sample paper of physics for class 12th, Exercises of Physics

Download sample paper of physics for class 12th and more Exercises Physics in PDF only on Docsity! Material downloaded from myCBSEguide.com. 1 / 18 CBSE SAMPLE PAPER-01 Class – XI PHYSICS (Theory) Time allowed: 3 hours, Maximum Marks: 70 General Instructions: 1. All the questions are compulsory. 2. There are 26 questions in total. 3. Questions 1 to 5 are very short answer type questions and carry one mark each. 4. Questions 6 to 10 carry two marks each. 5. Questions 11 to 22 carry three marks each. 6. Questions 23 is value based questions carry four marks. 7. Questions 24 to 26 carry five marks each. 8. There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions in five marks each. You have to attempt only one of the choices in such questions. 9. Use of calculators is not permitted. However, you may use log tables if necessary. 10. You may use the following values of physical constants wherever necessary: , , , , , 1. If one mass of one electron is 9.11 x 10-31 kg, then how many electrons would weigh in 1 kg? Ans. 9.11 x 10-31 x n = 1 kg Therefore, n = 1.1 x 1030 2. What do you understand by the term conservative force? Ans. Any force is called conservative force if, a) Work done against is independent of path. b) Work done in a closed path is zero. Material downloaded from myCBSEguide.com. 2 / 18 3. Give reason: “Liquid set in rotation comes to rest after some line”. Ans. The liquid comes to rest due to the viscous force, due to internal fluid friction between its different layers. 4. What is the number of degree of freedom of a molecule of a diatomic gas at room temperature? Ans. A molecule of diatomic gas possesses five degrees of freedom at room temperature which is due to translational motion and rotational motion. 5. What is the slope of stress-strain body within the elastic limit? Ans. Within elastic limit, the slope of stress-strain curve gives the value of modulus of elasticity of the given material. 6. If the velocity at the maximum height of a projectile is half its initial velocity of projection u, then find its range on the horizontal plane. Ans. cos = = 600 Horizontal range =R = 7. What fraction of its mechanical energy is lost in each bounce, if a ball bounces to 80% of its original height? Ans. Let the ball fall from height h then, Kinetic energy of ball at the time of just striking the ground = Potential energy of ball at height h, K = mgh Similarly, on rebounding the ball moves to a maximum height h’, then kinetic energy will be K’ = mgh’ Loss of kinetic energy K – K’ = mgh – mgh’ = mg (h –h’) = mg (h – 80/100h) = mgh x (0.2) Fractional loss in K.E. of ball in each re-bounce = K –K’/K Material downloaded from myCBSEguide.com. 5 / 18 b) The tension in the 12 kg. Ans. Let the downward linear acceleration of the cylinder be a. if M be the mass of the cylinder then Mg – 2T = Ma T = ½ m (g – a) ------------------------- (i) Torque = moment of inertia x angular acceleration 2Tr = Iα 2Tr = ½ mr2 x (a/r) T = ma /4 ----------------------------- (ii) From equation (i) and (ii), m a/4 = ½ m(g – a) ---------------------- (iii) Solving equation (i) and (iii) Substituting the value of a in equation (ii) we get, 12. An object weighing 70 kg is kept in a lift. Find its weight as a recorded by a spring balance when the lift a) Moves upwards with a uniform velocity of ms-1 b) Moves upwards with a uniform acceleration of 2.2 ms-2 c) Moves downwards with a uniform acceleration of 2.8 ms-2 d) Falls freely under gravity Material downloaded from myCBSEguide.com. 6 / 18 Or A body of mass 2 kg is at rest at a height of 10 m above the ground. a) Calculate its potential energy and kinetic energy after it has fallen through half