Work Energy Power

JEE Main level, single correct, medium

171 of 183

Question 171 of 183

A bob of mass mm is suspended at a point OO by a light string of length ll and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity v0v_0 at the point ' A ', the string becomes slack when, the bob reaches at the point ' D '. The ratio of the kinetic energy of the bob at the points BB and CC is ______ -.

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Answer: (B) 2

12mvA2=12mvB2+mgh⇒12 m(5 gℓ)=12mvB2+mgℓ2⇒5mgℓ2−mgℓ2=KEB⇒KEB=2mgℓ12mvC2=12mvD2+mgℓ2⇒KEC=12mgℓ+mgℓ2=mgℓ⇒KEBKEC=2\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{A}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgh} \\ & \Rightarrow \frac{1}{2} \mathrm{~m}(5 \mathrm{~g} \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mg} \frac{\ell}{2} \\ & \Rightarrow \frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}} \\ & \Rightarrow \mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell \\ & \frac{1}{2} \mathrm{mv}_{\mathrm{C}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{D}}^2+\mathrm{mg} \frac{\ell}{2} \\ & \Rightarrow \mathrm{KE}_{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell \\ & \Rightarrow \frac{\mathrm{KE}_{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2\end{aligned}

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