Question

In the figure shown, string $AB$ and $BC$ have masses $m$ and $2m$ respectively. Both are of the same length $l$. Mass of each string is uniformly distributed on its length. The string is suspended vertically from the ceiling of a room. A small jerk wave pulse is given at the end ${}^{\prime }{C}^{\prime }$. It goes up to upper end ${}^{\prime }{A}^{\prime }$ in time ${}^{\prime }{t}^{\prime }$. If $m=2\text{kg},l=\frac{9610}{1681}\text{m},g=10{\text{m/s}}^2,\sqrt{2}=1.4,\sqrt{3}=1.7$, then the value of ${}^{\prime }{t}^{\prime }$ is :-

• A. $\frac{620}{697}\text{s}$
• B. $\frac{434}{205}\text{s}$
• C. $2\text{s}$
• D. $0.2\text{s}$

Answer 1

The correct option is C $2\text{s}$

For $BC$, at distance $y$ above $C$:


$T={\mu }_{1}yg$
where mass per unit length of BC is ${\mu }_1=\frac{2m}{l}$
$\Rightarrow v=\sqrt{\frac{T}{{\mu }_1}}=\sqrt{gy}$
$\Rightarrow \frac{dy}{dt}=\sqrt{\frac{T}{{\mu }_1}}=\sqrt{gy}$
$\Rightarrow \int \frac{dy}{\sqrt{y}}=\sqrt{g}dt$
Integrating with limits of $y=0$ to $y=l$
$\Rightarrow 2\sqrt{l}=\sqrt{g}t$
$\Rightarrow t_{CB}=2\sqrt{\frac{l}{g}}$

For $AB$, tension at distance $y$ above $B$


$T={\mu }_{2}yg+2mg;\text{mass per unit length of AB is}{\mu }_2=\frac{m}{l}$
$\Rightarrow \frac{T}{{\mu }_2}=gy+2gl$
$\Rightarrow v^2=gy+2gl$
$[\because v=\sqrt{\frac{T}{{\mu }_2}}]$
$\Rightarrow v=\frac{dy}{dt}=\sqrt{gy+2gl}$
$\Rightarrow \int \frac{dy}{\sqrt{gy+2gl}}=dt$

Integrating with limits of $y=0$ to $y=l$
$\Rightarrow \frac{2}{g}\sqrt{gl+2gl}-\frac{2}{g}\sqrt{0+2gl}=t$
$\Rightarrow t_{BA}=2(\sqrt{\frac{3l}{g}}-\sqrt{\frac{2l}{g}})$
$\Rightarrow t_{BA}=2\sqrt{\frac{l}{g}}(\sqrt{3}-\sqrt{2})$
$\Rightarrow t_{BA}=2\sqrt{\frac{l}{g}}\frac{1}{(\sqrt{3}+\sqrt{2})}$


Therefore, Total time taken to travel from $C$ to $A$ is,
$t=t_{CB}+t_{BA}$
$t=2\sqrt{\frac{l}{g}}[1+\frac{1}{\sqrt{3}+\sqrt{2}}]$
$t=2\sqrt{\frac{l}{g}}[1+\frac{1}{1.7+1.4}]$
$t=2\sqrt{\frac{l}{g}}\left[\frac{4.1}{3.1}\right]$
$t=2\sqrt{\frac{9610}{1681\times 10}}\times \frac{41}{31}$
$t=2\times \frac{31}{41}\times \frac{41}{31}$
$\therefore t=2\text{s}$