A pendulum using a uniform wire of area A has a period T. Adding a mass M to the bob changes the period to T_M. If Y is the Young's modulus, then frac 1 Y is equal to:

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EEJ MAIN Physics QUESTION #7080

Question 1

A pendulum using a uniform wire of area $A$ has a period $T$. Adding a mass $M$ to the bob changes the period to $T_M$. If $Y$ is the Young's modulus, then $\frac{1}{Y}$ is equal to:

$[(\frac{T_M}{T})^{2}-1]\frac{A}{Mg}$✔️
$[(\frac{T_M}{T})^{2}-1]\frac{Mg}{A}$
$[1-(\frac{T_M}{T})^{2}]\frac{A}{Mg}$
$[1-(\frac{T}{T_M})^{2}]\frac{A}{Mg}$

Correct Answer Explanation

The time period of a pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, so $L \propto T^2$.

Initial length: $L = kT^2$
New length: $L + \Delta L = k T_M^2$
Strain: $\frac{\Delta L}{L} = \frac{T_M^2 - T^2}{T^2} = (\frac{T_M}{T})^2 - 1$
From $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$
Therefore, $\frac{1}{Y} = \frac{\Delta L/L}{Mg/A} = [(\frac{T_M}{T})^2 - 1]\frac{A}{Mg}$

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