Grade XI Physics Mock Test #GrowWithGreen
Mock Test Physics Questions General Instructions: (i) (ii) (iii) (iv) (v) (vi) (vii) The question paper comprises 4 Sections, A, B, C and D. You are to attempt all the sections. All questions are compulsory. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions. Question numbers 1 to 5 in section A are one-mark questions. Question numbers 6 to 12 are two-marks questions. Question numbers 13 to 24 are three-marks questions. Question numbers 25 to 27 are five-marks questions. Questions Q1. Find the vector product of the given vectors. Section A Discuss in brief whether the function is periodic or not. 1 Q2. Under what conditions can a real gas be approximated to an ideal gas? 1
Q3. Why are beams between railway tracks given an I shape? 1 Q4. A ping-pong ball and a cricket ball are moving in air with the same velocity. Which ball is easier to catch? In a collision of two objects, is momentum always conserved? Validate your answer with an example. 1 Q5. Draw a velocity time graph illustrating a uniformly accelerated motion. 1 Section B Q6. The cone of a loudspeaker of a music player vibrates in SHM at a frequency of 300 Hz where the amplitude at the centre of the cone is 6 10-4 m and x = A at t = 0. (A) Find the equation that describes the motion of the centre of the cone. (B) Find the position of the cone at t = 2.4 ms. An object experiences a simple harmonic oscillation along the x = axis and its position varies with time according to the equation Determine the amplitude, frequency and period of the motion. 2 Q7. What are thermodynamic state variables? Give an example of thermodynamic state variable. Why is that quantity called thermodynamic state variable? 2 Q8. Verify that, obtained from kinetic theory of gases, is dimensionally correct. 2 Q9. A ray of light is perpendicularly incident on the face of a right-angled prism ( n = 1.41). Find the largest value of angle θ so that the ray is totally reflected off the oblique face of the prism if it is immersed in air. 2
Q10. Two express trains A and B, each 400 m long, are running on straight parallel tracks heading in the same direction. B travelling with 40km/hr, which is just behind A, overtakes A which is travelling with a speed of 30km/hr. Calculate the time taken by B to overtake A and the distance covered by A in this time. 2 Q11. Establish and state the relationship between angular momentum and torque. 2 Q12. An artificial satellite of mass 500 kg orbits at a height of 750 km above the Earth s surface. If the radius of the earth is 6.4 10 6 m and the acceleration due to gravity at its surface is 9.81 m/s 2, then find a) the weight of a 5 kg object at the height of the satellite b) the centripetal acceleration experienced by the satellite 2 Section C Q13. Derive an expression for final velocities of two point objects colliding elastically with each other. A 15 g bullet is fired into a stationary block of wood ( m = 5 kg). The relative velocity of bullet inside the block is zero. The speed of the wood after collision is 0.5 m/s. Find the original speed of the bullet. 3 Q14. A simple pendulum of length 0.5 m has bob of mass 150 g. It is released from an angle of 15 o with the vertical. (A) Find the frequency of vibration assuming simple harmonic motion. (B) Find the speed of bob when it passes through the lowest point of oscillation. (C) Find the total energy stored in this oscillation. 3 Q15. The given figure shows the strain-stress curve for a given material. 3
Calculate (i) Young s modulus of the material (ii) Ultimate strength for the material (iii) stress corresponding to the fracture point Q16. The mass and radius of a planet X are half of the mass and radius of earth. Obtain the ratio of the escape speed from the surface of the planet to that from the surface of the earth. 3 Q17. Show that the path of a projectile is a parabola. Show position-time graph of an object if the object is moving with 1. zero acceleration 2. positive acceleration 3. negative acceleration 3 Q18. What is the total amount of heat energy required to convert 5 kg of ice at 20 C into steam at 100 C? The pressure stays constant throughout. [Specific heat of ice = 2100 J/kg/K 1, Latent heat of fusion of ice = 3.36 10 5 J/kg, Specific heat of water = 4200 J/kg/K, Latent heat of vaporization of water = 2.25 10 6 J/kg] 3 Q19. A car of mass 2500 kg and moving with a velocity of 15 m/s is suddenly brought to a halt. If the tire brakes start to cool off and if the temperature of the surrounding is 25 C, then what will be the change in entropy? Establish the relation,. 3 Q20. (i) Name the principle which helps us to determine the speed of efflux. 3
