Question

A small body of mass m attached to a string is rotated in a circular vertical path, when at the top tension in the string is zero. When the same body reaches the lowest position, tension in the string is

• A. 1 mg
• B. 3 mg
• C. 5 mg
• D. 6 mg

Answer 1

The correct option is D 6 mg

$\begin{array}{c}\text{from given data;}\\ \text{let length}=l\\ \text{mass}=m\\ \text{velocity at}A=v_n\\ \therefore \text{for}T=0\text{, at}A\end{array}$

$\frac{m{v}^2}{l}=mg$, or $m{v}^2=mgl$

So, from energy conservation at 3

$\frac{1}{2}m{v}_B^2=\frac{1}{2}m{v}_A^2+mg2l$

$m{v}_B^2=m{v}_A^2+4mg$

$m{v}_B^2=mgl+4mgl$ (from 1)

$\frac{m{V}_B^2}{l}=5mg$

$\therefore T_B=mg+5mg=6mg$

Ans D