JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 ______ -.
- A 1
- B 2
- C 4
- D 3
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}…
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