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A wire of resistance $3 \Omega$ is stretched to twice its original length. The resistance of the new wire will be
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Verified Answer
The correct answer is:
$12 \Omega$
Given, resistance, $R_{1}=3 \Omega$
If $l$ is the original length of the wire, then after stretching the length of wire is
$l^{\prime}=2 l$
From the relation, resistance of a wire,
$R=\rho \frac{l}{A}$
where, $\rho$ is the resistivity and $A$ is cross-sectional arca.
$\begin{aligned}
&R \propto \frac{l}{A} \\
&\Rightarrow \quad R \propto \frac{l^{2}}{V} \quad\left[\because A=\frac{\text { Volume }}{\text { Length }}\right]
\end{aligned}$
$\begin{aligned}
&\Rightarrow \quad \frac{R_{1}}{R_{2}}=\frac{l_{1}^{2}}{V} \times \frac{V}{l_{2}^{2}} \\
&\text { Here, } l_{1}=l, l_{2}=2 l \\
&\Rightarrow \quad \frac{3}{R_{2}}=\frac{l^{2}}{4 l^{2}} \\
&\text { or } \quad R_{2}=3 \times 4=12 \Omega
\end{aligned}$
If $l$ is the original length of the wire, then after stretching the length of wire is
$l^{\prime}=2 l$
From the relation, resistance of a wire,
$R=\rho \frac{l}{A}$
where, $\rho$ is the resistivity and $A$ is cross-sectional arca.
$\begin{aligned}
&R \propto \frac{l}{A} \\
&\Rightarrow \quad R \propto \frac{l^{2}}{V} \quad\left[\because A=\frac{\text { Volume }}{\text { Length }}\right]
\end{aligned}$
$\begin{aligned}
&\Rightarrow \quad \frac{R_{1}}{R_{2}}=\frac{l_{1}^{2}}{V} \times \frac{V}{l_{2}^{2}} \\
&\text { Here, } l_{1}=l, l_{2}=2 l \\
&\Rightarrow \quad \frac{3}{R_{2}}=\frac{l^{2}}{4 l^{2}} \\
&\text { or } \quad R_{2}=3 \times 4=12 \Omega
\end{aligned}$
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