Mechanics

91. A block of mass 2 kg, moving with the initial speed of 3 m/s comes to

A block of mass 2 kg, moving with the initial speed of 3 m/s comes to rest on a rough horizontal surface after travelling a distance of 3 m. The magnitude of the frictional force is :

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9 N
3 N
18 N
1 N
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We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1/2) * m * v_i^2 = (1/2) * 2 kg * (3 m/s)^2 = 9 J. Final KE = (1/2) * m * v_f^2 = (1/2) * 2 kg * (0 m/s)^2 = 0 J. Change in KE = 0 – 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180°) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) / (-3 m) = 3 N. The magnitude of the frictional force is 3 N.
– Work-Energy Theorem: Net Work Done = Change in Kinetic Energy.
– Initial Kinetic Energy = (1/2)mv_i^2.
– Final Kinetic Energy = (1/2)mv_f^2.
– Work done by friction = – f * d.
Alternatively, one could calculate the deceleration using kinematics (v_f^2 = v_i^2 + 2ad) and then find the force using Newton’s second law (F=ma). 0^2 = 3^2 + 2 * a * 3 => 0 = 9 + 6a => 6a = -9 => a = -1.5 m/s^2. The magnitude of acceleration is 1.5 m/s^2. The net force in the direction of motion is the frictional force (acting opposite to initial motion). F_net = m * a = 2 kg * (-1.5 m/s^2) = -3 N. The magnitude of the force is 3 N.

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92. If the block P as shown in the figure below were to be at rest, what s

If the block P as shown in the figure below were to be at rest, what should the magnitude of force F be ?

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5 N
6 N
8 N
10 N
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The magnitude of force F should be 10 N.
For block P to be at rest, the net horizontal force on it must be zero. The forces are the applied force F (left), the tension T from the string (right), and static friction f. The tension T is equal to the weight of block Q, T = m_Q * g = 2 kg * g. Assuming a standard value for gravitational acceleration, g = 10 m/s², the tension T = 20 N. Static friction f opposes the impending motion, with a maximum value f_max = μs * N = μs * m_P * g = 0.4 * m_P * g (where N is the normal force equal to the weight of P). Given the options (5, 6, 8, 10N) are less than T=20N, the impending motion is to the right due to tension. Friction must act to the left to help F balance T: F + f = T. The minimum force F required to prevent motion to the right occurs when friction is maximum and acts left: F_min = T – f_max = 20 – 0.4 * m_P * 10 = 20 – 4 * m_P. If we assume the mass of P is m_P = 2.5 kg, then f_max = 0.4 * 2.5 * 10 = 10 N. In this specific scenario, F_min = 20 N – 10 N = 10 N. This value matches option D and represents the minimum force needed to keep the block at rest against the pull of tension.
Without the mass of block P, the problem is technically underspecified for a unique value of F that keeps the block at rest (any F in the range [T – f_max, T + f_max] works). However, the presence of a specific value among the options suggests either an implicit assumption about m_P or that the question is asking for a significant value within the possible range, likely the minimum required force, which is calculable if a specific m_P value is assumed.

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93. A metallic bob X of mass m is released from position A. It collides el

A metallic bob X of mass m is released from position A. It collides elastically with another identical bob Y placed at rest at position B on a horizontal frictionless table. The angle AOB is 30°. How high does the bob X rise immediately after the collision?

