Mechanics

81. The free fall acceleration g increases as one proceeds, at sea level,

The free fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is

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Earth is a sphere with same density everywhere
Earth is a sphere with different density at the polar regions than in the equatorial regions
Earth is approximately an ellipsoid having its equatorial radius greater than its polar radius by 21 km
Earth is approximately an ellipsoid having its equatorial radius smaller than its polar radius by 21 km
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The correct answer is C. The primary reasons for the variation of the free fall acceleration ‘g’ with latitude at sea level are the Earth’s shape and its rotation.
The Earth is not a perfect sphere but is approximately an oblate spheroid, bulging at the equator and flattened at the poles. This means the distance from the center of the Earth is greater at the equator than at the poles. The gravitational acceleration ‘g’ is inversely proportional to the square of the distance from the center of the Earth. Therefore, points at the equator are further from the center, resulting in a weaker gravitational pull compared to the poles. Additionally, the Earth’s rotation creates a centrifugal force that opposes gravity, and this force is strongest at the equator and zero at the poles. This centrifugal effect further reduces the effective ‘g’ at the equator compared to the poles. Option C correctly states that the Earth is approximately an ellipsoid with the equatorial radius greater than the polar radius, which is the key geological reason for the variation in ‘g’.
The difference between the equatorial and polar radii is approximately 21 km. The average value of ‘g’ at sea level is about 9.80665 m/s². It is minimum at the equator (approx. 9.78 m/s²) and maximum at the poles (approx. 9.83 m/s²). The effect of Earth’s shape (distance from center) accounts for about two-thirds of the variation, while the effect of rotation (centrifugal force) accounts for about one-third.

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82. Pressure is a scalar quantity because

Pressure is a scalar quantity because

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it is the ratio of force to area and both force and area are vectors
it is the ratio of magnitude of force to area
it is the ratio of component of force (normal to area) to area
none of the above
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Pressure is defined as the force acting perpendicular to a unit area ($P = F_{\perp}/A$). Although force is a vector, pressure is a scalar quantity because it is defined based on the magnitude of the force component normal to the surface and the magnitude of the area.
Pressure at a point in a fluid acts equally in all directions (Pascal’s principle), which is characteristic of a scalar quantity.
Option A is incorrect because the definition combines vectors in a way that yields a scalar. Option B is incorrect because it doesn’t specify the component of force normal to the area. Option C accurately reflects the scalar nature of pressure by defining it using the magnitude of the normal force component per unit area.

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83. Which one of the following statements is not correct?

Which one of the following statements is not correct?

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In steady flow of a liquid, the velocity of liquid particles reaching at a particular point is the same at all points
Steady flow is also called streamlined flow
In steady flow, each particle may not follow the same path as taken by a previous particle passing through that point
Two streamlines cannot intersect each other
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The correct answer is C) In steady flow, each particle may not follow the same path as taken by a previous particle passing through that point. This statement is incorrect. In steady flow (or streamlined flow), the velocity of the fluid at any given point in space is constant over time. Consequently, the path followed by every fluid particle passing through a particular point is the same as the path followed by previous particles passing through that same point. These paths are called streamlines.
In steady flow, streamlines are fixed paths in space, and fluid particles follow these fixed paths.
Statement A is poorly phrased but likely intends to convey that velocity at a fixed point is constant over time, which is true for steady flow, though the “at all points” part is confusing. Statement B is correct; steady flow is also known as streamlined flow. Statement D is correct; streamlines cannot intersect because if they did, a fluid particle at the intersection would have two different velocity vectors simultaneously, which is physically impossible.

