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(B) Let the horizontal distance of the block from the pulley be $x$ and the length of the rope from the block to the pulley be $y$. Let the vertical h

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$\frac{v}{\sin 	heta}$

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(B) Let the horizontal distance of the block from the pulley be $x$ and the length of the rope from the block to the pulley be $y$. Let the vertical height of the pulley from the plane be $h$.From the geometry of the right-angled triangle formed,we have $x^2 + h^2 = y^2$.Differentiating both sides with respect to time $t$,we get:$2x \frac{dx}{dt} + 0 = 2y \frac{dy}{dt}$.Here,$\frac{dx}{dt}$ is the horizontal velocity of the block $(v_b)$ and $\frac{dy}{dt}$ is the rate at which the length of the rope decreases,which is equal to the velocity of the rope $v$.So,$x v_b = y v$.$v_b = \frac{y}{x} v$.From the figure,$\cos 	heta = \frac{h}{y}$ and $\sin 	heta = \frac{x}{y}$.Therefore,$\frac{y}{x} = \frac{1}{\sin 	heta}$.Substituting this into the expression for $v_b$,we get $v_b = \frac{v}{\sin 	heta}$.

$A$ block is dragged on a smooth plane with the help of a rope which moves with a velocity $v$ as shown in the figure. The horizontal velocity of the block is

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