Mechanics Archives

61. A solid disc and a solid sphere have the same mass and same radius. Wh

A solid disc and a solid sphere have the same mass and same radius. Which one has the higher moment of inertia about its centre of mass ?

The disc

The sphere

Both have the same moment of inertia

The information provided is not sufficient to answer the question

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The moment of inertia ($I$) for a solid disc about an axis through its center and perpendicular to its plane is $I_{disc} = \frac{1}{2}MR^2$. The moment of inertia for a solid sphere about an axis through its center is $I_{sphere} = \frac{2}{5}MR^2$. Given that both the disc and the sphere have the same mass ($M$) and radius ($R$), we compare the coefficients $\frac{1}{2}$ and $\frac{2}{5}$. Since $\frac{1}{2} = \frac{5}{10}$ and $\frac{2}{5} = \frac{4}{10}$, we have $\frac{1}{2} > \frac{2}{5}$. Therefore, $I_{disc} > I_{sphere}$. The solid disc has the higher moment of inertia.

– Moment of inertia depends on the mass distribution relative to the axis of rotation.
– Formulas for the moment of inertia of common shapes are standard results derived from integration.
– For objects of the same mass and radius, the object with more mass distributed further from the axis of rotation will have a higher moment of inertia. In the disc, all mass is at a distance up to R from the axis in a plane, whereas in the sphere, mass is distributed in a volume, including closer to the center.

– Moment of inertia is a measure of an object’s resistance to changes in its rotational motion.
– A higher moment of inertia means it is harder to start or stop the rotation.
– The formulas used are for axes passing through the center of mass, as specified in the question.

62. If an object moves at a non-zero constant acceleration for a certain i

If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

depends on its initial velocity.

is independent of its initial velocity.

increases linearly with time.

depends on its initial displacement.

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The distance covered by an object moving with constant acceleration is given by the kinematic equation: $s = ut + \frac{1}{2}at^2$, where $s$ is the distance (or displacement magnitude if direction is constant), $u$ is the initial velocity, $a$ is the constant acceleration, and $t$ is the time interval. This equation clearly shows that the distance covered ($s$) depends on the initial velocity ($u$), assuming $a \neq 0$ and $t > 0$.

– For motion with constant acceleration, the relationship between distance, initial velocity, acceleration, and time is described by standard kinematic equations.
– The equation $s = ut + \frac{1}{2}at^2$ explicitly includes the initial velocity $u$.

– Option B is incorrect because the term $ut$ makes the distance dependent on initial velocity.
– Option C is incorrect because the presence of the $\frac{1}{2}at^2$ term (with $a \neq 0$) means the distance increases quadratically with time, not linearly.
– Option D is incorrect; the distance covered (change in position) is independent of the starting position (initial displacement). Initial displacement affects the final position, but not the distance traveled during the interval.

63. When a ball bounces off the ground, which of the following changes sud

When a ball bounces off the ground, which of the following changes suddenly ?
(Assume no loss of energy to the floor)

Its speed

Its momentum

Its kinetic energy

Its potential energy

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The correct option is B) Its momentum. When a ball bounces off the ground, its velocity changes direction suddenly, leading to a sudden change in its momentum.

Velocity is a vector quantity, having both magnitude (speed) and direction. Momentum is mass multiplied by velocity (p = mv), so it is also a vector quantity. When a ball bounces, its downward velocity just before impact changes to an upward velocity just after impact. Even if the speed remains the same (in a perfectly elastic collision with no energy loss), the change in direction causes a significant change in the velocity vector, and thus a sudden change in the momentum vector.

Speed is the magnitude of velocity. In an idealized elastic collision with no energy loss to the floor, the speed of the ball remains the same before and after the bounce. Kinetic energy (KE = 1/2 mv²) depends on the square of the speed, so it also remains the same if there’s no energy loss. Potential energy (PE = mgh) changes gradually with height and is at a minimum (or zero) at the point of bounce; it does not change suddenly during the bounce event itself.

64. Let there be an object having some chemicals in it. It starts moving w

Let there be an object having some chemicals in it. It starts moving with a uniform velocity v and a chemical reaction starts happening. In this case, which of the following statement/s is/are correct ?

1. Chemical reactions happening in the system cannot change the velocity v of the center of mass of the object.
2. Chemical reactions happening in the system cannot change the kinetic energy of the particles inside with respect to the center of mass of object.

