A ring is allowed to roll down on an incline of $1$ in $10$ without slipping. The acceleration of its centre of mass is

9.8 m s^{- 2}

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4.9 m s^{- 2}

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0.98 m s^{- 2}

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0.49 m s^{- 2}

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Solution

The correct option is D $0.49 m s^{- 2}$
Consider $1$ in $10$ means $s i n θ = \frac{1}{10}$

$β = \frac{k^{2}}{r^{2}}$

For ring, $β = 1$

Acceleration of center of mass $= \frac{g sin θ}{(1 + β)}$
$\begin{matrix} = \frac{(9.8 m / s^{2} \times \frac{1}{10})}{(1 + 1)} = 0.49 m /^{2} \end{matrix}$

Hence, The Option $D$ is the correct answer.

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A ring is allowed to roll down on an incline of 1 in 10 without slipping. The acceleration of its centre of mass is

The correct option is D 0.49 ms−2Consider 1 in 10 means sinθ=110β=k2r2For ring, β=1Acceleration of center of mass =gsinθ(1+β) =(9.8m/s2×110)(1+1)=0.49m/2Hence, The Option D is the correct answer.

A ring is allowed to roll down on an incline of $1$ in $10$ without slipping. The acceleration of its centre of mass is