JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Rotational Mechanics

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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 - 3) / 10 s

, then the height of the top of the inclined plane, in metres, is ____. Take

g = 10 m s^{- 2}

Answer: 0.75

Solution

For rolling without slipping,

a = \frac{g s i n θ R ^{2}}{k ^{2} + R ^{2}}

. For disc:

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

, so

a_{d i sc} = \frac{g}{3}

. For ring:

k^{2} = R^{2}

, so

a_{r in g} = \frac{g 3}{4}

. Time to travel distance

s = h / sin 60°

t = \frac{4 h}{a 3}

. The time difference is

t_{r in g} - t_{d i sc} = \frac{2 h}{3 ^{1/4}} [\frac{- 1}{g} 3^{1/4} + \frac{2}{g 3}] = \frac{2 - 3}{10}

. Solving gives

h = 0.75

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer

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JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Rotational Mechanics

Solution

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