In the arrangement shown in the Fig, the ends $P$ and $Q$ of an unstretchable string move downwards with uniform speed $U$. Pulleys $A$ and $B$ are fixed. Mass $M$ moves upwards with a speed

Step 1: Define the total length of the string.

Let $\ell_P$ be the length of the string segment from pulley $A$ to end $P$, $\ell_Q$ be the length of the string segment from pulley $B$ to end $Q$, and $\ell$ be the length of each of the two inner segments connecting the pulleys to the mass $M$. Since the string is unstretchable, its total length $L$ is constant.

$$L = \ell_P + \ell_Q + 2\ell$$

Step 2: Relate the rates of change of the outer segments to the speed $U$.

The ends $P$ and $Q$ move downwards with a uniform speed $U$, which means the lengths of the outer segments are increasing at rate $U$.

$$\frac{d\ell_P}{dt} = U$$ $$\frac{d\ell_Q}{dt} = U$$

Step 3: Differentiate the total length equation with respect to time.

Since $L$ is constant, its time derivative is zero.

$$\frac{dL}{dt} = 0$$ $$\frac{d\ell_P}{dt} + \frac{d\ell_Q}{dt} + 2\frac{d\ell}{dt} = 0$$

Substitute the rates of change of the outer segments into the differentiated equation.

$$U + U + 2\frac{d\ell}{dt} = 0$$ $$\frac{d\ell}{dt} = -U$$

Step 4: Relate the rate of change of the inner segment length to the upward speed $v$ of mass $M$.

Let $v$ be the upward speed of mass $M$. The component of the velocity of mass $M$ along the direction of each inner string segment is $v \cos\theta$. This component represents the rate at which the length of the inner segment $\ell$ is decreasing.

$$\frac{d\ell}{dt} = -v\cos\theta$$

Step 5: Solve for the upward speed $v$.

Equate the two expressions obtained for $\frac{d\ell}{dt}$.

$$-U = -v\cos\theta$$ $$v\cos\theta = U$$ $$v = \frac{U}{\cos\theta}$$

Answer: The mass $M$ moves upwards with a speed of $\dfrac{U}{\cos\theta}$.