Kinematics Question 537

Question: A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at constant speed of 4 m/sec. The tension in the string is 6 N, when the stone is at (g = 10 m/sec2) [AIIMS 1982]

Options:

A) Top of the circle

B) Bottom of the circle

C) Half way down

D)None of the above

Show Answer

Answer:

Correct Answer: A

Solution:

$ mg=1\times 10=10N, $

$ \frac{mv^{2}}{r}=\frac{1\times {{(4)}^{2}}}{1}=16 $

Tension at the top of circle = $ \frac{mv^{2}}{r}-mg=6N $

Tension at the bottom of circle = $ \frac{mv^{2}}{r}+mg=26N $



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