Mechanics

111. Inhabitants are unaware of the speed of rotation of the planet Earth b

Inhabitants are unaware of the speed of rotation of the planet Earth because

  • the angular velocity is constant for each place on the Earth’s surface
  • the atmosphere rotates with the Earth
  • there are no nearby objects, either stationary or moving at a rate different from that of the Earth

Which of the above is/are the correct explanation(s)?

1 only
1 and 2 only
2 and 3 only
1, 2 and 3
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All three statements provide valid reasons why inhabitants are unaware of the Earth’s rotation speed.
Our perception of motion is relative. Since everything around us (atmosphere, land, buildings) is rotating with us at the same speed, and the rotation is constant, we don’t feel the motion or observe significant relative movement.
Statement 1: The constant angular velocity means there is no significant acceleration (except constant centripetal acceleration, which is small and contributes subtly to gravity variations but not the feeling of rotation speed) that we would feel as a change in motion. Our senses primarily detect changes in velocity (acceleration).
Statement 2: The atmosphere rotating with the Earth means there is no strong wind resistance or relative air movement caused by our speed, which would otherwise make the motion noticeable.
Statement 3: Without nearby stationary objects or objects moving at a different speed, we lack a visual reference frame to perceive our rapid motion. Distant celestial objects are too far away to provide this sense of rapid nearby movement.

112. A wooden box of mass 2 kg and dimensions (30 cm $\times$ 15 cm $\times

A wooden box of mass 2 kg and dimensions (30 cm $\times$ 15 cm $\times$ 10 cm) is placed on a table with sides 30 cm and 10 cm touching the tabletop. Which one of the following is the approximate pressure exerted on the table?

111.1 N/m$^2$
222.2 N/m$^2$
333.3 N/m$^2$
666.6 N/m$^2$
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The correct option is D.
Pressure is defined as force per unit area ($P = F/A$). The force exerted by the box on the table is its weight ($F = mg$). The area is the contact area between the box and the table.
Mass of the box $m = 2$ kg.
Assume acceleration due to gravity $g \approx 10$ m/s$^2$ for approximation, as is common in such problems.
Force (weight) $F = mg = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N}$.
The dimensions of the box are 30 cm $\times$ 15 cm $\times$ 10 cm.
The box is placed with sides 30 cm and 10 cm touching the tabletop. The contact area is $A = (30 \text{ cm}) \times (10 \text{ cm})$.
Convert area to square meters: $A = (0.30 \text{ m}) \times (0.10 \text{ m}) = 0.03 \text{ m}^2$.
Pressure $P = F / A = 20 \text{ N} / 0.03 \text{ m}^2 = 20 / (3/100) \text{ N/m}^2 = (20 \times 100) / 3 \text{ N/m}^2 = 2000 / 3 \text{ N/m}^2$.
$2000 / 3 \approx 666.67 \text{ N/m}^2$.
Looking at the options, 666.6 N/m$^2$ is the closest value.
If $g = 9.8$ m/s$^2$ was used, $F = 2 \times 9.8 = 19.6$ N. $P = 19.6 / 0.03 = 1960 / 3 \approx 653.33$ N/m$^2$. 666.6 is still the closest option, suggesting $g=10$ m/s$^2$ was intended.

113. All objects experience a buoyancy when they are immersed in a fluid. B

All objects experience a buoyancy when they are immersed in a fluid. Buoyancy is

a downward force
a downward pressure
an upward force
an upward pressure
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The correct option is C.
Buoyancy is an upward force exerted by a fluid on an immersed object, opposing the object’s weight.
Archimedes’ principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. This force acts vertically upwards through the centroid of the displaced fluid volume. Buoyancy is a force, not a pressure, and its direction is always opposite to the direction of gravity on the fluid (hence, upward in the case of earth’s gravity).

114. The energy possessed by a body due to its change in position or shape

The energy possessed by a body due to its change in position or shape is called

thermal energy
potential energy
kinetic energy
electric energy
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The correct answer is potential energy. Potential energy is the energy stored by an object due to its position or its shape.
– Potential energy is a form of stored energy.
– Gravitational potential energy depends on an object’s height above a reference point.
– Elastic potential energy is stored in a material when it is deformed (stretched or compressed), like a spring or a stretched rubber band.
– The definition provided in the question (“energy possessed by a body due to its change in position or shape”) precisely describes potential energy.
– Thermal energy is related to the internal energy of a system due to its temperature.
– Kinetic energy is the energy of motion.
– Electric energy is energy associated with the flow of electric charge or the presence of electric fields.

115. A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a s

A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?

2.15 m/s
1.89 m/s
1.51 m/s
2.51 m/s
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The correct answer is 1.89 m/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart.
– The system consists of the boy and the stationary cart.
– Before the jump, the boy has momentum (mass × velocity) and the cart has zero momentum.
– After the boy jumps onto the cart, they move together as a single system with a common velocity.
– According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned).
– Initial momentum = (mass of boy × velocity of boy) + (mass of cart × velocity of cart) = (52 kg × 2 m/s) + (3 kg × 0 m/s) = 104 kg m/s.
– Let the final velocity of the combined system (boy + cart) be ‘v’. The combined mass is 52 kg + 3 kg = 55 kg.
– Final momentum = (combined mass × final velocity) = 55 kg × v.
– By conservation of momentum: 104 kg m/s = 55 kg × v.
– v = 104 / 55 ≈ 1.89 m/s.
This is an example of a perfectly inelastic collision because the two objects stick together after colliding. Kinetic energy is not conserved in an inelastic collision, but momentum is always conserved in the absence of external forces.

