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Two identical masses are connected to a horizontal thin massless rod as shown in the figure. When their distance from the pivot is x, a torque produces an angular acceleration $α_{1}$ . If the masses are now repositioned so that they are at distance 2x each from the pivot, the same torque will produce an angular acceleration $α_{2}$ such that

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$α_{2} = 4 α_{1}$

$α_{2} = α_{1}$

$α_{2} = \frac{α_{1}}{2}$

$α_{2} = \frac{α_{1}}{4}$

answer is D.

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Detailed Solution

$\begin{array}{l} τ = (2 {mx}^{2}) α_{1} \\ τ = (2 \times m \times 4 x^{2}) α_{2} \end{array}$

Equating, $8 {mx}^{2} α_{1} or α_{2} = \frac{α_{1}}{4}$

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