JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Rotational Mechanics

A circular disc reaches from top to bottom of an inclined plane of length

l

. When it slips down the plane, if takes t s . When it rolls down the plane then it takes

(\frac{α}{2})^{1/2} ts

, where

α

is ____

Answer: 3

Solution

Sliding (frictionless) down length

l

a = g sin θ

, and

l = \frac{1}{2} a t^{2} ⟹ t = \frac{2 l}{g sin θ}

.

Rolling (disc,

k^{2} / R^{2} = 1/2

a^{'} = \frac{g sin θ}{1 + k ^{2} / R ^{2}} = \frac{2}{3} g sin θ

, and

l = \frac{1}{2} a^{'} t^{'2} ⟹ t^{'} = \frac{2 l}{a ^{'}} = \frac{3 l}{g sin θ} = \frac{3}{2} t

t^{'} = (\frac{α}{2})^{1/2} t

gives

\frac{α}{2} = \frac{3}{2} ⟹ α = 3

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer

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