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CBSE XII Sample / Mock 2015 : PHYSICAL EDUCATION with Answers

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PHYSICAL EDUCATION Class: XII MODEL QUESTION PAPER 2015 Max Time: 03 Hrs Max Marks: 70 Question paper consists of 26 questions. All questions are compulsory 01 Mark question must be answered in 10-20 words. 03 Marks question must be answered in 30-50 words. 05 Marks question must be answered in 75-100 words. Q1 Give one most important mechanical difference between walking and running. (1) Q2 How extrinsic motivation sometime may kill intrinsic motivation? (1) Q3 What principles should be followed for goal setting? (1) Q4 What do you understand by self-esteem? (1) Q5 which motor quality does a senior citizen lack who finds difficulty in tying the shoe laces while sitting on the chair? (1) Q6 In which conditions knock out tournaments are better than Round Robin? (1) Q7 What does the organizers intend by stating that, Only such students shall participate in the Intramurals who have not represented the school in any Football Championship in the past and minimum 10 substitutions shall be compulsory in a 90 min game . (1) Q8 Explain any two objectives of participation in adventure sports. (1) Q9 Enlist the equipments required for camping. (1) Q10 Calculate the Physical Fitness Index using short formula for a 12 year old boy having completed Harvard step test for a duration of 3 minute and a pulse rate of 54 beats for 1 to 1.5 minute. Q11 Why does the weight lifters diet include lots of protien? (1) (1) Q12 Explain why are the angle of release for shot-put, javelin and discus throws different? Q13 Friction is a necessary evil . Justify your answer with suitable examples from sport. (3) (3) Q14 Regular physical activities cannot stop the clock of ageing; but definitely it can slow the process. Justify. (3) ---2--- Q15 How can the minimum muscular strength for children assessed? (3) Q16 Explain the advantages of correct posture. (3) Q17 In sports such as Boxing and Wrestling, the players tend to lose weight sharply. Explain the pitfalls of dieting. Q18 Elaborate the various leadership qualities one inculcates by participating in adventure sports. (3) (3) Q19 Once upon a time, during an athletic meet in stadium, 8 girls were on starting line ready for the race. With the sound of pistol, all 08 girls started running. Hardly they had covered 10 to 15 metres, when accidently one girl slipped and fell. Due to pain the girl started crying. As soon as the other 07 girls heard her cry; all of them stopped running, stood for while, turned back, and ran towards her. Suddenly, the girls returned, pacified her, joined their hands together, lifted her, walked together and reached the finishing line. The officials were shocked to see such scene and unity. Quite a many eyes were filled with tears. Based on the passage above, answer the following questions: (1x3)(3) 1. What values do they teach? 2. What quality girls have shown by walking together? 3. What was so special about the race? Q20 Suggest various methods for flexibility training to improve the optimum flexibility. (5) Q21 Participation in sport results in all-round development of personality. Justify. (5) Q22 Recall the adaptive affects that take place in our cardiovascular system after engaging in exercise for a longer period. (5) Q23 Draw a circuit training plan for developing strength among school children. (5) Q24 Describe the procedure for administering Rikli and Jones Sr Citizen Fitness Test. (5) Q25 Suggest at what age children should be exposed to weight training and justify your answer. (5) Q26 While specifying all calculations, prepare a knock-out fixture for 21 teams. (5) Class XII Physics Sample Question Paper (Marking Scheme) Time Allowed: 3 Hours Maximum Marks: 70 01. P.E. = = p E Cos P.E. is maximum when Cos 02. 