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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 22 January · Morning Shift · Physics · Work, Power and Energy

A bob of mass m is suspended at a point O by a light string of...

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 ______ -. 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)

Options

  1. A1
  2. B2
  3. C4
  4. D3

Correct answer

B. 2

Step-by-step solution

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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