Rotational Motion — NEET UG practice

1. First find the mass of the wire using the definition of linear mass density (mass per unit length).

$$M = mL$$

2. Since the wire of length $L$ is bent into a full circle, use the circumference formula to get the radius of the ring.

$$L = 2\pi r \implies r = \frac{L}{2\pi}$$

3. For a ring, the moment of inertia about a diameter (an axis in its own plane through the centre) is half that about the axis perpendicular to its plane.

$$I_{cm} = \frac{1}{2}Mr^2$$

4. The axis $yy'$ shown is parallel to this diameter but shifted to the edge of the ring, so apply the parallel axis theorem to shift from the centre to the rim.

$$I_{yy'} = I_{cm} + Mr^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$$

5. Substituting $M = mL$ and $r = \frac{L}{2\pi}$ gives the final answer in terms of $m$ and $L$.

$$I_{yy'} = \frac{3}{2}(mL)\left(\frac{L}{2\pi}\right)^2 = \frac{3mL^3}{8\pi^2}$$

Hence, the answer is option B.

1. Angular momentum about a point can always be split into two parts — the part due to the motion of the centre of mass about that point, plus the part due to rotation about the centre of mass.

$$\vec{L} = \vec{r} \times \vec{P} + I_{cm}\vec{\omega}$$

2. Here the disc rotates about a fixed vertical axis through its centre $O$, so the centre of mass never moves, meaning its linear momentum is zero.

$$\vec{P} = 0$$

3. Since $\vec{P}=0$ holds for both points $A$ and $B$, the first term vanishes in each case, leaving only the spin part, which does not depend on which point you measure it about.

$$\vec{L}_A = \vec{L}_B = I_{cm}\vec{\omega}$$

4. Taking the ratio, the identical $I_{cm}\vec{\omega}$ terms cancel out exactly.

$$\frac{L_A}{L_B} = 1$$

Hence, the answer is option D.

1. For a solid sphere, the moment of inertia about an axis through its own centre is given by the standard formula.

$$I_{cm} = \frac{2}{5}mR^2$$

2. To find the moment of inertia of the whole system about the vertical axis through the centre of sphere $A$, apply the parallel axis theorem to sphere $B$, since its centre lies at a distance $(R+r)$ from this axis.

$$I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R+r)^2$$

3. Similarly, to find the moment of inertia about the vertical axis through the centre of sphere $B$, apply the parallel axis theorem to sphere $A$ instead, since now $A$'s centre lies at distance $(R+r)$ from this new axis.

$$I_B = \frac{2}{5}MR^2 + M(R+r)^2 + \frac{2}{5}mr^2$$

4. Subtracting the two expressions, the common terms $\frac{2}{5}MR^2$ and $\frac{2}{5}mr^2$ cancel out, leaving only the extra parallel-axis terms.

$$I_A - I_B = m(R+r)^2 - M(R+r)^2 = (m-M)(R+r)^2$$

Hence, the answer is option C.