Discussion Forum



1)

A 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 TM. If the young's modulus of the material of the wire is Y, then   $\frac{1}{Y}$ is equal to (g= gravitational acceleration)


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

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

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

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

Answer:

Option A

Explanation:

We  know that  time period 

$T=2\pi \sqrt{\frac{L}{g}}$

 When additional  mass M is added to its bob

 $T_{M}=2\pi \sqrt{\frac{L+\triangle L}{g}}$

 where   $\triangle L$     is increase in length

 we know that

 $Y=\frac{Mg/A}{\triangle L/L}=\frac{MgL}{A \triangle L}$

 $\Rightarrow$       $\triangle L=\frac{MgL}{AY}$

 $\Rightarrow$   $\therefore$                       $T_{M}=2\pi\sqrt{\frac{L+\frac{MgL}{AY}}{g}}$

 $\Rightarrow$    $\left(\frac{T_{M}}{T}\right)^{2}=1+\frac{Mg}{AY}$

 or         $\frac{Mg}{AY}=\left(\frac{T_{M}}{T}\right)^{2}-1$

 or    $\frac{1}{Y}=\frac{A}{Mg}\left[\left(\frac{T_{M}}{T}\right)^{2}-1\right]$