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

  $\left[1-{\left(\frac{T}{T_M}\right)}^2\right]\frac{A}{Mg}$
• Correct answer
  B

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

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

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

Solution

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

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

Now time period ${\mathit{\text{T}}}_{\text{M}}=2\pi \sqrt{\frac{l+\Delta l}{\mathit{\text{g}}}} ...\left(\mathrm{ii}\right)$       

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

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

$\Rightarrow \frac{T_m^2}{T^2}=1+\frac{∆l}{l}$
$\begin{array}{cc}1+\frac{F}{AY}={\left(\frac{T_m}{T}\right)}^2& \left[\mathrm{p}\mathrm{u}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{ }\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }∆l\right]\end{array}$
$\frac{1}{Y}=\left[{\left(\frac{T_m}{T}\right)}^2-1\right]\frac{A}{F}$
$F=Mg$
$\frac{1}{Y}=\left[{\left(\frac{T_m}{T}\right)}^2-1\right]\frac{A}{Mg}$

Hence, the correct answer is option: 'B'.