b
Apply \(W.E.T.\) from \(A\) to \(B\)

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

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

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

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

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

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

Hence option \((2)\) is correct