Select the correct answer using the code given below :

1 only

2 only

Both 1 and 2

Neither 1 nor 2

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The correct option is A) 1 only. Statement 1 is correct, while statement 2 is incorrect.

Statement 1: The velocity of the center of mass of a system can only be changed by external forces acting on the system. Chemical reactions happening *within* the object involve internal forces between constituent particles. Internal forces cannot change the motion of the center of mass of the system. Therefore, if the object is moving with uniform velocity (implying zero net external force or balanced external forces), a chemical reaction inside will not change the velocity of the center of mass.
Statement 2: Chemical reactions often involve changes in the internal energy of the system, which can include changes in the kinetic energy of the particles relative to the center of mass (manifesting as temperature changes). For example, an exothermic reaction releases energy, which increases the kinetic energy of the constituent particles. Therefore, chemical reactions *can* change the kinetic energy of the particles inside with respect to the center of mass of the object.

The total momentum of the system remains constant in the absence of external forces, which means the velocity of the center of mass remains constant if mass is also constant. The total energy of the isolated system is conserved, but this energy can be converted between different forms, including internal potential energy (chemical energy) and internal kinetic energy (thermal energy, kinetic energy of particles relative to CM).

65. A thin disc and a thin ring, both have mass M and radius R. Both rotat

A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their center of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true ?

The ring has higher kinetic energy

The disc has higher kinetic energy

The ring and the disc have the same kinetic energy

Kinetic energies of both the bodies are zero since they are not in linear motion

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The correct option is A) The ring has higher kinetic energy. The rotational kinetic energy is given by KE = (1/2) * I * ω², where I is the moment of inertia and ω is the angular velocity. Since both the ring and the disc have the same mass (M) and radius (R) and rotate at the same angular velocity (ω), the kinetic energy depends on their moments of inertia.

The moment of inertia of a thin ring about an axis through its center and perpendicular to its plane is I_ring = M * R². The moment of inertia of a thin disc about an axis through its center and perpendicular to its plane is I_disc = (1/2) * M * R². Since I_ring > I_disc (M*R² is greater than (1/2)*M*R²), and ω is the same for both, the rotational kinetic energy of the ring is higher than that of the disc (KE_ring = (1/2)*I_ring*ω² > KE_disc = (1/2)*I_disc*ω²).

Moment of inertia represents the resistance of an object to rotational motion; it depends on the mass distribution relative to the axis of rotation. For the same total mass and radius, the ring has mass distributed further from the axis (all at radius R) compared to the disc (mass distributed from center to R), resulting in a higher moment of inertia for the ring. Kinetic energy of a body rotating is non-zero unless its angular velocity is zero, so option D is incorrect.

66. A ball is released from rest and rolls down an inclined plane, as show

A ball is released from rest and rolls down an inclined plane, as shown in the following figure, requiring 4 s to cover a distance of 100 cm along the plane :
Which one of the following is the correct value of angle θ that the plane makes with the horizontal? (g = 1000 cm/s²)

θ = sin⁻¹ (1/9•8)

θ = sin⁻¹ (1/20)

θ = sin⁻¹ (1/80)

θ = sin⁻¹ (1/100)

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The correct value of angle θ that the plane makes with the horizontal is sin⁻¹ (1/80).

– The ball is released from rest, so initial velocity u = 0.
– The distance covered along the plane is s = 100 cm in time t = 4 s.
– The motion along the inclined plane is uniformly accelerated. The formula for distance is s = ut + (1/2)at².
– Substitute the values: 100 cm = (0)(4 s) + (1/2)a(4 s)².
– 100 = (1/2)a(16) => 100 = 8a.
– The acceleration down the inclined plane is a = 100 / 8 = 12.5 cm/s².
– The acceleration of an object rolling down an inclined plane (without slipping, but here it’s just about acceleration along the plane component of gravity) is a = g sin θ, where g is the acceleration due to gravity and θ is the angle of inclination.
– Given g = 1000 cm/s².
– So, 12.5 = 1000 sin θ.
– sin θ = 12.5 / 1000 = 125 / 10000 = 1 / 80.
– θ = sin⁻¹ (1/80).

