JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Rotational Mechanics

A cylinder is rolling down on an inclined plane of inclination

6 0^{\circ}

. It's acceleration during rolling down will be

\frac{X}{3}

m/s

^{2}

, where

x =

______. (use

g = 10

m/s

^{2}

Answer: 10

Solution

For a solid cylinder rolling down an incline of angle

θ

without slipping,

a = \frac{g sin θ}{1 + \frac{I}{M R ^{2}}} = \frac{g sin θ}{1 + \frac{1}{2}} = \frac{2}{3} g sin θ

With

θ = 6 0^{\circ}

g = 10 m/s^{2}

a = \frac{2}{3} (10) (\frac{3}{2}) = \frac{10 3}{3} = \frac{10}{3} m/s^{2}

Comparing with

a = \frac{x}{3}

gives

x = 10

.

Key confirmed: 10.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer

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