Trending ▼   ResFinder  

ICSE Prelims 2016 : Science (St. Francis School, Visakhapatnam)

112 pages, 0 questions, 0 questions with responses, 0 total responses,    0    0
Ethakoti Honey
St. Francis School (SFS), Madhurawada, Visakhapatnam
1st to 10th science and maths
+Fave Message
ProfileTimelineUploads
 Home > lovelyromana >   F Also featured on: School Page

Formatting page ...

EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES SOLUTION FOR EVERGREEN MODEL TEST PAPERS EVERGREEN MODEL TEST PAPER - 1 SECTION - I (40 MARKS) Answer 1. (a) No, the couple acting on a rigid body can cause only rotatory motion. It cannot produce translatory motion. (b) 1J = 1N 1m = 1 kgms 2 1m = 1 kgm2s 2 = 1000 g (100 cm)2s 2 = 107g cm2s 2 1J = 107 ergs (c) No, because a part of input is wasted : (i) in moving the parts of the machine. (ii) in overcoming friction between various parts of machine. (d) According to the question, L + W = 2T and E = T L = 2T W Mechanical Advantage = E L 2T W = E T W T W =2 E =2 (e) We know that, kinetic energy K = Load L p2 2m K1 2m p2 = 1 22 K2 2m1 p2 p1 = p2 , m1 = 5 g, m2 = 35 g m2 35 7 K1 = m = = K2 5 1 1 Here, Here, ratio is Answer 2. K1 : K2 = 7 : 1 (a) The first beaker will have water at a higher temperature because of hidden heat (latent heat) which is released by steam at 100 C when it condenses to form water at 100 C. (b) The image is enlarged and erect, therefore, the lens is convex and the measurement of O F will give the focal length of the lens. The course of rays is as given below. A L A F I O O (1) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) In the minimum deviation position, Angle of incidence [i] = angle of emergence [e] Angle of refraction [r1] = angle of refraction [r2] A 90 i e r2 r1 B (d) r < v as = C c vr > vv i.e., red light in the glass travels faster than violet light. v Normal (e) i A A B w ed + Yello Blue + R Re Yel d Blu low e D C Glass slab B C Prism ue Bl R Ye ed llow Answer 3. (a) It depends on the angle of incidence i, the thickness of the slab and the refractive index of the slab. (b) Given, Distance between Reema and the building (d) = 167 m Speed of sound (v) = 334 ms 1, t = ? Time interval t = 2 167 m 2d = 334 ms 1 = 1s v (c) The cloud cover traps the heat radiated by the ground at night between the ground and the cloud, making us feel warmer. On a clear night, no such trapping of heat takes place. (d) The specific latent heat of ice means that 1 kg of ice at 0 C absorbs 336000 J of heat to convert it into water at 0 C or 1 kg of water at 0 C will liberate 336000 J of heat to convert into ice at 0 C. (e) Given, R = 0.2 P = 10 kW = 10000 W V = 220 V Power loss in line = I2R P = VI I = P/V 2 P R = V 2 1000 W 0.2 = 413 W = 0.413 kW = 220 V Answer 4. (a) By using the magnetic effect of current, i.e., when a magnetic needle is placed on the wall, it will show deflection due to the concealed current carrying wire. (2) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) Effective resistance between A and E : 1 1 2 1 1 1 = + = + = R1 ( R AF +R FE ) R AE (3 + 3) 6 6 R1 = 3 Effective resistance of network AFEDC is given by R2 = R1 + RED = 3 + 3 = 6 R2 and RAD are in parallel. Effective resistance will be : 6 6 36 = =3 6+6 12 This 3 is in series with RDC , so effective resistance = 3 + 3 = 6 Let this effective resistance be R3. Now, R3, RAC and RABC are parallel, so effective resistance between A and C is given by = 1 1 1 + + R AC R ABC R3 = 1 R4 1 1 1 + + R AC R3 ( R AB + R BC ) [ RAB and RBC are in series] 1 1 1 1 = + + 6 R4 (3 + 3) 6 3 1 1 1 1 1 = + + = = R4 = 2 6 2 R4 6 6 6 (c) Steel or alcino is used for making permanent magnets because of their large values of coercivity. (d) Yes, in -decay, the element shifts two places left in the periodic table. In -decay, the element shifts one place right in the periodic table. (e) In a TV set, electrons are accelerated by about 5000 V and strike the screen due to which soft X-rays are produced which can damage our body. These X-rays are absorbed by a length of a few feet of air due to ionisation by collisions. Thus, it is advisable to watch TV from a distance of about 10 - 12 feet. SECTION - II (40 MARKS) Answer 5. (a) (i) An inclined plane is used by labourers to load heavy barrels on a truck. (ii) No, a single fixed pulley does not produce any multiplication of force. It is useful because it helps us to apply force in a more convenient direction. (b) Let a lady of mass 60 kg sit a distance X from the fulcrum. 1.5 m X 3.5 m 40 kg Child 30 kg Child 60 kg Lady According to the principle of moments, Clockwise moments = Anticlockwise moments (40 kg g 3.5 m) + (30 kg g 1.5 m) = 60 g x (140 + 45) kg m = 60 x kg m (3) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 185 60 x = 3.08 m x = (c) Weight of water lifted = 20 10 N = 200 N Work done = 200 N 10 m = 2000 J Power = Work done Time taken 2000 J = 10 s = 200 Js 1 Power = 200 W Hence, Answer 6. (a) A real, inverted and diminished image is formed between F2 and 2F2. A F2 B 2F1 O F1 B1 2F2 A1 A L (b) 45 M 45 B C N P The angle of incidence at N inside the prism is less than 42 , therefore, the ray is able to emerge along NP. (c) The absolute refractive index of a medium is given by, sin i c n = = sin r v where c = Speed of light in vacuum, v = Speed of light in medium Thus, for same angle of incidence c sin r sin i v sin r v = or i.e., v will be minimum for the medium in which r or angle of refraction r is minimum. Thus, the speed of light is minimum in medium [I]. Answer 7. (a) Let X1 and X2 be the distances of the two cliffs from the observer. Then, 2X1 = 340 ms 1 5s 2X1 = 1700 m X1 = 850 m 2X2 = 340 ms 1 7s = 2380 m X2 = 1190 m Distance between the two cliffs = X1 + X2 = 850 + 1190 = 2040 m (4) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) Dissimilarities : Sound waves Light waves (i) These are longitudinal waves (also known as mechanical waves) which need a medium to travel. (ii) Their speed in air is 332 ms 1. (i) These are transverse waves (also known as electromagnet waves) which do not need a medium to travel. (ii) Their speed is 3 108 ms 1. Similarities : Both sound waves and light waves obey the laws of reflection and refraction and also exhibit the phenomenon of interference. (c) (i) We know that the frequency of the string depends on the length of the string (l ), mass per unit length of the string (m) and tension of the string (T). 1 2l T m Hence, the tension and length are generally changed to bring about the desired tuning. (ii) 1. Figure (2) represents the vibration of the largest amplitude and figure (3) represents the vibration of the fundamental note. i.e., v = 2. For figure (1), l = 3 1 2 1 = l 2 3 v v 3v = Frequency, 1 = = 2 1 2l l 3 v 2v For figure (2), l = 2 2 = = 2 (frequency) l 2l 2 v v For fig. (3), frequency, 3 = = 2l 3 Hence, 1 : 2 : 3 = 3v 2v 1v : : =3:2:1 2l 2l 2l Answer 8. (a) When the salt crystal dissolves, its crystal lattice is destroyed. The process requires certain amount of energy, i.e., latent heat which is taken from water. In the second vessel, a part of the intermolecular bonds of the crystal has already been destroyed in crushing the crystal. Hence, less energy is required to dissolve the powder. Therefore, the water will be at a higher temperature in second vessel. (b) Given, m = 1 kg Initial temperature = 10 C In this case, heat energy is required in the following four steps : 1. Heat energy required to raise the temperature of ice from 10 C to 0 C = m C (ice) rise in temperature = 1 (2.1 103) [0 ( 10)] = 21 103 J 2. Heat energy required to melt ice at 0 C into water at 0 C = mLice = 1 (336 103) = 336 103 J 3. Heat energy required to increase the temperature of melted ice water from 0 C to 100 C = m C (water) rise in temperature = 1 (4.2 103) (100 0) = 420 103 J (5) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 4. Heat energy required for vaporization of water = mLsteam = 1 (2268 103) = 2268 103 J Total heat energy required = (21 + 336 + 420 + 2268) 103 = 3045 103 J = 3045 kJ (c) Let the thermal capacity of the vessel be C JK 1. Heat energy given by hot water = 40 4.2 (60 30) = 5040 J Heat energy taken by cold water = 50 4.2 (30 20) Heat energy taken by cold water = 2100 J Heat energy taken by vessel = C (30 20) = 10C J If there is no loss of heat energy, Heat energy given by hot water = Heat energy taken by cold water and vessel 5040 = 2100 + 10C 10C = 2940 C = 294 JK 1 Thus, thermal capacity of the vessel = 294 JK 1. Answer 9. (a) The electric fuse is a device used to limit the current in an electric circuit. It has the following characteristics : (i) Fuse has high resistance and low melting point. (ii) Fuse is made-up of the alloy of lead and tin. (iii) Fuse is always connected to the live wire of the circuit. (b) C B N S D B1 (i) N and S are strong horse shoe magnets. (ii) ABCD is a soft iron core armature. (iii) S1 and S2 are slip rings. (iv) B1 and B2 are carbon brushes. (c) Given, P = 100 W, V = 230 V, t = 20 60s Resistance of bulb, R = A S1 B2 S2 Load V2 (230 V)2 = = 529 P 100W Supply voltage V = 115 (115V)2 (20 60)s V 2 Therefore, energy produced by the bulb = t= = 30000 J 529 R (6) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 10. (a) Thermionic emission is the phenomenon of emission of electrons from the metal surface on heating suitably. Plate Filament B Plate Filament Directly heated cathode (b) (c) B Bakelite F F P base Bakelite C F F P base Indirectly heated cathode (i) Number of protons. (ii) An element becomes radioactive when : 1. its atomic number exceeds 82. 2. there is an imbalance of protons and neutrons as compared to a normal stable atom. (i) < < (ii) > > (iii) 235 92 U+1 0n 148 57 La + 85 35 Br + 31 0 n + 186 MeV (7) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 2 SECTION - I (40 MARKS) Answer 1. (a) We know that K.E. = 1 mv2. 2 (i) If mass m is doubled, the kinetic energy gets doubled. (ii) If velocity v is halved, the kinetic energy gets one-fourth. (b) Given : F = 25 kgf = 25 10 = 250 N, m = 0.5 tonne = 500 kg, a = ? F 250 = = 0.5 ms 2 m 500 (c) According to the principle of machine, Work done by the machine will be at the most equal to the work done on the machine, i.e., for a perfect machine (whose parts are weightless and frictionless) . Work done on the machine (input) = Work done on the machine (output) (d) Levers having a mechanical advantage, necessarily greater than one are levers of second order. The nutcracker is an example of a lever of second order. (e) Maximum potential energy at points B and C. Maximum kinetic energy at point A. Answer 2. We know that a = (a) Frequency of second harmonics = 392 Hz. Frequency of fourth harmonics = 784 Hz. (b) Necessary conditions for total internal reflection : (i) Ray of light must travel from optically denser medium to rarer medium. (ii) The angle of incidence in the optically denser medium must be more than the critical angle. (c) A 45 P 90 Q 45 45 B C S (i) Angle of incidence of light on its face i. (ii) The refracting angle of the prism. (iii) The refractive index of the prism. (iv) The wavelength of the light used. (e) Ultraviolet radiations are used : (i) to detect adulteration in ghee because of its property of fluorescence. (ii) to detect forgery of cheques. Answer 3. (d) (a) Time period = 1 Frequency (b) Amount of heat required to melt 1080 g of ice at 0 C into water = mL = 1080 g 80 calg 1 = 86400 cal. Amount of heat lost by 1080 g of water at 80 C = m C t Q = 1080 1 (80 0) C = 86400 cal. (8) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES This is precisely the amount of heat lost by 1080 g of water at 80 C to bring down its temperature to 0 C. So, the temperature of the mixture will be 0 C. (c) Water molecules posseses more potential energy. The heat given to melt the ice at 0 C is used up in increasing the potential energy of water molecules formed at 0 C. 1 V2 V2 R= R R P P i.e., the resistance of a lamp is inversely proportional to its power. Thus, a 40 W lamp has more resistance than a 60 W lamp. (e) It is not safe to touch switches with wet hands because a layer of water between fingers and a live point in a switch will conduct electricity into body and a person will receive shock. Answer 4. (d) We know that, P = (a) The loss of energy along the transmission lines is proportional to the square of current (H I2). Hence, the transmission of electrical energy is economical at high voltage and low current. V . If resistance is reduced to half, the current will become double, i.e., 2I. R (c) Look at the face of the coil, if the current around that face is in the clockwise direction, the face has south polarity. This can be tested by using a compass needle. (b) We know that I = N S Current in clockwise direction Current in anticlockwise direction (d) A neutron in the nucleus of the -emitter radioactive element is converted into a proton and an electron as 1 0 0 n H1 1 + 1 e + v The electron so created leaves the nucleus immediately. (e) Tungsten (melting point is 3655 K) is used as an electron emitter. It has a high work function of 4.52 eV. When it is heated upto a temperature of 2500 K, it emits electrons. When it is coated with suitable coating material, its work function decreases and it can emit electrons, at even low temperatures. e.g., Thoriated Tungsten, work function reduces to 2.6 eV and it emits electrons when heated upto 2000 K. SECTION - II (40 MARKS) Answer 5. (a) Consider a body of mass m at a height h from the earth s surface which falls freely under gravity, i.e., position A in figure. Let us now calculate the sum of kinetic energy and potential energy at different positions say at A, B and C. At point A, [ vA = 0 at A] K.E. = 0 P.E. = mgh Total energy = K.E. + P.E. = 0 + mgh = mgh ...(i) At point B, 1 K.E. = mvB2 2 From third equation of motion, A x B h h x C vB2 = vA2 + 2gx vB2 = 0 + 2gx ( vA = 0) (9) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 1 m 2gx = mgx 2 P.E. = mg (h x) K.E. = Hence, total energy = K.E. + P.E. = mgx + mg (h x) = mgh ...(ii) At point C, K.E. = 1 mv 2 2 C From third equation of motion, vC2 = vA2 + 2gh ( vA = 0) vC2 = 0 + 2gh 1 m 2gh = mgh 2 P.E. = mg 0 = 0 K.E. = and Hence, total energy = P.E. + K.E. = 0 + mgh = mgh ...(iii) Thus, it is clear from eq. (i), (ii) and (iii) that the total energy of a body remains constant at each point of motion. (b) (i) Force = mg = 80 kg 10 ms 2 = 800 N (ii) Work done = F S = 800 (50 0.20) m = 8000 J (iii) P.E. = Total work done = 8000 J (iv) Power = W 8000J = = 400 W t 20s 400 = 0.53 HP 750 400 Power in kW = = 0.40 kW 1000 Power in HP = Answer 6. A (a) C E O B D O F (b) This is due to the unequal refraction of rays diverging from the lower and upper ends of the sun. Rays from the lower edge have to go through a greater thickness of air than rays from the upper edge. So, the vertical diameter appears to be diminished in size, whereas the horizontal diameter remains unaltered. (c) The sounds from the different sources of a factory form a mixture of different frequencies and give rise to the worst of noise. Frequency of wave = 1012 Hz (d) Velocity = 3 108 ms 1 Wavelength = v = 3 108 1012 = 3 10 4 m = 3 10 2 cm These waves are heat waves or infrared waves that varies from 10 2 to 10 4 cm. (10) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (e) Focal length of a convex lens is determined by following methods : (i) Distant object method. (ii) Auxillary plane mirror method. Answer 7. (a) No, echoes and resonant vibration are two different phenomenon. Echo is caused due to reflection of sound from a distant obstacle while resonant vibrations are produced when the frequencies of vibrating sources become equal. (b) Free vibrations : When a body is disturbed from its mean position and left free, it vibrates with its natural frequency which is the characteristic of the body. These vibrations of the body are said to be free vibrations. Forced vibration : When a body is set into vibrations with the help of a strong periodic force and the body vibrates with the frequency of the periodic force, then the phenomenon is said to be forced vibration. (c) The function of wooden box in a sonometer is to increase the sound intensity by forced vibrations. It decreases the duration of emission of sound energy. (d) Fire is extinguished by the vaporization of water which lowers the temperature of the burning body. Further, the water vapour envelops the body keeping oxygen away. Hot water vaporizes more than cold water, as it is closer to its boiling point. The latent heat of vaporization being large, hot water will extinguish easily than cold water. (e) To prevent heat loss by radiation, radiations are reflected back to the calorimeter by the surrounding vessel. Answer 8. (a) Suppose the equal mass of water and iron be m and T be the rise in temperature. The amount of heat required in case of water will be, Qw = mCw T and amount of heat required by same mass of iron will be, QFe = mCFe T Dividing both, Qw C = w Q Fe CFe i.e., the ratio of amount of heat is equal to the ratio of specific heats. Putting the values of Cw and CFe , we have Qw 1 cal g 1 C 1 = =9 Q Fe 0.11 cal g 1 C 1 Hence, for the same rise in temperature, water requires about 9 times more heat than iron. (b) Specific latent heat of fusion of a solid substance is defined as the amount of heat energy required to change a unit mass of a substance at its melting point from solid to liquid state without any change in temperature. (c) Let C be the specific heat capacity of copper and m be the mass of ice present in the calorimeter. So, we have 100 C [ 2 ( 4)] + m 0.5 [ 2 ( 4)] = 520 200C + m = 520 ...(i) Also, 100C [2 ( 2)] + m 0.5 [0 ( 2)] + m 80 + m 1 (2 0) = 41540 400C + 83m = 41540 ...(ii) Solving eq. (i) and (ii), we have m = 500 g C = 0.1 cal g 1 C 1 Putting this value of C = 0.1, we have 400 0.1 + m = 520 m = 520 40 = 480 g (11) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Regulator Answer 9. Open switch (a) Fan Fuse L Fuse Closed switch Main fuse Live wire E Earth wire Neutral wire N L E N Power socket (b) Effective emf = 20 8 = 12 V Total resistance R = 3 + 1 = 4 12V = 3A 4 p.d. across R1 = IR1 = 3 A 3 = 9 V From the battery of emf 20 V, a p.d. of 9 V will fall across R1. p.d. across A and B = 20 9 = 11 V. Effective current I = (c) (i) The effective resistance will be shown as in fig. given below. Here, the resistance of arm EFGH = 10 + 10 + 10 = 30 will be in parallel to the resistance of arm EH = 10 . 10 A E 10 B 10 10 F Their effective resistance 10 H G 10 10 30 = 7.5 10 + 30 Total resistance between A and B = 10 + 7.5 + 10 = 27.5 (ii) The effective circuit will be shown as given below : Reff = E H F A G C Resistance of arm EHG = 10 + 10 = 20 will be parallel to resistance of arm EFG = 10 + 10 = 20 20 20 = 10 20 + 20 So, total resistance between A and C = 10 + 10 + 10 = 30 Effective resistance = Answer 10. (a) Advantages of filling the bulb with inert gas at low pressure : (i) It prevents bulb from blackening. (12) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) It doesn t allow the temperature of the bulb to go beyond and hence, prevent the filament from getting melted. (b) Oxide coated filament produces more electrons at a lower temperature. Its operating temperature is 1000 K. Thermions (c) The free electrons emitted by a metal on being heated are called thermions. Free electrons The electrons revolving in the outermost orbit which are weakly held by the nucleus are free electrons. (d) There is no difference between : (i) A -particle and an electron -particle is the name given to an electron emitted from the nucleus. (ii) A -particle and a helium nucleus -particle is the name given to the positively charged particle emitted from the nucleus. (e) Equations representing -decay : (i) 238 92 U (ii) 226 88 Ra 234 90 4 Th + 2 He 222 86 Rn + 4 2 He Equations representing -decay : (i) 24 11 Na (ii) 32 15 0 P 32 16 S + 1e 24 12 Mg + 0 1e (13) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 3 SECTION - I (40 MARKS) Answer 1. (a) Yes, a body can remain in rest position if it is acted by more than one external forces when their vector sum is zero. Or When the force applied is less than opposing force of friction. (b) It is the inherent property of the body due to which it opposes the change in its state of rest or motion. It depends only upon the mass of the body. (c) Yes, the CG of a body can be outside the body. The CG of a uniform ring is at its centre, a point which is not on the body. (d) The efficiency of a machine is the ratio of the work output to the work input. Mathematically, = Work output Work input (e) All machines follow the law : Resistance force (Load) Load arm = Effort force Effort arm. Answer 2. (a) Work done is given by the expression, W = FS cos . Work done is zero if, (i) displacement is zero. (ii) force and displacement are perpendicular to each other. 1 mv2, whereas momentum is given by p = mv 2 Now, dividing and multiplying K.E. with m we will get, (b) K.E. is given by the relation, K.E. = K.E. = K = p = 1 (mv)2 2 m 1 p2 2 m ( p = mv) 2 mK (c) Laws of refraction : (i) The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane. (ii) The ratio of sine of angle of incidence to the sine of angle of refraction is constant for a pair of given media. (d) Yes, the type of lens changes, if it is placed in a medium having a higher refractive index than that of the lens. (e) When the atmosphere is absent there will not be any scattering of light and no light will reach the eye. The sky will, therefore, appear dark. Answer 3. (a) Loudness depends upon the square of the amplitude. Hence, with the doubling of the amplitude, loudness will become four times more. (b) On being set into vibration, the stretched string forces the surrounding air to vibrate. This vibrating air, in turn affects our ear drum and produce an audible sound. (c) Conductors which do not obey Ohm s law of V I are called non-ohmic conductors. For example, Diode valve. (14) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Graph of current-voltage : Current n-o No ic hm Voltage (d) It is a process of connecting a metallic body of an appliance to the earth through a conductor. It protects the user of an appliance from electric shock. (e) Heat dissipated depends upon : (i) The square of current through the conductor. (ii) The resistance of the conductor. (iii) The time for which current is passed. Q = I2Rt Answer 4. (a) When a current carrying conductor, capable of moving freely, is placed in a magnetic field, it experiences a force (torque) and begins to move in a direction given by Fleming s left hand rule. (b) Alternating current changes in magnitude and its direction reverses periodically whereas direct current direction remains same. (c) According to the principle of calorimetry, When two objects exchange heat, the amount of heat lost by one of them is equal to the amount of heat gain by the other . In other words, Heat gained = Heat loss. (d) The land cools at a faster rate than water. This is because the specific heat capacity of water is more than land. For the same amount of heat lost, the fall in temperature of a given mass of land, therefore, will be more than that of an equal mass of water. (e) Two uses of cathode ray tube : (i) Used in television. (ii) Used to investigate the varying potential difference. SECTION - II (40 MARKS) Answer 5. (a) We know that momentum p of a body of mass m with a velocity is p = mv. Since m is constant for small velocities compared to light, Change in momentum p when a change in velocity v takes place, is given by, p = m v p v =m t t ...