A stone of mass 1 kg is tied to a string 2m long and is rotated at constant speed of 40 ms^{-1} in a vertical circle . The ratio of the tension at the top and at the bottom is [g=10m/s]

A)12/19

B)79/81

C) 19/12

D)81/79

The correct answer is option B) 79/81

Explanation::

Formula for tension in vertical circle along a string is \(T={mv^2\over r}+mg \cos\theta\)

For highest point \(\theta\) = \(π\)

\(\therefore T={mv^2\over r}-mg \)

For lowest point \(\theta =0\)

\(\therefore T={mv^2\over r}+mg\)

After simplification

Ratio =\(v^2-gr\over v^2+gr\)

Putting values 40 m/s^{2}, 10 m/s, 2m

\(={1600-20\over 1600+20}\)

\(={79\over81}\)