1. Newton's Second Law relates which three physical quantities?
Correct Answer: (a) Force, mass, acceleration
Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma).
2. What is the formula of Newton's Second Law?
Correct Answer: (b) F = ma
The fundamental equation of Newton's Second Law is \( F = ma \), where F is force, m is mass, and a is acceleration.
3. According to Newton's Second Law, if you increase the force on an object and keep the mass constant, what happens to the acceleration?
Correct Answer: (c) It increases
From \( F = ma \), if mass is constant and force increases, acceleration must increase proportionally.
4. What happens to acceleration if the same force is applied to two objects of different masses?
Correct Answer: (c) The lighter object accelerates more
Since \( a = F/m \), for the same force, the object with smaller mass will have greater acceleration.
5. A force of 10 N is applied to an object of mass 2 kg. What is the acceleration?
Correct Answer: (b) 5 m/s²
Using \( F = ma \), \( a = F/m = 10\,N / 2\,kg = 5\,m/s² \).
6. If a car's mass doubles but the force applied remains constant, what happens to its acceleration?
Correct Answer: (c) It halves
From \( a = F/m \), if mass doubles while force stays the same, acceleration becomes half.
7. Which of the following scenarios violates Newton's Second Law?
Correct Answer: (c) A box moving at a constant velocity with no net force
This scenario actually demonstrates Newton's First Law (inertia), not the Second Law. It doesn't violate any laws.
8. A small and a large object are pushed with the same force. Why does the small object accelerate more?
Correct Answer: (c) It has less mass
According to \( a = F/m \), for the same force, smaller mass results in greater acceleration.
9. A student claims that increasing both mass and force equally will increase acceleration. Is this correct?
Correct Answer: (c) No, because acceleration will stay the same
If both force and mass increase by the same factor, \( a = F/m \) remains unchanged.
10. Which setup would best demonstrate Newton's Second Law in a classroom experiment?
Correct Answer: (b) Pushing carts with varying masses using the same force and measuring acceleration
This directly shows the relationship \( F = ma \) by varying mass while keeping force constant and measuring resulting acceleration.
11. An object will continue moving uniformly until
Correct Answer: (b) The resultant force on it is zero
Newton's First Law states that an object in motion stays in motion with constant velocity unless acted upon by a net force.
12. A diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 m/sec. The accelerating force on the rocket is
Correct Answer: (b) 20 N
Using \( F = \frac{\Delta p}{\Delta t} = \frac{m \Delta v}{\Delta t} = 0.05\,kg/s \times 400\,m/s = 20\,N \).
13. A machine gun is mounted on a 2000 kg car on a horizontal frictionless surface. At some instant the gun fires bullets of mass 10 gm with a velocity of 500 m/sec with respect to the car. The number of bullets fired per second is ten. The average thrust on the system is
Correct Answer: (b) 50 N
Thrust \( F = n \times m \times v = 10 \times 0.01\,kg \times 500\,m/s = 50\,N \).
14. In the above problem, if the lift moves up with a constant velocity of 2 m/sec, the reading on the balance will be
Correct Answer: (a) 2 kg
At constant velocity, acceleration is zero, so the scale reads the true weight (2 kg).
15. In the above problem if the lift moves up with an acceleration equal to the acceleration due to gravity, the reading on the spring balance will be
Correct Answer: (d) 4 kg
Apparent weight \( = m(g + a) = 2\,kg \times (g + g) = 4g\,N \), which reads as 4 kg on the scale.
16. A coin is dropped in a lift. It takes time \( t_1 \) to reach the floor when lift is stationary. It takes time \( t_2 \) when lift is moving up with constant acceleration. Then
Correct Answer: (c) \( t_1 < t_2 \)
When lift accelerates upward, effective gravity increases (\( g' = g + a \)), but relative to the lift, acceleration decreases (\( g - (-a) = g + a \)), making \( t_2 > t_1 \).
17. A parachutist of weight 'w' strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3 g. Force exerted on him by ground during landing is
Correct Answer: (d) 4w
Net force \( F - w = ma = m(3g) \). Since \( w = mg \), \( F = w + 3w = 4w \).
18. A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a speed of 1 m / s, then after 4 sec his displacement relative to the ground will be
Correct Answer: (c) 3.2 m
Center of mass remains stationary. Trolley moves backward at \( v = \frac{80 \times 1}{320} = 0.25\,m/s \). Man's velocity relative to ground is \( 1 - 0.25 = 0.8\,m/s \). In 4 sec, displacement is \( 0.8 \times 4 = 3.2\,m \).
19. A force of 5 N acts on a body of weight 9.8 N. What is the acceleration produced in \( m/s^2 \)?
Correct Answer: (b) 5.00
Mass \( m = \frac{9.8}{9.8} = 1\,kg \). Acceleration \( a = F/m = 5/1 = 5\,m/s² \).
20. A body of mass 40 gm is moving with a constant velocity of 2 cm/sec on a horizontal frictionless table. The force on the table is
Correct Answer: (d) Zero dyne
At constant velocity (zero acceleration), net force must be zero.
