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Byju's Answer

Standard XII

Physics

Tension in a String

A stone of ma...

Question

A stone of mass $1 k g$ is tied to the end of a string of $1 m$ length. It is whirled in a vertical circle. If the velocity of the stone at the top be $4 m / s$ . What is the tension in the string (at that instant)?

6 N

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16 N

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5 N

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10 N

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Solution

The correct option is A $6 N$
$T + m g = \frac{{m v}^{2}}{r} T = \frac{{m v}^{2}}{r} - m g \frac{1 \times 4^{2}}{1} - 1 (10) = 16 - 10 = 6 N$

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Similar questions

Q. A stone of mass

1 k g

is tied of the end of a string

1 m

long. It is whirled in a vertical circle. If the velocity of stone at the top is

4 m / s

. What is the tension in the string at the lowest point?

Take

g = 10 m / s^{2}

A stone of mass 1 kg is tied with a string and it is whirled in a vertical circle of radius 1m. If tension at the highest point is 14N, then velocity at lowest point will be:

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. The tension in the string is

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stone at the end of

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

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A stone of mass 1kg is tied to the end of a string of 1m length. It is whirled in a vertical circle. If the velocity of the stone at the top be 4m/s. What is the tension in the string (at that instant)?

The correct option is A 6NT+mg=mv2rT=mv2r−mg1×421−1(10)=16−10=6N

A stone of mass $1 k g$ is tied to the end of a string of $1 m$ length. It is whirled in a vertical circle. If the velocity of the stone at the top be $4 m / s$ . What is the tension in the string (at that instant)?

The correct option is A $6 N$
$T + m g = \frac{{m v}^{2}}{r} T = \frac{{m v}^{2}}{r} - m g \frac{1 \times 4^{2}}{1} - 1 (10) = 16 - 10 = 6 N$