JEE Physics 2024 Question with Solution

Given: A solid cylinder rolls down an inclined plane of angle $60^\circ$ with $g = 10 \, \text{m/s}^2$.

Find: The value of $x$ if acceleration is written in the given form.

For rolling without slipping, the acceleration is

$$a = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}}$$

For a solid cylinder,

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

Substituting,

$$a = \frac{g \sin \theta}{1 + \frac{1}{2}}$$

Now use $g = 10$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$:

$$a = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{10\sqrt{3}}{3}$$

Thus the acceleration is $\frac{10\sqrt{3}}{3} \, \text{m/s}^2$.

From the solution working, the final concluded answer is $x = 10$.

Therefore, the numerical answer is $10$.

Given: A body rolls without slipping on an incline of angle $60^\circ$ and $g = 10 \, \text{m/s}^2$.

Find: The value of $x$.

Another standard form is

$$a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$$

For a solid cylinder,

$$\frac{k^2}{R^2} = \frac{1}{2}$$

So,

$$a = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2g \sin \theta}{3}$$

Substitute the values:

$$a = \frac{2 \times 10 \times \sin 60^\circ}{3}$$ $$a = \frac{2 \times 10 \times \frac{\sqrt{3}}{2}}{3} = \frac{10\sqrt{3}}{3} \, \text{m/s}^2$$

The extracted the solution contains an internal inconsistency in the comparison step, but both approaches conclude the final answer as $x = 10$.

Therefore, the numerical answer is $10$.