(i) p v = force, F and rate of change of velocity = acceleration, a. Hence, t t substituting the values in eq. (i), we will get Now, rate of change of momentum F = ma (b) The various positions where the weights are acting as shown in the fig. AB is the metre scale of mass W. A 10 cm 20 cm 30 cm E D F 100 g 50 g 50 cm C 100 cm B W Since the metre scale balances about the point F, sum of the clockwise moments must be equal to the sum of the anticlockwise moments. Taking moments about F, we get 100 EF = 50 DF + W CF 100 10 = 50 10 + W 30 (15) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES W 30 = 500 500 = 16.67 g 30 (ii) Class I W = (c) (i) Class III (iii) Class II (iv) Class I Answer 6. (a) (i) Initial velocity, u = 0 Final velocity, v = ? Distance S = h = 50 cm = 0.5 m Acceleration, g = 10 ms 2 Using the equation of motion, v2 u2 = 2gh v2 = u2 + 2gh v2 = 0 + (2 10 0.5) = 10 Hence, v = (ii) ( u = 0) 10 = 3.16 ms 1 Kinetic energy of the body = 1 mv2 2 1 10 10 = 50 J ( m = 10 kg) 2 (iii) Velocity of the body doesn t depend upon its mass, because the earth attracts all bodies with same acceleration due to gravity. (b) = 5400 = 5400 10 10 m K.E. = a = g 3 , in glass = ? 2 As frequency remains same, v =f 3 108 = f 54 10 8 f = 3 108 54 10 8 f = ( v = speed of light) 3 1016 54 = 300 1014 54 Hence, f = 5.5 1014 Hz ... (i) a = 3 2 a = c velocity in glass g g c Velocity of glass = a = g 3 108 2 = 2 108 3 Now, since v =f v = 2 108, f = 5.5 1014 Hz ... (ii) = v 2 108 = f 5.5 1014 = 0.36 10 6 = 3600 (16) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 7. (a) Yes, it is possible to burn a piece of paper using a convex lens in day light. The rays from the sun which are coming from infinity are made to pass through a convex lens. The lens bring them to focus at its focus. F (b) Given, = 5 Hz Therefore, T = 1 = 1 = 0.2 s 5 Time for 8 vibrations t = 8 0.2 = 1.6 s v = 340 ms 1 d =? During the given time, the sound wave travels the distance between the gun and the cliff twice. d = Hence, A v t 340 1.6 = = 272 m 2 2 (c) A B 2F F B F 2F AB = Object, A B = Virtual, erected, magnified image. Answer 8. (a) Clearly, from the fig., CD = DB = CB = 10 cm As AC = CE = 20 cm 1 AC = 10 cm 2 Clearly, AB = BC = CD = DE = BD = 10 cm R1 = R2 = R3 = R4 = R5 = 10 As R2 and R3 are in series, therefore, their combined resistance is 10 + 10 = 20 This combination is in parallel with R5. (i) Hence, resistance between points B and D is AB = DE = 1 1 3 1 20 = + = R= = 6.67 R 20 10 20 3 (ii) Now, resistances R1, R and R4 are in series, therefore, resistance across AB is R = R + R1 + R4 = 6.67 + 10 + 10 = 26.67 (b) The equivalent circuit : 3 2 1 X Y 3 2 (17) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Total resistance is given by 1 1 1 7 1 = + + = 5 1 5 5 R R= 5 = 0.71 7 (c) We know that P = Therefore, V2 , where V is constant. R 100 P R 25 = 100 = =4 25 P R100 25 i.e., the resistance of 25 W bulb is 4 times the resistance of 100 W bulb. When both bulbs are put in parallel, p.d. becomes same. Therefore, the ratio of power dissipated is R 100 P 25 V 2 /R 1 = 2 25 = R = V /R 100 4 P 100 25 Hence, 100 W will glow 4 times more than 25 W. Answer 9. (a) Given, mice = 30 g, mw = 200 g, T1 = 0 C, T2 = 30 C, T = ?, Lice = 80 cal g 1, C = 1 cal g 1 C 1. Heat lost by 200 g of water in cooling from 30 C to T C = 200 1 (30 T) = 200 (30 T) cal This heat is used by 30 g of ice in getting just completely melted and in raising the temperature of melted water to T C. Now, heat needed to just melt 30 g of ice = 30 80 = 2400 cal Heat needed to raise the temperature of 30 g water from 0 C to T C = 30 1 (T 0) cal = 30 T cal Thus, according to the principle of heat, we have 2400 + 30T = 200 (30 T) 2400 + 30T = 6000 200T 230T = 3600 T = 3600 / 230 = 15.65 C (b) Let C be the specific heat capacity of the liquid. Heat used in raising the temperature of the liquid from 25 C to 31 C = 5000 C (31 25) J = 5000 C 6 J ...(i) Heat produced by the electric heater in 120 s = 1000 watt 120 s = 120000 J ...(ii) From (i) and (ii), we have 5000 C 6 = 120000 C = 120000 = 4 J g 1 C 1 5000 6 Hence, heat capacity = mC = 5000 4 = 20000 J C 1 (18) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 10. Armature coil D Electromagnet A (a) N S C B Split ring Carbon brush Battery (b) (i) N Both -rays and cathode rays are electrons and hence, both are deflected by electric and magnetic field. N Both produce fluorescence when they strike against a fluorescent material. e N Both carry a negative charge and have the same value of . m (ii) (a) (b) Mg 2 4 12 ...... 4 2 He ........ (iii) Tungsten or Thoriated Tungsten, because it has high melting point and low work function. (19) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 4 SECTION - I (40 MARKS) Answer 1. (a) No, when the particle moves in a curved path, it posseses an acceleration and hence, some force must be acting on it. (b) By Newton s second law, F = ma or a = 1 F i.e., a m m a m (c) The moment of a force about an axis of rotation is the turning effect of the force about the axis of rotation. It is measured by the product of force and perpendicular distance between the axis of rotation and the line of action of the force. (d) The statement means that the effort required for the machine to overcome a load is less than the load. (e) An inclined plane is a simple machine consisting of a plane kept at a certain inclination to the ground. It acts as a force multiplier and has M.A. greater than one. Answer 2. (a) There is an interchange in the kinetic and potential energy of the simple pendulum. At the mean position, its energy is wholly kinetic and at the extreme position, it is wholly potential. The K.E. changes to P.E. when it moves from the mean position to the extreme position. (b) It states that energy can neither be created nor destroyed, it can be transformed from one form into another. (c) Optical fibre is a fine quality fibre of glass. Its diameter is of the order of 10 4 cm with refractive index of material being of order 1.7. Optical fibre is based on total internal reflection. (d) In both cases, the two surfaces are convex towards air. In case of double convex lens, the two surfaces may have any values for the two radii of curvature of the two surfaces, while in bi-convex lens, the radii of curvature of the two surface has the same value. Thus, every bi-convex is a double convex lens, but every double convex lens is not a bi-convex. (e) The splitting of white light into its seven constituent colours is called dispersion of light. Answer 3. (a) For hearing echoes, there should be atleast a distance of 17 m between the source of sound and the body from which sound is reflected. In small rooms, this is not the case. Hence, no echoes are heard. (b) The intensity of the sound is the quantity of energy, passing per second, through a unit area, held perpendicular to the direction of sound propagation. (c) Resistivity of a material is the resistance of a conductor of this material of unit length and unit cross-sectional area. Its unit is Ohm metre, m. (d) (i) The resistance of wire increases with increase in length of wire. R L (ii) The resistance of wire decreases with increase in thickness of wire. 1 A (e) The fuse protects an electric circuit by melting and breaking the circuit, whenever the current exceeds the predetermined limit in the circuit. Due to low melting point, the fuse wire melts and hence, breaks the circuit. R Answer 4. (a) Eddy currents, which gets generated within the core of the transformer, are undesirable since they result in wear and tear of the transformer and also in loss of energy. So these have to be minimised. If the core is laminated with insulation in between, the eddy currents are likely to be terminated, thus, saving some amount of energy. (20) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) The two ways through which the emf of an A.C. generator can be increased are as follows : (i) Increase the speed of rotation of the armature. (ii) Increase the strength of the magnetic field in which the armature is placed. (c) (i) Melting point of ice decreases with an increase in the pressure. (ii) Boiling point of water increases with an increase in the pressure. (d) Ice-cream draws latent heat of fusion for its melting which water doesn t draw. To melt, ice-cream needs latent heat of fusion, which is higher than the specific heat of water. (e) The two factors are as follows : (i) The temperature of the surface. (ii) The surface area and material of the surface. SECTION - II (40 MARKS) Answer 5. (a) (i) If the body is moving along a straight line with a uniform speed, no force acts on it because there is no acceleration. However, for a body moving in a circular path with uniform speed experiences a force called the centripetal force. This force only changes the direction of motion. (ii) No net force acts on such a body because there is no acceleration. (b) A body is said to be in equilibrium under the action of a number of forces, if the forces are not able to produce any change in the state of rest or of uniform motion or uniform rotation. For a body to be in equilibrium, the following conditions must be satisfied : (i) Vector sum of all the forces acting on the body must be zero, i.e., F = 0. (ii) Vector sum of all the torques of the different forces must also be zero, i.e., = 0. (c) Here, L = 200 kg, E = 1500 N Resistance R = 250 N L 200 10 4 = = = 1.33 3 E 1500 When the load is being lifted without acceleration, (i) M.A. = P = mg sin + F l 1500 = 200 10 sin + 250 2000 sin = 1250 P F h mg 1250 Hence, sin = = 0.625 2000 (ii) If V.R. is the velocity ratio of the inclined plane, then V.R. = (iii) Hence, % is given by = = 1 1 = = 1.6 sin 0.625 M.A. 100 V.R. 1.33 100 = 83.33% 1.6 (iv) Also, from the figure, 1 l = sin h (l = length of inclined plane, h = height of inclined plane) l 1.5 l = 1.6 1.5 = 2.4 m 1.6 = (21) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 6. (a) Given, Mass = 50 kg Height = 10 m K.E. = ? v=? K.E. of the stone when it reaches the ground = P.E. of the stone at the top of the ladder Hence, K.E. = P.E. = mgh = 50 9.8 10 = 4900 J Speed v of the stone just before hitting the ground. Initial velocity, u = 0, v = ?, h = 10 m, g = 9.8 ms 2 According to the equation of motion, v2 u2 = 2gh v2 0 = 2gh v = 2gh v = 2 9.8 10 = 14 ms 1 (b) If candle flame is placed in front of a thick plane glass mirror and is viewed obliquely, a number of images are seen. Out of these images, the second image is the brightest while others are of diminishing brightness. When a ray of light AB falls on the surface LM of the mirror, a small Eye part of light is reflected in the direction of BP, forming a faint P image of A, while a larger part of light is refracted along BB Q inside the glass and then strongly reflects as B C and then gets R S refracted along CQ in air and forms a virtual image A2, which is A the brightest image because it is due to light suffering a strong B E C D reflection. The reflected ray further keeps on forming multiple M L images of gradually decreasing brightness. P N A1 B C D A2 A3 Answer 7. A4 (a) A A F B B O Concave Lens AB = Object, A B = Image (b) In a prism, refraction of light takes place at two slant surfaces. The dispersion of white light occurs at the first surface of prism where its constituent colours are deviated through different angles. At the second surface, these split colours suffer only refraction and they get further separated. But in a rectangular glass block, the refraction of light takes place at the two parallel surfaces. At the first surface, although the white light splits into its constituent colours on refraction, but these split colours on suffering refraction at the second surface emerge out in the form of a parallel beam, which give an impression of white light. (c) Material medium is needed for the propagation of sound. On moon their is vacuum. Therefore, sound cannot propagate on the moon, due to this absence of material medium. Hence, the astronants can t hear each other. Answer 8. (a) Potential difference across 5 resistor = 10 V Potential difference across R2 resistor = 6 V (22) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Value of resistance R1 = 5 , I = ?, R2 = ? and V = ? By Ohm s law, the current through resistor of 5 is, 10 V (i) I = = = 2A 5 R1 (ii) Since R2 is connected in series with R1, I1 = I 2 = 2 A Hence, current through R2 is 2 A. 6 (iii) R2 = = 3 2 (iv) Since the resistors are in series. Total resistance = R1 + R2 = 5 + 3 = 8 Hence, Total V = Total resistance Current V = 8 2 = 16 V (b) Given, Power of motor 1 HP = 746 W Power of one bulb, V I = 220 0.2 = 44 W Therefore, power of 5 bulbs = 5 44 = 220 W Power of two fans = 2 80 = 160 W Total power of all electrical appliances = 746 + 220 + 160 = 1126 W Since, all appliances work for 5 hours, therefore, Energy consumed in one day = 1126 5 watt h = 5630 Wh Hence, energy consumed in one month = 30 5630 Wh Thus, Therefore, units consumed = ( R = V , where V = 6 V, I = 2 A) I 30 5630 = 168.9 units 1000 cost of energy = 0.20 168.9 = ` 33.78 (c) The fuse wire rated 20 A is thicker of two. The thicker wire would have a lower resistance than the thin wire. It will, therefore, produce less heat for a given current. It can, therefore, carry a much larger current through it before melting down. Answer 9. (a) Construction : A step-up transformer consists of two coil, namely primary and secondary coil. The number of turns in secondary coil must be more than primary coil. These coils are wound over a laminated core. This is made by taking thin rectangular sheets made by soft iron and dipping them in insulating point. These insulated sheets are stacked together to form a rectangular frame. This is known as laminated core of transformer. Secondary winding Primary winding Laminated core Working : The A.C. source which is connected to the primary coil produces a varying current to flow through this coil. This, therefore, produce a varying magnetic field whose lines of force pass through the secondary coil. Because the magnetic field produced is a varying one, the flux linked with the secondary coil is a changing one. Hence, an induced emf is produced in the secondary. The emf induced in the secondary is directly proportional to the number of turns in the secondary. Thus, (23) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES VS N = S VP NP The step-up transformer increases the voltage. (b) Diagram of cathode ray tube : Cylindrical anode Heating element 6V Electron beam 100 V Grid Screen Vertical plates Horizontal plates Answer 10. (a) (i) Heating curve : D B 80 C Melting 60 Liquid So lid Temp. ( C) 100 40 20 A O 30 60 90 120 150 180 210 240 Time (s) (ii) Cooling curve : A Temp. ( C) 100 Liquid C B 80 Freezing 60 40 Solid D 20 O 30 60 90 120 150 180 210 240 Time (s) (b) We know that the heat content of 1 kg of steam (100 C) is more than that of 1 kg of boiling water (100 C) by nearly 2268 kJ (the specific latent heat of vaporization of water). Each kilogram of steam can thus, give out nearly 2268 kJ of heat more than each kg of boiling water. This makes steam more effective than boiling water. (c) Let x -particles and y -particles be emitted. We know that -particles decreases the mass number by 4 and the atomic number by 2. Further, -particles increases the atomic number by 1 and mass number remains unaltered. 238 4x = 206 4x = 238 206 = 32 32 =8 4 Hence, 8 -particles are emitted. x= Considering atomic number 92 2x + y = 82 92 (2 8) + y = 82 y = 82 92 + 16 = 6 Hence, 6 -particles are emitted. (24) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 5 SECTION - I (40 MARKS) Answer 1. (a) A force is said to be conservative if the work done by it is dependent on the path followed between two points. Non-conservative forces are those forces for which the work done is dependent of the path connecting the two points. (b) This is so because near the hinge, the distance of the point of application of force from the axis of rotation becomes negligible, so the force applied increases. (c) Rotational effect of a force depends on two factors : (i) The magnitude of the force. (ii) Perpendicular distance of its line of action from the axis of rotation. (d) A single fixed pulley has its axis of rotation fixed and hence is called a fixed pulley. It is used to lift a water bucket from the well. It is used to change the direction of the effort. (e) (i) Class III (ii) Class II (iii) Class II (iv) Class III Answer 2. (a) As, p = 2mK . Since K, kinetic energy is the same for both bodies, momentum p m. Therefore, the heavier body will have more momentum. (b) (i) Electric energy to sound energy. (ii) Electric energy to mechanical energy. (iii) Chemical energy to electrical energy. (iv) Mechanical energy to electrical energy. (c) The lateral displacement depends upon : (i) angle of incidence. (ii) refractive index of the medium. (iii) thickness of the glass slab. No, emergent ray is parallel to the incident ray. There is no deviation. (d) The three characteristics are : (i) Virtual image. (ii) Upright or erect image. (iii) Diminished. Yes, when a concave lens is placed in the path of a converging beam, it produces a real image. (e) As we know, c = f c = 3 108 ms 1, f = 1012 Hz The wavelength of the wave = c 3 108 = 1012 f = 3 10 4 m The name of this wave is microwave. Answer 3. (a) No, we will not hear any audible sound on the surface of the moon. This is because sound requires a medium to propagate. Since there is no atmosphere on the surface of the moon, no sound will be heard. (b) Electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point. (c) Physical conditions remain same, the electric current flowing through a conductor is directly proportional to the potential difference across the two ends of the conductor. (25) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (d) The resistivity of a conductor does not depend upon its shape or size. Therefore, there will be no change in the resistivity of wire. (e) The rate of doing work, or the rate of dissipation of energy, is called electric power. It is denoted by P i.e., P = W V2 = t R SI unit is watt. Answer 4. (a) We wind an insulated copper wire around the magnet and pass alternating current through it. This A.C. current will change the polarities of the magnet rapidly. This will ultimately result in the loss of magnetic property of magnet. (b) The functioning of a transformer depends upon the principle of mutual induction. It is used to increase the alternating voltage. (c) Evaporation of water from the leaves is called transpiration. Transpiration helps plants to keep cool. During summer, when the temperature is high, a plant must transpire more to keep itself cool. More transpiration requires more leaves. So plants acquire more leaves in summer. (d) When we make the base thick, the capacity of cooking pass, its thermal capacity increases. Thus, it can import sufficient energy for cooking even at low temperature. (e) A chemical change is an extra nuclear phenomenon. It is concerned with the electrons alone. A nuclear reaction involves the nucleus and its constituents. In this reaction, new elements are produced. SECTION - II (40 MARKS) Answer 5. (a) (i) It is the inherent property of a body due to which it opposes the change in its state of rest or motion. It depends only upon the mass of the body. Mass (ii) Weight (a) Mass is the matter contained in the body. (a) Weight is the force with which a body is attracted towards its planet. (b) Weight is a vector qauntity and its value depends on the value of g . (b) Mass is a scalar quantity and its value remains same everywhere. (b) Let D be the position of the knife edge so that the metre scale and the various masses are balanced. Since the metre scale is balanced. Sum of the clockwise moments must be equal to the sum of the anticlockwise direction. O 10 cm E 20 cm F 90 cm G x C D 100 cm B 60 g 20 g 30 g 80 g Taking moments about D, we have 80 GD = 20 ED + 30 FD + 60 CD 80 (40 x) = 20 (40 + x) + 30 (30 + x) + 60x 3200 80x = 800 + 20x + 900 + 30x + 60x 1500 = 190x x = 1500 = 7.9 cm 190 Position of the mark on the metre rod where the knife edge should be placed is 50 + 7.9 = 57.9 cm. (26) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) It is the ratio of the load to the effort. M.A. = Load Effort Answer 6. (a) (i) At point A : Total mechanical energy = mg (x + y) = potential energy (ii) At point B : Total mechanical energy is sum of kinetic and potential energy. P.E. = mgy ...(i) 1 mv2 2 v2 u2 = 2gx K.E. = u =0 during free fall 2 v = 2gx As 1 m 2gx = mgx 2 P.E. + K.E. = mgy + mgx = mg (y + x) (iii) At point C, the body has only K.E., no P.E. K.E. = K.E. = ...(ii) 1 mv2 and v2 = 2g (x + y) 2 1 m 2g (x + y) = mg (x + y) ...(iii) 2 (iv) The law which is verified is conservation of energy. (b) The pond appears to be three-quarter of its actual depth due to the property of refraction of light. K.E. = Refractive index = Real depth Apparent depth 4 for water. 3 3 Eye So apparent depth is th of real depth. 4 D E A ray from bottom of the tank A on striking normally at B goes straight. B C Another ray from A striking the interface at C is refracted towards A CE and when produced back, it appears to come from A . When BD and CE enter the eye, these form image of A at A . Thus, the A depth appears to be A B instead of actual AB. (c) Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90 . Since = Answer 7. B (a) B A F A O F2 AB = Object A B = Image (b) The refractive index of glass is different for different colours of light. When white light is incident on the surface of a prism, different colours are deviated by different amounts as the deviation depends upon refractive index which is different for different colours. Of these, violet is deviated the most and red is deviated the least. Thus, all the colours come out of the prism from different points. (i) For the first echo, the sound takes a time of 2 seconds to reach the nearest cliff which is 640 m, therefore, (c) using the formula, v= S 640 = = 320 ms 1 t 2 (27) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) Let x be the distance of the second cliff from the person. Now, distance travelled by the second echo to the second cliff is, x =v t = 320 3.5 = 1120 m Therefore, the distance between the two cliffs is D = 640 + x = 640 + 1120 = 1760 m Answer 8. (a) Since 15 and 10 resistances are in parallel. 