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To the same height as that of position A on the other side in the same trajectory
To half the height as that of position A on the other side along the same trajectory
The same height at position A
It stops at position B
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Immediately after the collision, bob X stops at position B.
– The collision is described as **elastic**. In an elastic collision, both momentum and kinetic energy are conserved.
– The colliding bobs are **identical**, meaning they have the same mass (m).
– Bob Y is **at rest** at position B before the collision. Bob X has a certain velocity (V) just before the collision at B due to its swing from A.
– For a head-on elastic collision between two objects of equal mass where one object is initially at rest, the moving object comes to rest, and the stationary object moves off with the initial velocity of the moving object.
Let $v_{Xi}$ and $v_{Yi}$ be the initial velocities of bob X and Y respectively, and $v_{Xf}$ and $v_{Yf}$ be their final velocities after the collision.
Initial state: $v_{Xi} = V$, $v_{Yi} = 0$.
Conservation of Momentum: $m v_{Xi} + m v_{Yi} = m v_{Xf} + m v_{Yf} \implies m V + 0 = m v_{Xf} + m v_{Yf} \implies V = v_{Xf} + v_{Yf}$.
Conservation of Kinetic Energy: $\frac{1}{2} m v_{Xi}^2 + \frac{1}{2} m v_{Yi}^2 = \frac{1}{2} m v_{Xf}^2 + \frac{1}{2} m v_{Yf}^2 \implies \frac{1}{2} m V^2 + 0 = \frac{1}{2} m v_{Xf}^2 + \frac{1}{2} m v_{Yf}^2 \implies V^2 = v_{Xf}^2 + v_{Yf}^2$.
From the momentum equation, $v_{Yf} = V – v_{Xf}$. Substituting into the energy equation:
$V^2 = v_{Xf}^2 + (V – v_{Xf})^2 = v_{Xf}^2 + V^2 – 2Vv_{Xf} + v_{Xf}^2$
$0 = 2v_{Xf}^2 – 2Vv_{Xf} = 2v_{Xf}(v_{Xf} – V)$.
This gives two possible solutions for $v_{Xf}$: $v_{Xf} = 0$ or $v_{Xf} = V$.
If $v_{Xf} = V$, then $v_{Yf} = V – V = 0$, which means no collision occurred (they passed through each other, which is not the case).
Therefore, the physical solution is $v_{Xf} = 0$.
If $v_{Xf} = 0$, then $v_{Yf} = V – 0 = V$.
So, bob X stops ($v_{Xf}=0$) and bob Y moves off with velocity $V$ ($v_{Yf}=V$).
Since bob X stops at position B, it does not rise to any height immediately after the collision.

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94. Which one of the following about different frictional forces is

Which one of the following about different frictional forces is correct?

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”Kinetic
”Static
”Static
”Static
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The correct order of frictional forces from largest to smallest is Static friction > Kinetic friction > Rolling friction.
– **Static friction** is the force that opposes the start of motion between two surfaces in contact. It is generally the maximum friction force.
– **Kinetic friction** (or sliding friction) is the force that opposes the relative motion between two surfaces that are already sliding over each other. It is typically less than the maximum static friction.
– **Rolling friction** is the force that opposes the rolling motion of a wheel or ball over a surface. It is generally significantly less than kinetic or static friction for comparable loads and surfaces, which is why rolling is much easier than sliding.
The magnitude of frictional forces depends on the nature of the surfaces in contact and the normal force pressing the surfaces together. The coefficients of static, kinetic, and rolling friction usually follow the relationship $\mu_s > \mu_k > \mu_r$. The frictional force is proportional to the normal force ($F_f = \mu N$).

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95. Which one of the following statements regarding a simple pendulum is c

Which one of the following statements regarding a simple pendulum is correct? Simple pendulum has a time period independent of amplitude:

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only for small amplitudes because then the net force on its bob is independent of its displacement.
for any amplitude because the net force on the bob is always proportional to its displacement.
for any amplitude because the net force on the bob is independent of its displacement.
only for small amplitudes because then the net force on its bob is proportional to its displacement.
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The correct statement is D. A simple pendulum has a time period independent of amplitude only for small amplitudes because then the net force on its bob is proportional to its displacement.
The motion of a simple pendulum is approximately Simple Harmonic Motion (SHM) for small angular displacements (amplitudes). In SHM, the restoring force is directly proportional to the displacement from equilibrium (and directed towards equilibrium). For a simple pendulum, the restoring torque (or force component) is proportional to sin(theta), where theta is the angular displacement. For small theta, sin(theta) ≈ theta (in radians), which means the restoring force is proportional to the angular displacement and, consequently, the arc displacement. This proportionality is the condition for the period to be independent of amplitude.
The formula for the period of a simple pendulum, T = 2π * sqrt(L/g), is derived using the small angle approximation. For larger amplitudes, the period actually increases with amplitude because the approximation sin(theta) ≈ theta becomes less accurate, and the motion deviates from perfect SHM.

