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 $F=\frac{Y A \Delta l}{l}$
Substituting area $A=\pi \left(\frac{d}{2}\right)^{2}=\frac{\pi d^{2}}{4}$
So, we have $F=\frac{Y \pi d^{2} \Delta l}{4 l}$
$F∝\frac{d^{2} \Delta l}{l}$ $\left[\because \frac{Y \pi }{4} = c o n s t a n t\right] \, $
Hence, using equation (i), we have
$F_{A}∝\frac{d_{A}^{2} \Delta l_{A}}{l_{A}}$ ...(i)
$F_{B}∝\frac{d_{B}^{2} \Delta l_{B}}{l_{B}}$ ...(ii)
Form equations (i) and (ii), we get
$∴\frac{F_{A}}{F_{B}}=\frac{d_{A}^{2} \Delta l_{A}}{l_{A}}\times \frac{l_{B}}{d_{B}^{2} \Delta l_{B}}$
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 $F_{A}:F_{B}=8:1$