1 1 3+ 2 1 = + = 15 R 10 30 R =6 p.d. in parallel = Total current Resistance only in parallel =1 6=6V I1 (current flowing in 10 resistor) = 6 V = = 0.6 A 10 R1 I2 (current flowing in 15 resistor) = 6 V = = 0.4 A 15 R2 p.d. in wire AB = IR = 1 5 = 5 V Total p.d. = 6 + 5 = 11 V (b) Total internal resistance is, 1 1 1 2 = + = r 1.5 1.5 1.5 r = 1.5 = 0.75 2 emf of one cell = emf of all the cells in parallel E =2V Total resistance in the circuit = r + R = 0.75 + 4.25 = 5 2 V, 1.5 A B 2 V,1.5 4.25 I = E R+r I = Now, 2 = 0.4 A 5 P . Thus, the current flowing through a device is directly proportional to its V power. Thus, knowledge of the power voltage rating of a device helps us to know the current flowing through it. This, in turn, helps us to decide about the type of leads to be used for it. We use thick wires if the current flowing is more than 5 A. For currents less than 5 A, we use relatively thinner wires. (c) We know that P = VI or I = (28) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 9. (a) It works on the principle that when a coil rotates in a uniform magnetic field, a current is induced in the coil. Working : Let the coil abcd , initially in the horizontal position, be rotated anticlockwise i.e., the arm ab moves downwards and arm cd upwards. Due to the motion of the coil in the magnetic field, induced current is produced in the coil from d to c in arm cd and b to a in arm ab, according to Fleming s right hand rule. After half rotation, the arms of the coil interchange their positions. Arm ab comes to the right and arm cd to the left. During the second half of the rotation, current flows from a to b in the arm ab. The two half rings (R1 and R2) rotate with the coil and touch the two carbon brushes (B1 and B2) one by one. As a result, each carbon brush continues to have the same polarity (+ or ). Brush B2 always remain positive (+) and brush B1 remains negative ( ). The current so produced is a direct current. b c N S a d Electromagnet B1 B2 S1 S2 Load S1 and S2 = Slip rings B1 and B2 = Brushes abcd = Armature (b) Electrons which are present in the outermost orbit are loosely bound as they are weakly attracted by the nucleus, so they are called free electrons. Free electrons can move randomly but they don t have sufficient energy to leave the metal surface. Answer 10. (a) Effect of electric field (i) The electric field exerts a force on the electrons in a direction opposite to the electric field. (ii) The path of electron beam is parabolic. (iii) It changes both the kinetic energy and momentum of the electrons. (iv) Electric field exerts a force whatever be the direction of motion of electrons. (b) Effect of magnetic field (i) It exerts a force on the electron beam perpendicular to both the magnetic field and direction of motion. (ii) The path of electron beam is circular. (iii) It changes the momentum but has no effect on the kinetic energy of electrons. (iv) Magnetic field exerts a force if the electrons move in the direction other than that of magnetic field. (i) Here, m = 150 g, t1 = 30 C, t2 = 5 C Specific heat of water = 4.2 Jg 1 C 1 Specific heat of material = 0.4 Jg 1 C 1 (29) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Specific latent heat of fusion of ice = 336 Jg 1 Heat lost by 150 g of water = 150 4.2 25 = 15750 J Heat lost by 100 g of the vessel = 100 0.4 25 = 100 J Total heat lost = 15750 + 1000 = 16750 J Heat gained by ice in changing into water at 0 C = m 336 Heat gained by m g of water to in raising its temperature from 0 C to 5 C = m 4.2 5 = 21m Total heat gained by ice = m 336 21m = 357m But, heat gained = Heat lost 357m = 16750 16750 = 46.92 g 357 (ii) Calorimeter is usually made of copper. Hence, m = (30) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 6 SECTION - I (40 MARKS) Answer 1. (a) (i) 1 Newton is that force which when acting on a body of mass 1 kg, produces an acceleration of 1 ms 2 in it. (ii) S.I. unit of force is Newton (N) and C.G.S. unit is dyne. 1N = 105dyne (b) Axis of rotation is the line or point about which the object rotates. It may be within the object or outside the object. (c) Skidding takes place on a rainy day because the force of friction between the roads and the tyres considerably decreases and is not sufficient to supply the necessary centripetal force. (d) The mechanical advantage of second order levers is always more than 1. This is because the effort arm is always longer than the load arm of these levers. (e) The various applications of gears are as follows : (i) Gears are used in mechanical clock. (ii) Gears are used in transports like car, bus etc. Answer 2. (a) In case of a body moving along a circular path, the centripetal force is always along the radius while displacement in tangential. Hence, work done = F S cos 90 = 0 as angle between F and S is 90 . Thus, work done is zero. p2 (b) As, K.E. = and p is same for both. 2m 1 (m = mass) m Thus, the lighter body has more kinetic energy. Hence, K (c) A Q 60 30 Normal a 60 60 B 60 60 C (d) Every part of a lens forms a complete image. If the lower part of the lens is blackened, the complete image will be formed but its intensity will decrease. (e) Danger signals are red because the red colour is scattered less on account of its longer wavelength, since 1 I 4 . The red colour can be seen even from far of distance. Answer 3. (a) The intensity of the sound is the quantity of energy, passing per second, through a unit area, held perpendicular to the direction of sound propagation. (b) (i) Terminal potential difference : It is the potential of the cell when a current is drawn from it. In other words, it is the potential of the cell in a closed circuit. (ii) emf : It is the potential of the cell when no current is drawn from it. In other words, it is potential of the cell in an open circuit. (31) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) The internal resistance of a cell depends upon : (i) The nature of the electrolyte used. (ii) The area of the two electrodes and the distance of separation between them. (d) The fuse is placed in the live wire just after the electricity meter of the consumer. Fuse wires are generally made of an alloy of tin and lead and have a relatively low melting point and high resistance. (e) It means that if the bulb is lighted on a 220 V supply it consumes 60 W electrical power or 60 J of electrical energy converts into heat and light energy in 1 second. Answer 4. (a) The function of the split ring in the D.C. motor is to ensure that the current through the armature is reversed after half a revolution. This makes the armature always rotate in the same direction. (b) A transformer works when the magnetic flux linked with the primary and the secondary changes. In case of D.C., there is no change in the magnetic flux. Hence, a transformer does not work on D.C. (c) (i) Since the device is being used to step-up 12 V A.C. to 200 V A.C., it is called a step-up transformer. (ii) A transformer works on the principle of mutual induction. (d) The specific latent heat of vaporization of a given liquid is defined as the heat required to convert one kilogram of the liquid into vapour, at its boiling point, without any change in its temperature. (e) Free electrons leave the metal surface only if sufficient energy is supplied to the metal. This emission of electrons from metal surface is called the electron emission. SECTION - II (40 MARKS) Answer 5. (a) When a force acts on a body capable of rotation, it causes a rotation of the body either in the clockwise direction or anticlockwise direction. According to the principle of moments, when a number of forces act on body free to rotate about an axis and if no rotation occurs, then the sum of the clockwise moments must be equal to the sum of the anticlockwise moments. To illustrate the principle, let us consider a uniform rod AB of weight W, to be in equilibrium under the action of a number of forces, F1, F2, F3 and weight W acting at points D, E, B and C respectively, then according to the principle of moments. F3 F1 D A C E B W F2 Taking moments about A, Sum of clockwise moments = Sum of anticlockwise moments W AC + F2 AE = F1 AD + F3 AB If we take moments about C, then F2 CE + F1 CD = F3 CB (b) Here, ra = 2 cm, rb = 20 cm, Na = (i) We know that gear ratio = (ii) Also, (iii) Also, 100 r.p.s., Nb = 40. 60 ra Na 2 1 = = = rb Nb 20 10 rb Na 20 = r = = 10 Nb 2 a 100 60 Nb = 10 Nb = 100 1 = r.p.s. = 10 r.p.m. 600 6 ra 1 NA = = 10 rb NB NA = 1 40 = 4 10 (32) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 6. (a) Given, Mass of the rocket (m) Height (h) Acceleration due to gravity P.E. of the rocket = 3 106 kg = 25 km = 25 103 m = 10 ms 2 = mgh = 3 106 10 25 103 = 7.5 1011 J 1 2 1 mv = 3 106 103 103 2 2 = 1.5 1012 J (b) Let the mass of each body be m. Then K.E. of the first body, Now, K.E. of the rocket = K1 = 1 mv 2 2 ...(i) K2 = 1 1 9 mv 2 = m (3v)2 = mv 2 2 2 2 ...(ii) K.E. of the second body, Dividing (ii) by (i), we have 9 2 mv K2 9 2 = 1 = 2 K1 1 mv 2 Thus, the kinetic energy of the second body is nine times the K.E. of the first body. (c) Let the thickness of the glass block be t mm. According to the question, apparent depth of the postage stamp = (t 7) mm. We know, Refractive index of glass, r = Real thickness of glass block Apparent thickness of glass block t t 7 1.5t 10.5 = t 1.5 = 10.5 0.5 = 21 mm or 2.1 cm t = Answer 7. (a) F O The magnification of the image formed is unity (or 1). The image is real and inverted. (b) The clouds generally consists of dust particles and water droplets whose size is much larger than the wavelength of light coming from the sun and incident on them. There is thus very little scattering. Thus, the light received from the sun has all the colours of light. Thus, we receive almost white light and , therefore, the clouds are white. (c) It happens due to resonance. When a vehicle is driven, the piston of the engine makes in and out motion at a frequency depending upon its speed. The vibrations caused by the movement of piston are transmitted to all (33) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES parts of the vehicle. It is just possible that some parts of the vehicle, may have the natural frequency of to and for movement of piston at a certain speed of the vehicle. When this happens, then at the particular speed of the vehicle, that part starts vibrating vigorously due to resonance. Answer 8. (a) Given, I1 = 0.6 A, R1 = 2 , I2 = 0.3 A, R2 = 8 , r = ? E R+r Using I = (i) 0.6 = E E and (ii) 0.3 = r+2 8+ r 2 From the above two, we have, 0.6 A As same cell is used, E is same so, (i) = (ii) 8 0.6(2 + r) = 0.3(8 + r) 0.3r = 1.2 r= 4 0.3 A Substituting the value of r in (i), we will get E = 0.6 (2 + 4) = 3.6 V (b) The current in the ammeter B and C is inversely proportional to the value of resistance in the parallel branch. Therefore, reading of ammeter C 6 = =2 reading of ammeter B 3 Therefore, reading of ammeter C = 2 0.5 = 1.0 A Hence, reading of ammeter A A = (0.5 + 1.0) A = 1.5 A Let the total resistance of the circuit be R, therefore, R = 2 + RP, where 1 1 1 1 = + = 3 6 2 RP RP = 2 R=2+2= 4 Therefore, (c) Given, P1 = 40 W, I1 = 0.812 A, V1 = 220 V, P2 = 60 W, I2 = 0.272 A, V2 = 220 V Also, V = 220 V Now, R1 = V12 (220) 2 = = 1210 40 P2 and R2 = 2 V2 (220) 2 = = 806.7 P2 60 When the lamps are connected in series, then the total resistance becomes RS = R1 + R2 = 1210 + 806.7 = 2016.7 Hence, current through the series combination and hence, through each lamp I = 220 V = = 0.109 A 2016.7 RS (34) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 9. (a) (i) A.C. generator D.C. motor It is a device which converts mechanical energy to electrical energy. A.C. generator makes use of two separate slip rings. It works on the principle of electromagnetic induction. It is a device which converts electrical energy to mechanical energy. It makes use of one split ring or commutator. It works on the principle, force produced on a current carrying conductor placed in a magnetic field. (ii) 1. We know that, Power, P = IP EP 400 = IP 220 IP = 400 = 1.8 A 220 IP EP = IS ES 2. Now, IS = IP EP ES Thus, IS = 1.8 220 15 = 26.4 A (iii) No, transformer can t work in D.C. source, because it works on the principle of electromagnetic induction, which is only applicable for A.C. (b) (i) Given, A = 235, Z = 92 Atomic no. Z = No. of protons = No. of electrons = 92 Therefore, a neutral atom of uranium 235 has 92 electrons. (ii) The number of protons = Z = 92. The number of neutrons = A Z = 235 92 = 143 (iii) Isotopes of an element have the same number of protons but different number of neutrons, since their mass numbers are different. (iv) Uranium-238, is an isotope of Uranium-235, therefore, it has the same number of protons as that in Uranium-235, i.e., 92. Answer 10. (a) Mass of water = 200 g, Mass of ice = 40 g, t1 = 50 C, Final temperature = t2 = ? Now, we can say that, mL (ice) + mC (of water formed from ice) = mC (water) (40 336) + (40 4.2 t2) = 200 4.2 (50 t2) 13440 + 168 t2 = 840 (50 t2) 168 t2 + 840t2 = 42000 13440 1008t2 = 28560 t2 = 28560 = 28.33 C 1008 (35) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES D Vaporization E am Ste r 120 110 100 90 80 70 60 50 40 30 20 10 0 10 Wa te Temperature ( C) (b) Ice B A Melting C Time (sec.) (c) In heavy nuclei, the number of protons is very high. As a result of this, electrostatic respulsive forces between them also become very high. The attraction due to nuclear forces cannot contain the protons together in the nucleus. The presence of large number of neutrons separates the protons, thereby weakening the repulsion between them. (36) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 7 SECTION - I (40 MARKS) Answer 1. (a) The rate of change of momentum of an object is equal to the net force acting on it and it is in the direction of force. (b) Given, F = 6000 N, S = 200 m, W = ? We know that, W = FS = 6000 200 = 12 105 J = 1.2 106 J = 1.2 106 107 erg ( 1 J = 107 erg) 13 = 1.2 10 erg (c) (i) M.A. is always less than 1 as effort arm is always smaller than the load arm. (ii) It is called speed multiplier as load moves through a larger distance as to compared to effort. (d) Given, Load arm = 0.1 m, L = 100 kgf Effort arm = 2 0.1 m = 1.9 m, E = ? Taking moments about the pivot, Effort Effort arm = Load Load arm E 1.9 m = 100 kgf 0.1 m E = 100 kgf 0.1 m = 5.26 kgf 1.9 m (e) When the ray strikes normal to the surface of the glass block. Answer 2. (a) v = (b) 2 86 m 2d = = 344 ms 1 2.5 t s 5 Real image Virtual image (i) A real image is formed by actual intersection of rays. (i) A virtual image is formed by apparent intersection of rays when rays are produced in backward direction. (ii) It can be taken on the screen. (ii) It cannot be taken on the screen. (c) Fisherman is in air. So light travels from a rarer medium to a denser medium, thereby it bends towards the normal which appears to be coming from a longer distance. Therefore, to the fish under water, the fisherman looks taller. (d) The full image will be formed of less brightness. Angle of deviation ( ) (e) m Angle of incidence (i) (37) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 3. (a) The part of the spectrum beyond the red end is called infrared and beyond violet end is called ultraviolet . Infrared radiations are used to for therapeutic purpose and ultraviolet radiations for sterilization. (b) The loudness of the sound heard by a plucked wire is increased due to the forced vibration of the sounding box. (c) Heat required by 100 g of ice to melt into water at 0 C, Q1 = 100 80 = 8000 cal Heat lost by 300 g of water in cooling to 0 C, Q2 = 300 (25 0) 1 = 7500 cal Since Q2 < Q1, it follows that the whole of ice will not melt and final temperature will be 0 C. (d) No, water in the beaker will be heated to 100 C but will not boil as required latent heat for boiling will not be transferred from the bath to the beaker as both are at same temperature, 100 C. (e) A heating wire must have high resistance and high melting point while a fuse wire must have high resistance and low melting point. Answer 4. (a) If bulb is rated 500 W, 240 V, it means that bulb is lighted on a 240 V supply and it will consume 500 W electrical power. (b) In BCD, resistances BC and CD are in series but combination of BC and CD is in parallel with BD, 1 R BCD Hence, = RBCD = 1 1 3 + = (2 + 2) 2 4 4 3 Now, RAB, RBCD and RDE is series. R = RAB + RBCD + RDE = 2 + 4 +2 3 16 = 5.33 3 (c) Magnetic field lines will be in the form of concentric circles with the conductor at centre, lying in a plane perpendicular to the straight conductor. (d) Fast moving cathode ray possess high K.E. which produce considerable heat when stopped by metal. Tungsten was first melted by focussing cathode rays. R = (e) (i) In the emission of an electron from the nucleus, a neutron is converted into a proton. Thus, the number n of neutrons decreases and number of protons increases, so the ratio decreases. p (ii) In the emission of positron from a nucleus, a proton is converted into a neutron. Therefore, n ratio p increases. SECTION - II (40 MARKS) Answer 5. (a) Given, mass = 50 kg, length = 10 m K.E. of the stone when it reaches the ground = P.E. of the stone at the top of the ladder K.E. = mgh = 50 9.8 10 K.E. = 4900 J Speed v of the stone just before hitting the ground, According to the law of conservation of energy, P.E. at the top = K.E. when it reaches the ground 4900 J = 1 50 v2 2 (38) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 4900 = 196 25 v = 14 ms 1 v2 = (b) Principle of levers : Load Load arm = Effort Effort arm (c) (i) M.A. = no unit (ii) Weight = newton (iii) V.R. = no unit (iv) Mass = kg (d) The V.R. is always equal to the number of strands of the string used to support the load. (e) Take initial direction of ball as direction of positive x-axis. pi = Linear momentum of ball before striking the wall = 0.1 kg 10 ms 1 = 1 kgms 1 pf = Linear momentum of ball after rebounding = 0.1 kg ( 8 ms 1) = 0.8 kgms 1 Hence, change in linear momentum = pf pi = ( 0.8 1) = 1.8 kgms 1 Time taken for this change = 0.2 s Magnitude of force exerted by the wall on the ball = Magnitude of change in linear momentum 1.8 kgms 1 = =9N Time taken 0.2 Answer 6. c vr > vv v Hence, red light travels faster than violet light. (a) r < v, as = (b) A lens can be considered to be divided into several blocks, each functioning as prism. Incident rays which are parallel to the principal axis, diverge after refraction through the prisms forming the lens. These diverging rays appear to originate from what is called the virtual principal focus F of concave lens. The distance between the principal focus and the optical centre is called focal length. F (c) (i) Totally reflecting prism : A right-angled, isosceles prism, made of glass ( = 1.5), has a ray incident normally on one face, is known as a totally reflecting prism. (ii) Ultraviolet rays are absorbed by ozone layer, thus preventing us from the harmful effects of these rays. (iii) No, we cannot produce a temperature of 10000 K even if we focus the rays by a big lens. Answer 7. (a) A tone is a sound of single frequency whereas a note is a sound made-up of several tones emitted by a musical instrument. (b) This is due to the fact that the over tones produced by the two sources may be different. In other words, the quality of sound produced by two instruments of same fundamental frequency is different. (c) Empty vessels contain air so that they can vibrate freely. But when they are filled with water, their vibrations a heavily damped. (39) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (d) Let the specific latent heat of fusion of ice be L Jg 1 Heat lost by water = 100 g 4.2 Jg 1 C 1 (15 3) C Heat lost by calorimeter = 50 g 0.42 Jg 1 C 1 (15 3) C Total heat lost by water and calorimeter = [100 4.2 12 + 50 0.42 12] J Heat gained by ice at 0 C in changing to water at 0 C = (15 L) J Heat gained by this water (formed from the ice) in raising its temperature to 3 C = [15 4.2 (3 0)] J Total heat gained = [15 L + 15 4.2 3] J Heat gained = Heat lost 15 4.2 3 + 15L = 50 0.42 12 + 100 4.2 12 189 + 15L = 252 + 5040 L = 5103 = 340.2 Jg 1 15 Answer 8. (a) Heat required to melt ice = mL = 0.5 kg 336 103 Jkg 1 = 168 103 J Heat required to raise the temperature of ice from 10 C to 0 C = m C = 0.5 kg 2.1 103 Jkg 1 C 1 [0 ( 10)] C = 10.5 103 J Heat required to raise the temperature of water from 0 C to 100 C = m C 1 = 0.5 kg 4.2 103 Jkg 1 C 1 100 C = 210000 J Heat required to evaporate the water = mL = 0.5 kg 2268 103 Jkg 1 = 1134 103 J Total heat required = (10.5 + 168 + 210 + 1134) 103 J = 1522.5 103 J = 1.5225 106 J (b) The electric fuse is a device used to limit the current in an electric circuit. It has following characteristics : (i) Fuse has high resistance and low melting point. (ii) Fuse is made-up of alloy of lead and tin. (iii) Fuse is always connected to the live wire in a circuit. (c) Echo is a sound heard after reflection from a large obstacle, after the original sound from the given source dies off. Conditions for the formation of echo are : (i) The size of the obstacle must be large compared to the wavelength of the incident sound. (ii) The minimum distance between the source of sound and the reflector should be at least 17 m. (iii) The intensity or loudness of sound should be sufficient so that it can be clearly heard after reflection. Answer 9. (a) (i) According to question, three resistances are in parallel. 1 1 1 1 = + + R3 R R1 R2 1 1 1 1 = + + 12 R 4 2 1 1 1 1 Now, = R 2 12 4 6 1 3 2 1 = = R=6 12 12 6 (ii) PQ in series with PR, R1 = RPQ + RPR R1 = 2 + 2 = 4 (40) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 4 2 8 4 = = 3 4+ 2 6 = RAQ + R2 + RRB R1 || RQR , R2 = As Now, RAB 4 16 +2= 3 3 (b) Here, P = 100 W, V = 230 V, t = 20 60 s RAB = 2 + Resistance of bulb, R = V2 (230 V)2 = = 529 P 100 W Now, supply voltage = 115 V Energy dissipated by the bulb = V2 (115 V)2 (20 60)s t = = 30000 J R 529 (c) N S + N S K Bar magnet Answer 10. Solenoid (a) Nuclear forces are : (i) Charge independent. (ii) Short range forces. (b) (i) < < (ii) > > (c) An element becomes radioactive when : (i) its atomic no. exceeds 82. (ii) there is an imbalance of protons and neutrons as compared to normal stable atom. (d) (e) 235 92 85 1 U 01n 148 57 La + 35 Br + 30 n + Energy Isotopes (i) Isotopes are atoms of same elements, same atomic no. but different mass no. (ii) Isotopes have same chemical properties. Isobar (i) Isobars are atoms of different elements having same mass no. but different atomic no. (ii) Isobars have different chemical properties. (41) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 8 SECTION - I (40 MARKS) Answer 1. (a) Given, m = 250 g = 0.25 kg, v = 20 ms 1, p = ? Momentum of the ball p = mv = 0.25 20 ms 1 p = 5.0 kg ms 1 (b) 1 mv2 2 K.E. = p = mv Now, p2 2m K.E. = (c) Class II lever : A nut cracker, wheel barrow, oar of a boat etc. Effort B C A Fulcrum Load (d) Given, m = 5 kg, h = 2 m, g = 9.8 ms 2 Now, Work done = mgh = 5 2 9.8 = 98 J (e) Let the new velocity be v . According to the question, 1 1 m v 2 = m2v 2 2 1 1 2 m1 v 2 = m v12 2 v 2 = v = m 2 v 81m v 9 Answer 2. (a) If r = 90 , the corresponding angle of incidence is called critical angle. (b) A spectrum is said to be pure if there is no overlapping of one colour on another and the different colours have well defined boundaries and are distinctly visible. (c) i r e d = incident angle = refractive angle = emergence angle = lateral displacement i r r e Em d er ge nt ra y (42) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (d) A (i) A F2 B 2F1 F1 B O (ii) A A F2 B B F1 AB = Object, A B = Image Answer 3. (a) Quartz prism is used for obtaining spectrum of uv light because ordinary glass absorbs the ultraviolet radiation. Infrared radiation (b) Sound waves (i) They travel with the speed of light, i.e., 3 108 ms 1. (ii) These are transverse waves. (c) Velocity = (i) They travel with the speed of 332 ms 1. (ii) These are longitudinal waves. Distance v t ,d= 2 Time Velocity of sound, v = 340 ms 1, t = 0.1 s 340 0.1 = 17 m 2 (d) The longer and thicker pin is the earth pin. It is long so that the earth connection is made first. It ensures the safety of the user because if the appliance is defective, the fuse will blow off. Earth pin is thicker so that even by mistake it cannot be inserted with the live or neutral connections of the socket. (e) Advantages of electromagnet : (i) An electromagnet can produce a strong magnetic field. (ii) The strength of the magnetic field of an electromagnet can be varied by varying the current or the number of turns in the solenoid. d= Answer 4. (a) (i) Use of laminated core to reduce the losses due to eddy current. (ii) Use of core material to reduce the hysterisis loss. (b) Radioactivity is the process of spontaneous disintegration of atomic nuclei with the emission of radiation from the nuclei of the atoms. (c) Background radiation means the radiation present in the atmosphere at any place, even when there is no known radioactive source in the vicinity. (d) If we calculate the effective resistance R between A and C, the resistances between A and B and B and C are in series which are in parallel with the resistance between A and C. 1 1 1 1 1 + = + = R (R AB + R BC ) R AC (3 + 3) 3 (43) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES A 1 1 1 1 = + = R 6 3 2 R =2 m = 1 kg, Q = 8.4 kJ = 8400 J, Given, T 1 = 90 C, T2 = 100 C, C = ? Heat released by copper ball = mC (T2 T1) 3 3 (e) Let the specific heat capacity of copper be C. 8400 = 1 kg C (100 90) C C B 3 8400 10 C = C = 840 J kg 1 C 1 SECTION - II (40 MARKS) Answer 5. (a) Work is the product of force and the displacement of the point of application of force in the direction of force. Work = Force Displacement of point of application of the force in the direction of force W =F S The S.I. unit of work is Joule. Power is defined as the rate of doing work, i.e., Power (P) = Work done Time taken S.I. unit of power is Watt. Energy of a body is defined as its capacity to do the work and measured by total quantity of work it can do. Expression for kinetic energy : Let a body of mass m which is initially at rest, cover a distance S on applying a force F due to which its velocity becomes v. The kinetic energy of the body = Work done K.E. = F S ...