Q 11-04-163JEE MainJEE Main 2021 (25 Feb, Shift 1)Medium
A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string is in the ratio 5 : 1. The velocity of the bob at the highest position is ______ $\text{m s}^{-1}$. (Take $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 5
Let $v$ be the speed at the top. At the bottom $u^2 = v^2 + 4gl$.
$T_{min} = \dfrac{mv^2}{l} - mg$ (top), $T_{max} = \dfrac{mu^2}{l} + mg = \dfrac{mv^2}{l} + 5mg$ (bottom). With $l = 1$ m:
$$\frac{v^2 + 50}{v^2 - 10} = 5 \Rightarrow v^2 + 50 = 5v^2 - 50 \Rightarrow v^2 = 25 \Rightarrow v = 5\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics