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← JEE Main 2022 · 27 Jun · Shift 2
PhysicsSingle correct+4 / -1Question 35 of 90

JEE Main 2022 · 27 Jun · Shift 2, Physics question 35

A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed uu. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x(u2−gL)\sqrt{x\left(u^2-gL\right)}. The value of xx is-

Choose one option
Answer and worked solution

Official answer: B

Speed at the horizontal position. v2=u2−2gLv^2=u^2-2gL, directed vertically while uu was horizontal.

Change. ∣Δv⃗∣2=u2+v2=2(u2−gL)|\Delta\vec v|^2=u^2+v^2=2(u^2-gL), so x=2x=2.

(B)