A body of \(\mathrm{m} \mathrm{kg}\) 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 _______. (given, \(\mathrm{R}=14 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\) and \(\sqrt{2}=1.4\) )

\(\Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}}\)

\(\Rightarrow \mathrm{mg} \times\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{R}\right)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2-0\left\{\mathrm{v}_{\mathrm{A}}=0 \text { rest }\right\}\)

\(\Rightarrow \mathrm{mgR} \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2\)

\(\Rightarrow \sqrt{\mathrm{gR} \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=\mathrm{v}_{\mathrm{B}}\)

\(\Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}}=\mathrm{v}_{\mathrm{B}}\)

\(\Rightarrow 21.9=\mathrm{v}_{\mathrm{B}}\)

Hence option \((2)\) is correct