(ii) A water tank has a small hole in its side at a height 25 cm from the bottom. The height of the water column in the tank is 1 m. The air pressure inside the tank above the water column is 1.05 10 5 Pa. Calculate the velocity with which water will come out from the small hole. Q21. A rod of length 15 cm is kept along the principal axis of a concave mirror of focal length 20 cm. The closer end of the rod is at a distance of 25 cm from the mirror. Find the position of the image. In the following diagram, an object 'O' is placed 15 cm in front of a convex lens L 1 of focal length 20 cm and the final image is formed at I at a distance of 80 cm from the second lens L 2. Find the focal length of the lens L 2. Q22. A large number of spherical raindrops of similar size are falling with a terminal velocity of 0.13 m/s. They combine together and form a large spherical drop. The terminal velocity of the large drop is 1.17 m/s. Calculate the number of small raindrops. 3 Q23. A 2 kg block is initially at rest on a horizontal frictionless table and a horizontal force in the positive direction of x -axis is applied to the block. The force is given by, where x is in metre and the initial position of the block is x = 0. What is the kinetic energy of the block when it passes through x = 2 m? 3 Q24. A microscope of magnification 600 uses objective lens of focal length 0.42 cm. The length of the microscopic tube is 16.2 cm. Find the focal length of the eyepiece. Assume normal adjustment. 3 Section D Q25. A stone of mass m is whirling in a vertical plane with the help of an attached string of length l. Velocity of the stone at the lowermost point is u. Obtain general expressions for the velocity and tension in the string at any point. Also, find the minimum velocity at the 5 3
lowermost point to perform vertical circle. (i) Derive an expression for the maximum velocity with which a vehicle can take a turn on a circular banked track. (ii) A car is about to take a turn on a circular banked track of radius 3 m. The track is banked at a 45 angle. The coefficient of static friction between the track and the tyres is 0.11. What should be the maximum velocity of the car in order to avoid slip? Q26. What is Carnot theorem? Describe Carnot cycle and derive an expression for efficiency of the Carnot engine. (a) A scientist claims to have constructed an engine that has an efficiency of 55% when operated between the boiling and freezing points of water. Can his claim be true? Give reasons. (b) Prove that the efficiency of all reversible heat engines operating between two reservoirs is the same? 5 Q27. (a) State and prove the theorem of parallel axis. (b) What is the moment of inertia of uniform circular disc about an axis passing through it circumference and perpendicular to its plane? The mass of the disc is 250 gm and its radius is 25 cm. A cylinder measures 100 cm in length and has a radius of 20 cm. A sphere of radius 20 cm is cut out from the cylinder, such that the centre of the sphere lies at a distance of 30 cm from the central axis of the cylinder (as shown). By what distance does the centre of mass of the cylinder shift, after cutting out the sphere? 5
Grade XI Physics Mock Test Solutions #GrowWithGreen
Mock Test Physics Solutions 1. Vector product is defined as Section A The function is not periodic. With increasing time, the function decreases monotonically and tends to zero as time tends to infinity. Hence, the function never repeats. 2. A real gas can be approximated to an ideal gas when its density is very low. This can be brought at very low pressure and very high temperature. 3. The beams on railway tracks have been given an I-shape because: (i) The shape provides a larger cross-sectional area for bearing load to prevent buckling of the beam under load. (ii) The shape reduces the weight of the beam without sacrificing the strength. This also helps in reducing the cost. 4. The mass of a ping-pong ball is less than a cricket ball. Now, we know that the momentum transferred directly depends on mass and velocity of the body and inversely to the time in which the body is brought to rest. Since, mass of ping-pong ball is more than cricket ball, so it will be relatively easier to catch a ping-pong ball than a cricket ball.
