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Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 L$. Assuming, the resistivity and density of the material is unchanged, the resistance of the elongated wire is
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$45 \Omega$
Resistance of wire of length $L, R=5 \Omega$
If area of cross-section be $A$, then
$$
R=\rho \frac{L}{A} \quad(\rho=\text { resistivity of the wire })
$$
After applying external force new length becomes, $L^{\prime}=3 L$
$$
\begin{array}{ll}
\text { As, density, } \sigma & =\frac{m}{V}=\frac{m}{L \times A} \\
\Rightarrow \quad & A=\frac{m}{L \sigma} \\
\therefore & R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \times \frac{3 L \times L^{\prime} \sigma}{m}=\frac{3 \rho \sigma L \times 3 L}{m} \\
\Rightarrow & R^{\prime}=\frac{9 L^2 \rho \sigma}{m}
\end{array}
$$
Also, $\quad R=\rho \cdot \frac{L}{A}=\rho \frac{L L \sigma}{m}$
$=9 \times 5 \Omega=45 \Omega$
If area of cross-section be $A$, then
$$
R=\rho \frac{L}{A} \quad(\rho=\text { resistivity of the wire })
$$
After applying external force new length becomes, $L^{\prime}=3 L$
$$
\begin{array}{ll}
\text { As, density, } \sigma & =\frac{m}{V}=\frac{m}{L \times A} \\
\Rightarrow \quad & A=\frac{m}{L \sigma} \\
\therefore & R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \times \frac{3 L \times L^{\prime} \sigma}{m}=\frac{3 \rho \sigma L \times 3 L}{m} \\
\Rightarrow & R^{\prime}=\frac{9 L^2 \rho \sigma}{m}
\end{array}
$$
Also, $\quad R=\rho \cdot \frac{L}{A}=\rho \frac{L L \sigma}{m}$
$=9 \times 5 \Omega=45 \Omega$
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