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84. The acceleration due to gravity ‘g’ for objects on or near the surface

The acceleration due to gravity ‘g’ for objects on or near the surface of earth is related to the universal gravitational constant ‘G’ as (‘M’ is the mass of the earth and ‘R’ is its radius):

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G = g $ rac{ ext{M}}{ ext{R}^2}$
g = G $ rac{ ext{M}}{ ext{R}^2}$
M = $ rac{ ext{gG}}{ ext{R}^2}$
R = $ rac{ ext{gG}}{ ext{M}^2}$
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The correct option is B, g = G $\frac{\text{M}}{\text{R}^2}$.
The acceleration due to gravity ‘g’ on the surface of the Earth is caused by the gravitational force exerted by the Earth on an object. According to Newton’s Law of Universal Gravitation, the force (F) between the Earth (mass M) and an object (mass m) on its surface (distance R from the center, where R is the Earth’s radius) is given by F = G $\frac{\text{Mm}}{\text{R}^2}$, where G is the universal gravitational constant. This gravitational force is also the object’s weight, W, which is given by W = mg (by Newton’s second law, where ‘g’ is the acceleration due to this force). Equating the two expressions for the force (F = W), we get mg = G $\frac{\text{Mm}}{\text{R}^2}$. Dividing both sides by the mass of the object ‘m’ gives the relationship for the acceleration due to gravity: g = G $\frac{\text{M}}{\text{R}^2}$.
This formula shows that the acceleration due to gravity ‘g’ is independent of the mass of the object itself and depends only on the mass and radius of the planet (or celestial body) and the universal gravitational constant G. The value of ‘g’ varies slightly across the Earth’s surface due to factors like altitude, latitude, and local geological variations.

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85. Two forces, one of 3 newton and another of 4 newton are applied on a s

Two forces, one of 3 newton and another of 4 newton are applied on a standard 1 kg body, placed on a horizontal and frictionless surface, simultaneously along the x-axis and the y-axis, respectively, as shown below:
The magnitude of the resultant acceleration is:

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7 m/s²
1 m/s²
5 m/s²
√7 m/s²
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The correct option is C, 5 m/s².
Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $\sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ N. According to Newton’s second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N / 1 kg = 5 m/s².
The direction of the resultant force (and thus acceleration) would be at an angle to the x-axis, given by θ = atan(Fy/Fx) = atan(4/3). However, the question only asks for the magnitude of the acceleration. This problem illustrates how to find the resultant of perpendicular forces and apply Newton’s second law.

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86. Conservation of momentum in a collision between particles can be under

Conservation of momentum in a collision between particles can be understood on the basis of:

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Newton's first law of motion
Newton's second law of motion only
Both Newton's second law of motion and Newton's third law of motion
Conservation of energy
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Conservation of momentum for a system of particles is a direct consequence of Newton’s laws of motion. Specifically, when particles collide, the forces they exert on each other are internal forces within the system. Newton’s third law states that these forces are equal in magnitude and opposite in direction (action-reaction). Applying Newton’s second law (F = dp/dt) to each particle and summing over the system shows that the net internal force is zero, meaning the total momentum (sum of momenta of all particles) remains constant in the absence of external forces. Thus, conservation of momentum stems from both Newton’s second and third laws.
Conservation of momentum for a system relies on the principle of action-reaction (Newton’s third law) and the relationship between force and change in momentum (Newton’s second law).
Newton’s first law describes inertia. Conservation of energy is a separate fundamental principle, although in elastic collisions, both momentum and kinetic energy are conserved. In inelastic collisions, only momentum is conserved (assuming no external forces).

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87. A man is sitting in a train which is moving with a velocity of 60 km/h

A man is sitting in a train which is moving with a velocity of 60 km/hour. His speed with respect to the train is:

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10/3 m/s
60 m/s
infinite
zero
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The correct answer is D) zero.
The question asks for the man’s speed *with respect to the train*. Since the man is sitting *in* the train, he is at rest relative to the train. His position relative to any point inside the train does not change.
His speed with respect to an observer outside the train would be 60 km/hour, but the reference frame specified is the train itself. Relative velocity is calculated based on the difference in velocities of the two objects. If the man’s velocity relative to the ground is V_man_ground and the train’s velocity relative to the ground is V_train_ground, the man’s velocity relative to the train is V_man_train = V_man_ground – V_train_ground. Since the man is sitting still in the train, V_man_ground = V_train_ground. Thus, V_man_train = 0.

