Question 55 of 183

A body of $m \mathrm{~kg}$ slides from rest along the curve of vertical circle from point $A$ to $B$ in friction less path. The velocity of the body at $B$ is:

$\text { (given, } R=14 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2 \text { and } \sqrt{2}=1.4 \text { ) }$

A$16.7 \mathrm{~m} / \mathrm{s}$B$19.8 \mathrm{~m} / \mathrm{s}$C$10.6 \mathrm{~m} / \mathrm{s}$D$21.9 \mathrm{~m} / \mathrm{s}$

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Answer: (D) $21.9 \mathrm{~m} / \mathrm{s}$

Apply W.E.T. from A to B
$\begin{aligned}
& \Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}} \\
& \Rightarrow \mathrm{mg} \times\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{R}\right)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2-0\left\{\mathrm{v}_{\mathrm{A}}=0 \text { rest }\right\} \\
& \Rightarrow \mathrm{mgR} \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2 \\
& \Rightarrow \sqrt{\mathrm{gR} \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=\mathrm{v}_{\mathrm{B}} \\
& \Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}}=\mathrm{v}_{\mathrm{B}} \\
& \Rightarrow 21.9=\mathrm{v}_{\mathrm{B}}
\end{aligned}$