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 the material of the wire is Y then$$\frac{1}{Y}$$ is equal to: (g = gravitational acceleration)

Question

TuteeHUB · Accepted Answer

$$\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{Mg}{A}$$

TuteeHUB · Answer

$$\left[ 1-{{\left( \frac{{{T}_{M}}}{T} \right)}^{2}} \right]\frac{A}{Mg}$$

TuteeHUB · Answer

$$\left[ 1-{{\left( \frac{T}{{{T}_{M}}} \right)}^{2}} \right]\frac{A}{Mg}$$

TuteeHUB · Answer

$$\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{A}{Mg}$$

1.	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 the material of the wire is Y then\[\frac{1}{Y}\] is equal to: (g = gravitational acceleration)
A.	\[\left[ 1-{{\left( \frac{{{T}_{M}}}{T} \right)}^{2}} \right]\frac{A}{Mg}\]
B.	\[\left[ 1-{{\left( \frac{T}{{{T}_{M}}} \right)}^{2}} \right]\frac{A}{Mg}\]
C.	\[\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{A}{Mg}\]
D.	\[\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{Mg}{A}\]
Answer» D. \[\left[ {{\left( \frac{{{T}_{M}}}{T} \right)}^{2}}-1 \right]\frac{Mg}{A}\]

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 the material of the wire is Y then\[\frac{1}{Y}\] is equal to: (g = gravitational acceleration)

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