Kinematics Question 539

Question: A 2 kgstone at the end of a string 1 m long it’s whirled in a vertical circle at a constant speed. The speed of the stone is 4 m/sec. The tension in the string will be 52 N, when the stone is [AIIMS 1982]

Options:

A) At the top of the circle

B) At the bottom of the circle

C) Halfway down

D) None of the above

Show Answer

Answer:

Correct Answer: B

Solution:

$ mg=20N $ and $ \frac{mv^{2}}{r}=\frac{2\times {{(4)}^{2}}}{1}=32N $ .

It’s clear that 52 N tension will be at the bottom of the circle..

Because we know that $ {T _{\text{Bottom}}}=mg+\frac{mv^{2}}{r} $



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