JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2025
- Medium
- MCQ
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity \(v_0\) 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 \(B\) and \(C\) is ______ -.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{A}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgh} \\ & \Rightarrow \frac{1}{2} \mathrm{~m}(5 \mathrm{~g} \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mg} \frac{\ell}{2} \\ & \Rightarrow \frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}} \\ & \Rightarrow \mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell \\ & \frac{1}{2} \mathrm{mv}_{\mathrm{C}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{D}}^2+\mathrm{mg} \frac{\ell}{2} \\ & \Rightarrow \mathrm{KE}_{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell \\ & \Rightarrow \frac{\mathrm{KE}_{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2\end{aligned}\)
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