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Standard XII

Physics

Pendulum in Non Inertial Frame

A pendulum ma...

Question

A pendulum made of a uniform wire of cross-sectional area A has time period. T. When and additional mass M is added to its bob, the time period changes to $T_{M}$ . If the Young's modulus of the material of the wire is Y, then $\frac{1}{Y}$ is equal to (g = gravitationl acceleration)

[1 - {(\frac{T}{T_{M}})}^{2}] \frac{A}{M g}

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[{(\frac{T_{M}}{T})}^{2} - 1] \frac{M g}{A}

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[- 1 {(\frac{T_{M}}{T})}^{2} - 1] \frac{A}{M g}

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[{(\frac{T_{M}}{T})}^{2} - 1] \frac{A}{M g}

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Solution

The correct option is B $[{(\frac{T_{M}}{T})}^{2} - 1] \frac{M g}{A}$
Watch the next video for the solution.

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A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to T_M. If the Young’s modulus of

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A pendulum made of a uniform wire of cross-sectional area A has time period. T. When and additional mass M is added to its bob, the time period changes to TM. If the Young's modulus of the material of the wire is Y, then 1Y is equal to (g = gravitationl acceleration)

The correct option is B [(TMT)2−1]MgAWatch the next video for the solution.

A pendulum made of a uniform wire of cross-sectional area A has time period. T. When and additional mass M is added to its bob, the time period changes to $T_{M}$ . If the Young's modulus of the material of the wire is Y, then $\frac{1}{Y}$ is equal to (g = gravitationl acceleration)

The correct option is B $[{(\frac{T_{M}}{T})}^{2} - 1] \frac{M g}{A}$
Watch the next video for the solution.