Question 75 of 118

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 $\dfrac{1}{Y}$ is equal to:
($g=$gravitational acceleration)

A$\left[1-{\left(\dfrac{T}{{T}_{M}}\right)}^{2}\right]\dfrac{A}{Mg}$B$\left[{\left(\dfrac{{\text{T}}_{\text{M}}}{\text{T}}\right)}^{2}-1\right]\dfrac{A⁡}{Mg⁡}$C$\left[{\left(\dfrac{{\text{T}}_{\text{M}}}{\text{T}}\right)}^{2}-1\right]\dfrac{\text{Mg}}{A⁡}$D$\left[1-{\left(\dfrac{{T}_{M}}{T}\right)}^{2}\right]\dfrac{A}{Mg}$

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Answer: (B) $\left[{\left(\dfrac{{\text{T}}_{\text{M}}}{\text{T}}\right)}^{2}-1\right]\dfrac{A⁡}{Mg⁡}$

Initial length $=l$. Time period $\text{T}=2\pi \sqrt{\dfrac{l}{\text{g}}}...(i)$. After suspending mass $M,$. Youngs modulus $\text{Y}=\dfrac{\text{Stress}}{\text{Strain}}$$=\dfrac{\dfrac{F}{A}}{\dfrac{∆l}{l}}=\dfrac{Fl}{∆lA}$

Change in length of wire $\Delta l=\dfrac{\text{F}l}{\text{AY}}$

Now time period ${\text{T}}_{\text{M}}=2\pi \sqrt{\dfrac{l+\Delta l}{\text{g}}}...(ii)$

$\dfrac{\text{T}}{{\text{T}}_{\text{M}}}=\dfrac{2\pi \sqrt{\dfrac{l}{\text{g}}}}{2\pi \sqrt{\dfrac{\left(l+\Delta l\right)}{\text{g}}}}$ $\left[∵\dfrac{\text{i}}{\text{ii}}\right]$

$\dfrac{{\text{T}}^{2}}{{\text{T}}_{\text{M}}^{2}}=\dfrac{l}{l+\Delta l}$

$\Rightarrow \dfrac{{T}_{m}^{2}}{{T}^{2}}=1+\dfrac{∆l}{l}$
$\begin{matrix}1+\dfrac{F}{AY}={\left(\dfrac{{T}_{m}}{T}\right)}^{2} & \left[puttingvalueof∆l\right]\end{matrix}$
$\dfrac{1}{Y}=\left[{\left(\dfrac{{T}_{m}}{T}\right)}^{2}-1\right]\dfrac{A}{F}$
$F=Mg$
$\dfrac{1}{Y}=\left[{\left(\dfrac{{T}_{m}}{T}\right)}^{2}-1\right]\dfrac{A}{Mg}$