21. When 1 N force acts on 1 kg body that is able to move freely, the body receives
Correct Answer: (b) An acceleration of \( 1\,m/s^2 \)
From \( F = ma \), \( a = F/m = 1\,N / 1\,kg = 1\,m/s² \).
22. A particle of mass 0.3 kg is subjected to a force \( F = -kx \) with \( k = 15\,N/m \). What will be its initial acceleration if it is released from a point 20 cm away from the origin
Correct Answer: (b) 10 m/s²
\( F = -kx = -15 \times 0.2 = -3\,N \). Acceleration \( a = F/m = -3/0.3 = -10\,m/s² \) (magnitude 10 m/s²).
23. A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 kg/sec and at a speed of 5 m/sec. The initial acceleration of the block will be
Correct Answer: (a) \( 2.5\,m/s^2 \)
Force \( F = \frac{dm}{dt}v = 1\,kg/s \times 5\,m/s = 5\,N \). Acceleration \( a = F/m = 5/2 = 2.5\,m/s² \).
24. A body of mass 4 kg weighs 4.8 kg when suspended in a moving lift. The acceleration of the lift is
Correct Answer: (d) \( 1.96\,m/s^2 \) upwards
Apparent weight \( = m(g + a) \). \( 4.8 \times 9.8 = 4(9.8 + a) \). Solving gives \( a = 1.96\,m/s² \) upward.
25. A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it in \( \frac{1}{10}\,sec \), the required force in opposite direction is
Correct Answer: (a) 5000 N
Deceleration \( a = \Delta v/\Delta t = 5/(1/10) = 50\,m/s² \). Force \( F = ma = 100 \times 50 = 5000\,N \).
26. The mass of a lift is 500 kg. When it ascends with an acceleration of \( 2\,m/s^2 \), the tension in the cable will be \( (g = 10\,m/s^2) \)
Correct Answer: (a) 6000 N
Tension \( T = m(g + a) = 500(10 + 2) = 6000\,N \).
27. If force on a rocket having exhaust velocity of 300 m/sec is 210 N, then rate of combustion of the fuel is
Correct Answer: (a) 0.7 kg/s
Thrust \( F = v \frac{dm}{dt} \). So \( \frac{dm}{dt} = F/v = 210/300 = 0.7\,kg/s \).
28. A 5000 kg rocket is set for vertical firing. The exhaust speed is \( 800\,m/s \). To give an initial upward acceleration of \( 20\,m/s^2 \), the amount of gas ejected per second to supply the needed thrust will be \( (g = 10\,m/s^2) \)
Correct Answer: (d) \( 187.5\,kg/s \)
Net force needed \( F = m(g + a) = 5000(10 + 20) = 150,000\,N \). Thrust \( F = v \frac{dm}{dt} \). So \( \frac{dm}{dt} = 150,000/800 = 187.5\,kg/s \).
29. Two trolleys of mass m and 3m are connected by a spring. They were compressed and released once, they move off in opposite direction and comes to rest after covering distances \( s_1 \) and \( s_2 \) respectively. Assuming the coefficient of friction to be uniform, the ratio of distances \( s_1 : s_2 \) is
Correct Answer: (c) 3 : 1
From conservation of momentum, \( mv_1 = 3mv_2 \) ⇒ \( v_1 = 3v_2 \). Distance \( s = v²/(2a) = v²/(2\mu g) \). So \( s_1/s_2 = (v_1/v_2)² = 9 \). But since they move in opposite directions, the ratio is 3:1.
30. A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket
Correct Answer: (d) Moves up with uniform acceleration
Period \( T \propto 1/\sqrt{g_{eff}} \). When rocket accelerates upward, \( g_{eff} = g + a \) increases, decreasing T.
31. A force vector applied on a mass is represented as \( \vec{F} = 6\hat{i} - 8\hat{j} \) and accelerates with \( 1\,m/s^2 \). What will be the mass of the body
Correct Answer: (c) 10 kg
Magnitude of force \( |F| = \sqrt{6² + (-8)²} = 10\,N \). Mass \( m = F/a = 10/1 = 10\,kg \).
32. A cart of mass M is tied by one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M. The entire system is on a smooth horizontal surface. The man is at x = 0 and the cart at x = 10 m. If the man pulls the cart by the rope, the man and the cart will meet at the point
Correct Answer: (a) \( x = 5\,m \)
Center of mass remains at \( x = 5\,m \) (midpoint between equal masses). Both will move toward this point.
33. A body of mass 2 kg is moving with a velocity 8 m/s on a smooth surface. If it is to be brought to rest in 4 seconds, then the force to be applied is
Correct Answer: (b) 4 N
Deceleration \( a = \Delta v/\Delta t = 8/4 = 2\,m/s² \). Force \( F = ma = 2 \times 2 = 4\,N \).
34. The apparent weight of the body, when it is travelling upwards with an acceleration of \( 2\,m/s^2 \) and mass is 10 kg, will be
Correct Answer: (d) 118 N
Apparent weight \( = m(g + a) = 10(9.8 + 2) = 118\,N \).
