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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

from Mht-cet 2019
in Physics
reopened | 98 views

+1 vote

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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