(i) = ma S ...(ii) But from third equation of motion, 2aS = v2 u2 v2 u2 2 Since the body is initially at rest, so u = 0 v2 aS = 2 aS = ...(iii) v2 in eq. (ii), we have 4 mv 2 K.E. = 2 1 K.E. = mass (velocity)2 2 Expression for potential energy : Let a body of mass m is lifted through a height h . The P.E. is equal to work done to lift the body against the force of gravity, i.e., P.E. = Work done = Weight of body Height = mgh Substituting aS = (44) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (i) Given, m = 2 kg, u = 0, h1 = 10 m, h2 = 5 m Gravitational potential energy of the ball = mgh P.E. = 2 9.8 5 = 98 J (ii) K.E. of the ball at 5 m height = P.E. at the top P.E. at halfway mark. K.E. = mgh1 mgh2 = mg (h1 h2 ) = 2 kg 9.8 (10 5) m = 98 Joule Answer 6. (b) (a) Force required to accelerate the gas out of cylinder is given by, F = Here, m = 12 kg, t = 1 m (v u ) t 1 5 min. = 90 s, u = 0, v = 90 = 25 ms 1 18 2 Avg. force on the gas = 12 kg (25 0) ms 1 = 3.3 N 90 s According to third law, Avg. force on gas = Equal reaction force exerted on the cylinder Avg. force exerted on the cylinder = 3.3 N. (b) The ratio of the useful work done by the machine to the total work put into the machine is called the efficiency of the machine. Given, V.R. = 25, L = 330 kgf, E = 0.165 KN = 16.5 kgf (i) M.A. = L 330 = = 20 16.5 E M.A. 20 100 = 100 = 80% V.R. 25 (c) When the tyre of the bicycle is pumped up, the work is done in compressing the air which increases the internal energy of the air. Hence, temperature of the pump barrel rises. (ii) Efficiency, = (d) Let the temperature of hot water be K. According to the question, Heat lost by water = Heat gained by water at 283 K m Cw ( 293) = 3m Cw (293 283) 293 = 3 10 Answer 7. = 30 + 293 = 323 K. (a) Angle of incidence at the air-glass interface i = 0 Angle of emergence at the glass-air interface e = 0 C 45 45 45 A 45 B (45) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) Dispersion of light is the phenomenon of splitting of white light into its constituent colours. A White light R Radiowaves Visible light UV - rays X - rays - rays Infrared radiations C B (c) G O Y B V I Microwave R Answer 8. (a) Let the required time be t min. So, the heat produced by heater in t min. = 7000t J Heat absorbed by water = m C T = 5 kg 4200 J kg 1 C 1 (47 22) C = 525000 J Heat generated by heater = Heat absorbed by water 7000t = 525000 525000 = 75 min. 7000 (b) Forced vibration : The free vibrations of a body eventually die out due to air resistance and friction. So, to keep the amplitude of the vibrations same, an external periodic force has to be applied. The vibrations which take place under the influence of external periodic force are called forced vibrations. For example, When a tuning fork is struck against a rubber pad and stem is placed against the top of the table, a loud sound is heard because the table is set into forced vibrations and vibrates with frequency of the applied periodic force of the tuning fork. Resonance : It is a special case of forced vibrations in which the frequency of the applied periodic force coincides with the natural frequency of vibrations of the body. This case of forced vibrations is known as the resonance. The body vibrates with a large amplitude and a loud sound is heard. For example, When a troop crosses a suspension bridge, the soldiers are asked to break steps because in steps, the frequency of the movement of soldiers becomes exactly equal to the frequency of the suspension bridge. Therefore, due to resonance, the bridge starts vibrating with a larger amplitude and it may even collapse. (c) A percussion instrument is one in which the tone is produced by percussing or tapping sharply or striking. e.g., drum and tabla. Factor : Size and nature of membrane tension. t = Answer 9. (a) Potential difference between two points is equal to the amount of work done in moving a unit charge from one point to another. p.d. (V) = Work done Quantity of charge transferred S.I. unit of p.d. is Joule/Coloumb or Volt. (b) Given, E = 1.08 V, r = 0.5 , V = 0.81, I = ? (i) Potential difference across the internal resistance = E V = 1.08 0.81 = 0.27 V (46) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) Current in the circuit, I = (iii) External resistance, R = E V 0.27 = = 0.54 A r 0.5 0.81 V = = 1.5 0.54 I (c) Principle of calorimetry : Calorimetry deals with the measurement of heat. The copper vessel which is generally used in such measurement is called calorimeter. When a hot body is mixed or kept in contact with a cold body, there is a transfer of heat from the body at a higher temperature to the body at lower temperature till both the bodies attain the same temperature, i.e., heat lost by the hot body is equal to the heat gained by the cold body provided there is no heat loss to the surrounding. This principle of calorimetry is based on the law of conservation of energy. Heat lost by hot body = Heat gained by the cold body. Answer 10. (a) Specific latent heat of vaporization of a substance is the amount of heat required to convert the unit mass of the liquid into vapours at its boiling point without any change in its temperature. Specific heat of vaporization (L) = Heat supplied Mass of the substance Q = mL Given, m = 5 g, Ts = 100 C, mw = 200 g, Ls = 2260 Jg 1, Tw = 10 C Specific heat capacity of water Cw = 4.2 Jg 1 C 1 Let the final temperature of water be t C According to the principle of calorimetry, Heat loss by steam = Heat gained by water msLs + ms Cw (Ts t) = mw Cw (t Tw) ms [Ls + Cw (Ts t)] = mw Cw (t Tw) 5[2260 + 4.2 (100 t)] = 200 4.2 (t 10) Solving this equation, we will get (b) (i) (ii) t = 25.32 C 1. -particles are fast moving electrons emitted from the nucleus of the atom. 2. -particles affect the photographic plate and cause fluorescence on striking a fluorescent material. 14 6 0 C 14 7 N + 1 e + v (iii) A cathode ray tube is used to convert signal into a visual signal on the deflecting plates. e.g., for checking the waveforms. (iv) Types of thermionic diodes : 1. The diode with directly heated cathode. 2. The diode with directly heated cathode. (47) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 9 SECTION - I (40 MARKS) Answer 1. (a) Given, m = 20 kg, retarding force F = 20 kgf = 20 10 = 200 N, u = 150 ms 1, v = 0, t = ? We know that, a = 200 N F = = 10 ms 2 20 kg m v = u + at t = v u 0 150 = = 15 sec. a 10 1 mv2 2 K.E. Kinetic energy (b) We know that, kinetic energy, K.E. = Mass (m) Velocity (v) (c) Inclined plane, M.A. = Mass of the body, m = (d) Length of inclined plane (l ) Height of inclined plane (h ) w 22 = g 9.8 m = 2.24 kg Weight of the particle at a point where g = 4.9 ms 2 w = 22 N 4.9 = 11 N 9.8 Mass will remain same = 2.24 kg. (e) Yes, when water falls from a height its potential energy changes to kinetic energy i.e., its velocity is maximum at the foot of the fall which suddenly reduces to zero. Hence, kinetic energy of water gets converted into heat energy which raises the temperature. Answer 2. (a) ( m = 10 10 6 g, v = 10 ms 1) Power of the bee, P = Fv = (mg) v = 10 10 6 9.8 10 = 9.8 10 4 W 2 We know that, 1 2 = 1 (b) w g = g w a = g a w (c) -rays, UV rays, yellow light, infrared rays, microwaves. (d) Yes, if refractive index of the material of the lens ( g) is less than the refractive index of the material of the medium ( m), the convex lens will behave like a concave lens. (48) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (e) (i) (ii) (iii) (iv) Angle of incidence. Angle of prism. Refractive index of material. Wavelength of light incident. Answer 3. = 1.1 m (a) Given, Displacement v = 300 Hz v = v = 300 Hz 1.1 m = 330 ms 1 0.8 m Distance 1.1 m (b) Resistances R1, R2, R3 and R4 are in series, so their equivalent resistance R = R1 + R2 + R3 + R4 Current I = E ( R1 + R 2 + R 3 + R 4 ) + r As resistances R1, R2 and R3 are in parallel, so 1 1 1 1 = + + R R2 R3 R1 R = R1R 2 R 3 R1R 2 + R 2 R 3 + R 3 R1 E E = R + r R 1R 2 R 3 +r + + R R R R R R 2 3 3 1 1 2 (c) Given, bulb 1, P1 = 110 W, V1 = 220 V I = For bulb 2, P2 = 55 W, V2 = 110 V, We know that, P = R1 =? R2 V2 R 2 2 R1 P2 V1 55 W 220 V = =2 R2 = P1 V2 110 W 110 V R1 : R 2 = 2 : 1 (d) Colour codes of the connecting leads : Live Brown Neutral Blue Earth Green or Yellow (e) The strength of an electromagnet can be increased by : (i) increasing the number of turns. (ii) increasing the current strength through the solenoid. Answer 4. Core (a) N Primary coil (NP) Secondary coil (NS) NP < NS (49) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) Final steady state temperature of the water in which ice is added will be lower that the other as ice will take 80 calg 1 of additional heat from the water in the glass to melt. (c) Amount of heat needed to convert 5 kg of water into steam = mwCw (T2 T1) + mwLv = 5 kg 4200 Jkg 1 C 1 (100 30) C + 5 kg 2250 103 J kg 1 = (1470 + 11250) 103 J = 12720 103 J = 1.2720 107 J (d) Electrons. Three properties of cathode rays : (i) Cathode rays are negatively charged particles. (ii) Cathode rays travel in a straight line. (iii) Cathode rays cause the emission of when they are stopped by certain metals. (e) 235 92 27 12 U + 01n Mg 27 13 Sb + 133 51 99 41 Nb + 4 01n +Q 0 Al + 1 e SECTION - II (40 MARKS) Answer 5. (a) According to the law of conservation of energy, energy can neither be created nor be destroyed. It can be transformed from one form to other form. Thus, the total energy of the system remains conserved. Given, m = 1000 kg, g = 9.8 ms 2, h = 150 m, t = 1s, P = ? According to the question, Power generated = 60% of energy gained from falling water. 60 mgh 100 t 0.6 1000 kg 9.8ms 2 150 m = 1s = 8.82 105 W. (b) According to the principle of lever, Power, P = Effort Effort arm = Load Load arm E AB = L AC AB L = E AC AB M.A. = AC M.A. < 1 AB < AC (c) Given, L = 400 kgf, E = 20 kgf, M.A. = ?, V. R. = ?, = ?, dE = 2.66, dL = 10 cm = 0.1 m Load (L) 400 kgf = = 20 Effort (E) 20 kgf d 2.66 V.R. = E = = 26.6 0.1 dL M.A. 20 = 100 = 100 26.6 V.R. = 75% M.A. = (50) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (d) Since the bodies have equal momentum, m1u1 = m2u2 u1 m2 u = m 2 1 Maximum height attained by the body is, For mass m1, u2 2g u12 h1 = 2g mass m2 , h2 = h = h1 h2 h1 h2 Answer 6. u12 2g u12 u12 2g = 2 = 2 u2 2g u2 2 m2 = 2 m1 Here, Load (L) Effort (E) M.A. = 8, L = 300 kgf, M.A. = (a) M.A. = E=? L 300 kgf = E E 300 kgf 300 kgf E = = M.A. 8 E = 37.5 kgf The man will pull the rope in the downward direction by applying an effort of 37.5 kgf while tension in the rope will pull him in the upward direction by a force of 37.5 kgf. Thrust of the man on the ground = Weight of the man Tension T = 80 kgf 37.5 kgf = 42.5 kgf. (b) Let x be the initial distance of the boy from the cliff and v is the speed of the sound. 2d v 2x v = 2.5 2( x 80) 2 = v v = x 80 t = 2.5 = 2x v ... (i) ... (ii) Comparing equations (i) and (ii), we have 2x = x 80 2.5 2x = 2.5x 200 200 0.5 x = 400 m Substituting the value of x in eq. (ii), we have Lens v = x 80 = 400 80 = 320 ms 1 x = Mirror (c) S * f (51) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 7. (a) Let the temperature of the mixture be . So, according to the principle of calorimetry, Heat lost by one substance = Heat gained by other substance m1C1 ( 1 ) = m2C2 ( 2) m1C1 1 m1C1 = m2C2 m2C2 2 = m1C1 1 + m2C 2 2 m1C1 + m2C 2 Hence, it is assumed that there is no loss in energy. (b) Let the equilibrium temperature of the calorimeter and its content s = t C According to the principle of mixtures, Heat lost by 10 g of water at 85 C = Heat gained by 10 g of ice 10 g 4.2 Jg 1 C 1 (85 ) C = (10 g 2.1 Jg 1 C 1) [0 ( 10)] C + (10 g 336 Jg 1) + (10 g + 4.2 Jg 1 C 1) ( 0) C 42 (85 ) = 3570 + 42 3570 42 = 3570 + 42 84 = 0 Hence, = 0 C (c) In very hot weather, to maintain the human body temperature at 37 C, due to rise in atmospheric temperature the body perspires. When the sweat evaporates, it takes the heat required for evaporation from the body thereby cools it and maintains the temperature. Answer 8. X Real depth w Apparent depth OC = Real depth O C = Apparent depth a (a) a C b = B Y i r O i Coin O (b) Mirage is an optical illusion caused due to total internal reflection of light in sandy deserts or in some extended surface like a black tarred road in very hot weather. In hot summer days, sandy land becomes very hot during the day time. The air in contact with the ground becomes hot and its density is reduced whereas the density of air at higher level remains unaffected. Therefore, the layers of air near the ground are warmer than the air at the upper level. Thus, the successive upper layers are denser than those below them. When the rays of light from the top of a tree travels from a denser to a rarer layer, it bends away from a normal and thereby at a stage the angle of incidence becomes greater than critical angle, the ray suffer total internal reflection and reach the eyes of the observer. Observer sees an inverted image of the trees and concludes that there is a water near tree. (c) For an echo of a simple sound to be heard, the minimum distance between the speaker and the walls should be 17 m. Hence, if the room is small, the echo cannot be heard. (d) Infrared radiations UV radiations (i) Spectrum of infrared radiations can be obtained by rock salt prism. (i) Spectrum of UV can be obtained by passing the radiation through a quartz prism. (ii) The infrared region extends (ii) The UV radiations have from wavelength range 7500 to 107 . wavelength range from 4000 to 135 . (52) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 9. Musical note (a) Noise (i) It is pleasant, smooth and agreeable to ear. (i) It is harsh, discordant and unpleasing to the ear. (ii) It is produced by periodic (ii) It is produced by an vibrations. irregular succession of disturbances. (b) When the motorcycle is driven at a particular speed, sometimes its frame is found to vibrate violently. This happens when the natural frequency of vibration of the frame becomes equal to the frequency of the engine is driven at that particular speed. E = 1.5 V (c) Given, E = 1.5 V, r = 1 , R1 = 4 , R2 = 20 (i) Current in the circuit I = E E = R+r (R1 + R 2 ) r I = 1.5 V 1.5 = = 0.06 A (4 + 20) + 1 25 R =4 R2 = 2 1 (ii) Potential difference across resistor R1 V1 = IR1 = 0.06 A 4 = 0.24 V Potential difference across resistor R2 V2 = IR2 = 0.06 A 20 = 1.20 V (iii) Potential difference across the cell, VC VC = V1 + V2 = 0.24 + 1.20 VC = 1.44 V (iv) Voltage drop across the internal resistance V = Ir = 0.06 A 1 = 0.06 V (d) In staircase wiring double pole type or dual switches are used to switch off or on from two different places in a staircase. a S1 c S2 b b c S1 S2 a b c a b c a Bulb Bulb A.C. mains A.C. mains S1 a b b S1 S2 c c S2 a b c a b c a Bulb Bulb A.C. mains A.C. mains To operate an electric bulb independently from two different positions, a pair of switches is required which has 3 terminals a, b and c. One is to connect the live wire from the supply line. If on operating the switch S1 and S2, terminals ab and ba, ac and ca are connected, the bulb gets lighted. On the other hand, if ab and ca or ac and ba are connected, the circuit will not be complete. Hence, bulb will not be lighted. Answer 10. (a) Amount of electrical energy consumed daily = Energy consumed by (a heater + lamp) = 500 W 2 h + 100 W 5 h = 1000 Wh + 500 Wh = 1500 Wh = 1.5 kWh (53) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Total unit of electrical energy consumed in the month of July = 1.5 kWh 31 days = 46.5 kWh Cost of electrical energy consumed in the month of July at the rate of ` 3.25 = 46.5 kWh ` 3.25 = ` 151.13 (b) A.C. generator is based on the principle of electromagnetic induction. b c N S a d B1 S1 B2 S2 Load N and S abcd S1 and S2 B1 and B2 = Electromagnet = Armature coil = Slip rings = Brushes (c) Horizontal plates Electron gun Vertical plates A1 F S A2 Screen Grid + H.T. (1000 V) Cathode ray tube works on the principle of (i) thermionic emission (ii) deflection of an electron beam (iii) fluorescence produced by the electrons on striking the screen. Electron gun consists of an electron emitting hot cathode which is indirectly heated by a filament F, to a low tension battery. The electrons are accelerated with the help of anodes A1 and A2 which are in H.T. (1000 V). The accelerated electron beam passed through the horizontal and vertical deflecting plates. It deflects the electron and the path of the electron beam becomes parabolic in both X and Y dimension. This deflected electron beam strikes the screen produced of zinc sulphide. Hence, it comes fluorescence on the screen. This is the working CRT. (d) When -particle is emitted from a nucleus of an element, the atomic number reduces by 2 and in case of emission of -particle, atomic number of new nucleus increases by 1. (54) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 10 SECTION - I (40 MARKS) Answer 1. (a) Force is a physical cause that changes or tends to change the state of rest or state of motion of the body. It can start or stop or change the speed and direction of body. Its S.I. unit is Newton, N. (b) When we walk on ground the force of friction between the sole of our shoe and the ground acts opposite to the direction of our motion. Due to the force of friction, the sole of shoe wore out. (c) Whenever a body is moving in a circular path with a uniform speed, its velocity is continuously changing due to change in its direction. The body thus possesses acceleration, which is called centripetal acceleration. The force which produces this acceleration is called centripetal force. (d) It is defined as the ratio of the velocity of the effort (VE) to the velocity of the resistance force (Load) (VL). Mathematically, V.R. = VE Velocity of effort = V L Velocity of load (e) The two reasons are : (i) There is always some friction between the pulley wheel and string. (ii) The string used in this system is never inextensible and massless. Answer 2. (a) Work done while taking a catch is negative because the force opposes the motion or tries to stop a moving body. W.D. = F S W.D. = ve (b) We know that, the rate of doing work is known as power. Power = Work Time Moreover, Work = Force Displacement Now, Power = ...(i) Force Displacement Time F S T P = Since S = Velocity = v t P = F .v (c) Yes, when a concave lens is placed in the path of a converging beam, it produces a real image. (d) Infrared rays, UV rays, X-rays and rays. (e) Infrared rays are not scattered much in fog. They can, therefore, penetrate to a greater extent through it. Answer 3. (a) Bats have special types of wings. When they fly they produce ultrasonic waves. These waves are received by the ears of the bat after they have been reflected by the object. The ears of the bats are so sensitive and trained that they not only get the information of distance of the obstacle but also that of the reflecting surface. (b) The subjective property of sound waves, related to its frequency, is known as its pitch. The name of the similar subjective property of light related to its wavelength, is the colour of light. (55) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES O hm ic Current (c) Conductors which obey Ohm s law (V I) are called ohmic conductors. For example, Silver. The I-V graph is as follows : Voltage (d) Nichrome is used for making wires to prepare heating coils. The reason is that its melting point and resistivity are high. (e) An electrical appliance is said to have a power of one Watt, if a current of one ampere flows through it under a potential difference of one Volt. Answer 4. (a) (i) It concentrates the lines of magnetic field through the galvanometer coil. (ii) It makes the field radial. (b) Mechanical energy gets converted into electrical energy. This phenomenon taking place is electromagnetic induction. (c) When ice melts at 0 C on heating, its volume decreases and converts into water at 0 C. Now, as water at 0 C is heated further it starts contracting till it has its minimum volume at 4 C. Thereafter, on heating, the volume of water begins to increase. (d) The specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of that substance through 1 C. Its unit is J kg 1 C 1. (e) The two characteristics are as follows : (i) Thermionic emitters must have high melting point. (ii) They should have a low work function. SECTION - II (40 MARKS) Answer 5. (a) Linear acceleration is defined as the rate of change of linear velocity whereas centripetal acceleration is defined as the rate of change of velocity, in circular motion. This acceleration acts towards the centre of the circular path along its radius. (b) (i) Let the mass of the scale be x g acting at centre C of the scale PQ. Taking the moments about the knife edge, we get Clockwise moment = Anticlockwise moment x gf 10 cm = 35 gf 15 cm 35 15 525 = = 52.5 g 10 10 (ii) The force on the knife edge = 35 gf + 52.5 gf = 87.5 gf (iii) The force of 87.5 gf will act in the downward direction. (c) No. of pulleys in the block and tackle = 5 M.A. = 4, L = 5N, E = ? Total resistance R = ?, = ? Now, we know that velocity ratio of a pulley block and tackle is equal to the total number of pulleys. V.R. = 5, M.A. = 4 Also, x = M.A. 4 100 = 100 = 80% V.R. 5 L M.A. = E % Efficiency = (56) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES E = Also, L 5 = = 1.25 N M.A. 4 L + R = V.R. E = 5 1.25 = 6.25 R = 6.25 L = 6.25 5 = 1.25 N Answer 6. (a) Given, h = 15 m, m = 1 kg The loss in P.E. of the block in moving down = mgh = 1 kg 9.8 ms 2 15 m = 147 J X 1 1 mv2 = 1 v2 2 2 K.E. = P.E. Gain in K.E. on reaching Y = But 15 m 1 1 v2 = 147 2 v2 = 294 v = (b) Y 294 = 17.14 ms 1 Initial mass = 5 kg Final mass = 10 kg Initial velocity = 10 ms 1 10 = 5 ms 1 2 1 Initial kinetic energy = mv2 2 1 = 5 10 10 = 250 J 2 Final velocity = 1 1 mv2 = 10 5 5 = 125 J 2 2 So, ratio of initial to final K.E. is 2 : 1. Final kinetic energy = (c) Given, air = 6600 = 6600 10 10 m, c = 3 108 ms 1, = 4 3 (i) From relation v = 3 108 c = = 4.54 1014 Hz 6600 10 10 air c From the relation = v Frequency of light in air vair = (ii) c 3 108 Speed of light in water v = = 43 v = 2.25 108 ms 1 (iii) Since the frequency of light remains unchanged in refraction, so vwater = vair = 4.54 1014 Hz. Now, speed of light in water v = water water water = v water = 2.25 108 4.54 1014 = 4.956 10 7 m or 4956 water = 6600 air = = 4950 43 water (57) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 7. A (a) L A F2 B F1 B O AB = Object, A B = Image (b) Both in the morning and in the evening, i.e., at sunrise and at sunset, when the sun is near the horizon, sunlight has to travel a much longer distance of the atmosphere. As a result, the light travelling from the sun loses most of the blue light due to scattering because its wavelength is small. The red light of long wavelength is scattered a little and reaches the observer. As a result, the sun appears to be red. (c) When the frequency of the forced vibrations is equal to the natural frequency of a body nearby, then the body vibrates with large amplitude. This phenomenon is known as resonance. Conditions of resonance : (i) The two bodies should be close enough for energy transfer. (ii) The frequency of the forced vibration must be equal to the natural frequency of the body. Answer 8. (a) (i) Alloys have high resistivity and high melting point as compared to pure metals, therefore, they are used as heating elements. (ii) In case of series arrangement, if one electrical appliance fails in the circuit the entire circuit will be switched off. Thus, series arrangement is not used in household electrical circuits. (iii) Resistance to the flow of current through a wire is due to the opposition offered by the atoms of the conductor to the electrons. If the area of cross-section is large the electrons have more space to move. This decreases the number of collisions taking place. Hence, resistance gets reduced, thus resistance is inversely proportional to the cross-sectional area of the conductor. (b) (i) A three-pin plug is used to supply electricity to any electrical appliance whose body is earthed. The third big pin of the plug helps us to do this earthing. The user of the electrical appliance is then protected against accidential electrical shocks. (ii) Thick wires are used in high power instruments like electric iron, electric heater, heating rod, electric radiator because for the given V1. P I Thus, the higher is the power of the electric appliance, larger is the current drawn by it and for given voltage V1. 