JEE Mains · Physics · STD 11 - 13. oscillations
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 \(\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{{Mg}}{A}\;\;\;\;\;\;\)
- B \(\;\left[ {1 - {{\left( {\frac{{{T_M}}}{T}} \right)}^2}} \right]\frac{A}{{{M_g}}}\)
- C \(\;\left[ {1 - {{\left( {\frac{T}{{{T_M}}}} \right)}^2}} \right]\frac{A}{{{M_g}}}\)
- D \(\;\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{{M_g}}}\)
Answer & Solution
Correct Answer
(D) \(\;\left[ {{{\left( {\frac{{{T_M}}}{T}} \right)}^2} - 1} \right]\frac{A}{{{M_g}}}\)
Step-by-step Solution
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…
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