JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 27)

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 $$\mathrm{F}_2$$.

The value of $$\frac{F_1}{F_2}$$ is _________.
Answer
16

Explanation

$$\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$$

$$\begin{aligned} & \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \\ & \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}} \end{aligned}$$

$$Y ~\& V$$ is same for both the wires

$$\begin{aligned} & \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \\ & \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \\ & \Delta \ell_1=\Delta \ell_2 \\ & \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \\ & \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16 \end{aligned}$$

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