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96. A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45

A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45 m and rises to a height of 0.20 m. If it was in touch with the floor for 0.1 s, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g = 10 m s⁻²)

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1.0 N
6.0 N
3.0 N
5.0 N
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The correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N.
First, calculate the velocity of the ball just before impact (v₁) and just after impact (v₂). Using energy conservation (v² = 2gh), v₁ = sqrt(2 * 10 * 0.45) = 3 m/s (downward) and v₂ = sqrt(2 * 10 * 0.20) = 2 m/s (upward). Let’s take upward as positive. v₁ = -3 m/s, v₂ = +2 m/s. The change in momentum of the ball is Δp = m(v₂ – v₁) = 0.1 kg * (2 – (-3)) m/s = 0.1 * 5 = 0.5 kg m/s (upward). The average net force on the ball during contact is F_net_on_ball = Δp / Δt = 0.5 Ns / 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N – mg. So, 5 N = N – (0.1 kg * 10 m/s²) = N – 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton’s third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.
The net force applied by the ball on the floor is essentially the reaction force to the average normal force exerted by the floor on the ball during the collision. Gravity also acts on the ball during the contact time, but the primary force during bouncing is the large normal force from the floor. The calculation involving the change in momentum accounts for the effect of all forces during the contact period, including gravity.

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97. Escape speed from the Earth is close to 11.2 km s⁻¹. On another planet

Escape speed from the Earth is close to 11.2 km s⁻¹. On another planet whose radius is half of the Earth’s radius and whose mass density is four times that of the Earth, the escape speed in km s⁻¹ will be close to :

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11.2
15.8
5.6
7.9
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The correct option is A. The escape speed from the new planet will be close to 11.2 km s⁻¹, the same as Earth’s.
The escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM/R). The mass M can be expressed in terms of density (rho) and volume (V = 4/3 * pi * R^3) as M = rho * (4/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4/3 * pi * R^3) / R) = sqrt(8/3 * pi * G * rho * R²) = R * sqrt(8/3 * pi * G * rho). This shows that v_e is proportional to R * sqrt(rho).
Let Earth’s radius, density, and escape speed be R_e, rho_e, and v_e_e respectively. So, v_e_e ∝ R_e * sqrt(rho_e) = 11.2 km/s. For the new planet, R_p = R_e / 2 and rho_p = 4 * rho_e. The escape speed from the new planet is v_e_p ∝ R_p * sqrt(rho_p) = (R_e / 2) * sqrt(4 * rho_e) = (R_e / 2) * 2 * sqrt(rho_e) = R_e * sqrt(rho_e). Thus, v_e_p is proportional to the same value as v_e_e, meaning v_e_p = v_e_e = 11.2 km/s.

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98. A uniform meter scale of mass 0.24 kg is made of steel. It is kept on

A uniform meter scale of mass 0.24 kg is made of steel. It is kept on two wedges, W₁ and W₂, in a horizontal position. W₁ is at a distance of 0.2 m from one of its ends, while W₂ is at distance of 0.4 m from the other end. If the force on the scale is N₁ due to W₁ and N₂ due to W₂, then : (take g = 10.0 m s⁻²)

