Question

A pendulum 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. $[1-{\left(\frac{T_M}{T}\right)}^2]\frac{A}{Mg}$
• B. $[1-{\left(\frac{T_M}{T}\right)}^2]\frac{Mg}{A}$
• C. $[{\left(\frac{T_M}{T}\right)}^2-1]\frac{A}{Mg}$
• D. $[{\left(\frac{T}{T_M}\right)}^2-1]\frac{Mg}{A}$

Answer 1

Answer: (A)

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

Since we know that

$T\propto \sqrt{l}\Rightarrow \frac{T_M}{T}=\frac{\sqrt{l+\Delta l}}{\sqrt{l}}\Rightarrow {\left(\frac{T_M}{T}\right)}^2=\frac{l+\Delta l}{l}=1+\frac{\Delta l}{l}\Rightarrow \frac{\Delta l}{l}={\left(\frac{T_M}{T}\right)}^2-1.........\left(I\right)$

$\because Y=\frac{\text{stress}}{\text{strain}}\Rightarrow \frac{1}{Y}=\frac{\text{strain}}{\text{stress}}=\frac{\left(\frac{\Delta l}{l}\right)}{\left(\frac{Mg}{A}\right)}\Rightarrow \frac{1}{Y}=\left[\frac{\Delta l}{l}\right]\frac{A}{Mg}\Rightarrow \frac{1}{Y}=[{\left(\frac{T_M}{T}\right)}^2-1]\frac{A}{Mg}\text{From}\left(I\right)$