A pendulum made of a uniform wire of cross sectional area A has...

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]$<br /><br />${\text{T}}^{2}{\text{T}}_{\text{M}}^{2} = \dfrac{l}{l + \Delta l}$<br /><br />$⇒ \dfrac{{T}_{m}^{2}}{{T}^{2}} = 1 + \dfrac{∆ l}{l}$<br />$\begin{cases}1 + \dfrac{F}{AY} ={\left(\dfrac{{T}_{m}}{T}\right)}^{2} & \left[puttingvalueof ∆ l\right]\end{cases}$<br />$\dfrac{1}{Y} =\left[{\left(\dfrac{{T}_{m}}{T}\right)}^{2} - 1\right]\dfrac{A}{F}$<br />$F = Mg$<br />$\dfrac{1}{Y} =\left[{\left(\dfrac{{T}_{m}}{T}\right)}^{2} - 1\right]\dfrac{A}{Mg}