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.
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 $K = \frac{3}{4}$.
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 $I = \mu \ell^2 = \frac{m_1 m_2}{m_1 + m_2}\ell^2$.
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.
The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} – 6\hat{k}$ at $(2, 0, -3)$, about the point $(2, -2, -2)$ is given by
(2018)
1. $-7\hat{i} - 8\hat{j} - 4\hat{k}$
2. $-4\hat{i} - \hat{j} - 8\hat{k}$
3. $-8\hat{i} - 4\hat{j} - 7\hat{k}$
4. $-7\hat{i} - 4\hat{j} - 8\hat{k}$
View Answer
The relative position vector is $\vec{r} = (2-2)\hat{i} + (0 - (-2))\hat{j} + (-3 - (-2))\hat{k} = 2\hat{j} - \hat{k}$. Torque is $\vec{\tau} = \vec{r} \times \vec{F} = (2\hat{j} - \hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k}) = -7\hat{i} - 4\hat{j} - 8\hat{k}$.