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88. The following figure represents the velocity-time graph of a moving ca

The following figure represents the velocity-time graph of a moving car on a road:
Which segment of the graph represents the retardation?

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AB
BC
CD
None
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In a velocity-time (v-t) graph, acceleration is represented by the slope of the graph. Retardation (or deceleration) occurs when the acceleration is in the opposite direction to the velocity, causing the object to slow down. This corresponds to a negative slope in the v-t graph, assuming positive velocity.
– Segment AB has a positive slope, indicating positive acceleration (velocity increasing).
– Segment BC is horizontal, indicating zero slope and thus zero acceleration (constant velocity).
– Segment CD has a negative slope, indicating negative acceleration. Since the velocity is positive during this segment, a negative acceleration means retardation or deceleration (velocity decreasing).
The slope of a velocity-time graph represents acceleration. A negative slope indicates deceleration or retardation if the velocity is in the positive direction.
If the velocity were negative and the slope were positive, it would also represent slowing down (velocity increasing towards zero from a negative value). In this graph, velocity is positive throughout the plotted segments.

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89. Which one of the following statements is not correct?

Which one of the following statements is not correct?

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If the velocity and acceleration have opposite sign, the object is slowing down
If the velocity is zero at an instant, the acceleration should also be zero at that instant
If the velocity is zero for a time interval; the acceleration is zero at any instant within the time interval
If the position and velocity have opposite sign, the object is moving towards the origin
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Statement B is not correct. If the velocity of an object is zero at a particular instant, its acceleration at that instant does not necessarily have to be zero. A classic example is an object thrown vertically upwards: at the peak of its trajectory, its instantaneous velocity is zero, but the acceleration due to gravity is still acting on it and is non-zero (approximately 9.8 m/s² downwards).
Acceleration is the rate of change of velocity. Zero velocity at an instant only means the object is momentarily stopped; it does not imply that the velocity is not changing at that instant.
Statement A is correct: if velocity and acceleration have opposite signs, the object is slowing down (e.g., positive velocity, negative acceleration). Statement C is correct: if velocity is zero for a time interval, the object is stationary, meaning its velocity is constant (zero), and thus its acceleration is zero. Statement D is correct: if position and velocity have opposite signs (e.g., positive position, negative velocity means moving towards origin from the positive side; negative position, positive velocity means moving towards origin from the negative side), the object is moving towards the origin (assuming origin is at position=0).

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90. A brass ball is tied to a thin wire and swung so as to move uniformly

A brass ball is tied to a thin wire and swung so as to move uniformly in a horizontal circle. Which of the following statements in this regard is / are true?

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  • 1. The ball moves with constant velocity
  • 2. The ball moves with constant speed
  • 3. The ball moves with constant acceleration
  • 4. The magnitude of the acceleration of the ball is constant

Select the correct answer using the code given below:

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1 only
1 and 3
1, 2 and 4
2 and 4 only
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In uniform circular motion, the speed (magnitude of velocity) is constant, but the velocity itself changes direction constantly. The acceleration (centripetal acceleration) is directed towards the center and its direction also changes constantly, but its magnitude is constant.
Statement 1 is false because the direction of velocity changes constantly, so velocity is not constant. Statement 2 is true because “uniformly” implies constant speed. Statement 3 is false because the direction of centripetal acceleration is constantly changing. Statement 4 is true because the magnitude of centripetal acceleration (v²/r) is constant when speed (v) and radius (r) are constant.
For uniform circular motion, the speed is constant, but there is always an acceleration directed towards the center of the circle, known as centripetal acceleration. This acceleration changes the direction of the velocity vector, thus keeping the object moving in a circle.

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