Moment of Inertia - NEET Physics Questions
Question 11: easy

Assertion (A): Inertia and moment of inertia are same quantities.


Reason (R): Moment of inertia represents the capacity of a rigid body to oppose its state of oscillatory motion.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Inertia (mass) measures resistance to translational motion, while moment of inertia measures resistance to rotational motion. They are distinct quantities. Moment of inertia opposes changes in a body's state of \(\text{rotational}\) motion, not oscillatory motion. Therefore, both Assertion (A) and Reason (R) are false.

Question 12: easy

Assertion (A): For the purpose of calculation of moment of inertia, body’s mass can be assumed to be concentrated at its centre of mass.


Reason (R): Moment of inertia of a rigid about an axis passing through its centre of mass is zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Moment of inertia depends critically on the distribution of mass relative to the axis of rotation, so mass cannot generally be assumed concentrated at the center of mass (A is false). Also, the moment of inertia of a rigid body about an axis passing through its center of mass is generally not zero (e.g., a disc has \(I = \frac{1}{2}MR^2\)). Thus, (R) is false. Both statements are incorrect.

Question 13: easy

Assertion (A): If two different axes are at same distance from the centre of mass of a rigid body then moment of inertia of the given rigid body about both the axes will always be equal.


Reason (R): According to perpendicular axis theorem \(\text{I} = \text{I}_{\text{cm}} + \text{Md}^2\) where symbols have their usual meaning.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A) is false. Moment of inertia depends on both the distance and the orientation of the axis. Reason (R) is false. The given formula is for the parallel axis theorem, not the perpendicular axis theorem.

Question 14: easy

A light rod of length $\ell$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:

(2016 – II)

1. $(m_1 + m_2)\ell^2$
2. $\sqrt{m_1 m_2}\ell^2$
3. $\frac{m_1 m_2}{m_1 + m_2}\ell^2$
4. $\frac{m_1 + m_2}{m_1 m_2}\ell^2$
View Answer

The moment of inertia about the center of mass uses reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, resulting in $I = \mu \ell^2 = \frac{m_1 m_2}{m_1 + m_2}\ell^2$.

Question 15: easy

The ratio of the radius of gyration of a thin uniform disc about an axis passing through it centre and normal to its plane to the radius of gyration of the disc about its diameter is:

(2022)

1. $1:\sqrt{2}$
2. $2:1$
3. $\sqrt{2}:1$
4. $4:1$
View Answer

Radius of gyration about center normal axis is $k_1 = R/\sqrt{2}$ and about diameter is $k_2 = R/2$. The ratio $k_1/k_2$ simplifies to $\sqrt{2}:1$.

Question 16: easy

From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a $90^\circ$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times $MR^2$. Then the value of ‘K’ is:

(2021)

1. $\frac{7}{8}$
2. $\frac{1}{4}$
3. $\frac{1}{8}$
4. $\frac{3}{4}$
View Answer

Since mass is uniformly distributed, removing a $90^\circ$ sector removes $1/4$ of the mass, leaving $3/4$ of the mass at distance $R$. Thus $I = \frac{3}{4}MR^2$, making $K = \frac{3}{4}$.

Question 17: moderate

The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is:

(2006, 2005)

1. $\frac{1}{2}MR^2$
2. $MR^2$
3. $\frac{2}{5}MR^2$
4. $\frac{3}{2}MR^2$
View Answer

Using the parallel axis theorem, $I = I_{cm} + Md^2$. Here $I_{cm} = \frac{1}{2}MR^2$ and $d = R$, so $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.

Question 18: easy

The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is:

(2004)

1. $2:1$
2. $\sqrt{5}:\sqrt{6}$
3. $2:3$
4. $1:\sqrt{2}$
View Answer

For disc about in-plane tangent, $I_d = \frac{5}{4}MR^2 \implies k_d = \frac{\sqrt{5}}{2}R$. For ring about in-plane tangent, $I_r = \frac{3}{2}MR^2 \implies k_r = \sqrt{\frac{3}{2}}R$. Ratio is $\sqrt{5}:\sqrt{6}$.

Question 19: moderate

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be:

(2003)

1. $\frac{K^2+R^2}{R^2}$
2. $\frac{K^2}{R^2}$
3. $\frac{K^2}{K^2+R^2}$
4. $\frac{R^2}{K^2+R^2}$
View Answer

Rotational kinetic energy $E_{rot} = \frac{1}{2}MK^2\omega^2$ and total energy $E = \frac{1}{2}M(K^2+R^2)\omega^2$. The fraction is $E_{rot}/E_{total} = \frac{K^2}{K^2+R^2}$.

Question 20: easy

A circular disc is to be made by using iron and aluminium so that it acquired maximum moment of inertia about geometrical axis. It is possible with:

(2002)

1. Aluminium at interior and iron surround to it.
2. Iron at interior and aluminum surround to it.
3. Using iron and aluminium layers in alternate order.
4. Sheet of iron is used at both external surface and aluminium sheet as internal layers.
View Answer

Moment of inertia is $I = \int r^2 dm$. To maximize $I$, denser material (iron) should be placed at the outer periphery and lighter material (aluminium) at the interior.