JEE Main 2025 — Rotational Motion Question with Solution

Let's analyze the problem step-by-step.

Energy conservation:

When the sphere rolls without slipping, its gravitational potential energy converts into both translational and rotational kinetic energy. The energy conservation equation is given by:

$$mgL\sin \theta =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$

where:

$L\sin \theta$ is the vertical height,

$I$ is the moment of inertia, and

$\omega$ is the angular speed.

Moment of Inertia and Rolling Condition:

For a solid sphere, the moment of inertia about its center is:

$$I=\frac{2}{5}m{r}^2$$

Since the sphere rolls without slipping, the linear speed $v$ and the angular speed $\omega$ are related by:

$$v=r\omega \Rightarrow \omega =\frac{v}{r}$$

Total Kinetic Energy:

Substituting the rolling condition into the rotational kinetic energy gives:

$\begin{array}{rl}K& =\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{v}{r}\right)}^2\\ & =\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)\frac{v^2}{r^2}\\ & =\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2\\ & =\frac{7}{10}m{v}^2\end{array}$

Finding $v^2$ in Terms of $L$ and $\sin \theta$:

Equate the initial potential energy to the final kinetic energy:

$$mgL\sin \theta =\frac{7}{10}m{v}^2$$

Solving for $v^2$:

$$v^2=\frac{10}{7}gL\sin \theta$$

Apply to the Two Cases:

For $\theta =30^{\circ }$, since $\sin (30^{\circ })=\frac{1}{2}$:

$$v_1^2=\frac{10}{7}gL\left(\frac{1}{2}\right)=\frac{5}{7}gL$$

For $\theta =45^{\circ }$, since $\sin (45^{\circ })=\frac{1}{\sqrt{2}}$:

$$v_2^2=\frac{10}{7}gL\left(\frac{1}{\sqrt{2}}\right)$$

Calculate the Ratio $\frac{v_1^2}{v_2^2}$:

$\begin{array}{rl}\frac{v_1^2}{v_2^2}& =\frac{\left(\frac{5}{7}gL\right)}{\left(\frac{10}{7}gL\frac{1}{\sqrt{2}}\right)}\\ & =\frac{5}{10}\sqrt{2}\\ & =\frac{\sqrt{2}}{2}\end{array}$

Expressing this ratio as $v_1^2:v_2^2$, we have:

$$v_1^2:v_2^2=1:\sqrt{2}$$

Conclusion:

According to the options given, the correct answer is:

Option C: $1:\sqrt{2}$.