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N₁ = 1.6 N and N₂ = 0.8 N
N₁ = 0.8 N and N₂ = 1.6 N
N₁ = 0.6 N and N₂ = 1.8 N
N₁ = 1.8 N and N₂ = 0.6 N
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The correct option is C. Calculating the forces using equilibrium conditions yields N₁ = 0.6 N and N₂ = 1.8 N.
For a uniform meter scale of mass 0.24 kg, its weight (W = mg = 0.24 * 10 = 2.4 N) acts at its center of mass, which is at the 0.5 m mark from either end. W₁ is at 0.2 m from one end (say, the 0m mark), and W₂ is at 0.4 m from the other end (the 1m mark), placing it at 1.0 – 0.4 = 0.6 m from the 0m mark. For equilibrium, the sum of upward forces equals the sum of downward forces (N₁ + N₂ = W = 2.4 N), and the sum of torques about any point is zero. Taking torques about the point W₁ (at 0.2m), the weight W (at 0.5m) creates a clockwise torque W * (0.5 – 0.2) = 2.4 * 0.3 = 0.72 Nm. The force N₂ (at 0.6m) creates an anticlockwise torque N₂ * (0.6 – 0.2) = N₂ * 0.4 Nm. Setting the sum of torques to zero: 0.72 – 0.4 N₂ = 0, which gives N₂ = 0.72 / 0.4 = 1.8 N. Substituting N₂ into the force equation: N₁ + 1.8 = 2.4, which gives N₁ = 2.4 – 1.8 = 0.6 N.
This problem is a classic example of applying the conditions for static equilibrium: zero net force and zero net torque. Choosing the pivot point for calculating torques at one of the unknown force locations simplifies the calculation by eliminating that force from the torque equation.

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99. Which one of the following statements is true ?

Which one of the following statements is true ?

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The force of gravity of the Earth on the Moon is greater than the force of gravity of the Moon on the Earth.
The force of gravity of the Moon on the Earth is greater than the force of gravity of the Earth on the Moon.
The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude and are in the same direction.
The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude but are in opposite directions.
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According to Newton’s Law of Universal Gravitation, the force of gravity exerted by the Earth on the Moon is equal in magnitude to the force of gravity exerted by the Moon on the Earth. This is an application of Newton’s Third Law of Motion (for every action, there is an equal and opposite reaction). These forces are always attractive and act along the line joining the centers of the two bodies, thus they are in opposite directions.
– Newton’s Law of Universal Gravitation states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them (F = G * m1 * m2 / r^2). This formula is symmetrical with respect to m1 and m2, showing the force is the same magnitude for both.
– Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. The force Earth exerts on the Moon is the action, and the force Moon exerts on the Earth is the reaction.
– The forces are attractive, meaning Earth pulls the Moon towards it, and the Moon pulls the Earth towards it. Therefore, the forces are in opposite directions.
Although the forces are equal in magnitude, their effects are different due to the difference in mass. The Earth’s mass is much greater than the Moon’s mass. Therefore, the acceleration of the Moon towards the Earth (due to the Earth’s pull) is much greater than the acceleration of the Earth towards the Moon (due to the Moon’s pull, which causes tides).

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100. Which one of the following is an example of Second Class Lever ?

Which one of the following is an example of Second Class Lever ?

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A pair of scissors
A bottle opener
A cricket bat
A bow and arrow
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Levers are classified into three types based on the relative positions of the fulcrum (pivot point), the load (resistance), and the effort (applied force).
– Class 1 Lever: Fulcrum is between the effort and the load (e.g., see-saw, scissors, pliers).
– Class 2 Lever: Load is between the fulcrum and the effort (e.g., wheelbarrow, bottle opener, nutcracker). Class 2 levers always provide mechanical advantage (>1).
– Class 3 Lever: Effort is between the fulcrum and the load (e.g., fishing rod, forceps, cricket bat, tweezers). Class 3 levers always have mechanical advantage (<1), prioritizing range of motion or speed over force multiplication.
– A pair of scissors consists of two Class 1 levers joined at the pivot (fulcrum).
– A bottle opener uses the lip of the bottle cap as the fulcrum. The opener is used to lift the cap (load) by applying force (effort) at the end of the handle. The load (cap) is lifted between the fulcrum and the effort. Hence, it is a Class 2 lever.
– A cricket bat has the player’s hands applying effort between the pivot (often lower hand near the handle end) and the point of impact with the ball (load). Hence, it is a Class 3 lever.
– A bow and arrow is not typically classified as a simple lever system.

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