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KARNATAKA ICSE SCHOOLS ASSOCIATION ICSE STD. X Preparatory Examination 2025 Subject: PHYSICS (SCIENCE PAPER 1) Maximum Marks: 80 Time Allowed: Two hours Date: _________ ANSWER KEY SECTION A (Attempt all questions from this Section.) Question 1 Choose the correct answers to the questions from the given options. (Do not copy the question, write the correct answers only.) i) (b)0.347 N ii) (b) 1: 2 iii) (c)Photocell iv) (a)A machine acts as a force multiplier as well as a speed multiplier simultaneously v) (b) Red vi) (a) C1< C2 < C3 vii) (b) zero viii) (a) 0.6 m ix) (c)Potential of earth and neutral wire is always the same x) (d)Assertion is false but reason is true. xi) (b) 5R/4 xii) (a)Power xiii) (c)67200J xiv) (b) S2 is louder and has a lower pitch xv) (c) 2 beta and one alpha particle Question 2 i) (a) (b) (c) (d) (e) (f) [15] blue thicker inwards kinetic energy class II speed ii) Velocity ratio =4 Efficiency =60% Efficiency =Mechanical advantage /velocity ratio Mechanical advantage = 4 =2.4 [1] [1] 1 iii) Tuning fork B starts vibrating and a loud sound is heard [1] Resonance [1] The air column of B starts vibrating with the frequency of fork A. Since the frequency of these vibrations is the same as the natural frequency of the fork B. Then fork B starts vibrating under resonance. Question 3 i) (a) kilogram [1] (b)W =F S Cos [1] ii) Let the real depth of the water tank =x cm Apparent depth =(x-80 ) cm =Real depth /Apparent depth = [1] 4x -320 =3x x=320 cm [1] iii) Let R1 and R2 be the value of two resistances Rs=15 ohm Rp = ohm R1+ R2= 15 --------------(equ.1) =10/3 On solving R1 R2 =50 Ohm R2 = Substituting in equ.(1) R1 + = 15 R12 -15 R1 +50 =0 (R1-10) (R1-5) =0 R1= 10 ohm [1] R1+R2 =15 10 +R2 =15 Therefore R2= 5 ohm [1] 2 iv) Secondary coil Turn ratio n< 1 n= = [1] So current in the secondary coil is more . As more current is passing in the secondary coil ,in order to reduce its resistance and therefore reduce the loss of heat energy in the coil ,secondary coil is thick. [1] v) (a)Moment of couple =Force x distance between forces. 40 x 5 =200 Nm [1] (b) No, there is no change in the moment of couple because the moment of couple is independent of the reference point. [1] vi) Inside the nucleus, 0 a) 0n1 1P1 +-1 e b) Isobars [1] [1] vii) (a) Q1=Q2 m1 c1 =m2c2 t = = [1] c1:c2 =3:2 [1] (b) The amount of heat energy required to raise the temperature of one gram of copper by 10 C is 0.4 J [1] Section B Question 4 i) (a) Convex lens [1] (b) Ray diagram [2] [Position of object between F1 and 2F1 & Magnified, real and inverted image is formed beyond 2F2] ii) Angle of minimum deviation =370 Angle of incidence corresponding to minimum deviation =460 For deviation = i1+ i2 -A [1] [1] 51=40 + X 62 X=51+ 22 = 730 [1] 3 iii) (a) convex lens (b) f=20 cm v=60 cm = [1] = = -1/u = [1] =2/60=1/30 u= ! "# magnification = m= = [1] [1] =-2 Question 5 i) (a)$ = &'( ) Sin C = [1] * (b) Red light -more than 45 [2] 0 Blue light -less than 45 ii) (a) Scattering of light [1] (b) Scattering is the process of absorption and then remission of light energy by the dust particles and air molecules present in air [1] (c) Size of the air molecules should be smaller than the wavelength of incident light. [1] iii) (a) Lateral displacement [1] (b) The perpendicular distance between the path of the emergent ray and the direction of incident ray. [1] (c) More is the refractive index of the medium, more is the lateral displacement. [1] (d) Lateral displacement increases. [1] Question 6 i) (a) one kilowatt hour =3.6 10 6 J [1] (b) Zero [1] Centripetal force on the body at any instant is directed towards the centre of the circular path and displacement is tangent to the circular path Displacement is normal to the direction of force. So, work done is zero 4 [1] ii) Solution: Let m be the mass of the ball and height of the ball after rebound be H Initial Potential energy of the ball =mgh =m 10 10 =100m Since 100m is the total energy, then the reduced energy is =100m (40% of 100m) =100m - [1] 100m =100m - 40m =60 m [1] This is the energy of the ball when it bounces back. Therefore 60m =m g H H= - = =6m [1] Hence, the ball can bounce back 6 m high iii) The force acts in vertically downward direction -the force is called weight or gravitational force [1] The force acts in vertically upward direction at A. The force is called effort. [1] When the wheel barrow is in equilibrium. Effort x effort arm =Load x load arm E .30 + 60234 = .10 + 90 2678 30 34 E= 9- = 33.33 kgf : [1] [1] Question 7 i) (a) P2 is used to change the direction of effort to a convenient direction [1] (b) [1] (c) L = T+T =2T ; < = =2 9- =2 E=25 kgf [1] ii) Let the width of river be x cm When the boy stands on the bank of river V=2d/t V = -------------------------(equ. 1) When the boy moves 170m away from the bank Distance of boy from cliff = x+ 170 5 = Now V= = > . ? 2 ? -----------------------------------(equ.2) Comparing equations (1) and (2) we get = [1] ? 14x =12x +2040 x=1020 m Width of the river = x= 1020 m Substituting the value of x, we get V= = 340m/s [1] [1] iii) (a) The frequency of the note produced in (b) is 2 times that produced in (a) ( Frequency of note in (a) is n or [1] (b) Figure (b) [1] Pitch is directly proportional to the frequency of sound note. The frequency of note in fig.(b) is more than the frequency of note in (a) iv) a) @ < A < B [1] b) B < A < @ [1] Question 8 i) (a) Semiconductor (Graphite, Germanium ..) [1] (b) Specific resistance decreases with increase in temperature [1] (c) Constantan/Manganin [1] ii) The number of rotations of the coil in one second(speed of rotation of the coil) [1] Eddy current loss lamination of iron core prevents the formation of eddy current [2] Copper loss or heating in the coil -by taking thick wire for the coil. Hysterisis loss - the core is made of soft iron [any two] iii) (a) P=V2/R or R= V2/P = [1] =40 [1] 6 (b) Energy consumed by oven in joules in 1/2 hour. =P t = 1000J/s 1800 s = 1800,000 J [1] (c) E =P t t= C 9DE = F =15 h [1] Question 9 i) -To increase its heat capacity, it imparts sufficient heat energy at a slow rate to the food for its proper cooking. [1] -It keeps the food warm for a longer period [1] Melting point of ice decreases. [1] ii) (a) 1.Clockwise 2.Anticlockwise [1] [Draw the diagram and mark the direction of induced current] (b) Lenz s law /Fleming s Right hand rule [1] (c) Mechanical energy spent in moving the magnet is transferred into electrical energy [1] iii) Solution: Mass of water = 104g Temperature of copper =30 0 C Mass of copper =42 g Final temperature =10 0 C By the principle of calorimetry Heat lost by the hot body =Heat gained by the cold body mw cw(Tw-T) + mc cc (Tw-T)=mice L + micecw (T-Tice) (104 4.2 (30 -10) + 42 0.4 (30 -10) =m 336 + m 4.2 ( 10 -0) ? Mass of ice m = = [2] [1] : ? ? = 24 g [1] End 7
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