Q 11-05-041NEETJEE MainTop questionMedium
A body of mass $m$ moves in a vertical circle on a string. The difference between the tensions in the string at the lowest and highest points is
Answer: (D) $6mg$
Bottom: $T_1 = mg + \dfrac{mv_1^2}{r}$. Top: $T_2 = \dfrac{mv_2^2}{r} - mg$. Energy: $v_1^2 = v_2^2 + 4gr$.
$$T_1 - T_2 = 2mg + \frac{m(v_1^2 - v_2^2)}{r} = 2mg + 4mg = 6mg$$
This holds whatever the speed, as long as the string stays taut.
Solution by Sreeraj P, M.Sc Physics