Question

Two wires of length $l$, radius $r$ and length $2l$, radius $2r$ respectively having same young's modulus are hung with a weight $mg$ as shown in figure. The net elongation is

• A. $\frac{3mgl}{\pi r^{2}Y}$
• B. $\frac{2mgl}{3\pi r^{2}Y}$
• C. $\frac{3mgl}{2\pi r^{2}Y}$
• D. $\frac{3mgl}{4\pi r^{2}Y}$

Answer 1

The correct option is C $\frac{3mgl}{2\pi r^{2}Y}$


Tension in both wires will be same.

Let elongation in wire 1 be $=\Delta l_1$
$\Delta l_1=\frac{mgl}{\pi r^{2}Y}$ $[\because \Delta l=\frac{Fl}{AY}]$
$[\because A=\pi r^2]$
Let elongation in wire 2 be $=\Delta l_2$
$\Delta l_2=\frac{mg\times \left(2l\right)}{\pi (2r{)}^{2}Y}=\frac{mgl}{2\pi r^{2}Y}$
[$\because$ given length of wire 2 $=2l$, radius of wire $2=2r$]
So, net elongation
$(\Delta l{)}_{net}=\Delta l_1+\Delta l_2=\frac{mgl}{\pi r^{2}Y}+\frac{mgl}{2\pi r^{2}Y}$
$\Rightarrow \Delta l_{net}=\frac{3mgl}{2\pi r^{2}Y}$