No, momentum is not always conserved. For example, momentum is not conserved when external forces (along with internal forces) act on colliding objects. 5. Section B 6. (A) We have ω = 2πf = 2 3.14 300 rad/s = 1884 rod/s The motion is described as X = A cos (ωt) => X = (6 10-4 ) cos (1884t) (B) At t = 2.4 10-3 s, the above equation gives x = (6 10-4 m) cos [(1884) 2.4 10-3 ] = (6 10-4 m) cos (4.5216 rad) = (6 10-4 m) ( 0.1896) x = 1.138 10-4 m The standard equation of motion for an oscillatory motion is So, comparing this equation with the given equation, we get amplitude Angular frequency, w = 3π rad/s
So, frequency, And time period is 7. Macroscopic variables (such as pressure, volume, temperature, mass) used for describing the equilibrium state of a thermodynamic system completely are called state variables. The values of state variables depend only on the given state of a system. Internal energy is one of the examples of the thermodynamic state variables because it describes the state of a system in terms of the total kinetic and potential energy of individual molecules forming the system. It does not depend upon the path taken by the system to bring about a change in the internal energy. Any change in internal energy depends only on the initial and final states and not on the path taken by the system to bring about a change. 8. According to the kinetic theory of gases, L.H.S:- R.H.S:- Hence, we see that is dimensionally correct. 9. Let the point of incidence of the ray on face AC be D. According to the figure,
X = 2 θ c (Alternate angles) Y = 90 θ c (Normal to surface AC) In triangle DEC, we have X + Y + θ = 180 Thus, 2 θ c + 90 θ c + θ = 180 Or, θ = 180 (90 + θ c ) = 90 θ c Thus, the minimum value of θ c (for total internal reflection to take place) limits the maximum value of θ. n = Refractive index of prism material with respect to air 10. The relative velocity of train B with respect to train A is given as V BA = V B V A V BA = 40 (30) Thus, V BA = 10 km/hr Now, the total time taken by B to overtake Here, the total distance travelled would be the sum of the lengths of the two trains which would be 400 m + 400m = 800 m = 0.8 km So, we have Thus, t = 0.08 hr So, during this time, the total distance covered by A is d = (V A t) So, by using appropriate values, we get d = (30 0.08) d = 2.4 km
Thus, total distance travelled d = 2400 m 11. For a rigid body, the angular momentum is given as Now, by differentiating the above equation with respect to ' t ', we get Here, angular acceleration Now, we know that the torque is given as By substituting this value in equation (2), we can arrive at a relation between angular momentum and torque. Thus, So, the torque acting on a (rigid) body is defined as the rate of change of angular momentum exhibited by the body. 12. a) The variation of g due to height ( h ) is given as Thus, g' = 8.78 m/s 2
Now, the weight would be given as W = mg' So W = 5 8.78 or W = 43.9 N b) The centripetal acceleration (a) can be calculated using the following formula for centripetal force So
Thus The centripetal acceleration experienced by the satellite will be a = 7.86 m/s 2 Section C 13. We know that linear momentum and kinetic energy are conserved in an elastic collision, m 1 v 1 + m 2 v 2 = m 1 u 1 + m 2 u 2 m 2 ( v 2 u 2 ) = m 1 ( u 1 v 1 )... (1) Also,
On dividing equation (2) by (1), Now, on solving, we obtain: Similarly, Mass of bullet m b = 15g = 0.015 kg Mass of wood m = 5 kg Initial speed of bullet = u Initial speed of wood v w = 0 Final speed of bullet-plus-wood combination = v = 0.5 m/s Applying liner conservation principle of momentum, we get, m b u + m 0 = ( m b + m ) v 0.015 u = (0.015 + 5) (0.5)
14. (A) Frequency of the pendulum, f = => f = => f = 0.7046 Hz (B) Let us assume the lowest point as the zero level of gravitational potential energy. According to the law of conservation of energy, we have energy at top = energy at bottom KE top + PE top = KE bottom + PE bottom 0 + mg ( L L cos ) =