the height. b) Find the velocity at this instant. Ans. a. When the lift is moving upwards with a uniform velocity 5 ms-1, the reaction R or the pressure on the base is R = mg = 70 x 9.8 N = 686 N b. When lift is moving upwards with a uniform acceleration of 2.2 ms-2, the reaction R’ or the pressure on the bass increase and is given by R’ = m (g + a) = 70(9.8 + 2.2) N = 840 N c. When the lift descends with a uniform acceleration of 2.8 ms-2, the reaction R’’ is given by R’’ = m (g – a) = 70(9.8 – 2.8) N = 490 N d.When the lift falls freely under gravity , the reaction R’’’ is given by R’’’ = m (g – g) =0 The object appears to have become weightless. Or Total energy at B = kinetic energy + potential energy = 0 + mgh = 2 x 9.8 x 10 = 198 J It descends half the height, it loses potential; energy which is given by = ½ mgh = 98 J Its potential energy at C = (196 – 98) = 98 J Material downloaded from myCBSEguide.com. 7 / 18 The loss of potential energy = gain in kinetic energy = 196 – 98 = 98 J Kinetic energy = ½ mv2 ½ x 2 x v2 = 98 v2 = 98 = 7√2 m/s 13. If the potential energy of a spring when stretched through a distance ‘a’ is 25 J, then what is the amount of work done on the same spring so as to stretch it by an additional distance ‘5a’? Ans. P. E = = 25 Additional distance of 5a becomes 6a Substituting ‘a’ we get = 900 J Additional work done = 900 – 25 = 875 J 14. If a copper plate has an area of 250 cm2 at 00 C, then calculate the area of this plate at 600. Given: The coefficient of linear expansion of copper = 1.7 x 10-50C-1 Ans. A0 = 250 cm2 β = 2 α = 2 x 1.7 x 10-5 = 3.4 x 10-50C-1 ΔT = (60 – 0) = 600 A = A0 (1 + βΔT) = 250(1 + 3.4 x 10-5 x 60) = 250(1 + 0.00204) = 250.51 cm2 Area of copper plate at 600 = 250.51 cm2. 15. If a progressive wave and a stationary wave have frequency 300 Hz and the same wave velocity 360 m/s, then calculate, (i) The phase difference between two points on the progressive wave which are 0.4 m apart. Material downloaded from myCBSEguide.com. 10 / 18 Let A and B be two points on velocity – time graph corresponding to the instants t1 and t2. As the motion is uniform hence AA1 = BB1 = v Area under v-t graph between t1 and t2 = area ABB1A1 = AA1 x A1B1 = v(t2 – t1) Velocity is defined as v(t2-t1) = x2-x1 Area ABB1A1 = (x2-x1) Hence displacement of a particle in time interval (t2-t1) is numerically equal to the area under velocity- time graph between the instant t1 and t2. 19. If two bodies of different masses m1 and m2 are dropped from two different heights a and b, give the ratio of time taken by the two bodies to drop through these distance? Ans. Let t1 and t2 are the time taken by two bodies of masses m1 and m2 to drop from heights ‘a’ and ‘b’. Using equation of motion h = ut + ½ at2 u = 0 and a =g t1:t2 = 20. What will be the energy shared in the spring at the instant when the 10 kg mass has acceleration 12 m/s2 if two masses 10 kg and 20 kg are connected by a massless spring. A force of 200 N acts a 20 kg mass? Material downloaded from myCBSEguide.com. 11 / 18 Ans. Since F = ma F = 10 x 12 = 120 N F = k = 2400 Energy stored in the spring E = ½ kx2 = 3J 21. In the diagram given below, a tangential force of 2 kg wt is applied round the circumference of the flywheel with the help of a string and mass arrangement. Now, if the radius of the wheel is 0.1 m, find the acceleration of the mass. Assume that the moment of inertia of a solid fly wheel about its axis is 0.1 kg m-2. Ans. Let ‘a’ be the linear acceleration of the mass and ‘T’ the tension in the spring. It is clear that mg – T = ma ----------------------- (i) Let the angular acceleration of the flywheel be ‘α’. The couple applied to the flywheel is I α = TR ------------------------- (ii) The linear acceleration α and angular acceleration are related to each other as a = Rα ------------------------- (iii) Material downloaded from myCBSEguide.com. 12 / 18 From equation (i), (ii) and (iii) Mg – Iα/R = m R α It is given that m = 2kg, R=0.1m and I=0.1kgm2. Substituting these values we get, rad s-2 = 16.7 rad s-2 22. If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth, then a. Determine the height of the satellite above the earth’s surface. b. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed with which it hits the surface of the earth Ans. a. In this problem vu =1/2 vu = h = R = 6400 km b. By conservation of ME. = 8 km/s 23. Pranav was climbing the stairs to meet his friend. On his way he saw a person was hypnotizing a lady with a pendulum moving to and fro. The lady was giving all her gold Material downloaded from myCBSEguide.com. 15 / 18 Derive the expression for excess pressure inside: a) A liquid drop . b) A liquid bubble. c) An air bubble. Ans. Mass of the diamond = 175 g Density = 3.5 gcm-3 Volume = = 50cm3 If the original volume of the diamond were V, then V = 50 + ΔV Where ΔV is the increase in volume under the pressure during its formation, Bulk modulus = B = Substituting (V – 50) for ΔV and the values of P and B, we have V = 40V – 2000 39V = 2000 V = 51.28 cm3 - can be calculated as, Now adding this value to the present value giving V = 51.25 cm3. The difference is only in the second decimal place, less than 0.06%. Hence the original volume of the diamond must have been equal to 51.3cm3. Or (a) Let r =radius of a spherical liquid drop of Centre O, T = surface tension of the liquid. Let Pi and Po be the value of pressure inside and outside the drop. Material downloaded from myCBSEguide.com. 16 / 18 Excess pressure inside the liquid drop = Pi -Po Let r be the increase in its radius due to excess pressure. It has one free surface outside. Increase in surface area of the liquid drop = 4π(r + r)2 - 4πr2 = 4 [r2 + ( r)2 + 2r r – r2] = 8 r r Increase in surface energy of the drop is W = surface tension x increase in area = T x 8 r r W = force due to excess of pressure x displacement ---------- (i) = Excess pressure x area of drop x increase in radius = (Pi – Po) 4 r2 r ------------ (ii) From equation (i) and (ii) we get, (Pi –Po) x 4 r2 r = T x 8 r r Pi – Po = (b) Inside the liquid Bubble: A liquid bubble has air both inside and outside it therefore it has two free surfaces. Increase in surface area = 2[4 (r+ r)2 - 4 r2] = 2 x 8 r r = 16 r r W = T x 16 r r ----------------------- (1) W = (Pi – Po) 4 r2 x r ---------------- (2) From equation (1) and (2) (Pi-Po) x 4 r2 x r = T x 16 r r Pi-Po = (c) Inside an air bubble: Air bubble is formed inside the liquid, thus air bubble has one free surface inside it and liquid is outside. r = radius of the air bubble, r = increase in its radius due to excess of pressure (Pi-Po) inside it, T = surface tension of the liquid in which bubble is formed Increase in surface area = 8 r r W = T x 8 r r W = (Pi-Po) x 4 r2 r (Pi-Po) x 4 r2 r =T x 8 r r Pi-Po = . Material downloaded from myCBSEguide.com. 17 / 18 26. A SHM is expressed by the equation x = A cos (ωt + φ) and the phase angle φ =0. Draw graphs to show variation of displacement, velocity and acceleration for one complete cycle in SHM. Or If two tuning forks A and B give 9 beats in 3 seconds and a sound with a closed column of air 15 cm long and B with an open column 30.5 cm long, then calculate their frequencies Ans. Let x = A cos (ωt + φ) and if phase angle φ is zero x = A cosωt, v = -Aω sin ωt a = v = -Aω2 cosωt = - ω2x The values of ‘x’, ‘v’ and ‘a’ at different times over one complete cycle as follow: Using the given data, x – t, v– t and a – t graphs are plotted as shown below: