ln the arrangement shown in figure the ends P and Q of an non-stretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed

 

As P and p fall down, the length / decreases at the rate of U m/s.

From the figure, ${\mathrm{l}}^2 = {\mathrm{b}}^2+{\mathrm{v}}^2$

Differentiating with respect to time

$2\mathrm{l} \times \frac{\mathrm{dl}}{\mathrm{dt}} = 2\mathrm{b}\times \frac{\mathrm{db}}{\mathrm{dt}}+2\mathrm{y}\times \frac{\mathrm{dy}}{\mathrm{dt}}(\mathrm{As} \frac{\mathrm{db}}{\mathrm{dt}}=0, \frac{\mathrm{dl}}{\mathrm{dt}} = \mathrm{U})$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = \left(\frac{\mathrm{l}}{\mathrm{y}}\right)\times \frac{\mathrm{dl}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = \left(\frac{1}{\cos \mathrm{\theta }}\right)\times \mathrm{U} = \frac{\mathrm{U}}{\cos \mathrm{\theta }}$