Question

a mass M is revolving in a vertical circle at the end of a string of length 20 CM by how much year does the tension of the string at the lowest point exceed the tension at the top most point.

Answer 1

Let tension $T_1$ at the topmost point is given by,
$T_1+mg=\frac{mv_1^2}{l}$.....(1)
here tension and weight of body acting towards centre of circle and balancing centrifugal force acting outward. here l is length of string.

similarly, The tension $T_2$ at the lowest point is given by,
$T_2-mg=\frac{mv_2^2}{l}$ .....(2)
here, tension acting inward while balancing centrifugal force and weight of body acting outward.

$T_2-T_2=\frac{mv_2^2}{l}+mg-\frac{mv_1^2}{l}+mg\\\\=\frac{m}{l}(v_2^2-v_1^2)+2mg$

we know,from law of kinematics equation,
$v_1^2=v_2^2-2g(2l)$

$T_2-T_1=\frac{m}{l}\{2g(2l)\}+2mg\\\\=6mg$

hence, answer is 6mg

Answer 2

Answer:

6mg

Explanation:

This is the shortest method

Pls mark this as brainliest answer