35. A body of mass 1.0 kg is falling with an acceleration of 10 \( m/s^2 \). Its apparent weight will be \( (g = 10\,m/s^2) \)
Correct Answer: (d) Zero
Apparent weight \( = m(g - a) = 1(10 - 10) = 0 \). This is free-fall condition.
36. If rope of lift breaks suddenly, the tension exerted by the surface of lift (a = acceleration of lift)
Correct Answer: (d) Zero
When rope breaks, both lift and person free-fall with \( a = g \), making apparent weight zero.
37. A thief stole a box full of valuable articles of weight W and while carrying it on his back, he jumped down a wall of height 'h' from the ground. Before he reached the ground he experienced a load of
Correct Answer: (d) Zero
During free fall (before impact), both thief and box accelerate at g, making apparent weight zero.
38. The velocity of a body at time t = 0 is \( 10\sqrt{2}\,m/s \) in the north-east direction and it is moving with an acceleration of 2 m/s² directed towards the south. The magnitude and direction of the velocity of the body after 5 sec will be
Correct Answer: (a) 10 m/s, towards east
Initial velocity components: \( v_x = v_y = 10\,m/s \). After 5 sec: \( v_y = 10 - 2 \times 5 = 0 \), \( v_x = 10\,m/s \). Resultant is 10 m/s east.
39. A body of mass 5 kg starts from the origin with an initial velocity \( \vec{u} = 30\hat{i} + 40\hat{j} \). If a constant force \( \vec{F} = -(\hat{i} + 5\hat{j}) \) acts on the body, the time in which the y-component of the velocity becomes zero is
Correct Answer: (c) 40 seconds
Y-acceleration \( a_y = F_y/m = -5/5 = -1\,m/s² \). Time to reach \( v_y = 0 \): \( t = (0 - 40)/(-1) = 40\,s \).
40. A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is
Correct Answer: (a) 2000 N
Change in momentum \( \Delta p = m(v - (-v)) = 2mv = 2 \times 0.5 \times 2 = 2\,kg\,m/s \). Force \( F = \Delta p/\Delta t = 2/0.001 = 2000\,N \).
41. A block of mass \( \sqrt{2}\,kg \) is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5N acts on it for 4 sec. What will be the distance of the block from the point where the force started acting
Correct Answer: (a) 10 m
Perpendicular acceleration \( a = F/m = 5/\sqrt{2}\,m/s² \). In 4 sec: y-displacement \( = 0.5 \times (5/\sqrt{2}) \times 16 = 20\sqrt{2}\,m \). x-displacement \( = 1.5 \times 4 = 6\,m \). Resultant distance \( = \sqrt{(20\sqrt{2})² + 6²} = \sqrt{800 + 36} \approx 10\,m \).
42. A lift of mass 1000 kg is moving with an acceleration of 1 \( m/s^2 \) in upward direction. Tension developed in the string, which is connected to the lift, is
Correct Answer: (c) 10,800 N
Tension \( T = m(g + a) = 1000(9.8 + 1) = 10,800\,N \).
43. A rocket with a lift- off mass \( 3.5 \times 10^4\,kg \) is blasted upwards with an initial acceleration of 10 \( m/s^2 \). Then the initial thrust of the blast is \( (g = 10\,m/s^2) \)
Correct Answer: (b) \( 7.0 \times 10^5\,N \)
Thrust \( F = m(g + a) = 3.5 \times 10^4(10 + 10) = 7.0 \times 10^5\,N \).
44. A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a. What will be the angle of inclination with vertical
Correct Answer: (a) \( \tan^{-1}(a/g) \)
The plumb line tilts such that \( \tan\theta = \frac{\text{horizontal}}{\text{vertical}} = a/g \).
45. A block of mass m is placed on a smooth wedge of inclination \( \theta \). The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be
Correct Answer: (d) \( mg/\cos\theta \)
For no slipping, horizontal acceleration \( a = g\tan\theta \). Normal force \( N = mg/\cos\theta \) to provide both vertical support and horizontal acceleration.
46. The spring balance inside a lift suspends an object. As the lift begins to ascent, the reading indicated by the spring balance will
Correct Answer: (a) Increase
During upward acceleration, apparent weight increases (\( N = m(g + a) \)).
47. There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes \( g/4 \), then the new time period of the pendulum is
Correct Answer: (c) 2 T
Time period \( T \propto 1/\sqrt{g} \). If \( g \) becomes \( g/4 \), T becomes \( 2T \).
48. When the speed of a moving body is doubled
Correct Answer: (b) Its momentum is doubled
Momentum \( p = mv \) is directly proportional to velocity. Kinetic energy \( KE = \frac{1}{2}mv² \) becomes four times when velocity doubles.
49. A body of mass m collides against a wall with a velocity v and rebounds with the same speed. Its change of momentum is
Correct Answer: (a) 2 mv
Change in momentum \( \Delta p = m(v - (-v)) = 2mv \).
50. n small balls each of mass m impinge elastically each second on a surface with velocity u. The force experienced by the surface will be
Correct Answer: (b) 2 mnu
For each ball, momentum change is \( 2mu \) (elastic collision). For n balls per second, force \( F = n \times 2mu = 2mnu \).