= 0.5778 m/s (C) Total energy at the bottom is also the total energy throughout the oscillation. So, E total = 0.025 J 15. (i) It is evident from the given graph that for stress 8 10 7 N/m 2, strain is 4 10 3. Young s Modulus, Hence, Young s modulus for the given material is 2 10 10 N/m 2. (ii) Ultimate strength of a material is the maximum stress that the material can sustain without crossing the elastic limit. It is evident from the given graph that this material has the ultimate strength equal to 16 10 7 N/m 2 or 1.6 10 8 N/m 2. (iii) The fracture point exists beyond the yield strength. It can be observed in the graph that stress for the fracture point is approximately equal to 14 10 7 N/m 2 or 1.4 10 8 N/m 2. 16. Escape speed from the surface of a planet is given by the relation, Let ME and RE b e the mass and radius of the earth respectively. Hence, acceleration due to gravity for the earth is given as
It is given that mass and radius of a planet X is half of the mass and radius of earth. Hence, the expression for acceleration due to gravity for planet X is given as Escape velocity from the surface of the earth, Escape velocity from the surface of planet X, Hence, the required ratio is 1:1. 17. Suppose an object is thrown upward at an angle θ with the horizontal with an initial velocity of v. The acceleration acting on the object is due to gravity only, which is acted vertically downwards. Hence, the expression for acceleration vector can be written as This equation suggests that the acceleration is directed along the y -axis only. Hence, the two components of acceleration can be written as a x = 0 a y = g.(1) The motion of the projectile is represented in the given figure as
Where, the components of initial velocity are given as The horizontal range of the projectile is given as The height reached by the projectile in time t is given as Substituting the value of t in equation (4), we obtain
This is the equation of a parabola. Hence, the path of a projectile is parabola. (1) An object moving with zero acceleration possesses uniform motion i.e., moves with a constant velocity. Its position-time graph is a straight line as represented in the given figure. (2) An object moving with a positive acceleration is said to have accelerating motion. Its position-time graph is a curve with positive curvature as represented in the given figure. (3) An object moving with a negative acceleration is said to have decelerating motion. Its position-time graph is a curve with negative curvature as represented in the given figure.
18. Now, the following process takes place So, the total heat energy required will be Q t = Q 1 + Q 2 + Q 3 + Q 4 = mc i ΔT i + m L i + mc w ΔT w + m L w Here, m = 5 kg ΔT i = 0 C ( 20 C) = 20 C ΔT w = 100 C 0 C = 100 C c i specific heat of ice c w specific heat of water L i latent heat of fusion L w latent heat of vaporization
So, Q t = (5 2100 20) + (5 3.36 10 5 ) + (5 4200 100) + (5 2.25 10 6 ) = 2.1 10 5 + 16.8 10 5 + 21 10 5 + 112.5 10 5 Thus, the total heat energy required will be Q t = 1.524 10 7 J 19. Now, we know that the change in entropy is given as Here, the heat energy (Q) is transferred by the change in kinetic energy of the system Or, So, Here, m = 2500 kg V = 15 m/s and T = 25 C = 25 + 273 = 298 K Thus, by using the values, we get, Thus, the change in entropy will be ΔS = 943.8 J/K First law of thermodynamics is given as Where, = Amount of heat supplied to a system = Change in internal energy of the system = Work done = p
= Specific heat capacity at constant volume = Specific heat capacity at constant pressure Specific heat is given by the relation, At constant pressure p, specific heat is given as 20. (i) The speed of efflux from the side of a container is given by the application of the Bernoulli s principle. (ii) It is given that, Height of the small hole from the bottom = 25 cm = 0.25 m
Height of the water column = 1 m Hence, depth at which small hole is located, h = 1 0.25 = 0.75 m Water density, ρ = 10 3 kg/m 3 Air pressure inside the tank above the water column, P = 1.05 10 5 Pa Atmospheric pressure, P a = 1.01 10 5 Pa Acceleration due to gravity, g = 9.8 m/s 2 Speed of efflux, v, is given as Hence, the speed of efflux of water is 3.94 m/s. 21. The radius of curvature of the mirror,. Thus, the closer end of the rod is situated between radius and focus of the mirror. So, the image will be formed at the same place as that of the object. Now, distance of farther end from the pole = (25cm + 15cm) = 40 cm Now, using the mirror formula, we get v = 40 cm As per the figure, The image formed by lens L 1 is at I'. Therefore, using lens formula
As per the parameters given in the question So, the image distance will be Now, this image is acting as an object for the lens L 2. We can again use the lens formula and other parameters given in the question and question figure to find the focal length of lens L 2. Here, So, the focal length of the lens L 2 = 40 cm. 22. Number of small rain drops = n Radius of each rain drop = r Radius of the larger rain drop = R Terminal velocity of small drop, v = 0.13 m/s Terminal velocity of the larger drop, v = 1.17 m/s Volume of the larger rain drop is equal to the volume of n small raindrops.
Terminal velocity is given by the relation, Hence, the ratio can be written as Equation (1) reduces to Hence, the number of raindrops is 27. 23. Using work energy theorem W (all force) = k f k i W (T) + W (N) + W (Mg) + W ( f ) = k f k i (1) W (N) = 0 [Since displacement is perpendicular to N and Mg] W (Mg) = 0 k i = 0 [Initially at rest] k f =?
Given that, F x = (3 x 2 ) N Plugging these values in equation (1), we get, 0 + 0 + 3.33 = k f = 0 k f = 3.33 J 24. Magnification of microscope in normal adjustment
f e = 1.60 cm 25. The given situation can be represented as Section D M and P are the lowermost and topmost points respectively. T P is the tension in the string when the stone is at point P. From the figure, it can be written that MQ = OM OQ = l l cos θ = l (1 cos θ ) Velocity at point M = u Velocity at point N = v N v N can be obtained by using the relation, Equation (1) represents the velocity of the stone at any point N on the vertical circle. At point N at equilibrium,
Equation (2) represents the tension in the string when the stone is at any point N on the vertical circle. Tension in the string should be non-zero at the topmost point P to perform a complete circle. Hence, in order to perform a complete vertical circle, the minimum velocity required at the lowermost point is. (i) Free body diagram of a car taking a turn on a circular track is represented in the given figure.
θ = Angle of bank N = Normal reaction on the car f = Frictional force between the tyres and road m = Mass of the car g = Acceleration due to gravity At equilibrium, the net vertical force on the car is zero. Again, the net horizontal force on the car is also zero. Where, v = Velocity of the car R = Radius of the circular track = Centripetal force required for the circular motion The relation for frictional force is Equation (1) reduces to
Substituting the values of N and f in equation (2), we obtain (ii) It is given that, R = 3 m θ = 45 = 0.11 g = 9.8 m/s 2 Maximum allowed speed of the car is given by the relation, 26. Carnot's theorem states that No heat engine working at a given temperature has greater efficiency than that of a reversible engine working at the same temperature under same conditions. Working between the two given temperatures of the hot ( T 1 ) and cold ( T 2 ) reservoirs, efficiency of no engine can be more than that of a Carnot engine. Efficiency of the Carnot engine is independent of the nature of the working substance For a Carnot cycle
The sequence of steps constituting one cycle of a Carnot engine is called a Carnot cycle.it consists of four steps Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression The p-v curve describing the carnot cycle is shown as: The initial pressure,volume and temperature of the ideal gas enclosed in the cylinder P 1,V 1 and T 1. From the curve it is observed that the gas first expands isothermally to the state having pressure, volume and temperature to be P 2,V 2 and T 1. From this state it compressed isothermally to the state having pressure,volume and temperature to be P 3,V 3 and T 2.From this state it is compressed adiabatically to the state having pressure,volume and temperature to be P 1,V 1 and T 1. The efficiency of the carnot engine can be found by calculating the work done in individual cycle. Isothermal expansion of gas Heat absorbed ( Q 1 ) by the gas from the reservoir is the work done. ( W 1 2 ) by the gas,