4 7 = -1, i.e. = ( ) ( ) 0 (180 ) yellow Violet ( ) ( ) 03. The capacitor (1) 04. (1) 05. Cm (t) is the frequency modulated wave. (1) = 10-2 A 06. The current through the potentiometer wire = ( ) Potential drop per unit length of the potentiometer wire = = Balancing length (l) = 240 cm (given) = X (1) V= l= X (= 60 mV) ( ) (1) 07. Polarization Two Uses: Polaroids can be used in sunglasses, window panes, photographic cameras, 3D movie cameras (Any Two) ( + ) OR Diffraction; Condition: Size of the obstacle sharpness should be comparable to the wavelength of the light falling. (1+1/2) Any application (1/2) 08. E = Energy of the photon = h = ( ) = Wave length of the moving electron = = ( ) Momentum of the electron = p = ( ) = kg =2X kg ( ) 09. Rydberg formula for the wavelengths of spectral lines in hydrogen spectrum is =R( - The short wavelength limit = R( R= ) ( ) for the Lyman series would be )=R ( ) The short wavelength limit = R( - )= = = for the Balmer series, would be ( ) 4 X 913.4 = 3653.6 ( ) 10. a) LAN, WAN, Internet (1) b) Three (1) 11. Let the radius of each drop be r. The capacitance C of each drop is kr, where k is a constant. Also q = CV, V = 900 volt ( ) charge on each drop = q = (kr X 900) C ( ) Total charge on all the eight drops = Q = 8 q =7200 kr ( ) Let R be the radius of the large drop. Then R3 = 8 X r3 R = (8) r = 2r Capacitance ( ) of the large drop = kR = 2kr Potential of the large drop = = ( ) volt = 3600V ( ) 12. With key K2 open, the current I in the galvanometer is given by I = and When K2 is closed, the equivalent resistance, say R , of the parallel combination of S is given by = The total current, say ( ) drawn from the battery would now be = This current gets subdivided in the inverse ratio of S and G, would now be given by = ; Hence the current through = ( ) = = ( ) But = = = Or n RS + n S or (n-1) RS = R = RS + R ( ) +S - (n-1) S or (n-1) RS + ( ) This is the required expression When R S , we have = (n-1) S ( ) 13. We know that F is an attractive (-ve) force when the currents I1 and I2 are like currents i.e. when the product I1I2 is positive. Similarly F is a repulsive (+ve) force when the currents I1 and I2 are unlike currents, i.e. when the product I1I2 is negative. (1/2) Now F (I1 I2 ), when d is kept constant and F when I1I2 is kept constant. The required graphs, therefore, have the forms shown below. (1/2) (1+1/2+1/2) OR (i) P- Paramagnetic ( Q-Ferromagnetic materials ( (ii) (1) 14.The current, I, leads the voltage, V, by an angle tan Here and tan where = = = =1 = 450 = 500 (1/2) L = 100 , R = 400 ( ) The power factor becomes unity when = 00. Hence we need to adjust C to a new value C where = = 100 (1/2) ( ) Thus, phase angle is 450 with the current LEADING the voltage. To make power factor as unity we need to have Xc also equal to 100 ohms. For this C needs to have a value of 10 F. (1/2) We, therefore need to put an additional capacitor of (10-2), i.e, 8 F in parallel with the given capacitor (1/2) 15. Discussion on the observed inconsistency (1) Discussion on Removing the contradiction through the concept of an additional current, called the displacement current . (1 +1) (Pages 270 and 271 of NCERT Book, Part I) 16. Focal length of the convex lens = m (1/2) Let v be the position of the image, I, of the object formed by the convex lens alone. We then have = v = 150cm (1/2) Hence the distance of the image (formed by the convex lens alone ) from the convex mirror would be (150-50) cm ,i.e.,100 cm. This distance equals the radius of curvature of the convex mirror. (1/2) Hence focal length of the convex mirror equals100/2, i.e., 50 cm. ( ) Radius of curvature of the convex mirror R = LI LM ( ) (1/2) 17. Calculation of fringe width, the normal set up (1 ) Calculation of fringe width, in the changed set up (1) Observing that ( ) (Reference : NCERT Book, Part II, pages 363, 364) 18. Initial number of radioactive atoms, per unit volume, in the blood streams of persons A and B are ( /V) and ( /V ) respectively. ( ) After a time nT (T = Half life), these numbers would get reduced by a factor 2n. ( ) Hence N1= ( ) . And N2 = ( ) . ( ) = or ( ) =V ( ) Additional volume of blood needed by person B is V- =V V =( )V ( ) 19. The equivalent gates in the two cases are the OR gate and the AND gate respectively. (1/2 + ) (i) A combination of three NAND gates, connected in the manner shown, would be equivalent to an OR gate. (Fig 14.46 (b) Page 511) (1) (i) A combination of three NOR gates connected in the manner shown would be equivalent to an AND gate. (Fig 14.48 (b) page 512) and plot a graph between E (photon energy in eV) 20. One can calculate the values of and (1) (in nm-1). (1) The resulting straight line graph can be used to nm-1 (1/2) ( in nm-1) corresponding to E = 1eV (i) read the value of (1/2) , corresponding to (ii) read the value of = (iii) We have E = The slope of the graph (after appropriate adjustment of the units) would equal hc. Since h is known, one can calculate c. (1) 21. The equation of the (amplitude) modulated signal is sin This can be rewritten as sin Where = = sin (1) sin = modulation index sin sin + 2 sin cos (1/2) sin (1/2) cos (1/2) These are the three sinusoidal waves present in the amplitude modulated signal. The frequencies of these three waves are (1/2) and 22. (i) Heavy doping makes the depletion region very thin. This makes the electric field of the junction very high, even for a small reverse bias voltage. This in turn helps the Zener diode to act as a voltage regulator . (1) (ii) When operated under reverse bias, the photodiode can detect changes in current with changes in light intensity more easily. (1) (iii). The photon energy, of visible light photons varies from about 1.8 eV to 3 eV. Hence for visible LED s, the semiconductor must have a band gap of 1.8 eV. (1) 23. (i) Expression for the force in vector form (1) - Statement of Flemings left hand rule (ii) (iii) (1) Adaptation to different situations and flexible and adjustable attitude (1) - Sharing excitement in classroom learning with family members Avoiding unnecessary arguments in conflicting situations in everyday life (1) 24. (a) Statement of Gauss s law Simple Proof (1) (2) (Pages 33 and 34 of NCERT Book I) (b) Set be the uniform density of negative charge. We then have (1/2) Taking a sphere of radius r (centred at the nucleus) as the Gaussian surface, we have (1/2) Where Q is the net charge enclosed by the Gaussian surface. Now Q = (+Ze) + ( - . = Ze Ze ( ) ) = Ze ( 1- (1/2) ) Substituting this value of Q, we get (1- E(r) = = ( ) - (1/2) ) OR Derivation of the expression for (page 28 NCERT Book I) For r >> a, = (3) Thus, the graph has the form shown. (1) Derivation of the expression for torque (Page 31, NCERT Book I) (1) 25. Statement of Faraday s law of e-m induction (1) Derivation of the expression for induced emf (2) Justification on the basis of the concept of Lorentz s force (2) (Page 212 + page 213 NCERT Part I) OR Applying Kirchoff s loop rule to obtain expressions for (i) (ii) Current flowing in the circuit Inductive reactance of L ( (1/2) Finding expression for instantaneous power, Pi, the graph has the form shown. (1) (1) 26. (a) Explaining the use of the phenomenon of total internal reflection in (i) an optical fibre (NCERT Page 322 fig 9.16 + Explanation) (1/2 +1) (ii) a prism that inverts an image without changing its size. (NCERT Page 322 fig 9.15 9(c) + Explanation) (1/2 +1) (b) (1/2) The angle of incidence at the face AC = i But i = A (Angle between two lines is the same as the angle between their perpendiculars) Also = /2 A = /2 i (1/2) The minimum value of I, so that there is total internal reflection at the face AC, equals ic where ic = sin The maximum value of max (1/2) corresponding to the minimum value of i (= ic) is therefore, = /2 ic = /2 sin (1/2) OR Statement of Huygen s principle (1) Absence of back wave (Page 354 NCERT book part II) Drawing the refracted wave front (1) ( Obtaining Snell s law of refraction (Fig 10.5 Page 357 NCERT book part II) (

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