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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Correct Answer

(D) \(\;\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{{M_g}}}\)

Detailed explanation

As we know, time period, \(T=2 \pi \sqrt{\frac{\ell}{g}}\)

\( When\, additional \,mass \mathrm\,{M}\) is added then

\(\mathrm{T}_{\mathrm{M}}=2 \pi \sqrt{\frac{\ell+\Delta \ell}{\mathrm{g}}}\)

\(T_{\frac{M}{T}}=\sqrt{\frac{\ell+\Delta \ell}{\ell}}\) or \(\left(\frac{T_{M}}{T}\right)^{2}=\frac{\ell+\Delta \ell}{\ell}\)

or, \(\left(\frac{T_{M}}{T}\right)^{2}=1+\frac{M g}{A y}\left[\because \Delta \ell=\frac{M g \ell}{A y}\right]\)

\(\therefore \frac{1}{y}=\left[\left(\frac{T_{M}}{T}\right)^{2}-1\right] \frac{A}{M g}\)

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