Physics - W.E.P. Question with Solution | TestHub

The chain (length $\pi R$, mass per unit length $\lambda$) initially fills the semicircular arch; its CM is at height $\frac{2R}{\pi }$ above the diameter. When it has fully emerged at B it hangs vertically, CM at depth $\frac{\pi R}{2}$ below B.

$$\frac{1}{2}(\lambda \pi R)v^2=\lambda \pi Rg\left(\frac{2R}{\pi }+\frac{\pi R}{2}\right)$$

$$v=\sqrt{2gR\left(\frac{2}{\pi }+\frac{\pi }{2}\right)}$$

Ans. (C)