JEE Advanced · Physics · 8. Rotational Motion

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60 °$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2 - \sqrt{3}) / \sqrt{10} s$ , then the height of the top of the inclined plane (in meters) is __________. Take $g = 10 m s^{- 2}$ .

A 0.85
B 0.75
C 0.7
D 71

Verified Solution

Answer & Solution

Correct Answer

(B) 0.75

Step-by-step Solution

Detailed explanation

a_{c} = \frac{g \sin θ}{1 + \frac{I_{c}}{M R^{2}}}

a_{r i n g} = \frac{g \sin θ}{2}

a_{r i n g} = \frac{2 g \sin θ}{3}

\frac{h}{\sin θ} = \frac{1}{2} (\frac{g \sin θ}{2}) t_{1}^{2} \Rightarrow t_{1} = \sqrt{\frac{4 h}{g \sin^{2} θ}} = \sqrt{\frac{16 h}{3 g}}

\Rightarrow \frac{h}{\sin θ} = \frac{1}{2} (\frac{2 g \sin θ}{3}) t_{2}^{2} \Rightarrow t_{2} = \sqrt{\frac{3 h}{g \sin^{2} θ}} = \sqrt{\frac{4 h}{g}}

\Rightarrow \sqrt{\frac{16 h}{3 g}} - \sqrt{\frac{4 h}{g}} = \frac{2 - \sqrt{3}}{\sqrt{10}}

\sqrt{h} [\frac{4}{\sqrt{3}} - 2] = 2 - \sqrt{3}

\sqrt{h} = \frac{(2 - \sqrt{3}) \sqrt{3}}{(4 - 2 \sqrt{3})} = \frac{\sqrt{3}}{2} \Rightarrow h = \frac{3}{4} = 0.75 m

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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