Q115 / 150 · MHT CET 2011 · Physics
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 $F_{R}$ respectively they get equal increase in their lengths. Then the ratio $F_{A} / F_{B}$ should be
- A$1: 2$
- B$1: 1$
- C$2: 1$
- Correct answerD$8: 1$
Solution
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 $\quad F=\frac{Y \pi d^{2} \Delta l}{4 l}$
$$
F \propto \frac{d^{2} \Delta l}{l}\left[\because \frac{Y \pi}{4}=\text { constant }\right]
$$
Hence, using Eq. (i), we have
$$
\begin{array}{l}
F_{A} \propto \frac{d_{A}^{2} \Delta l_{A}}{l_{A}} \\
F_{B} \propto \frac{d_{B}^{2} \Delta l_{B}}{l_{B}}
\end{array}
$$
From Eqs. (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 $\quad F_{A}: F_{B}=8: 1$
Hence, the correct answer is option: 'D'.