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Physics Question on Current electricity

The resistivity of the material of a potentiometer wire is 5×10−6Ω m5 \times 10^{-6} \Omega\, m and its area of cross section is 5×10−6  m25 \times 10^{-6} \; m^2. If 0.2 A0.2\, A current is flowing through the wire, then the potential drop per metre length of the wire is

A

0.1 Vm−10.1\, Vm^{-1}

B

0.5 Vm−10.5\, Vm^{-1}

C

0.25 Vm−10.25\, Vm^{-1}

D

0.2 Vm−10.2\, Vm^{-1}

Answer

0.2 Vm−10.2\, Vm^{-1}

Explanation

Solution

Given, i=0.2 Ai=0.2\, A Resistivity, ρ=5×10−6Ω−m\rho=5 \times 10^{-6} \Omega- m and A=5×10−6m2,l=1 mA=5 \times 10^{-6} m ^{2}, l=1\, m We know that, V=iRV=i R ⇒V=iρlA[∵R=ρlA]\Rightarrow V=\frac{i \rho l}{A}\left[\because R=\frac{\rho l}{A}\right] ⇒V=0.2×5×10−6×15×10−6\Rightarrow V=\frac{0.2 \times 5 \times 10^{-6} \times 1}{5 \times 10^{-6}} ⇒V=0.2 Vm−1\Rightarrow V=0.2\, Vm ^{-1}