Current Electricity

JEE Main level, single correct, medium

60 of 380

Question 60 of 380

A uniform metallic wire carries a current 2A2A, when 3.4V3.4V battery is connected across it. The mass of uniform metallic wire is 8.92×10−3kg8.92\times {10}^{-3}kg, density is 8.92×103kgm−38.92\times {10}^{3}kg{m}^{-3} and resistivity is 1.7×10−8Ω−m1.7\times {10}^{-8}\Omega -m The length of wire is :

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Answer: (B) l=10ml=10m

Given here, I=2A,V=3.4V,m=8.92×10−3kg,I=2A,V=3.4V,m=8.92\times {10}^{-3}kg,
d=8.92×103kgm−3andρ=1.7×10−8Ω−md=8.92\times {10}^{3}kg{m}^{-3}and\rho =1.7\times {10}^{-8}\Omega -m
From Ohm's law, resistance is R=VI=3.42=1.7ΩR=\dfrac{V}{I}=\dfrac{3.4}{2}=1.7\Omega
As resistance of wire is given by R=ρlAR=\dfrac{\rho l}{A}, where, AA is area and ll is length of wire.
⇒1.7=ρlA\Rightarrow 1.7=\dfrac{\rho l}{A} ⇒l=1.7Aρ\Rightarrow l=\dfrac{1.7A}{\rho }
Now, volume=md=8.92×10−38.92×103=10−6m3=\dfrac{m}{d}=\dfrac{8.92\times {10}^{-3}}{8.92\times {10}^{3}}={10}^{-6}{m}^{3}
Thus,
l2=1.7Aρ×l=1.7(volumel)ρ×l=1.7ρ(10−6)=1.71.7×10−8×10−6{l}^{2}=\dfrac{1.7A}{\rho }\times l =\dfrac{1.7\left(\dfrac{\text{volume}}{l}\right)}{\rho }\times l =\dfrac{1.7}{\rho }\left({10}^{-6}\right) =\dfrac{1.7}{1.7\times {10}^{-8}}\times {10}^{-6}
Hence, l=10ml=10m

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