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 $\mathrm{P}$, an extension of $\Delta l$ is produced. The force which is required to produce same extension in $\mathrm{Q}$ is $\mathrm{F}_{2}$.
The value of $\frac{\mathrm{F}{1}}{\mathrm{~F}{2}}$ is
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Answer: (16)
Solution:
$\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\Delta \ell / \ell}=\frac{F \ell}{A \Delta \ell}$
$\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}}$
$\mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}}$
$\Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^{2} \mathrm{Y}}$
$\mathrm{Y} \backslash & \mathrm{~V}$ is same for both the wires
$\Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^{2}}$
$\frac{\Delta \ell_{1}}{\Delta \ell_{2}}=\frac{\mathrm{F}{1}}{\mathrm{~A}{1}^{2}} \times \frac{\mathrm{A}{2}^{2}}{\mathrm{~F}{2}}$
$\Delta \ell_{1}=\Delta \ell_{2}$
$\mathrm{F}{1} \mathrm{~A}{2}^{2}=\mathrm{F}{2} \mathrm{~A}{1}^{2}$
$\frac{F_{1}}{F_{2}}=\frac{A_{1}^{2}}{A_{2}^{2}}=\left(\frac{4}{1}\right)^{2}=16$
