Question

As shown in the figure, a chain of mass $m$ is placed on a smooth quarter circular portion of radius $R$ . End $A$ is tied with a wedge while the remaining chain is free, then the minimum work required to be done by the external agent to make the chain horizontal keeping point $A$ fixed, is

• A. $m\text{gR}$
• B. $m\text{g}\frac{2\text{R}}{\pi }$
• C. $m\text{g}\sqrt{2}\frac{2\text{R}}{\pi }$
• D. None

Answer 1

Answer: (B)

Center of mass of the chain is at $\frac{2R}{\pi }$ from horizontal
As shown in figure .......
Height gain by center of mass is $\frac{2\text{R}}{\pi }$



$\text{Wg}=-m\text{g}\left(\frac{2\text{R}}{\pi }\right)$
${\mathrm{W}}_{\text{ext. }}=-\mathrm{Wg}=\left(m\mathrm{g}\frac{2\mathrm{R}}{\pi }\right)$
Alternate solution



${\mathrm{U}}_{\mathrm{i}}=\int -(\mathrm{dm})\mathrm{gy}$
$=\text{g}\int \text{R cos}\theta \left(\frac{m}{\frac{\pi }{2}}\right)\text{d}\theta$
$=-\frac{2\text{Rg}m}{\pi }{\int }_0^{\frac{\pi }{2}}cos\theta \text{d}\theta$
$=-\frac{2\text{Rg}m}{\pi }[\text{sin}\theta {]}_0^{\pi /2}$
$=-\frac{2m\text{gR}}{\pi }$
${\text{U}}_{\text{f}}=0\Rightarrow \Delta \text{U}={\text{w}}_{\text{ext}}=\frac{2m\text{gR}}{\pi }$