The resistivity of the material of a wire of length 15 m, area of cross-section \( 6\times10^{-7}\,\mathrm{m^2} \) and resistance \( 10\,\Omega \) will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4\times10^{-7}\,\Omega\,\mathrm{m} \)
The resistance of a uniform wire is \( R = \rho L/A \), so the resistivity is
\( \displaystyle \rho = \frac{R\,A}{L} \)
\( R = 10\,\Omega \)
\( A = 6\times10^{-7}\,\mathrm{m^2} \)
\( L = 15\,\mathrm{m} \)
\( \displaystyle \rho = \frac{10 \times 6\times10^{-7}}{15} = \frac{6\times10^{-6}}{15} = 4\times10^{-7}\,\Omega\,\mathrm{m} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
