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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

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Best answer

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/s2, 10 m/s, 2m 

\(={1600-20\over 1600+20}\) 

\(={79\over81}\)

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