Question 241: Properties of Matter

Question

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 _________.

Accepted Answer

$$\mathrm{Y}=\frac{	ext { Stress }}{	ext { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$$
$$\begin{aligned}
& \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \
& \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \
& \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}}
\end{aligned}$$
$$Y ~\& V$$ is same for both the wires
$$\begin{aligned}
& \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \
& \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} 	imes \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \
& \Delta \ell_1=\Delta \ell_2 \
& \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \
& \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}ight)^2=16
\end{aligned}$$

Properties of Matter

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