116. A 5 N force is defined when a mass of 10 kg is accelerated with

A 5 N force is defined when a mass of 10 kg is accelerated with

5.0 cm/s²
0.5 m/s²
0.5 cm/s²
5.0 m/s²
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This question requires the application of Newton’s Second Law of Motion, which states that Force (F) equals mass (m) times acceleration (a), or F = ma. We are given the force (F = 5 N) and the mass (m = 10 kg) and asked to find the acceleration (a) with which the mass is accelerated by this force. Rearranging the formula, we get a = F/m. Plugging in the given values: a = 5 N / 10 kg. Recalling that 1 Newton (N) is defined as 1 kg⋅m/s², the units work out correctly: a = (5 kg⋅m/s²) / 10 kg = 0.5 m/s². Checking the options, 0.5 m/s² corresponds to option B.
Newton’s Second Law (F=ma) is fundamental in relating force, mass, and acceleration. Ensure consistent units (SI units like Newtons, kilograms, and meters per second squared).
Option A (5.0 cm/s²) is 0.05 m/s², Option C (0.5 cm/s²) is 0.005 m/s², and Option D (5.0 m/s²) is significantly larger than the calculated value. It’s important to be careful with unit conversions (cm to m).

117. A mass M is dragged by a pulley on a horizontal plane by a force anti-

A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is

zero
positive
infinite
negative
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Work done (W) is calculated as the dot product of force (F) and displacement (d): W = F ⋅ d = |F| |d| cos(θ), where θ is the angle between the force vector and the displacement vector. The problem states that the force is “anti-parallel” to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180°)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative.
Work done is negative when the force and displacement are in opposite directions (anti-parallel).
Work done is positive when the force and displacement are in the same direction (parallel, θ=0°, cos(0°)=1). Work done is zero when the force is perpendicular to the displacement (θ=90°, cos(90°)=0).

118. The time period of a 1 m long pendulum approximates to

The time period of a 1 m long pendulum approximates to

6 s
4 s
2 s
1 s
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The time period of a 1 m long pendulum approximates to 2 s.
The time period ($T$) of a simple pendulum for small oscillations is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.
Given length $L = 1$ m. Taking the standard value of $g \approx 9.8$ m/s² and $\pi \approx 3.14$: $T = 2 \times 3.14 \times \sqrt{\frac{1}{9.8}} \approx 6.28 \times \sqrt{0.102} \approx 6.28 \times 0.319 \approx 2.005$ s. A common approximation for calculation purposes is sometimes $\pi^2 \approx g$, which gives $T = 2\pi \sqrt{\frac{1}{\pi^2}} = 2\pi \times \frac{1}{\pi} = 2$ s. Both calculations yield a value very close to 2 seconds.

119. What is the dimension of gravitational constant?

What is the dimension of gravitational constant?

ML³T⁻²
M⁻¹L³T⁻²
M²L⁻²T⁻²
M²L⁻¹T⁻²
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The dimension of the gravitational constant ($G$) is M⁻¹L³T⁻².
The gravitational constant $G$ appears in Newton’s Law of Universal Gravitation, which states that the force ($F$) between two masses ($m_1$, $m_2$) separated by a distance ($r$) is given by $F = G \frac{m_1 m_2}{r^2}$.
We can determine the dimensions of $G$ by rearranging the formula: $G = \frac{F r^2}{m_1 m_2}$. The dimensions of Force ($F$) are [MLT⁻²], distance ($r$) are [L], and mass ($m$) are [M]. Substituting these dimensions: $[G] = \frac{[MLT⁻²] [L]^2}{[M][M]} = \frac{[ML³T⁻²]}{[M²]} = [M^{1-2} L³ T⁻²] = [M⁻¹L³T⁻²]$.

120. Which one of the following statements is true for a simple harmonic os

Which one of the following statements is true for a simple harmonic oscillator ?

Force acting is directly proportional to the displacement from the mean position and is in same direction.
Force acting is directly proportional to the displacement from the mean position and is in opposite direction.
Acceleration of the oscillator is constant.
The velocity of the oscillator is not periodic.
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The correct answer is (B) Force acting is directly proportional to the displacement from the mean position and is in opposite direction.
A simple harmonic oscillator (SHO) is defined by the condition that the restoring force acting on the oscillating body is directly proportional to its displacement from the equilibrium (mean) position and is always directed towards the equilibrium position. This is mathematically represented as F = -kx, where F is the force, x is the displacement, and k is the force constant. The negative sign indicates the force is in the opposite direction to the displacement.
Option A is incorrect because the force is in the opposite direction to the displacement (restoring force). Option C is incorrect because the acceleration (a = F/m = -kx/m) is proportional to the displacement and thus varies; it is not constant. Option D is incorrect because in SHM, the motion is periodic, meaning all properties like velocity, displacement, and acceleration repeat over time.
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