1 R i.e., if resistance is low, power will be more and for low resistance, area of cross-section will be more. P Answer 9. (a) (b) (i) At end X - North polarity will be formed. (ii) The strength of the electromagnet formed can be increased by : 1. increasing the number of turns of the winding. 2. increasing the current through it. A.C. (i) Its magnitude changes continuously. (ii) Its direction reverse every half-cycle. (iii) It is more dangerous and fatal when someone gets a shock than D.C. (iv) A transformer can be used in A.C. D.C. (i) Its magnitude remains constant. (ii) Its direction always remains same. (iii) It is less dangerous and fatal, during any shock. (iv) Transformer doesn t work in D.C. (58) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (i) Ice has the highest specific latent heat of fusion of 336000 Jkg 1. Thus, when small pieces of hail melt, they absorb large amount of heat energy from the surroundings. This leads to a decrease in atmospheric temperature. (ii) Water has the highest specific heat capacity of 4.2 Jg 1 C 1. Thus, water absorbs large amount of heat from the roads, but its own temperature does not rise much. Thus, on the whole, the roads get cooled. Answer 10. (c) (a) (b) (c) (i) Work function of a metal is the minimum amount of energy required by an electron to just escape from the metal surface so as to overcome restraining forces. (ii) Rate of emission of thermions depends on the following factors : 1. The nature of the metal surface, i.e., lower the work function of the metal greater is the rate of emission of electrons from its surface. 2. The surface area of metal, i.e., larger the surface area of metal, the emission of electron is greater. (i) Radioactivity is the process of spontaneous disintegration of the atomic nuclei with the emission of particles from within the nuclei of atoms. (ii) If a radioactive substance is oxidized, no change takes place in the nature of its radioactivity. If the phenomenon of radioactivity be due to orbital electrons, it would have been affected but it is the property of nucleus. That is why the nature of radioactivity of the oxidized substance is unaffected. (i) The precautions that must be taken while handling a radioactive source are : 1. Handle radioactive materials with long lead tongs. 2. Put on special lead lined aprons and lead gloves. (ii) Two possible sources of background radiation are internal sources and external sources. Internal sources - Potassium (K-40), Carbon (C-14) and Radium present in our body. External sources - Cosmic rays from high altitudes and local terrestrial -rays from the radioactive rocks in the earth s crust. (59) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 11 SECTION - I (40 MARKS) Answer 1. (a) Centripetal force Centrifugal force (i) The force which acts on an object towards the centre of a circle to produce centripetal acceleration and so keeps the object moving in a circle. (i) The force which does not act on the object moving in the circle but it is equal and opposite force to centripetal force. It is a reaction force. (ii) It is a real force. (ii) It is not a real force. (b) A couple is always formed by two equal and opposite forces. The resultant force is therefore, always zero. Hence, no linear motion can be produced by a couple. (c) Torque is a vector quantity. Its direction is along the axis of rotation, i.e., perpendicular to the plane of rotation. (d) The M.A. of third order lever is always less than 1 because they have effort in between the load and fulcrum so that the effort arm is always shorter than load arm. These types of levers are used to gain in speed. (e) (i) Class I (ii) Class I (iii) Class III (iv) Class II Answer 2. (a) In order to balance the load on his head, the coolie applies a force on it in the upward direction, equal to its weight. His displacement is along the horizontal direction. Thus, the angle between force F and displacement is 90 . Therefore, work done W = FS cos . W = FS cos 90 = 0 ( cos 90 = 0) (b) A stretched bow posseses potential energy on account of a change in its shape. To shoot an arrow ; the bow is released. The potential energy of the bow is converted into kinetic energy of the arrow. (c) Optical fibre is a fine quality fibre of glass or quartz. Its diameter is of the order of 10 4 cm with refractive index of material being of the order of 1.7. Optical fibre is based on the phenomenon of total internal reflection. It is used in telephone and transmitting wires. (d) The focal length of a mirror does not depend upon the nature of the medium in which it is placed whereas the focal length of a lens depends upon the medium in which it is placed. Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change. (e) Danger signals are red because the red colour is scattered less on account of its longer wavelength. Since I 1 4 . The red colour can be seen from far of distance. Answer 3. (a) The intensity of the sound is the quantity of energy, passing per second, through a unit area, held perpendicular to the direction of sound propagation. (b) The subjective property of sound waves, related to its frequency, is known as its pitch. The name of the similar subjective property of light, related to its wavelength, is the colour of light. (c) When a given metalic wire is doubled on itself, its length is reduced to half, but its area of cross-section gets doubled. So, the resistance of the wire become one-fourth. (d) (i) Connected devices are switches and fuse to live wire. (ii) Fuse : It is a safety device which is used to limit current in an electric circuit. Switch : Its main function is either to connect or to disconnect an electric appliance in an electric circuit. (60) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (e) MCB stands for Miniature Circuit Breaker. This is a device that automatically switches off the mains supply if the current drawn exceeds the upper safe limit. It protects the entire circuit and appliances against short circuiting. Answer 4. (a) Whenever magnetic field through a coil is changed, an emf is produced in the coil. The emf lasts as long as the magnetic field through the coil changes. (b) (i) Electrical energy is converted into mechanical energy. (ii) It is based on the principle that a current carrying conductor when placed in a magnetic field experiences a force. (c) (i) It concentrates the lines of magnetic field, through the galvanometer coil. (ii) It makes the field radial. (d) Specific heat capacity : The specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of that substance through 1 C. Its S.I. unit is Jkg 1 C 1. (e) Alpha-particle are nuclei of helium He4. When it absorbs electrons it will become a singly charged helium atom. On absorbing two electrons, it will become a helium atoms. SECTION - II (40 MARKS) Answer 5. (a) It is inherent property of a body due to which it opposes the change in its state of rest or motion. It depends only upon the mass of the body. (b) (i) The moments of the weight of the metre rule about 0 cm mark = 10 gf 50 cm = 500 gfcm clockwise. (ii) To make it horizontal by applying the least force, the force must be applied at the farthest point i.e., at B. Let F be the force applied upwards. Then taking moments about 0 cm mark F 100 = 50 10 F = 5 gf A minimum force of 5 gf must be applied at B in the upward direction. (c) Given, E = 80 N, dE = 0.15 m, g = 10 ms 2, L = 10 10 = 100 N, dL = 10 cm = 0.10 m, M.A. = ?, V.R. = ?, work input = ?, work output = ? By definitions, 100 Load = = 1.25 80 Effort d 0.15 (ii) V.R. = E = = 1.5 dL 0.10 (i) M.A. = (iii) Work input = E dE = 80 N 0.15 m = 8 1.5 = 12 J (iv) Useful work output = L dL = 100 N 0.10 m = 10 J output 10 (v) Efficiency, = = = 0.833 = 83.3% input 12 Answer 6. (a) Given, Mass Height K.E. v = 50 kg = 10 m =? =? (61) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES K.E. of the stone when it reaches the ground = P.E. of the stone of the top of the ladder. mgh = 50 9.8 10 Energy when it reaches the ground = 4900 J Speed v of the stone just before hitting the ground. Initial velocity, u = 0 Final velocity, v = ? Now, we know, h 10 m, g = 9.8 ms 2 v2 u2 = 2gh v2 0 = 2gh 2gh v = (b) v = 1 2 9.8 10 = 14 ms . Initial mass = 5 kg Final mass = 10 kg Initial velocity = 10 ms 1 Final velocity = 10 2 = 5 ms 1 Initial K.E. = 1 2 mv2 = 1 2 5 10 10 = 250 J Final K.E. = 1 2 mv2 = 1 2 10 5 5 = 125 J So, the ratio of initial to final K.E. is 2 : 1. (c) Since the ray is diverged, therefore the lens is a concave lens. The image formed is virtual, erect and diminished. A A B B Answer 7. (a) In a prism, the refraction of light takes place at two slant surfaces. The dispersion of white light occurs at the first and surface of prism where its constituent colours deviated through different angles. At the second surface, these split colours suffer only refraction and they get further separated. But in a rectangular glass block, the refraction of light takes place at the two parallel surfaces. At the first surface, although the white light splits into its constituent colours on refraction, but these split colours on suffering refraction at the second surface emerge out in form of a parallel beam, which give an impression of white light. (b) Here, t1 = 5 s. Let d be the distance of the hill from the man and v be the velocity of the sound. Then t = 2d v 2d 5v or d= ...(i) v 2 When the man moves, through a distance 310 m towards the wall, then d = d 310 and t = 3 s 5 = 2( d 310) v 3v (d 310) = 2 Subtracting (ii) from (i), we have Hence, 3 = ...(ii) 5v 3v 2v = =v 2 2 2 v = 310 ms 1 d d + 310 = (62) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 8. (a) The circuit diagram is given alongside. If total resistance between A and B is R, then 1 5 1 1 = + = 2 3 R 6 6 R = = 1.2 . 5 Now, total resistance of the circuit is RT = R + 0.3 = 1.2 + 0.3 = 1.5 Hence, current in the circuit is V 1.5 I = = =1A 1.5 RT (b) (i) The resistances 1 and 2 are in series. 1 +2 =3 This combination is in parallel with the resistor of 3 . 1 1 1 or = + 3 3 R 3 R = = 1.5 2 Total resistance in the circuit R = R + v R = (1.5 + 0.5) = 2.0 3 B A 2 + Key 1.5 V r = 0.3 V 2 = = 1 A. 2 R (iii) emf of the cells = 4 V Total internal resistance of the cells = 1 (ii) I = 4 = 1.6 A 2.5 (c) Given, P1 = 40 W, I1 = 0.182 A, V1 = 220 V P2 = 60 W, I2 = 0.272 A, V2 = 220 V Also, V = 220 V Ammeter reading = V12 220 220 R1 = P = = 1210 . 40 1 V22 (220) 2 = = 806.7 . Now, R2 = P2 60 When the lamps are connected in series, then the total resistance becomes RS = R1 + R2 = 1210 + 806.7 = 2016.7 Hence, current through the series combination and hence, through each lamp. I = V 220 = = 0.109 A RS 2016.7 Primary winding Answer 9. (a) The labelled diagram is given alongside. 200 V 15 V Secondary winding The name of the device is step down transformer. (63) Core e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) (i) Every 1 kg of ice at 0 C absorbs 336000 J of heat energy to form water at 0 C. As ice can extract 336000 J of heat energy, more than water at 0 C, therefore, it cools the bottled drinks more effectively. (ii) Water has the highest specific heat capacity of 4.2 J g 1 C 1 and its temperature does not rise beyond 100 C. Thus, it can absorb large amount of heat from the working engine, when is then radiated through the radiator. This effectively controls the temperature of the engine thereby increasing its efficiency. (c) Heat Temperature (i) Heat is the energy of transit. (i) Temperature is the fundamental quantity which determines the direction of flow of heat. (ii) Its S.I. unit is Joule. (ii) Its S.I. unit is Kelvin. Answer 10. (a) (i) The atomic no. is 90 2 = 88 and the mass no. is 223 4 = 219 (ii) The required scheme of decay is as below : 223 90 219 Th 88 Ra + 4 2 He + Energy (iii) The difference in the mass of the reactants, and the products is converted into energy according to Einstien s mass energy relation E = mc2. (b) Cathode Cylindrical anode S 6V Low tension battery Electron beam 1000 V Horizontal plates (i) When a hotter filament is used, we have an increase in the number of electrons emitted. (ii) When the anode voltage is increased, we have an increase in the energy of the emitted electrons. (64) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 12 SECTION - I (40 MARKS) Answer 1. (a) One newton is that force which when acting on a body of mass 1 kg, produces an acceleration of ms 2 in it. S.I. unit of force is Netwon (N) and C.G.S. unit is dyne. 1 N = 105 dyne. (b) We know that, Moment of force = Force Perpendicular distance. For a given force, the torque will be maximum, if perpendicular distance is maximum. The force should be, therefore, applied in a direction perpendicular to the plane of the gate. (c) Weight of the body of mass 12 kg = mg = 12 10 ( m = 12 kg, g = 10 ms 2) = 120 N In gravitational units Weight = 12 kgf Force = mg = 120 N (d) (i) Rotating the wheel of a cycle by applying force on the pedal with the help of chain. (ii) Lifting a car with a jack. (e) In an ideal machine, the work output is equal to the work input. The efficiency of an ideal machine is 100%. Efficiency = M.A. V.R. In an ideal machine, ratio = 1 : 1. Answer 2. (a) We know that rate of doing work is known as power. Work Time Moreover, Work = Force Displacement Now, equation (i) becomes Power = Power = P = Since ...(i) Force Displacement Time F S t S = Velocity = v t P = F .v . (i) Work done is positive as the bucket moves in the direction of force applied by the man. (ii) Work done is negative, as frictional force acts opposite to the direction of motion of the body. (c) If ray is incident perpendicular to the surface separating the two media, it passes undeviated, i.e., i = 0, r = 0 and angle of deviation is also zero. (d) Hold the given piece of glass over some printed matter. (i) If the letters appear magnified, the given piece is a convex lens. (ii) If the letters appear diminished, the given piece is a concave lens. (e) The image of an extended source, as formed by a concave lens, has the following characteristics : (i) Smaller in size than the source itself. (ii) A virtual image and an erect image. (b) (65) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 3. (a) X-rays and -rays both have frequencies higher than that of violet light. X-rays are used in X-ray crystallography. -rays are used in radio theraphy. (b) Loudness depends upon the square of the amplitude of the wave, therefore, when the amplitude of wave is doubled the loudness becomes four times. (c) Nichrome is used for making wires to prepare heating coils. The reason is that its melting point and resistivity are high. (d) Earthing is a safety device which is used to prevent the shocks due to short-circuiting and leakages. Earthing means that the metal body of the appliance is connected to a thick copper wire which is buried deep in the ground. This prevents the user from getting shocks. (e) The rate of doing work or the rate of dissipation of energy is called electric power. It is denoted by P, i.e., P = W t = V2 R Its S.I. unit is Watt. Answer 4. (a) The loss of energy along the transmission lines is proportional to the square of current. Hence, the transmission of energy is economical at high voltage and low current. (b) The force F depends on : (i) Current through the conductor, I . (ii) Strength of Magnetic field, B . (iii) Length of the conductor, l . It is expressed as F = B.I.l. (c) (i) It concentrates the lines of magnetic field through the galvanometer coil. (ii) It makes the field radial. (d) The specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of that substance through 1 C. Its unit is J kg 1 C 1 . (e) Water is often used to cool the radiators of the engines in motorcars and trucks etc., as it can draw large amount of heat without much rise in its own temperature. SECTION - II (40 MARKS) Answer 5. (a) Linear acceleration is defined as the rate of change of linear velocity whereas centripetal acceleration is defined as the rate of change of velocity in circular motion. This acceleration acts towards the centre of the circular path along its radius. (b) Consider two wheels A and B having NA and NB as the number of teeth in them respectively to form a gear system. The wheel which is closer to the source of power is called the driver wheel and the other wheel is known as driven wheel. If the driven wheel B is rotated in the anticlockwise direction, the driving wheel A rotates in the clockwise direction. Let A be the driver wheel and let it rotate in the clockwise direction as shown. In this case, the driven wheel is slow but the driving wheel is fast. Let in a given time the wheel A complete A revolutions and wheel B complete B revolutions. Since the total number of teeth passing through a given point is the same for both driver and driven wheel. NA A = NB B B NA = ...(i) NB A Since width of the teeth is the same in both the wheel. 2 rB 2 rA = , NB NA B A where rA and rB are the radius of the two wheels A and B. NA rA = ...(ii) NB rB (66) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES From (i) and (ii), we get Clearly, if NB > NA, then rB > rA and A > B. (c) Given, Load = Weight of water, L = 6.0 10 N, Effort, E = 70 N. Now, M.A. = Answer 6. L 60 = = 0.86 E 70 (a) The boy has to overcome the force of gravity. Hence, force of gravity on the boy F = mg = 40 9.8 = 392 N. Total distance covered S = 50 10 = 500 cm = 5 m. (i) Work done by the boy in climbing = force distance W = 392 5 = 1960 J W 1960 = = 392 W 5 t (b) Here, mass of water lifted = 500 kg, S = h = 80 m, t = 10 s. (i) Work done by the pump = F S = mgh = 500 10 80 J = 4 105 J (ii) Power developed = 4 105 J = 40 kW. 10 Useful power = 100 Input power (ii) Power at which it is working = (iii) 40 = 40kW 100 Input power 40 100 = 100 kW 40 (c) When a glass block is placed over a mark on a paper, the mark seems to rise due to refraction of light from denser medium to rarer medium at the plane surface separating the two media. Input power = A B I Answer 7. D (a) O C Glass slab B B F2 A F1 A O AB = Object, A B = enlarged, upright and virtual image. The object should be placed in between the optical centre and focus of the lens. (b) Both in the morning and evening i.e., at sunrise and sunset, when the sun is near the horizon, sunlight has to travel a much longer distance of the atmosphere. As a result, the light travelling from the sun loses most of the blue light due to scattering because its wavelength is small. The red light of long wavelength is scattered a little and reaches the observer. As a result, the sun appears to be red. (67) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) (i) The quality of a give note is determined by the overall effect of the harmonics present in it. The harmonics are multiples of the fundamental or basic frequency of the note . Depending on the conditions under which vibrations are taking place, sometimes we get one set of harmonics and sometimes another set. The quality of the two notes will, therefore, differ even though their fundamental frequencies may be the same. (ii) Pure note is a sound or vibration of single frequency. Answer 8. (a) (i) From the fig., it can be seen that R2 and R3 are in parallel, therefore, the total resistance is 1 1 1 5 1 1 = + = + = 8 12 24 R3 R2 RP RP = 24 = 4.8 . 5 Also, R1 and RP are in series. Equivalent resistance of the circuit R will be R = RP + R1 = 4.8 + 7.2 =12 . (ii) The p.d. in the circuit is 6 V. Let the current in the circuit be I and the equivalent resistance is 12 , then from V = IR. 6 = I 12 I = 0.5 A (iii) Since R1 = 7.2 and I = 0.5 A Hence, potential difference across R1 is V1 = IR1 = 0.5 7.2 = 3.6 V Also, the p.d. of 6V 3.6 V = 2.4 V is maintained across the parallel combination of R2 and R3. Hence, p.d. across R2 = 2.4 V and across R3 = 2.4 V. (b) Heat capacity of calorimeter mC = 50 J C 1 Mass of liquid M = 1.0 kg Specific heat capacity of liquid C = 450 Jkg 1 C 1 Rise in temperature (T2 T1 ) = 10 C Current I = 2.0 A t = 10 min = 10 60 s = 600 s Total heat energy required = Heat energy required by the calorimeter + Heat energy required by the liquid = (mC + MC ) (T2 T1) = [50 + 1.0 450] 10 = 5000 J (i) If R is the resistance of coil, heat energy produced in the coil = I2Rt = (2)2 R 600 = 2400 R Joule If there is no loss of heat, Heat energy required = Heat energy produced 5000 = 2400 R R = (ii) 5000 = 2.08 2400 p.d. across the coil, V = IR = 2 2.08 = 4.16 V Assumption : No loss of heat. (68) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 9. Electromagnet Permanent magnet (i) They are temporary magnets and produce magnetic field only for the time current passes through them. (i) Its magnetic field is permanent except when rough handled or heated. (a) (ii) The strength of the magnetic field can be (ii) Once made its magnetic field strength made large by increasing the no. of turns cannot be changed. and increasing the current through it. (iii) During its operation, its strength can be (iii) During its operation, its strength cannot be varied. varied. (iv) Its polarity can be reversed during its (iv) Its polarity cannot be reversed. operation. (v) It has large no. of applications. (v) It has only a few applications. (b) The specific heat capacity of water is very high i.e., 4200 Jkg 1 C 1. It is about 5 times as high as that of sand. So, water takes a very long time to get heated up and an equally long time to get cooled. Thus, there is a large difference in temperature between the land and the sea due to which land and see breezes are formed. (c) When we make the base thick, the capacity of cooking pass, its thermal capacity increases. Thus, it can import sufficient energy for cooking even at a low temperature. Answer 10. (a) Mass of the hammer = 1.0 kg Velocity of the hammer = 50 ms 1 1 1 mv2 = 1 (50)2 = 1250 J 2 2 Kinetic energy transferred to the iron nail = Half of K.E. possessed by the hammer Therefore, K.E. of the hammer = Therefore, K.E. transferred = 1250 = 625 J. 2 mass of the nail (m) = 200 g = 200 10 3 kg Specific heat capacity of the nail (C) = 452 Jkg 1 C 1 Rise in temperature of the nail ( T) = ? By the expression of specific heat, we have Q = m C T Now, Therefore, (b) rise in temperature is T = Q 625 = = 6.9 C 200 10 3 452 m C Effect of electric field Effect of magnetic field (i) The electric field exerts a force on the electrons in a direction opposite to the electric field. (i) It exerts a force on the electron beam perpendicular to both the magnetic field and direction of motion. (ii) The path of electron beam is parabolic. (ii) The path of the electron beam in the magnetic field is circular. (iii) It changes both the K.E. and momentum of the electrons. (iii) It only changes the momentum of electrons and not K.E. (iv) Electric field exerts a force whatever be the direction of the motion of electrons. (iv) Magnetic field exerts a force on the electrons moving in the direction different from that of magnetic field. (c) X-rays are produced when fast moving electrons strike a metal of high atomic weight and high melting point. (69) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 13 SECTION - I (40 MARKS) Answer 1. (a) Water comes out of the cloth in the form of fine drops. Initially, the cloth along with water drops remains at rest. When the cloth is shaken, it is set into motion, but the drops of water try to maintain the inertia of rest. Thus, the drops are separated from the cloth. (b) Magnitude of torque = Force Perpendicular distance of the line of action from the turning point. Unit of torque = Unit of force Unit of distance = Newton metre = Nm (c) Mass = 60 kg K.E. = 750 J Velocity = ? 