JEE Main 2024 — 4 April, Shift 2 — Question 40

Apply W.E.T. from A to B

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

$\Rightarrow \operatorname{mgR} \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} \operatorname{mv}_{\mathrm{B}}^{2}$ $\Rightarrow \sqrt{\operatorname{gR} \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=\mathrm{v}_{\mathrm{B}}$ $\Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}}=\mathrm{v}_{\mathrm{B}}$ $\Rightarrow 21.9=\mathrm{v}_{\mathrm{B}}$