– The acceleration due to gravity along an inclined plane is g sin θ, and perpendicular to the plane is g cos θ. The net force along the plane determines the acceleration.
– The value of g is given as 1000 cm/s², which is equivalent to 10 m/s².

67. Consider the following velocity and time graph: Which one of the follo

Consider the following velocity and time graph:
Which one of the following is the value of average acceleration from 8 s to 12 s?

8 m/s²

12 m/s²

2 m/s²

-1 m/s²

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The correct value of average acceleration from 8 s to 12 s is -1 m/s².

– Average acceleration is defined as the change in velocity divided by the time interval.
– From the given velocity-time graph, the velocity at t = 8 s is approximately 10 m/s.
– From the graph, the velocity at t = 12 s is approximately 6 m/s.
– The time interval is Δt = 12 s – 8 s = 4 s.
– The change in velocity is Δv = v(12s) – v(8s) = 6 m/s – 10 m/s = -4 m/s.
– Average acceleration = Δv / Δt = (-4 m/s) / (4 s) = -1 m/s².

– A negative acceleration indicates that the velocity is decreasing.
– In a velocity-time graph, the slope of the line represents instantaneous acceleration. For average acceleration over an interval, we consider the endpoints of the interval.

68. A planet has a mass M₁ and radius R₁. The value of acceleration due to

A planet has a mass M₁ and radius R₁. The value of acceleration due to gravity on its surface is g₁. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?

g₁

2g₁

g₁/2

g₁/4

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The acceleration due to gravity on the surface of planet 2 is g₁/2.

The acceleration due to gravity (g) on the surface of a planet is given by the formula g = GM/R², where G is the gravitational constant, M is the mass of the planet, and R is its radius.

For planet 1, the acceleration due to gravity on its surface is g₁ = GM₁/R₁².
For planet 2, the mass M₂ = 2M₁ and the radius R₂ = 2R₁.
We can calculate the acceleration due to gravity on planet 2’s surface (g₂) using the same formula:
g₂ = G * M₂ / R₂²
Substitute the values for M₂ and R₂ in terms of M₁ and R₁:
g₂ = G * (2M₁) / (2R₁)²
g₂ = G * (2M₁) / (4R₁²)
g₂ = (2/4) * (GM₁/R₁²)
g₂ = (1/2) * (GM₁/R₁²)
Since g₁ = GM₁/R₁², we have g₂ = g₁/2.

69. The time period of oscillation of a simple pendulum having length L an

The time period of oscillation of a simple pendulum having length L and mass of the bob m is given as T. If the length of the pendulum is increased to 4L and the mass of the bob is increased to 2m, then which one of the following is the new time period of oscillation?

T/2

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The new time period of oscillation is 2T.

The time period (T) of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob (m) does not affect the time period of a simple pendulum.

Let the initial time period be T₁ = T, with length L₁ = L and mass m₁ = m. The formula is T₁ = 2π√(L/g).
When the length is increased to L₂ = 4L and the mass is increased to m₂ = 2m, the new time period T₂ is calculated using the formula, considering only the change in length:
T₂ = 2π√(L₂/g) = 2π√((4L)/g) = 2π * √4 * √(L/g) = 2π * 2 * √(L/g) = 2 * (2π√(L/g)).
Since T₁ = 2π√(L/g), the new time period T₂ = 2 * T₁. Thus, the new time period is 2T.

70. Which one of the following statements about energy is correct ?

Which one of the following statements about energy is correct ?

Energy can be created as well as destroyed.

Energy can be created but not destroyed.

Energy can neither be created nor destroyed.

Energy cannot be created but can be destroyed.

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The statement that energy can neither be created nor destroyed is a fundamental principle known as the Law of Conservation of Energy (also the First Law of Thermodynamics). This law states that the total energy of an isolated system remains constant over time. Energy can be transformed from one form to another (e.g., potential energy to kinetic energy, chemical energy to heat and light), but the total amount of energy in the universe is conserved.

– Law of Conservation of Energy: Energy cannot be created or destroyed.
– Energy can be transformed between different forms.
– The total energy in a closed system remains constant.

In the context of mass-energy equivalence (E=mc²), mass can be converted into energy and vice versa, for example, in nuclear reactions. However, this is still considered a transformation of mass into energy, not the creation or destruction of the fundamental entity of “mass-energy”. In many common physical and chemical processes, mass conservation and energy conservation can be treated separately.