Rotational Motion — JEE Main practice

1. Immediately after one string is cut, the rod is acted upon by only two external forces:

• Tension $T$ acting upward at the remaining end.
• Weight $mg$ acting downward through the centre of mass.

2. Let the centre of mass accelerate downward with acceleration $a$.

Applying Newton's second law for translational motion,

$$mg-T=ma.$$

3. The tension also produces a torque about the centre of mass.

Since the tension acts at a distance

$$\frac{l}{2},$$

the torque is

$$\tau=T\cdot\frac{l}{2}.$$

4. The moment of inertia of a uniform rod about its centre is

$$I=\frac{1}{12}ml^2.$$

Using

$$\tau=I\alpha,$$

we obtain

$$T\cdot\frac{l}{2} = \frac{1}{12}ml^2\alpha.$$

5. Immediately after the string is cut, the end attached to the remaining string behaves as the instantaneous pivot.

Hence, the linear acceleration of the centre of mass and the angular acceleration are related by

$$a=\alpha\cdot\frac{l}{2},$$

or

$$\alpha=\frac{2a}{l}.$$

6. Substitute this value of $\alpha$ into the torque equation:

$$T\cdot\frac{l}{2} = \frac{1}{12}ml^2 \left(\frac{2a}{l}\right).$$

Simplifying,

$$T=\frac{ma}{3},$$

or equivalently,

$$ma=3T.$$

7. From the translational equation,

$$mg-T=ma.$$

Replacing $ma$ by $3T$,

$$mg-T=3T.$$

Therefore,

$$mg=4T,$$

giving

$$\boxed{T=\frac{mg}{4}}.$$

8. Hence, the correct answer is Option C.

1. The pulley consists of a thin rim and two identical rods.

For a thin rim of mass $M$ and radius $R$,

$$I_{\text{rim}}=MR^2.$$

2. Each rod has

$$\text{mass}=M,\qquad \text{length}=2R.$$

The moment of inertia of one rod about its centre is

$$I=\frac{1}{12}ML^2.$$

Substituting $L=2R$,

$$I_{\text{one rod}} = \frac{1}{12}M(2R)^2$$ $$= \frac{1}{3}MR^2.$$

Since there are two rods,

$$I_{\text{rods}} = 2\times\frac{1}{3}MR^2$$ $$= \frac{2}{3}MR^2.$$

3. Therefore, the total moment of inertia of the pulley is

$$I = I_{\text{rim}} + I_{\text{rods}}$$ $$= MR^2+\frac{2}{3}MR^2$$ $$= \frac{5}{3}MR^2.$$

4. Let the linear acceleration of the blocks be $a$.

Since the string does not slip,

$$a=R\alpha,$$

where $\alpha$ is the angular acceleration of the pulley.

5. Apply Newton's second law to the two blocks.

For the block of mass $M$,

$$Mg-T_1=Ma.$$

For the block of mass $m$,

$$T_2-mg=ma.$$

6. The torque on the pulley is produced by the difference in tensions:

$$(T_1-T_2)R=I\alpha.$$

Using

$$I=\frac{5}{3}MR^2, \qquad \alpha=\frac{a}{R},$$

we obtain

$$T_1-T_2=\frac{5}{3}Ma.$$

7. Add the two linear equations:

$$(Mg-T_1)+(T_2-mg) = (M+m)a.$$

Rearranging,

$$(M-m)g-(T_1-T_2) = (M+m)a.$$

8. Substitute

$$T_1-T_2=\frac{5}{3}Ma.$$

Then,

$$(M-m)g = (M+m)a+\frac{5}{3}Ma.$$

Simplifying,

$$(M-m)g = \left(\frac{8}{3}M+m\right)a.$$

9. Hence,

$$\boxed{ a = \frac{(M-m)g} {\left(\frac{8}{3}M+m\right)} }.$$

10. Therefore, the correct answer is Option B.

1. The axis of rotation is the common tangent passing through the point of contact of the two spheres.

For each sphere, the axis is parallel to a diametric axis and is at a distance equal to its radius from the centre.

2. Apply the Parallel Axis Theorem:

$$I=I_{\mathrm{CM}}+Md^2,$$

where

• $I_{\mathrm{CM}}$ is the moment of inertia about the centre,
• $M$ is the mass of the sphere,
• $d$ is the perpendicular distance from the centre to the tangent axis.

3. For a uniform solid sphere,

$$I_{\mathrm{CM}}=\frac{2}{5}Mr^2.$$

Since

$$d=r,$$

the moment of inertia about the tangent becomes

$$I_{\mathrm{tangent}} = \frac{2}{5}Mr^2+Mr^2$$ $$= \frac{7}{5}Mr^2.$$

4. For the first sphere,

$$M_1=5\ \mathrm{kg}, \qquad r_1=10\ \mathrm{cm}=0.1\ \mathrm{m}.$$

Therefore,

$$I_1 = \frac{7}{5}\times5\times(0.1)^2$$ $$= 7\times0.01$$ $$= 0.07\ \mathrm{kg\cdot m^2}.$$

5. For the second sphere,

$$M_2=10\ \mathrm{kg}, \qquad r_2=20\ \mathrm{cm}=0.2\ \mathrm{m}.$$

Hence,

$$I_2 = \frac{7}{5}\times10\times(0.2)^2$$ $$= 14\times0.04$$ $$= 0.56\ \mathrm{kg\cdot m^2}.$$

6. The total moment of inertia is the sum of the moments of inertia of the two spheres:

$$I_{\mathrm{total}} = I_1+I_2$$ $$= 0.07+0.56$$ $$= 0.63\ \mathrm{kg\cdot m^2}.$$

7. Therefore, the required moment of inertia is

$$\boxed{0.63\ \mathrm{kg\cdot m^2}}.$$

Hence, the correct answer is Option D.