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(D) Let the distance of end $A$ from the corner $O$ be $x$ and the distance of end $B$ from the corner $O$ be $y$.Since the rod is of constant length

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$v \cot 	heta$

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(D) Let the distance of end $A$ from the corner $O$ be $x$ and the distance of end $B$ from the corner $O$ be $y$.Since the rod is of constant length $L$,we have the relation: $x^2 + y^2 = L^2$.Differentiating both sides with respect to time $t$,we get:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.Given that the end $A$ is pulled to the left with velocity $v$,we have $\frac{dx}{dt} = -v$ (since $x$ is decreasing).Let the downward velocity of end $B$ be $v^{\prime}$,so $\frac{dy}{dt} = -v^{\prime}$ (since $y$ is decreasing).Substituting these values into the differentiated equation:$2x(-v) + 2y(-v^{\prime}) = 0$$-xv - yv^{\prime} = 0$$yv^{\prime} = -xv$Since we are looking for the magnitude of the downward velocity $v^{\prime}$,we have:$v^{\prime} = \frac{x}{y} v$.From the geometry of the right-angled triangle formed by the rod,the wall,and the floor,we have $\cot 	heta = \frac{x}{y}$.Therefore,$v^{\prime} = v \cot 	heta$.

$A$ thin uniform rod of length $L$ is resting against a wall and the floor as shown in the figure. Its lower end $A$ is pulled towards the left with a constant velocity $v$. Then the downward velocity $v^{\prime}$ of the other end $B$ when the rod makes an angle $\theta$ with the floor is

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