A circular disc reaches from top to bottom of an inclined plane of length 'L' when it slips down the plane, it takes time '$\mathrm{t}_{1}$'. When it rolls down the plane, it takes time $\mathrm{t}_{2}$. The value of $\frac{t_{1}}{t_{2}}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ will be \_\_\_\_.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Rotational Mechanics

A circular disc reaches from top to bottom of an inclined plane of length 'L' when it slips down the plane, it takes time '

t_{1}

'. When it rolls down the plane, it takes time

t_{2}

. The value of

\frac{t _{1}}{t _{2}}

\frac{3}{x}

. The value of

x

will be ____.

Answer: 4.5

Solution

Sliding down a frictionless incline:

a_{1} = g sin θ

. A solid disc rolling without slipping:

a_{2} = \frac{g sin θ}{1 + I / m R ^{2}} = \frac{g sin θ}{1 + \frac{1}{2}} = \frac{2}{3} g sin θ

. Since

L = \frac{1}{2} a t^{2} \Rightarrow t \propto 1/ a

\frac{t _{1}}{t _{2}} = \frac{a _{2}}{a _{1}} = \frac{2}{3}

. Equating to

3/ x

\frac{2}{3} = \frac{3}{x} \Rightarrow x = \frac{9}{2} = 4.5

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Numerical answer

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