Step 2 2 Adiabatic expansion of gas Work done by the gas, Step 3 4 Isothermal compression Heat released ( Q 2 ) by the gas to the reservoir is the work done ( W 3 4 ) on the gas by the environment. Step 4 1 Adiabatic compression Work done on the gas, Total work done, W
Efficiency (η) of a Carnot engine, From the adiabatic processes in equations (2) and (4), we get On putting the values of equation (6) in equation (5), we get
a. The efficiency of a real engine must be less than the efficiency of a carnot engine, operating between two temperatures. Thus, Efficiency, E of a carnot engine operating between 0 C (273 K) and 100 C (373 K) Hence, the claimed efficiency of 55% for this real engine is impossible. b. Let us consider two reversible heat engines operating between the same reservoirs. Let us assume that heat engine 1 is more efficient than heat engine 2. Now, each engine is supplied with the same amount of heat Q H. The work produced by 1 is W 1, while by 2 is W 2. We assume that the efficiency of engine 1, η 1 is greater than the efficiency of engine 2, η 2 i.e., η 1 > η 2. Now, let reversible engine 2 be made to operate as a refrigerator. This refrigerator receives an input work of W 2 and rejects heat Q H to the hot reservoir. We can say that this heat is being received by engine 1, since it receives Q H amount of heat from the reservoir, thereby eliminating any net heat exchange with the hot reservoir. Thus, we have a combined heat engine that produces work while exchanging heat with a single reservoir. This is in direct contradiction to the Kelvin-Planck statement of the second law of thermodynamics, proving that the assumption η 1 > η 2 was incorrect.
27. (a) Statement: The moment of inertia of a body about an axis, parallel to the axis passing through its center of mass, is equal to the sum of the moments of inertia of the body about the axis passing through its centre of mass and the product of the mass and the square of distance between the two parallel axes. Proof: Consider a particle of mass m at P. Let d be the perpendicular distance between parallel axes YY and. Let GP = x M.I of the particle about YY = m ( x + d ) 2 M.I of the whole of lamina about YY, However, Σ mx 2 = I g and Σ md 2 = (Σ m ) d 2 = M d 2 Where M (= Σ m ) is the mass of the lamina. Also, Σ 2 mxd = 2d Σ mx ΣI = I g + M d 2 + 2 d Σ mx The lamina will balance itself about its centre of gravity. Therefore, the algebraic sum of the moments of the 'weight of consistent particles' about the centre of gravity G should zero.
Σ mg x = 0 g Σ mx = 0 Σ mx = 0 [ g 0] From (equation (i), I = I g + M d 2 (b) According to theorem of parallel axis, the moment of inertia will be I' = I + MR 2...(1) for a circular disc, I = MR 2 So, the moment of inertia of a circular disc about the said axis will be I' = 2MR 2 Here M = mass of the disc = 250 g = 0.25 kg R = radius of the disc = 25 cm = 0.25 m So, I' = 2[0.25] [0.25] 2 Thus, the moment of inertial will be I' = 3.125 10 2 kg. m 2 Let the mass of the cylinder be M. Density of the cylinder Mass of the sphere = Volume Density Mass of the remaining part of the cylinder is. Let the x axis be the longitudinal axis of the cylinder. The origin is at the centre of the mass of the cylinder. Hence, the original centre of the mass of the cylinder lies at (0, 0). Let x 1 be the position of the centre of the mass of the sphere. Thus, x 1 = 30
Let x 2 be the coordinate of the centre of the mass of the remaining cylinder. Thus, we have: Hence, the position of the centre of the mass shifts 10.9 cm to the right of its original position.