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CBSE Class 10 Science Exam 2014 Term 1 Paper

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Question : 15 of 36
Marks: +1, -0
The resistance of a wire of 0.01cm0.01 \text{cm} radius is 10Ω10 \Omega. If the resistivity of the material of the wire is 50×10−8ohm50 \times 10^{-8} \text{ohm} metre, find the length of the wire.
Solution:  
Given : Resistance of a wire, R=10ΩR=10 \Omega
Radius of wire,
r=0.01cm=0.01×10−2mr=0.01 \text{cm}=0.01 \times 10^{-2} \text{m}
Resistivity, ρ=50×10−8Ωm\rho=50 \times 10^{-8} \Omega \text{m}
Area of cross-section of wire A,
=πr2=3.14×(0.01×10−2)2=\pi r^2=3.14 \times (0.01 \times 10^{-2})^2
=3.14×0.01×0.01×10−4=3.14 \times 0.01 \times 0.01 \times 10^{-4}
=3.14×10−8m2=3.14 \times 10^{-8} \text{m}^2
As,
R  =ρ  lAR\;=\rho \;\frac{l}{A}
l  =  R×Aρ=  10×3.14×10−850×10−8l\;=\;\frac{R \times A}{\rho}=\;\frac{10 \times 3.14 \times 10^{-8}}{50 \times 10^{-8}}
  =0.628m\;=0.628 \text{m}
or
l  =  R×Aρ=  10×3.14×10−850×10−8l\;=\;\frac{R \times A}{\rho}=\;\frac{10 \times 3.14 \times 10^{-8}}{50 \times 10^{-8}}
  =0.628m\;=0.628 \text{m}
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