1 mv2 2 1 750 = 60 v2 2 v2 = 25 K.E. = v = 1 25 = 5 ms . (d) It is defined as a sloping surface which behaves like a simple machine having mechanical advantage greater than one. 1 1 M.A. = = sin h (e) A lever works on the principle of moments according to which moment of load about the fulcrum is equal to the moment of effort about the fulcrum, i.e., Load Load arm = Effort Effort arm Answer 2. (a) A total reflecting prism is a right-angled prism made of glass with the other two angles each equal to 45 . It is used in periscope and binoculars. (b) (i) Light must travel from a denser medium to a rarer medium. (ii) The angle of incidence must be greater than the critical angle for the pair of media. (c) (i) A real image can be obtained on a screen but a virtual image cannot be obtained on a screen. (ii) A real image is formed by the actual intersection of refracted rays while a virtual image is formed when the refracted rays do not actually meet but they meet when they are produced backwards. (d) (i) Ultraviolet radiations are of shorter wavelengths as compared to visible light. (ii) Ultraviolet radiations are chemically more active than that of visible light. (e) Infrared radiations are used for photography in fog because they are not scattered by fog and can penetrate appreciably through it. Answer 3. (a) The amplitude of a wave is defined as the maximum displacement of medium particles on either side of its mean position. The S.I. unit is metre (m). (b) Resonance is a special case of forced vibration, when the body vibrates under the influence of an external periodic force, with a very large amplitude. (c) The resistivity of a substance is the resistance of a wire of unit length and unit area of cross-section. Its S.I. unit is Ohm m. (d) The various appliances as well as circuits are connected in parallel so that they work at the same voltage and also if there is a short circuiting in one distribution circuits, its fuse will blow off without affecting the electric supply in the other circuits. (70) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (e) The wire marked 16 A will be thicker as its resistance should be low so that heat produced is small when heavy current is passed through it. Answer 4. (a) When an insulating conducting wire is wound in the form of a cylindrical coil and its length is greater than the diameter, it behaves like a magnet on passing direct current through it. This is called a solenoid. (b) The causes are as follows : (i) Heating in the coils. (ii) Eddy current in the core. (iii) Hysteresis loss in the core. (iv) Magnetic field link loss. (c) (i) Whenever there is a change in magnetic flux linked with a coil, an emf is induced. (ii) The magnitude of the emf induced is directly proportional to the rate of change of magnetic flux linked with the coil. (d) Since specific heat of water is high, the fall in temperature of the juice bottles immersed in water will be lesser. As a result, the juice does not freeze. (e) (i) Both of them are electromagnetic waves. (ii) Both the gamma X-rays travel with speed 3 108 ms 1 in vacuum or air. SECTION - II (40 MARKS) Answer 5. W = mgh = V gh = 30000 10 3 103 9.8 45 [ Mass (m) = volume (V) density ( )] (a) W 13230000 = = 22050 W = 22.05 kW t 10 60 (b) A renewable source of energy must be capable to provide adequate amount of useful energy in a convenient form over a long period of time. The energy harnessed by flowing water, wind, tides and biogas, are directly or indirectly derived from solar energy and can be used forever as the earth continues to receive heat and light from the sun. It is a renewable form of energy and is inexhaustible. It doesn t cause pollution. (c) (i) Given, V.R. = 5, = 80%, L = 10 kgf, h = 2 m, t = 10 s, g = 10 ms 2. P = M.A. = ?, P = ?, = M.A. 100 5 80 M.A. = 5=4 100 So, T 80% = (ii) M.A. = M.A. V.R. L E E T T P = L = 5T L 10 10 E = = = 25 N M.A. 4 T 25 N 2 m E h = =5W 10 s t Answer 6. (a) B (i) L Load, L B X A (ii) Convex lens (iii) Simple microscope A O X L1 (71) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) 45 30 30 20 60 60 30 30 60 45 75 75 (i) (c) 70 (iii) (ii) (i) Radiowaves, infrared, visible light, x-rays, -rays. (ii) 3 sin 60 2 = 1 sin 45 2 a = l = 3 2 = 60 30 45 1.5 = 1.225 Answer 7. (a) (i) A Displacement B Time (ii) When an artist plays a guitar, the strings of the guitar execute forced vibrations. (iii) Conditions to hear a distinct echo : 1. The echo should reach the ears at least 0.10 s after the original sound. 2. The reflecting surface in air should be at a minimum distance of 17 m from the listener. (b) According to the question, Let the distance between John and hill be x m, then 2x 5 2( x 320) v = 3 v = and ...(i) ...(ii) Equating (i) and (ii) 2x 5 3x 2x x 2( x 320) 3 = 5x 1600 = 1600 = 800 m = 2x 2 800 = = 320 ms 1 5 5 (c) A - Ammeter, B - Voltmeter, C - Unknown resistance, D- Variable resistance, E - Battery, F - Key v = I V (72) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 8. (a) (i) If key K is closed, 4 and 8 resistances are in parallel. Req. = I = 4 8 8 = (4 + 8) 3 30 V 10 V = = = 3.75 A 8 8 R eq. 3 (ii) If key K is open, 8 resistance will be in the circuit. Therefore, I = V 10 = = 1.25 A R 8 Energy consumed = n Power (in kW) Time (hrs) (b) Energy consumed by 6 bulbs = 6 100 8 = 4.8 kWh 1000 Energy consumed by a heater = 1 2 0.5 = 1.0 kWh Energy consumed by 5 fans = 5 50 6 = 1.5 kWh 1000 Total Energy consumed = (4.8 + 1.0 + 1.5) kWh = 7.3 kWh Cost of energy consumption for a month of 30 days = 7.3 kWh ` 3.50 30 days = ` 25.55 kWh 30 days = ` 766.50 (c) (i) 1. When magnet is dropped into the coil sudden deflection is observed in the galvanometer. 2. On increasing the number of turns of the coil, the deflection of the pointer will increase. (ii) As the magnet approaches the coil, the magnetic field lines linked with the coil increases, an emf is induced in the coil which would oppose motion of north pole of the magnet towards the coil. It is because north pole will be developed at this end of the coil to oppose the motion of the magnet. The direction of current flowing through the coil, when the magnet is dropped, will be anticlockwise. When magnet will be inside the coil, magnetic field lines linked with the coil will be constant so the induced emf will be zero. (iii) Lenz s law : It states that in case of electromagnetic induction, the direction of induced emf is such that it always opposes the motion which produced it. Answer 9. (a) According to the problem and the given parameters : Amount of heat given out while 400 g of water at 30 C is cooled and converted into ice at 2 C. = (0.4 kg 4200 Jkg 1 K 30 K) + (0.4 kg 336000 Jkg 1) + (0.4 kg 2100 Jkg 1 K 2K) = 50400 J + 134400 J + 1680 J = 186480 J (b) Assuming no loss of heat, according to the principle of mixture, Heat energy given out by body 1 = Heat energy taken by body 2 m1C1 (t1 t 3 ) = m2C2 (t3 t2) m1C1t1 m1C1t3 = m2C2t3 m2C2t2 m1C1t1 m2C2t2 = m2C2t3 + m1C1t3 t3 = m1C1t1 + m2 C2 t2 (m1C1 + m2 C2 ) (73) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) b c N S a d Electromagnet B1 B2 S1 S2 Load abcd = armature Field magnets : To produce strong magnetic field. Carbon brushes : To pass current from armature coil to the external resistance. Answer 10. (i) Work function of a metal is the minimum amount of energy required by an electron to just escape from the metal surface so as to overcome restraining forces. (ii) Rate of emission of thermions depend on the following factors : 1. The nature of the metal surface i.e., lower the work function of the metal greater is the rate of emission of electrons from its surface. 2. The surface area of metal i.e., larger the surface area of metal emitting the electrons greater is the rate of emission of electrons. (b) (i) Radioactivity is the process of spontaneous disintegration of the atomic nuclei with the emission of the particles from within the nuclei of atoms. (ii) If a radioactive substance is oxidized, no change takes place in the nature of its radioactivity. If the phenomenon of radioactivity be due to orbital electrons it would have been affected but it is the property of nucleus. That is why the nature of radioactivity of the oxidized substance is unaffected. (c) A small amount of nuclear fuel (Uranium - 235) can produce a tremendous amount of energy. Once the nuclear energy is loaded into a nuclear power plant, it continues to release energy for several years and can carry out the fission reaction in a controlled way to produce electricity. A high standard of protection for the persons working with the power plant and also for the environment is needed. (a) (74) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 14 SECTION - I (40 MARKS) Answer 1. (a) The force applied to a body is directly proportional to the rate of change in momentum which it produces in the body. (b) The power depends on the time for which the work is done, while the amount of work done does not depend on time. (c) Zero, when a body completes a circular path its displacement is zero and hence, the work done by the body is zero. (d) (i) Mechanical advantage = Velocity Ratio. (ii) Mechanical advantage < Velocity Ratio. (e) (i) Yes, e.g., Centre of gravity of a hollow sphere lies at its centre where there is no material. (ii) Given, F = 525 V, and a = 3.5 ms 2. Applying the formula, F = ma, we have m= 525 F = = 150 kg. 3.5 a Answer 2. (a) A lens of short focal length has more power. (b) The three characteristics of image formed by a concave lens : (i) The image will be virtual. (ii) The image will be upright. (iii) The image will be diminished. 4 a = (c) Given, w 3 1 Applying the formula = a , we have w 1 3 = . 4 4 3 (d) The optical fibres are mainly used in long distance communication of light signals, in testing the internal parts of human body, and in sending the laser beam inside the body. (e) Angle of incidence at face AC = 45 . Angle of incidence at face BC = 0 . w a = Answer 3. (a) The velocity of a wave in a medium depends on the density and the elasticity of the medium. (b) (i) UV rays : They are used in producing vitamin D in plants. (ii) X-rays : They are used for diagnostic purposes for detecting bone fractures. (c) f = 250 Hz But T = 1 1 = = 0.004 sec. f 250 (d) Resonance : It is a special case of the forced vibration when frequency of applied periodic force is equal to the natural frequency. Forced vibration : The vibration which take place under the influence of an external periodic force is called the forced vibration. (e) The subjective property related to frequencies is called pitch. The subjective property related to wavelength is called colour of light. (75) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 4. (a) The reason for drop in temperature is that as the common salt melts, it takes heat from melting ice itself. Thus, the temperature of ice drops. (b) 10125 = 4.5 L (c) L = 10125 = 2250 Jg 1 4.5 (i) (ii) 1. 2. Current Volt (d) Electrical to mechanical energy. A D.C. motor works on the principle that when a current carrying coil is placed in a magnetic field, it experiences a torque which tends to rotate it. (e) The speed of -rays is equal to the speed of light in vacuum, i.e., 3 108 ms 1. -rays are not deflected by electrostatic field as they are electromagnetic waves having no charge. SECTION - II (40 MARKS) Answer 5. (a) (i) Velocity ratio, Distance moved by the effort Distance moved by the load 5m = =1 5m W 400 M.A. = = = 0.89. 450 P V.R. = (ii) (iii) Efficiency, M.A. 100 % = V.R. 0.89 100 % = 89% = 1 (iv) Energy gained by the load = Load Displacement = (400 5) J = 2000 J Effort Displacement (v) Power developed by the boy = Time 450 5 = = 225 W. 10 (vi) To change the direction of the force applied in a convenient direction. (b) Let the body is brought to rest by a retarding force F after travelling a distant S. From relation, v2 = u2 + 2aS. v2 2S Retarding force F = ma retardation a = =m mv 2 v2 = 2S 2S Work done, W =F S Hence, mv 2 1 = S= mv2 2S 2 1 kinetic energy = mv2 2 (76) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) The acceleration produced on a body of given mass is directly proportional to force applied on it. The acceleration produced by a given force is inversely proportional to the mass of a body. Answer 6. Angle of deviation (a) Angle of incidence First, the angle of deviation produced by a prism decreases with the increase in angle of incidence. For a certain angle of incidence, the angle of deviation becomes minimum. On further increasing the angle of incidence, the angle of deviation increases. (b) Here, i = 60 , r = 45 sin i According the Snell s law, = sin r or = sin 60 3 2 = = sin 45 1 2 1.732 3 2 = = 1.22. 1.414 2 (c) If a blackened bulb is moved beyond the red end, there is a rapid rise in temperature. This shows existence of some kind of radiation producing the heating effect beyond the red end of spectrum. If the radiations beyond the violet end of the spectrum, the silver chloride solution becomes dark brown or black. This shows existence of some kind of radiation producing the chemical effect beyond the violet end of the spectrum. Answer 7. (a) The frequency depends upon : (i) Length of the string. (ii) Tension in the string. (iii) Mass per unit length of string. (b) According to the problem, For first echo, v = d1 = v t1 2d1 d1 = 2 t1 340 4 = 680 m. 2 For second echo, v t2 2d2 d2 = 2 t2 340 6 d2 = = 1020 m. 2 v = The distance between two cliffs = d1 + d 2 = 680 m + 1020 m = 1700 m (c) (i) All heated bodies such as heated iron ball, flame and fire are examples of infrared radiations. (ii) The name of the ray is -ray. Answer 8. (a) Mass of lead = 0.5 kg Let the specific heat capacity of lead is S . (i) Heat given out to cool 0.5 kg of lead at 327 C to 27 C = [0.5 5 (327 27)] J (77) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES According to question, [0.5 S (327 27)] J = 22500 0.5 S 300 = 22500 22500 = 150 cal kg 1 C 1. 300 0.5 (ii) Specific heat in Joules = 150 4.2 J kg 1 C 1. = 630 J kg 1 C 1. (b) Effect of pressure on melting point of ice : The melting point of ice is lowered by an increase of pressure because it contracts on melting. Effect of pressure on boiling point of water : The boiling point of water increases with the increase in pressure and decreases with decrease in pressure. (c) (i) 1g of steam at 100 C contains more heat. (ii) 1g of water at 0 C contains more heat. S = Answer 9. (a) (i) When the N-pole of the magnet is moved to the right, the current flows in the coil. There is a change in the magnetic flux linked with the coil. As a result emf is induced across the ends which causes induced current to flow in the coil. Thus, the galvanometer shows deflection. (ii) Anticlockwise. (iii) When the coil is moved away from N, the galvanometer needle deflects to left side. (iv) When both the coil and the magnet are moved at the same speed, there is no change in the magnetic flux linked with the coil. So, the galvanometer needle does not deflect. (b) (i) 1. The length of the conductor. 2. Intensity of magnetic field. 3. Magnitude of current flowing. (ii) When conductor is placed parallel to the direction of magnetic field, no force acts on it and it does not move. (c) (i) Heating in the coils. (ii) Eddy current in the core. (iii) Hysteresis loss in the core. (iv) Magnetic field link loss. Answer 10. (a) (i) A chemical change is always associated with the emission or absorption of heat. There is no effect of physical conditions in a nuclear change. (ii) By coating the screen with fluorescence material such as zinc sulphide. (b) 238 92 4 U 2 He 234 30 Th 234 90 Th 234 91 Pa 0 1 e (c) The rate of emission depends on : (i) Material or nature of the surface. (ii) The temperature to which the surface is heated. (d) The -particle is composed of 2 protons and 2 neutrons. They are Helium nuclei. (78) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 15 SECTION - I (40 MARKS) Answer 1. (a) The body will weigh more in vacuum. The air will exert a buyont force on the body resulting in the apparent loss of weight. (b) In S.I. system, it is kilogram force (kgf) which is equal to the force required to produce an acceleration of 9.8 ms 2 in a body of mass 1 kg. (c) 1. In a microphone. 2. When a simple pendulum moves from its mean position to its extreme position. (d) M.A. has no unit as it is the ratio of two similar quantities that is load and effort. (e) Work (energy) = 220 kJ t = 55 sec. Power = Answer 2. Energy 220 = = 4 kW. Time 55 (a) (i) The image is brighter and sharper as there is 100% reflection. (ii) The presence of moisture and dust on the glass does not affect the clarity of image. (b) When light travels from diamond to air, at an angle of incidence of 24 , the corresponding angle of refraction is 90 . (c) Concave lens Inc ide nt r ay Emergent ray F1 F2 (d) (i) Infrared radiation is beyond the spectrum of red. (ii) Ultraviolet radiation is beyond the spectrum of violet end. (e) Given, Power, P = 2.0 D P = 1 f 2.0 = 1 f f = 1 2 f = 0.5 m = 50 cm Answer 3. (a) (i) The sound produced in an organ pipe is due to vibration of Air Column. (ii) The sound produced in a violing is due to vibrations in the string. (b) (i) Loudness of the musical sound decreases. (ii) Pitch of the musical sound decreases. (c) (i) Application for high specific heat capacity of water for cooling is used in car radiators as coolant. (ii) In hot water bottles for maintaining temperature. (d) According to the problem, Heat capacity We know that, Mass, m = Sp. Heat capacity m= 93.75 = 125 g 0.75 (79) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (e) (i) Latent heat. (ii) Specific heat capacity. Answer 4. (a) S.I. unit of electrical power is Watt. Electrical energy is measured in Kilowatt hour. (b) It works on the principle of electromagnetic induction. If D.C. supply is given, transformer will not work. (c) The strength of the magnetic field can be changed by changing the current or the number of turns in the solenoid. (d) The current rating of a fuse is 2 A means that maximum of 2 A current can flow through it. (e) Radioactive radiation causes fluorescence on zinc sulphide. SECTION - II (40 MARKS) Answer 5. (a) (i) Work done normal to the direction of the gravitational force, Now, F = 20 kg = 20 10 = 200 N. W = Fd cos = 200 25 cos 90 = 200 25 0 =0J (ii) Work done parallel to the direction of the gravitational force, W = Fd cos W = 200 25 cos 0 ( F = 200 N, d = 25 m) = 200 25 1 = 5000 J (b) Let a machine overcome a load L by applying an effort E, in time t, the displacement of effort dE and displacement by load is dL. Then, Work input = Effort Displacement of effort = E dE Work output = Load Displacement of load = L dL By definition, Efficiency = = But L E and dE dL Work output Work input LE L dL = dE dL E dE = M.A. = V.R. Efficiency = M.A. V.R. (80) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) E T T Load, L Answer 6. (a) Here, u = 12.5 cm, f = 25 cm, v = ? A 1 1 1 = f v u Now, A 1 1 1 + = v 12.5 25 B B 12.5cm 1 1 1 1 2 1 F1 O F2 = = = v 25 12.5 25 25 25 cm v = 25 cm i.e., the image is virtual as shown in the figure. (b) In a prism, the refraction of light takes at two slant surfaces. The dispersion of white light occurs at first surface of prism where its constituent colours are deviated through different angles. But in rectangular glass block, the refraction of light takes place at two parallel surface. (c) (i) A print appears to be raised when a glass block is placed over it. (ii) A tank of water appears shallower than its actual depth. Answer 7. (a) The vibration of blade die away after sometime due to air resistance. The frequency of vibration of the blade can be lowered by : (i) increasing the vibrating length of the blade. (ii) increasing the thickness of the blade. (b) Given, = 1 3 m, f = 996 Hz. (i) Velocity, v = f = 1 3 996 = 332 ms 1. (ii) Since the medium is same, the velocity of sound will remain the same. Now, = 332 ms 1 , f = 1328 Hz v =f 332 v = = 0.25 m 1328 f (c) To produce a loud sound. When the string is set into vibration, forced vibrations are produced in a large volume of air inside the sound chamber. Thus, a loud sound is produced. = Answer 8. (a) Let the specific heat capacity of the solid = S Jg 1 C 1. Heat lost by the solid in dropping from 100 C to 40 C. (81) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES = 200 S (100 40) = 200 S 60 = 12000 S Heat gained by water in increasing temperature from 10 C to 40 C = 80 4.2 (40 10) = 80 4.2 30 Now, Heat lost = Heat gain 12000S = 80 30 4.2 S = ( C = 4.2 Jg 1 C 1) 80 30 4.2 12000 S = 0.84 Jg 1 C 1 (b) When water evaporates after shower, it absorbs large amount of heat from the surrounding due to high specific heat capacity of water thus, causing a drop in temperature. (c) Heat is the cause while temperature is the effect. Heat is the energy of transit while temperature determines the direction of flow of heat. Answer 9. (a) The labelled diagram is as shown alongside. Total internal resistance = 4 0.1 = 0.4 (i) Total resistance = 1.6 + 0.4 + R1 R = (2 + R1) (ii) Total emf = (2 + 2 + 2 + 2) V = 8 V Total emf Total resistance Now, according to the problem, we can say, (iii) Here, I = I = (iv) 2V 2V 2V 0.1 0.1 0.1 0.1 A 0.6 R1 E R1 + 2 2 = 2V 8 R1 + 2 2R1 + 4 = 8 2R1 = 4 R1 = 2 Potential difference across R1 = IR = 2 2 = 4 V. (b) Armature B C N S Field magnets A D Brushes Commutator S1 K S2 V (i) The field magnet is marked as NS. (ii) The armature coil is mounted on an axle, marked as ABCD. (82) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (iii) S1 and S2 is known as commutator. In this case, electrical energy is converted into mechanical energy. (c) (i) Coloumb - Electric charge. (ii) Kilowatt hour - Electrical energy. (iii) Kilowatt - Electric power. Answer 10. (a) - radiations compared to and are most energetic and so have smaller collisions with the atoms of the medium they pass through. Thus, they have the least ionising power but more penetrating power. (b) (i) Atomic no. and mass no. of radium are 88 and 229 respectively. 