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Current Electricity question

2023 · 25 Jan · Shift 1 · Q57
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  5. /2023 · 25 Jan · Shift 1 · Q57

Current Electricity question

2023 · 25 Jan · Shift 1 · Q57

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92 ×\times× 10 −3^{-3}−3 kg, density is 8.92 ×\times× 10 3^{3}3 kg/m 3^33 and resistivity is 1.7 ×\times× 10 −8 Ω^{-8}~\Omega−8 Ω-m\mathrm{m}m. The length of wire is :
  1. A
    l=100l=100l=100 m
  2. B
    l=6.8l=6.8l=6.8 m
  3. C
    l=5l=5l=5 m
  4. D
    l=10l=10l=10 m
View written solutionFree

Correct answer: D

  1. Find the resistance of the wire

Given current I=2 AI=2\,\text{A}I=2A and voltage V=3.4 VV=3.4\,\text{V}V=3.4V.

By Ohm’s law,

R=VI=3.42=1.7 ΩR=\frac{V}{I}=\frac{3.4}{2}=1.7\,\OmegaR=IV​=23.4​=1.7Ω
  1. Find the volume of the wire using mass and density

Given mass

m=8.92×10−3 kgm=8.92\times 10^{-3}\,\text{kg}m=8.92×10−3kg

and density

ρm=8.92×103 kg/m3\rho_m=8.92\times 10^{3}\,\text{kg/m}^3ρm​=8.92×103kg/m3

So volume is

Vw=mρm=8.92×10−38.92×103=10−6 m3V_w=\frac{m}{\rho_m}=\frac{8.92\times 10^{-3}}{8.92\times 10^{3}}=10^{-6}\,\text{m}^3Vw​=ρm​m​=8.92×1038.92×10−3​=10−6m3
  1. Relate resistance with length and area

For a wire,

R=ρlAR=\rho\frac{l}{A}R=ρAl​

where resistivity is

ρ=1.7×10−8 Ω m\rho=1.7\times 10^{-8}\,\Omega\,\text{m}ρ=1.7×10−8Ωm

Also, volume of wire is

Vw=AlV_w=A lVw​=Al

Hence,

A=VwlA=\frac{V_w}{l}A=lVw​​

Substitute into resistance formula:

R=ρlVw/l=ρl2VwR=\rho\frac{l}{V_w/l}=\rho\frac{l^2}{V_w}R=ρVw​/ll​=ρVw​l2​

So,

l2=RVwρl^2=\frac{R V_w}{\rho}l2=ρRVw​​
  1. Substitute the values
l2=(1.7)(10−6)1.7×10−8l^2=\frac{(1.7)(10^{-6})}{1.7\times 10^{-8}}l2=1.7×10−8(1.7)(10−6)​ l2=102=100l^2=10^2=100l2=102=100

Therefore,

l=10 ml=10\,\text{m}l=10m
  1. Check options
  • A: 100 m100\,\text{m}100m ✗
  • B: 6.8 m6.8\,\text{m}6.8m ✗
  • C: 5 m5\,\text{m}5m ✗
  • D: 10 m10\,\text{m}10m ✓

Therefore, the correct option is D.

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