A 2 kg stone at the end of a string 1 m long is 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

Q. A

1 k g

stone at the end of

1 m

long string is whirled in a vertical circle at a constant speed of

4 m s^{- 1}

. The tension in the string is 6 N when the stone is

A stone of mass m at the end of a string of length I is whirled in a vertical circle at a constant speed. The tension is the string will be maximum when the stone is

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A 2 kg stone at the end of a string 1 m long is 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

The correct option is B At the bottom of the circle Using mg=20 N and mv2r=2×(40)21=32 N Because we know that TBottom=mg+mv2r=52 N It is clear that tension at the bottom of the circle will be 52 N

A $2 kg$ stone at the end of a string $1 m$ long is 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

The correct option is B At the bottom of the circle
Using $m g = 20 N$ and $\frac{m v^{2}}{r} = \frac{2 \times (40)^{2}}{1} = 32 N$
Because we know that $T_{Bottom} = m g + \frac{m v^{2}}{r} = 52 N$
It is clear that tension at the bottom of the circle will be $52 N$