233 90 229 88 0 Ra 1 + 229 89 X 239 89 (ii) Th 0 X 1 + 229 88 Ra + 4 2 He 229 90 Th (c) E = mc2, where m is the loss in mass, c is the speed of light in vacuum and E is the amount of energy. (d) The three major precautions are : (i) Source should be kept in a thick walled lead container when not in use. (ii) Person should wear special protective clothing while working with a radioactive lead source. (iii) Radioactive source should be handled with a pair of lead tongs. (83) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 16 SECTION - I (40 MARKS) Answer 1. (a) (b) (c) (d) (e) (i) When velocity v is almost equal to c, then p = (mv) (ii) When v << c, then p = m v (i) Velocity ratio will not change for the machine of a given design. (ii) Velocity ratio is the ratio of velocity of effort to the velocity of the load. We know that, W =F S Work done by force is zero when : (i) there is no displacement. (ii) displacement is normal to the direction of the force. (i) A single movable pulley. (ii) A single fixed pulley. (i) Lever of class II. (ii) Given, FA = 40 cm, AB = 60 cm Load L = 50 N Load arm = FA = 40 cm Effort arm = FB = FA + AB = 40 + 60 = 100 cm M.A. = Effort arm 100 = = 2.5 Load arm 40 Answer 2. (a) (i) Light energy converted into electrical energy. (ii) The efficiency of conversion of solar energy to electrical energy is low. (b) We can say that, Loss of potential energy = mgh = 0.2 10 (10 6) = 0.2 10 4 = 8 Joule (c) (i) Class I lever. (ii) Class II lever. (d) (i) Rarer Denser Mirror (ii) This is known as principle of reversibility of the path of light. (e) Real depth Apparent depth (ii) The scattering of light by air molecules is responsible for the blue colour of sky. (i) Refractive index = Answer 3. (a) When acoustic resonance takes place, the natural frequency of vibration of body is equal to the frequency of external force, so the body vibrates with large amplitude and thus, conveys more energy to the ears. Hence, a loud sound is heard. (84) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) (c) (d) (e) (i) (ii) (i) (ii) (i) (ii) (i) (ii) Loudness of sound will change if its amplitude changes. Quality of a musical sound changes when its waveform changes. Frequency is inversely proportional to the length of air column. decibel (dB). North pole. The strength of the magnetic field of an electromagnet can be changed by changing the current. Fuse is used to protect electronic circuit from overloading and short circuiting. Overheating of fuse wire due to passage of excess current. Answer 4. (a) The heat capacity of a body is the amount of heat energy required to raise its temperature by 1 C or 1 K. Its S.I. unit is JK 1. (b) (i) Copper is used as a material of calorimeter. (ii) Copper is used because : (a) Copper is a good conductor of heat so the vessel soon acquires the temperature of its contents. (b) Copper has low specific heat capacity so the heat capacity of the calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its content is negligible. (c) (i) Tungsten coated with barium, caesium or strontium oxide, is used as an electron emitter metal. (ii) It is used because its work function is low, of only 1 eV and needs to be heated to only 1000 K to emit a good supply of electrons. (d) (i) The radioactive materials must be placed in the thick lead containers. (ii) The person handling the radioactive materials must wear lead aprons and lead gloves. (e) The resistance of a wire depends on : (i) length of wire. (ii) area of cross-section of wire. SECTION - II (40 MARKS) Answer 5. (a) (i) Product of mass and velocity of the body. (ii) Rate of change of momentum is directly proportional to the force, i.e., mv mu F t (iii) When mass is constant. (b) (i) p = mv Here, m = 50 kg and p = 3000 kg ms 1. 3000 = 50 v Now, (c) (i) (ii) M.A. = 3000 50 v = 60 ms 1 v = K.E. = M.A. = 3000 3000 p2 = = 90000 J 2 50 2m Effort arm Load arm l , where l is the length of the inclined plane and h is the height of the inclined plane. h l h (85) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (iii) 1. Single movable pulley is never frictionless. 2. Single movable pulley is never weightless. Answer 6. (a) (b) (c) (i) The incident ray, refracted ray and normal at point of incidence lie on the same plane. The ratio of sine of angle of incidence to the sine of angle of refraction is constant for given pair of media and for the given wavelength of light. (ii) i + e = A + d (i) 1. By using thermopile which shows the deflection in the galvanometer or blackened bulb thermometer shows the rise in temperature. 2. By using silver chloride solution which turns violet and then brown and then black in presence of these radiations. (ii) Muscular theraphy or for taking photographs in fog. (i) Concave lens can form this image. (ii) A A F B O B AB = Object, A B = Image Answer 7. (a) (i) SONAR is based on reflection of sound waves. (ii) Speed = 340 ms 1 t = 1.8 sec Distance Speed = 2 Time S 340 = 2 1.8 S = 340 1.8 = 306 m 2 (b) (i) Resonance. (ii) It happens when the natural frequency of air column becomes exactly equal to the frequency of the vibrating tuning fork. (c) (i) 1. Amplitude : The maximum displacement of a particle from its mean position. 2. Frequency : The number of vibrations produced by a wave in one second. (ii) The vibration is produced in air of the sound box when strings on it are made to vibrate. Since, surface area of air in the sound box is large, the forced vibration of air causes a loud sound. Answer 8. (a) (i) Terminal voltage < emf. emf is equal to terminal voltage when no current is drawn. (ii) Fuse is used to limit the current in an electric circuit. Its purpose is to provide safeguards to the circuit and the appliances connected in that circuit from being damaged. (iii) Fuse wire is made-up of a material of high resistivity and low melting point. So that it may easily melt due to overheating when current in excess of the prescribed limit passes through it. (b) (i) W = I2 Rt. (ii) 220 V. (iii) In parallel. (c) (i) According to the problem, equivalent resistance in parallel RP is, 1 1 1 = + R2 R1 RP 1 1 1 = + 12 8 RP (86) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 1 1 1 = + 12 8 RP 3+ 2 1 = 24 RP RP = 24 = 4.8 5 Equivalent resistance R = 7.2 + 4.8 = 12 (ii) Current = V 6 = = 0.5 A 12 R (iii) Potential difference across 7.2 = 7.2 0.5 = 3.6 V Answer 9. (a) (i) Amount of heat energy required by 1 kg mass of a solid to change the state from solid to liquid at constant temperature without any rise in temperature. (ii) Water has the highest specific heat capacity. (iii) Heat absorbed or given out by a body depends on mass of the body and specific heat capacity of the body. (b) (i) Heat capacity of A is less than B. (ii) 1. Kinetic energy of the molecules changes. 2. Potential energy of molecules changes. (c) Heat gained by ice = mL + mCt Q = (50 336) + (50 4.2 t) ( m = 50 g, L = 336 5 Jg 1, C = 4.2 Jg 1 C 1) Now, we can say, Heat lost by liquid = mL C (T t) mL = 300 g, C = 2.65 Jg 1 C 1 T = 30 C Lost heat = 300 2.65 (30 t) According to the principle, Heat gained = Heat lost Hence, (50 336) + (50 4.2 t) = 300 2.65 (30 t) 23850 16800 = 210 t + 795 t t = 7.01 C Answer 10. (a) The three main parts of cathode ray tube : The electron gun. The deflection system. The fluorescent screen. Radioactivity is the process of spontaneous emission of , and radiations from the nuclei of the atoms during their decay. (ii) The radioactive material after its use in known as nuclear waste. (iii) The nuclear waste must be stored in thick casks and buried in the deep underground stores away from the populated area and must be sealed. (i) (ii) (iii) (b) (i) (87) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) (i) b c N S a d Electromagnet B1 B2 S1 S2 Load A. C. Generator abcd = armature, NS = Electromagnet (ii) The voltage of A.C. can be stepped-up and stepped-down by the use of transformer. Thus, reducing loss of electrical energy as heat. (88) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 17 SECTION - I (40 MARKS) Answer 1. (a) Let a constant force F acting on a body displaces it by a distance S in its own direction. Then, Work, W = F S Power, P = W P = t F S t Now, we can say S t Hence, P = F.v Thus, Power = Force Velocity According to the problem, 1 kWh = 1 kW 1 h = 1000 W 3600 sec. = 3.6 106 J The energy can neither be created nor be destroyed. It may be transformed from one form to another form, but the total energy of the system remains constant. In a single movable pulley, when a force is applied at the free end in the upward direction, then M.A. of the pulley is greater than one. Here, it is given Load arm = 25 cm We can say, Effort arm = (150 25) = 125 cm ( Total length of crowbar = 150 cm) v = (b) (c) (d) (e) M.A. = Answer 2. Effort arm 125 = =5 25 Load arm (a) The refractive index of a medium is related to the speed of light as follows : = Speed of light in vacuum or air Speed of light in medium (b) When angle of deviation is minimum, then angle of incidence = angle of emergence. (c) The distance between the optical centre and principal focus of a lens is called the focal length of the lens. (d) Given, v = 40 cm, u = 20 cm I v = , we have O u 40 I 2 = = 20 O 1 Hence, the ratio between the image and the object is 2 : 1. (e) The various colours of the spectrum in order of their deviations are red, orange, yellow, green, blue, indigo and violet. Now, applying the formula, m = Answer 3. (a) The physical quantity whose unit is decibel is loudness. The other unit is bel. (b) The frequency of the sound emitted due to vibration in air column depends on the length of the air column. (c) Here, Heat energy released = m Latend heat of vap. of steam. m = 5 kg and L = 2268 kJkg 1. Heat energy released = 5 2268 = 11340 kJ = 11.34 106 J (89) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (d) Steam at 100 C condenses to water losing its latent heat of vaporization of 2268 Jg 1. Hence, we can say that 1 g of steam releases 2268 J heat more than 1 g of water, when both at 100 C. Thus, to heat a building, steam pipes are more effective that hot water pipes. (e) It means that 175 Joules of heat energy is required to raise the temperature of a copper calorimeter through 1K. Answer 4. (a) The strength of an electromagnet can be increased by : (i) increasing the current through the solenoid. (ii) increasing the number of turns. (b) Energy released by the battery in 1 hour E =V I t = 12 80 1 = 960 Wh Again, E =P t t = ( V = 12 V, I = 80 A) 960 E = = 16 h 60 P (c) (i) 1 HP = 0.746 kW (ii) 1 Watt = 0.001 kW (d) According to the Fleming s right hand rule, It will deflect towards left. (e) The properties of a thermionic emitter are as follows : (i) The melting point should be high. (ii) The work function should be low. SECTION - II (40 MARKS) Answer 5. (a) Newton s second law of motion states that the rate of change of momentum of a body is directly proportional to the net force acting on it and takes place in a direction in which the force acts. Expressing of force : Let a body of mass m is moving with velocity u . Its velocity becomes v after time t , under the action of a uniform force f . Initial momentum of the body = mu Final momentum of the body = mv Change in momentum in time t = mv mu mv mu Rate of change in momentum = t (v u) =m t Rate of change in momentum = ma [ v u = a] t Now, according to Newton s second law of motion, the rate of change of momentum of a body is directly proportional to the force applied (F). Rate of change of momentum F. F ma In S.I. system, F = kma where k =1 (as a is constant) Thus, F = ma Hence, force can be measured as the product of mass and acceleration. (90) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) We have, The radius of the driving wheel, rA = 18 cm The radius of the driven wheel, rB = ? No. of rotation of driving wheel, nA = 30 r.p.m. No. of rotations of driven wheel, nB = 150 r.p.m. No. of teeth on driving wheel, NA = 100 No. of teeth on driven wheel, NB = ? Then, NA 100 = NB NB Here, no. of teeth on driven wheel (NB) is calculated by, (i) The gear ratio = NB = 100 30 NA nA = = 20 150 nB 100 5 NA = = 20 1 NB Hence, gear ratio = 5 : 1 (ii) The no. of teeth on the driven wheel NB = 20 (iii) The radius of the driven wheel is given by Therefore, rB = rA nA 18 30 = = 3.6 cm nB 150 (c) When a body is thrown vertically upward, Then, v2 = u2 2gh 0 = u2 2gh u2 = 2gh 1 Now, u = 20 ms , g = 10 ms 2 (20)2 = 2 10 h h = [ v = 0] 400 = 20 m 20 Hence, P.E. at height h = 20 m is mgh = 0.20 10 20 = 40 J Answer 6. (a) Initial Now, heat developed Now, P.E. = mgh = m 10 50 = 500 m Q = mCt Q = mgh = 500 m 500 m = m 4200 t 500 = 0.12 C 4200 (b) We know that if a body of mass m is moving with a velocity v , then p Momentum, p = mv or v = ... (i) m 1 and kinetic energy, K = mv2 ... (ii) 2 From equations (i) and (ii), we have t = 2 p2 1 p K= m = 2m m 2 p2 , the K.E. 2m of lighter body will be more than that of heavier body as it is inversely proportional to the mass of the same momentum. As, the bodies have the same momentum but different masses. So according to the relation K = (91) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) As the metre rule is uniform, so its weight will act at its mid-point, i.e., 50 cm mark. 50 40 5 3.5 m 0 cm 3.5 m 75 25 10 100 cm 0 25 gf W 10 gf Taking the moments about 40 cm, we have Clockwise moment = W 10 cm + 10 gf 35 cm = 10 W cm + 350 gf cm Anticlockwise moment = 25 gf 35 cm = 875 gf cm According to the principle of moments, we have Clockwise moment = Anticlockwise moment 10 W cm + 350 gf cm = 875 gf cm 10 W cm = (875 350) gf cm 525 gf = 52.5 gf. 10 Hence, weight of metre rule = 52.5 gf. W = Answer 7. (a) Image formed is of the same size as the object. (i) (ii) (iii) (b) (i) (c) (ii) (i) (ii) (iii) Magnification = Size of image Size of object m =1 Image is real and inverted. Distance between object and optical centre of lens is equal to the focal length of the lens. 1. If seen at an angle from the normal, then the image of the letters will not be in the same place. 2. Letter V will be raised to the maximum because the light rays coming from it will bend more away from the normal. X-rays and gamma rays have shorter wavelength than violet light. When light passes from one medium to other, it deviates from its original path, this phenomenon is known as refraction. Velocity of light in vacuum (c) 1. = Velocity of light in medium (v ) sin i 2. = sin r Refractive index of medium I and II are equal. Answer 8. (a) If the distance of aeroplane from radar is x metre, then the distance covered by wave = 2x. Distance Time 2x 3 108 = 0.02 10 3 2x = 0.06 105 x = 0.03 105 = 3000 m x = 3 km (b) (i) AB represents melting of ice. CD represents vaporization of water. (ii) From the graph, the following observations are made : (a) Melting point of ice is 0 C. (b) Boiling point of water is 100 C. Velocity = (92) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) CD is longer than AB, which indicates that latent heat of vaporization is much more as compared to latent heat of fusion. (c) (i) 1. Heat lost by steam = Mass latent heat of fusion Q1 = m1 2268 2. Heat gained by ice = Mass latent heat of fusion. Q2 = m2 336 (ii) Q1 = Q2 m1 2268 = m2 336 m2 2268 m1 = 336 = 6.75 Answer 9. (a) P = 1500 W, V = 250 V. (i) We know Energy Time Energy = P T = 1500 60 60 60 E = 3240 105 J P = 1500 W, = 1.5 kW t = 60 hrs. E = 1.5 60 = 90 kWh Cost of electrical energy = 90 2.5 = ` 225 The split ring as a commutator in a D.C. motor. After every half rotation split rings exchange their position thus, the direction of the rotating coupling remains unchanged and the coil continues to rotate in the same direction. In moving coil galvanometer, soft iron core is used as : 1. It makes magnetic field radial. 2. It intensifies the magnetic field. 6 and 3 are in parallel. So, p.d. across the two will be same. So, 6 0.5 = 3 I 3 I = = 1A 3 Reading of ammeter, C = 1 A. Ammeter A reads total current of the circuit. Reading of ammeter A = 1 + 0.5 = 1.5 A 6 and 3 are in parallel. If their equivalent resistance in R1, then (ii) Now, (iii) (b) (i) (ii) (c) (i) (ii) P = VI P I = V 1500 I = =6A 250 P = 1 1+ 2 3 1 1 1 = + = = = 3 6 6 2 R1 6 R1 = 2 Now, R1 and 2 are in series. Hence, Total resistance R = R1 + 2 = 2 + 2 = 4 Answer 10. (a) (i) Since -particles are less energetic than -particles and -rays, so they suffer more collisions with the atoms when passing through in air. As they have high ionising power but less penetrating power, their energy is soon exhausted. (93) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) Since -rays are the highest energetic rays, so they have a smaller rate of collision with the atoms of the medium through which they pass. Thus, they have less ionising power but more penetrating power. (b) (i) The given out particles are electrons. Electrons are negatively charged. (ii) The amount of charge in each electron is 1.6 10 19 C. (iii) The voltage used to heat the filament is 6 V. (iv) The spot shifts towards the right or left along a horizontal line on applying the voltage between X-X plates. (c) The isotopes which exhibit the property of radioactivity are called radioisotopes. A few common radioisotopes are Iodine - 131, Phosphorus - 33, Carbon - 14, Cobalt - 60, Sulphur - 35 and Sodium -24. (94) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 18 SECTION - I (40 MARKS) Answer 1. (a) Mass Weight (i) It is the quantity of matter contained in a body. (i) It is a force which a body is pulled towards the centre of the earth. (ii) It varies from place to place on the surface of earth as well as in other place in universe. (ii) It is constant throughout the universe. Acceleration (b) No work done when a body moves in a circular path, it does not work against the centripetal force, since force is normal to the direction of motion. Work done = F. d cos 90 = 0. (c) Weight of the scale is greater than the weight of mass M. Load Load arm = Effort Effort arm. Principle of moment is applied to the lever. (d) Acceleration is inversely proportional to mass at constant force. Mass (e) (i) Distance covered by car in 5 second = Area of triangle 1 1 base height = 5 20 2 2 Distance = 50 m (ii) Acceleration of car = Slope of the graph = = Height 20 = Base 5 Acceleration = 4 ms 2 Answer 2. Angle of deviation (a) The velocity ratio of a pair of gears is defined as the ratio of the no. of rotations per unit time of the driving gear to the no. of rotations per unit time of the driven gear. (b) min Angle of incidence (95) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (c) Real image Virtual image (i) It is formed when the rays of light actually meet to focus at the image position. (i) It is formed when the rays of light appear to come from where the image seems to be. (ii) It can be obtained on a screen. (ii) It cannot be obtained on a screen. (d) The focal length of a lens depends on the : (i) radii of curvature of lens surfaces. (ii) refractive index of the material of lens. (e) Deviation produced by a prism is the angle between the incident ray on the prism and the emergent ray produced. Answer 3. (a) A large sound box contains large amount of trapped air. This starts vibrating with forced vibrations. As large volume of enclosed air vibrates, a loud sound is produced. 27 = 9 . 3 Now, these three wires are connected in parallel. (b) Resistance of each wire = 1 1 1 1 1 = + + = R 9 9 9 3 R=3 (c) Earth Live Neutral Plug (d) It is a device which is connected in series with the circuit to safeguard the circuit from an excessive current and to limit the current to be passed through the circuit. (e) A switch is always connected to the live wire of the circuit so that when it is off, no current flows through the appliance. If switch is connected in the neutral wire, the circuit remains on even when it is off . Answer 4. (a) Step-up transformer Step-down transformer (i) It has more no. of turns in the secondary coil than in the primary coil. (i) It has less no. of turns in the secondary coil than in the primary coil. (ii) Step-up transformer is used to increase the (ii) Step-down transformer is used to decrease the amplitude of an alternating emf. amplitude of an alternating emf. (b) Amount of heat released = mSt + mL = 5 4.2 (20 0) + 5 336 42 20 + 1680 10 Amount of heat released = 420 + 1680 = 2100 J (c) The resistance of wire depends on following factors : (i) Material of conductor. (ii) Length of conductor. (iii) Area of cross-section of wire. (iv) Temperature of wire. (any two) =5 (96) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (d) -particle on being emitted leaves the mass of nucleus unchanged but increase its atomic no. by 1. It is shown by 0 1 e. (e) The equation is as follow : 238 92 U 206 82 0 Pb + 8 4 2 He + 6 1e 8 -particles and 6 -particles are emitted. SECTION - II (40 MARKS) Answer 5. (a) Case I : Force = 140 N, distance = 50 cm = 0.5 m. Moment of force needed to open the nut = Force Perpendicular distance = 140 N 0.5 = 70 Nm. Case II : Let the length of handle needed by L m, then Moment of force needed to open the nut = Force Perpendicular distance = 50 N Lm = 50 L Nm Comparing (i) and (ii), we have 50 L = 70 L = ...(i) ...(ii) 70 = 1.4 m 50 Hence, 1.4 m long lever will be needed. (b) Single movable pulley (applying an effort equal to half the load. Since the pulley acts as a force multiplier, efforts required to overcome a given load is reduced). Fixed pulley T T Movable pulley T Effort (E) Load L (c) (i) Velocity Ratio = V.R. = 4 Efficiency = 90% Mechanical advantage = Mechanical advantage = (ii) 4 90 = 3.6 100 Load Effort ( M.A. = V.R. ) M.A. = L E Effort = 300 = 83.3 N 3.6 E= (97) L M.A. e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Answer 6. (a) The figure shows a uniform metre rule AB which is balanced horizontally at O. O 60 cm 0 cm A C 40 cm 100 cm B 10 gf W gf Let W gf be the weight of the metre rule. Then, as metre rule is uniform, its weight will act at its mid-point, i.e., 50 cm marks. (i) The weight W gf of the metre rule produces an anticlockwise moment about O. So, in order to balance it, 10 gf weight must be suspended at the end B to produce clockwise moment about O. (ii) Anticlockwise moment about O = W OC = W 10 cm = 10 W Clockwise moment about O = 10 gf OB = 10 gf 40 cm = 400 gf cm. According to the principle of moments, we have Anticlockwise moment = Clockwise moment 10 W gf cm = 400 gf cm W = 400 = 40 gf. 10 Hence, the weight of the metre rule is 40 gf. (b) Given, F = 10 kgf = 100 N and displacement S = 0.5 m. (i) The displacement is in the direction of force, i.e., = 0 On putting the values in the formula, W = FS cos We get, W = 100 0.5 cos 0 = 100 0.5 1 = 50 J (ii) Given, = 60 W = 100 0.5 cos 60 = 100 0.5 1 = 25 J 2 = 90 W = 100 0.5 cos 90 = 100 0.5 0 = 0 (c) The rate of doing work is called power. It is measured by the ratio of total work done in total time. Work Power = Time (iii) Given, Answer 7. (a) (i) PQN is known as critical angle. (ii) R N Air Q Water X P N (iii) If PQN i.e., critical angle is increased, then total internal reflection will occur. (98) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) (i) (ii) (iii) (c) (i) The above phenomenon is termed as Resonance. Frequency of loud sound in air column is equal to that of tuning fork. Unit of loudness is Decibel. Microwaves : It is used in transimitting TV programs and also used in radar to detect the presence of air crafts. (ii) UV rays : It is used for sterilizing purpose. (iii) Infrared radiation : It is used for therapeutic purpose by doctors and during wars they are used as signals. Answer 8. (a) (i) Heat energy required = Mass Specific heat capacity Rise in temperature Heat capacity t M Q = Heat capacity t Heat energy required = 966 15 = 14490 J Heat energy required = M Heat capacity Mass 966 = = 483 Jkg 1K 1 2 (b) (i) Specific latent heat of vaporization is the quantity of heat required to change a liquid at its boiling point into its vapour state at constant temperature. (ii) Principle of calorimetry states that when a hot substance is kept in contact with cold substance, then heat is transferred from hot substance to the cold substance till both attain the same temperature. Hence, Heat lost by hot substance = Heat gained by the cold substance. (c) Water is used in hot water bottle for fomentation and as universal coolant because water itself does not cool quickly and stores a large amount of heat energy for long period due to its high specific heat capacity. (ii) Specific heat capacity = Answer 9. (a) (i) I = 0.3 A, V = 6 V Total resistance of circuit 1 1 1 = + [R = Total resistance of circuit] R 60 R R = (ii) 60 R R + 60 p.d. = Current Resistance Resistance R = p.d. V = Current I Substituting the values, we will get (iii) 6.0 60 R = 60 + R 0.3 R = 30 R = Hence. 