Answer:
Option C
Explanation:
We know that
young's modulus (Y)$=\frac{Fl}{A\triangle l}\Rightarrow\triangle l=\frac{Fl}{AY}$
where, A= area of cross section of wire
$\triangle$ l= change in the length of the wire,
l= length of the wire
and F= applied force
According to the question, F, l and Y are same for both wires, i.e,
$\triangle l \propto \frac{1}{A}$ ......(i)
But m= $\rho$ V
where, $\rho$ = density, V= volume
m= $\rho$ Al [$\because$ V= Al]
$m \propto A$ .......(ii)
From Eqs. (i) and (ii), we get
$\frac{\triangle l_{1}}{\triangle l_{2}}=\frac{A_{2}}{A_{1}}=\frac{m_{2}}{m_{1}}$
