Mechanical Properties Of Solids Question 6
Question 6 - 2024 (29 Jan Shift 2)
Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio $4: 1$ and force $F _1$ is applied to $P$, an extension of $\Delta l$ is produced. The force which is required to produce same extension in $Q$ is $F _2$.
The value of $\frac{F _1}{F _2}$ is
Show Answer
Answer: (16)
Solution:
$Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\Delta \ell / \ell}=\frac{F \ell}{A \Delta \ell}$
$\Delta \ell=\frac{F \ell}{AY}$
$V=A \ell \Rightarrow \ell=\frac{V}{A}$
$\Delta \ell=\frac{FV}{A^{2} Y}$
$Y \& V$ is same for both the wires
$\Delta \ell \propto \frac{F}{A^{2}}$
$\frac{\Delta \ell _1}{\Delta \ell _2}=\frac{F _1}{A _1^{2}} \times \frac{A _2^{2}}{F _2}$
$\Delta \ell _1=\Delta \ell _2$
$F _1 A _2^{2}=F _2 A _1^{2}$
$\frac{F _1}{F _2}=\frac{A _1^{2}}{A _2^{2}}=\left(\frac{4}{1}\right)^{2}=16$
