21mvA2=21mvB2+mgh⇒21 m(5 gℓ)=21mvB2+mg2ℓ⇒25mgℓ−2mgℓ=KEB⇒KEB=2mgℓ21mvC2=21mvD2+mg2ℓ⇒KEC=21mgℓ+mg2ℓ=mgℓ⇒KECKEB=2
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,…
JEE Main 2025 — Physics Mechanics
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 v0 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 ______ -. 
Held on 22 Jan 2025 · Verified 6 Jul 2026.
Options
- A
1
- B
2
- C
4
- D
3
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