60 R R + 60 60 30 1800 = = 20 90 30 + 60 p.d. = I R = 0.3 20 = 6 V R = V 6 = = 0.2 A R 30 (b) Given, V = 110 V, I = 8 A, t = 2 h Now, we can say, Energy consumed in two hours = VI t Energy consumed = 110 8 2 Wh and current in R = (99) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES = 1760 Wh 1760 kWh = 1.76 kWh 1000 If cost of energy = ` 2.50 per kWh Then total cost = 1.76 2.50 = ` 4.40 (i) North pole compass needle is deflected towards West. When current flows from South to North direction. It is deflected towards East, when current flows from North to South. The phenomenon is magnetic effects of currents. (ii) Electrical energy change into mechanical energy and heat energy. = (c) Answer 10. (a) (i) Filament Anode Horizontal plates 6V battery Screen Grid 1000 V Electron Gun Deflecting plates are of two types : Horizontal deflecting plate : It deflects the electrons in X -X plane that strikes the screen. Vertical deflecting plates : It deflects the electrons in Y - Y plane that strikes the screen. (b) Thermionic emission is the emission of electrons from the surface of a metal when thermal energy is supplied to it. It is used in diode valves and cathode ray tube. Tungsten is a good thermionic emitter. (c) The three common properties of -rays and cathode rays : (i) They are negatively charged particles. They have same charge as that of an electron. (ii) Both of them affect photographic plates. (iii) They can produce fluorescence when strike a screen containing zinc sulphide. (100) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 19 SECTION - I (40 MARKS) Answer 1. (a) Class II levers have M.A. greater than 1. The effort arm can be increased or the load arm can be decreased. (b) Let the mass of A and B each be m . Their height are 20 m and 30 m. So, P.E. of A = m g 20 and P.E. of B = m g 30 PA 20 PB = 30 (c) S.I. unit of force is Newton. 1 N is that force which is exerted on a body of mass 1 kg to produce an acceleration of 1 ms 2. (d) Infrared radiations have a high penetrating power and thus can penetrate thick columns of fog, smoke etc. That is why, they are preferred over ordinary visible light for taking photographs in fog. (e) The energy of a body is its capacity to do work. Its S.I. unit are in Joule and erg. Energy = Power Time Answer 2. (a) As the mechanical advantage increases with the increase in effort arm, so a lesser effort is required to operate a hand pump with longer handle. That is why the handle of an ordinary pump is made longer. (b) Diamond is usually cut with its faces in such a manner that once a ray of light enters into it, it suffers the total internal reflection many of times at various faces before it has any chance to emerge out. A (c) 45 45 B C Q Object P P1 Image Q1 (d) Given, Velocity of light is diamond = 121000 kms 1 = 1.21 108 ms 1 Velocity of light in air = 3 108 ms 1 Refractive index of diamond = Velocity of light in air Velocity of light in diamond 3 108 = 2.48 1.21 108 (e) Since the rock-salt prism does not absorb infrared radiations, whereas a glass prism absorbs them. Therefore, infrared radiations can be obtained by a rock-salt prism. = Answer 3. (a) (i) Amplitude : Amplitude is the maximum displacement of a particle from its mean position. Its S.I. unit is metre (m). (101) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) Frequency : Frequency of a wave is defined as the number of complete wavelengths traversed by the wave in one second. Its S.I. unit is Hertz (Hz). (b) Object is placed at a distance greater than twice the focal length. A F B 2F O F B 2F A Here, AB represents the object and A B is its image. Since the two rays starting from A after passing through the lens, actually meet at the point A , thus A is the real image of point A. The image A B is between F and 2F on the other side of the lens, and its nature : (i) real (ii) inverted (iii) diminished (c) The entire incident light is reflected back into the denser medium, whereas in the case of reflection from a plane mirror some part of light is absorbed. V (d) We know that, slope = tan . B Also, slope of the graph = R (resistance). Since slope of graph B is greater than slope of graph A, therefore, A combination of resistance for slope B is more than slope A. Hence, A represents parallel. B represents series. I (e) Of the three connecting wires of a household circuit : (i) The ground wire and the neutral wire are at the same potential. (ii) The switch is always connected in live wire. Answer 4. (i) An -particle will change into a singly ionised Helium (He+), when it absorbs one electron. (ii) An -particle will change into the Helium atom (He), when it absorbs two electrons. (b) Let mass of hot water = m, mass of cold water = 3 m, Let the initial temperature of hot water = t C Initial temperature of cold water = 10 C Final temperature of the mixture = 20 C Heat lost by hot water = Heat gained by cold water mC (t 20) = 3 mC (20 10) t 20 = 3 10 t = 50 C Hence, the initial temperature of hot water is 50 C. (c) Since the effective resistance of the combination of two individual resistance is less than one of the two resistance, they must be connected in parallel. (a) Now, 1 1 1 = + R1 R2 R 1 1 1 = + 6 15 R2 + R1 = 15 5 2 1 1 1 = = 6 15 30 R2 R2 30 R2 = 10 3 (d) The specific latent heat of vaporization of steam is 2260 Jg 1. Therefore, 1 g of steam at 100 C contains 2260 J of heat energy more than the boiling water at 100 C. We can say when steam condenses, it gives out 2260 J of heat more than the water of same mass at same temperature. Hence, burn caused by steam is more severe. R2 = (102) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (e) The two important properties of a metal of good thermionic emitter : (i) It must have a high melting point. (ii) It must have a low work function. SECTION - II (40 MARKS) Answer 5. (a) (b) (c) (i) Kilowatt hour is the commonly used unit of electrical energy, equal to 1000 Wh or 3.6 MJ. 1 kWh = 1 kilowatt 1 h = 1000 Js 1 3600 sec. = 3.6 106 J = 3.6 MJ (ii) Measurement of work done when the force is applied at an angle to the direction of displacement. In such a case, the force is resolved into two rectangular components, the horizontal component and the vertical component. W = FS cos (where, S = displacement, F = Force) (i) 1. Photosynthesis in green leaves : The light energy from the sun is absorbed by green plants and they transform it into chemical energy. 2. Charging of battery : Electrical energy is transformed into chemical energy. (ii) Relationship between M.A., V.R. and . M.A. = V. R. where, M.A. = Mechanical advantage V.R. = Velocity ratio = Efficiency (i) T Effort T Load Block and Tackle pulley system with a V. R. 3. (ii) The lower block of this pulley system should be of negligible weight to obtain greater efficiency from the system. Suppose the weight W of the lower block of this pulley system is not negligible, then for a frictionless block, we have Load + W = 3E Load = 3E W Now, But Load 3E W = Effort E W M.A. = 3 3E 3d V. R. = = 3. d M.A. = (103) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES M.A. V.R. 3 W E W = =1 3 3E < 1 or 100% Efficiency = Hence, for greater efficiency, the lower block should be of negligible weight. Answer 6. 2.6 = 1.3 m 2 Moment of force needed to open the gate = Force Perpendicular distance = 80 1.3 = 104 Nm. (i) The force required will be least if it is applied at the farthest from the hinges, i.e., width of the gate (2.6 m) Moment of force = Force Perpendicular distance 104 Nm = Force 2.6 m (a) Given, Force F = 80 N and perpendicular distance = 104 = 40 N 2.6 Hence, the required least force = 40 N. (ii) The force should be applied at the free end of the door. Force = (b) Given, m = 1000 kg, t = 2 min. = 120 sec., u = 36 kmh 1 = 36 1000 = 10 ms 1 3600 1000 = 20 ms 1. 3600 According to the work energy theorem, Work done = Change in K.E. 1 m (v2 u2) W = 2 1 = 1000 [(20)2 (10)2] 2 1 = 1000 [400 100] 2 1 = 1000 300 2 = 150000 J = 1.5 105 J and v = 72 kmh 1 = 72 Power = 1.5 105 W = = 1.25 103 W t 120 (c) The perpendicular distance between emergent ray and original ray produced forward is called lateral displacement. Answer 7. (a) (i) In the given fig., ABC is a right-angled prism with a ray of light PQ incident normally on the side AB of the prism. It passes undeviated into the prism and strikes at the other side AC at an angle of incidence equal to 45 . Since C = 42 , therefore, since the incident ray being greater than C = 42 , hence, forth the ray will reflected totally and emerges through the base normally. A P Q 45 B R 45 S C (ii) The angle of deviation shown by the ray is 90 . (104) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) (c) (i) X-rays are used to study the structure of crystals. X-rays are also used for detecting fractures in bone. (ii) On increasing the wavelength of light, the speed of light in glass will increase. A (i) N A Apparent depth Real depth I B B Eye Air Q Water N O Object In the above diagram, an object is placed at the bottom of water, the incident ray OA is incident normally on the water-air interface and it emerges out normally. Now, the incident ray OB of i is refracted by r as BB . When this refracted ray is drawn backward, it produce the image I which is above the real object O. Hence, this image which is seen by our eye is apparently seen shallower than the real depth of the object. Thus, apparent depth is less than its real depth. (ii) Refractive index of water (a w) is related to the real depth and apparent depth of a column of water by the following relation : a w = Real depth Apparent depth Answer 8. (a) (i) Let the initial position of the man be A and the second position be B. For an echo : 2d t where, v = velocity of sound. d = distance between the observer and cliff. t = time after which the echo is heard. Case I : When the man is standing at position A : 2d v = t 2x v = 3v = 2x 3 3v 2x = 0 Case II : While the man is standing at position B : v = v = ...(i) 2d 2[ x 82.5] = t 2.5 2.5v = 2x 165 2.5v 2x = 165 Subtracting (ii) from (i), we have 0.5v = 165 ...(ii) 165 = 330 ms 1 0.5 Substituting v in eq. (i), we have 3(330) 2x = 0 2x = 990 v = x = 990 = 495 m. 2 The distance of the cliff from the initial position of the man is 495 m. (ii) Velocity of sound = 330 ms 1. (105) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (b) (i) Specific heat capacity of a substance is the quantity of heat required to raise the temperature of a mass of substance through 1 C or by 1 K. Q m where, m = mass, Q = heat, = change in temperature. The S.I. unit of specific heat capacity is Jkg 1 C 1 (ii) The heat supplied to a substance during the change of state does not cause any rise in temperature because the heat is used up in increasing the potential energy of the molecules of the substance. For example, when a solid changes into liquid without any change in temperature, the potential energy of the molecules as the distance between the molecules on an average increases. (i) Specific heat capacity, C = 500 Jkg 1 C 1 Specific heat capacity, C = (c) Change in temperature, = 80 C Heat, Q = 800 J Q = mC 800 = m 500 80 m = 0.02 kg = 20 g Mass of the substance = 20 g Q = mL (ii) 1600 = 0.02 L L = 1600 = 8 104 Jkg 1 0.02 Answer 9. (a) (i) Earthing of an electrical appliance means that the metal body is connected to a thick copper wire, which is buried deep in the earth and at its end is a copper plate surrounded by a mixture of charcoal and common salt. Earthing of appliance is essential because of the following reasons : 1. When the live wire of faulty appliance comes in direct contact with its metallic body due to break of insulation due to constant use / over loading / short circuiting, the whole appliance is at the same potential as that of the live wire. A person touching it will get a shock. Thus, when earthed the current will flow to the earth and hence, no shock will be received by the person. 2. The earthing also saves the appliance from being damaged. (ii) If alternating current is used instead of D.C., the electromagnet will not be magnetised properly due to change in polarity of the A.C. during each cycle and hence, the electric bell will not ring. (b) (i) The strength of an induced current depends on the rate of change of magnetic flux through the coil and the number of turns in the coil. (ii) When a solenoid which is carrying current is freely suspended, it comes to rest along the North-South direction. This is so because a solenoid which is carrying current behaves like a bar magnet. (c) (i) Total resistance of the circuit = R R = R1 + R2 + r E = 1.5 V = 4 + 20 + 1 = 25 Current in the circuit, I = = r = 1.0 E Total R 1.5 = 0.06 A. 25 (106) 4.0 20.0 e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) Potential difference across the 4 resistor, V1 = IR1 = 0.06 4 = 0.24 V (iii) Voltage drop when the current is flowing, V = Ir = 0.06 1 = 0.06 V (iv) p.d. across the cell is the total p.d. in the external circuit i.e., the sum of the p.d., across the 4.0 resistor and the p.d. across the 20.0 resistor. V2 = IR2 = 0.06 20 = 1.2 V p.d. across the cell = V1 + V2 = 0.24 + 1.2 = 1.44 V Answer 10. (a) Functions of the three main part of cathode ray tube : (i) Electron Gun : The electron gun contains a number of electrodes and produces a narrow beam of high speed electrons. (ii) Deflecting System : The deflecting system comprises of two pairs of deflecting plates which deflects the electron in X-X plane and Y-Y plane respectively. It spreads the electron, so that it reaches every part of the screen. (iii) Fluorescent Screen : The fluorescent screen is coated with fluorescent material. When the electron beam falls over it, a bright spot of light is produced. (b) (i) The two factors on which the magnitude of an induced emf in the secondary coil depends are as follows: (i) emf increases with the increase in number of turns. (ii) Area of cross-section, orientation between two coils. (ii) (c) 24 24 and Na11 are called isobars. Mg12 (i) 1. When an element emits -particle, its atomic no. decreases by two units. 2. When an element emits a -particle, its atomic no. increases by one unit. (ii) -particles and -particles are deflected an electric or magnetic field because they are charged particles whereas gamma rays don t have any charge. Hence, they cannot be deflected by electric or magnetic charge. (107) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES EVERGREEN MODEL TEST PAPER - 20 SECTION - I (40 MARKS) Answer 1. (a) (i) (ii) (i) (ii) 1 kgf is the force due to gravity on a mass of 1 kg. 1 kgf = 9.8 N. (b) The forces which act on bodies without being physically touched are called the non-contact forces. It decreases with the increase in separation and increases with decrease in separation. These forces vary inversely as the square of distance of separation. (c) According to the principle of moment of force F W1l1 = W2l2 2m l2 = ? m1l1 = m2l2 30 2 = 40 l2 30 kg (d) (e) l2 40 kg 30 2 = =1.5 m from fulcum. 40 (i) It is the turning effect of a force. The moment of a force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation. (ii) Clockwise. (i) Potential energy. (ii) Because of change of potential energy of spring into the kinetic energy of the ball. Answer 2. (a) (b) (c) (d) (e) (i) Light energy converted into electrical energy. (ii) The efficiency of conversion of solar energy to electrical energy is low. Loss of P.E. = mg h = 0.2 10 (10 6) = 0.2 10 4 = 8 Joule (i) Refractive index is defined the ratio of the speed of light in vacuum to the speed of light in the medium. (ii) Ray 2. Look at the printed piece of paper through both the lenses one after the other, holding the lens close to the paper. The lens through which letters appear to be magnified is convex lens. (i) e = i. (ii) The deviation suffered by the incident ray is minimum. Answer 3. (a) (b) (c) (d) (i) (ii) (i) (ii) (i) (ii) The phenomenon of splitting of white light into its constituent colours is known as dispersion of light. Red, because of its large wavelength. Pitch of a note or quality of a sound. Loudness. Frequency is inversely proportional to the length of air column. Decibel (dB). Power = 1000 kVA = 1000 kW t =2h (i) Electric energy consumed = 1000 kW 2 h = 2000 kWh (ii) 2000 kWh = 3.6 106 2000 = 7.2 109 J (108) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 20 (e) The equivalent circuit is shown alongside. P In series, RS = R1 + R2 = 10 + 10 = 20 . Now, in parallel, 1 RP 3 2 1 1 + 20 5 4 +1 5 1 = = 20 20 4 RP = 4 equivalent resistance = 3 + 2 + RP =3+2+4=9 1 RP Hence, = Q 5 = Answer 4. (a) (i) A.C. generator produces alternating voltage : It converts mechanical energy into electrical energy. (ii) A generator works on the principle of electromagnetic induction. (b) Heat capacity Specific heat capacity (i) It doesn t depend on the mass of the body. (i) It depends on the mass of the body. (ii) Its unit is JK 1 . (ii) Its unit is Jkg 1K 1. Heat capacity = Q S.H.C. = Q m (c) Heat energy given by hot solid of specific heat capacity C = 60 C (100 25) = 60 C 75 = 4500 C Now, Heat energy taken by water = 150 4.2 (25 20) = 150 4.2 5 ( C of water = 4200 Jkg 1K 1 = 4.2 Jg 1K 1) Now, according to the problem 4500 C = 150 4.2 5 C = 150 4.2 5 = 0.7 Jg 1K 1 4500 (i) 3 108 ms 1. (ii) Screen is coated with fluorescent mateiral such as zinc sulphide containing a trace of manganese. (e) Sources of background radiation are : (i) Naturally occurring radioisotopes. (ii) Radiation produced in the emission from nuclear power plants. (d) SECTION - II (40 MARKS) Answer 5. (a) (b) (c) (i) Constant speed in a circular path. (ii) Centripetal force. The direction of the force is along the radius towards the centre. (i) 1. Class II lever : Here, load is between effort and the fulcrum. 2. Class III lever : Here, effort is between load and fulcrum. (ii) Because inclined plane is a simple machine and acts as a force multiplier. Hence, less effort is needed. (i) We know, W = mg 400 W m = = = 40 kg. g 10 (109) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES Now, K.E. = v2 = 2K.E. 2 500 = 40 40 ( K.E. = 500) 100 10 = = 5 ms 1 2 4 1. The driving gear has more number of teeth than driven gear. 2. For a gain in torque, the number of teeth in the driven gear should be more than the number of teeth in the driver gear. (ii) 1 2 1 mv = 40 v2 2 2 v = Answer 6. (a) (i) Critical angle is the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90 . (ii) The refractive index and critical angle c are related as = (iii) Remains the same. (b) (i) A 1 sin c P Emergent ray Q 45 45 B C (ii) 180 is the angle of deviation. (iii) Periscope. (c) (i) Beyond 2F1. (ii) A F2 B 2F1 2F1 A Answer 7. (a) B O F1 (i) Resonance is a special case of the forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is called Resonance. (ii) Forced vibration Resonance 1. The amplitude of vibration is usually small. 2. The vibrations of the body are not in phase with the external periodic force. (b) 1. The amplitude of vibration is very large. 2. The vibrations of the body are in phase with the external periodic force. (i) Let d1 and d2 is the distance of the two cliffs from the observer, then 2d 1 2d1 = v 330 2d1 = 990 d1 = 495 m Similarly, for the second cliff, 3 = 4 = 2d 2 330 (110) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES 4 330 = 660 m 2 Distance between the two cliffs = 495 + 660 = 1155 m No echo will be heard when the distance between source and reflector is 10 m because it is less than the minimum required distance of 17 m. Damped vibrations. When a slim branch of tree is pulled and then released. The amplitude of vibrations decreases due to the frictional force which the surrounding medium exerts on the body vibrating in it. Body will stop vibrating. d2 = (ii) (c) (i) (ii) (iii) (iv) Answer 8. (i) Terminal voltage < emf i.e., E > V in closed circuit. emf is equal to the terminal voltage when no current is drawn. (ii) Fuse is used to limit the current in an electric circuit. Its purpose is to provide safeguards to the circuit and the appliances connected in that circuit from being damaged. (iii) Fuse wire is made up of a material of high resistivity and of low melting point so that it may easily melt due to overheating when current in excess of the prescribed limit passes through it. (b) (i) Energy spent = (Current)2 Resistance Time E = I2 Rt. (ii) 220 V is supplied to our houses. (iii) Bulbs are connected in parallel in buildings. (c) (i) In parallel, (a) 1 1 1 R P = R1 + R 2 1 1 1 R P = 8 + 12 1 3+2 5 = = RP 24 24 24 RP = = 4.8 5 Hence, equivalent resistance = 4.8 + 7.2 = 12 . 6 V = = 0.5 A 12 R (iii) Potential difference across 7.2 = 7.2 0.5 = 3.6 V (ii) Current = Answer 9. (a) (i) Q = mass (m) specific heat capacity (C) fall of temperature ( T). (ii) 25 K. (iii) Average kinetic energy remains same. Since temperature remains same. (b) Water (100 C) Steam (100 C) 0 C D Te m p. Ch an ge Temperature Phase Change A Phase B Change t1 t2 t3 t4 Time (Sec.) (i) AB represents the change of phase when ice at 0 C changes to water at 0 C. (although time increases 0 to t1). (111) e-mail : epildelhi@gmail.com EVERGREEN MODEL TEST PAPERS ICSE 100% SUCCESS IN PHYSICS-10 EVERGREEN SERIES (ii) BC represents the change in temperature from 0 C to 100 C when water increases its temperature and time reaches from t1 to t2. (iii) CD represents the change of phase when water at 100 C changes to steam at 100 C (no change in temp.) but time changes from t2 to t4 as latent heat of vaporization of steam is 2268 Jg 1. (c) According to the principle of calorimetry : Heat lost by water = Heat gained by ice to melt + Heat gained by water to reach 10 C Heat lost by water = m 4.2 (60 10) = 210 m Heat gained by ice to melt, mL = 40 336 = 13440 J Now, heat gained by water to raise its temperature to 10 C is = 40 4.2 10 = 1680 J So, 210 m = 13440 + 1680 m = 15120 = 72 g 210 Answer 10. (a) (i) EP NP NS ES NS > NP for a step-up transformer. (ii) (iii) (iv) (b) (i) (ii) The core losses are : 1. Hysteresis loss 2. Eddy current loss It works on the principle of electromagnetic induction. If D.C. is supplied instead of A.C., the transformer will not work. The emitted particle is -particle. The required equation : 14 0 X14 6 Y 7 + e 1 (iii) Oxidation does not alter the radioactivity since the nucleus remains the same. (c) In a cathode ray tube, if the anode plate is absent and the filament circuit is switched on, then the emission of electron takes place. As a result the cathode becomes positively charged and holds the emitted electrons close to it. These thermions surround the cathode in the form of a cloud. This collection of large no. of electrons near the cathode is called space charge. (112) e-mail : epildelhi@gmail.com

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

Formatting page ...

 

  Print intermediate debugging step

Show debugging info


 

 

© 2010 - 2025 ResPaper. Terms of ServiceContact Us Advertise with us

 

lovelyromana chat