Question

Two wires of same material and length are stretched by the same force their masses are in the ratio 3:2 find the ratio of their elongation?

Answer 1

Given:

Two wires of same material and length are stretched by the same force their masses are in the ratio 3:2.

To find:

Ratio of their elongation.

Calculation:

Let the masses be 3m and 2m;

Let initial length of both the wires be l and elongation be ∆l

Let Young's modulus be $\gamma$.

$\gamma=\dfrac{Fl}{(Area)(\Delta l)}$

$=>\gamma=\dfrac{F{l}^{2}}{(Area\times l)(\Delta l)}$

$=>\gamma=\dfrac{F{l}^{2}}{(Volume)(\Delta l)}$

$=>\gamma=\dfrac{F{l}^{2}}{(\dfrac{M}{\rho})(\Delta l)}$

$=>\Delta l \propto \dfrac{1}{M}$

Required ratio :

$\therefore \dfrac{\Delta l_{1}}{\Delta l_{2}}= \dfrac{M_{2}}{M_{1}}$

$=> \dfrac{\Delta l_{1}}{\Delta l_{2}}= \dfrac{M_{2}}{M_{1}}$

$=> \dfrac{\Delta l_{1}}{\Delta l_{2}}= \dfrac{2m}{3m}$

$=> \dfrac{\Delta l_{1}}{\Delta l_{2}}= \dfrac{2}{3}$

$=> \Delta l_{1} : \Delta l_{2}= 2:3$

So, final answer is :

$\boxed{\sf{\Delta l_{1} : \Delta l_{2}= 2:3}}$