Question

Two wires $A$ and $B$ are of the same materials. Their lengths are in the ratio $1:2$ and diameters are in the ratio $2:1$ . When stretched by force $F_A$ and $F_B$ respectively, they get an equal increase in their lengths. Then the ratio $\frac{F_A}{F_B}$ should be

• A. $1:2$
• B. $1:1$
• C. $2:1$
• D. $8:1$

Answer 1

Answer: (D)

Force $F=\frac{YA\Delta l}{l}$
Substituting area $A=\pi {\left(\frac{d}{2}\right)}^2=\frac{\pi d^2}{4}$
So, we have $F=\frac{Y\pi d^2\Delta l}{4l}$
$F\propto \frac{d^2\Delta l}{l}$ $\left[\because \frac{Y\pi }{4}=\text{ constant}\right]$
Hence, using equation (i), we have
$F_A\propto \frac{d_A^2\Delta l_A}{l_A}$ ...(i)
$F_B\propto \frac{d_B^2\Delta l_B}{l_B}$ ...(ii)
Form equations (i) and (ii), we get
$\therefore \frac{F_A}{F_B}=\frac{d_A^2\Delta l_A}{l_A}\times \frac{l_B}{d_B^2\Delta l_B}$
As $\Delta l_A=\Delta l_B=\Delta l$
Therefore $\frac{F_A}{F_B}=\frac{d_A^2}{d_B^2}\times \frac{l_B}{l_A}$
Putting $\frac{d_A}{d_B}=\frac{2}{1}$
And $\frac{l_B}{l_A}=\frac{2}{1}$
We get $F_A:F_B=8:1$