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Standard XII

Physics

Tension in a String

A stone of ma...

Question

A stone of mass $0.2 k g$ is tied to one end of a thread of length $0.1 m$ whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is $52 N$ , then velocity of the stone will be:

4 m / s

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5 m / s

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6 m / s

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7 m / s

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Solution

The correct option is B $5 m / s$
$T - m g = \frac{{m v}^{2}}{r} T = \frac{{m v}^{2}}{r} + m g 52 = 0.2 (10) + \frac{0.2 (v^{2})}{0.1} 2 v^{2} = 52 - 2 = 50 v = 5 m / s e c$

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A stone of mass 0.2kg is tied to one end of a thread of length 0.1m whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is 52N, then velocity of the stone will be:

The correct option is B 5m/sT−mg=mv2rT=mv2r+mg52=0.2(10)+0.2(v2)0.12v2=52−2=50v=5m/sec

A stone of mass $0.2 k g$ is tied to one end of a thread of length $0.1 m$ whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is $52 N$ , then velocity of the stone will be:

The correct option is B $5 m / s$
$T - m g = \frac{{m v}^{2}}{r} T = \frac{{m v}^{2}}{r} + m g 52 = 0.2 (10) + \frac{0.2 (v^{2})}{0.1} 2 v^{2} = 52 - 2 = 50 v = 5 m / s e c$