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,

Q: 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)

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

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

JEE MAIN 2015, Diffcult

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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