Statement-1:A pair of upright metersticks, with their lower ends against a wall, are allowed to fall to the floor. One is bare, and the other has a heavy weight attached to its upper end. The stick to hit the floor first is the weighted stick.
Statement-2: The torque acting on weighted stick is more than the bare stick.

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Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement-1.

Statement-1 is false, statement-2 is true.

Statement-1 is true, statement-2 is true and statement-2 is the Correct explanation for statement-1.

Statement-1 is true, statement-2 is false.

answer is D.

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Detailed Solution

$\begin{array}{l} α_{1} = \frac{mg \frac{l}{2} \sin θ}{m \frac{l^{2}}{3}} = \frac{3 gsin θ}{2 l} \\ α_{2} ≃ \frac{Mglsin θ}{{Ml}^{2}} = \frac{gsin θ}{l} \end{array}$

Since angular acceleration for the weighted stick is less, it will hit the floor later. Also, torque acting on weighted stick is more.

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