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A wire of length 15 m and uniform cross-section $6.0 \times 10^{-7}\ \text{m}^2$ has a measured resistance of 5.0 $\Omega$. The resistivity of its material is
A$2.0 \times 10^{-7}\ \Omega\text{m}$
B$1.25 \times 10^{6}\ \Omega\text{m}$
C$5.0 \times 10^{-8}\ \Omega\text{m}$
D$4.5 \times 10^{-6}\ \Omega\text{m}$
Answer & Solution
Correct answer: A. $2.0 \times 10^{-7}\ \Omega\text{m}$
1. From $R = \rho l / A$, solve $\rho = RA/l$.
2. Numerator: $RA = 5.0 \times 6.0\times10^{-7} = 3.0\times10^{-6}$.
3. $\rho = 3.0\times10^{-6}/15 = 2.0\times10^{-7}\ \Omega\text{m}$.
4. Option B inverts the formula ($l/(RA)$); the others mis-handle the powers of ten.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.25_
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