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, 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.
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Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
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