Verified Solution • 2025–26 Session moderate

QJEEM21Phde4d: JEE Main 2021 Physics - Class 12 JEEM21Phde4d Solution: In the experiment of Ohm's law, a potential difference of 5.0 V …

In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :

Multiple Choice Options

  • A 3.9
  • B 8.4
  • C 7.5
  • D 3

Step-by-Step Verified Solution

V=I×ρlAV = I \times \rho {l \over A}

⇒ρ=VAIl=π4Vd2Il\Rightarrow \rho = {{VA} \over {Il}} = {\pi \over 4}{{V{d^2}} \over {Il}}

Δρρ=2Δdd+ΔVV+ΔII+Δll{{\Delta \rho } \over \rho } = {{2\Delta d} \over d} + {{\Delta V} \over V} + {{\Delta I} \over I} + {{\Delta l} \over l}

=2(0.015)+0.15+0.012+0.110= 2\left( {{{0.01} \over 5}} \right) + {{0.1} \over 5} + {{0.01} \over 2} + {{0.1} \over {10}}

⇒\Rightarrow Δρρ=0.039=3.9%{{\Delta \rho } \over \rho } = 0.039 = 3.9\%

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