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Question

Physics Question on Electric Current

The external diameter of a 314 m314\, m long copper tube is 1.2 cm1.2\, cm and the internal diameter is 1 cm1 \,cm. Calculate its resistance if the specific resistance of copper is 2.2×10−8Ω−m2.2\times {{10}^{-8}}\Omega -m

A

5.0×10−2Ω5.0\times {{10}^{-2}}\Omega

B

4.4×10−2Ω4.4\times {{10}^{-2}}\Omega

C

3.14×10−2Ω3.14\times {{10}^{-2}}\Omega

D

0.2 Ω0.2\,\Omega

Answer

0.2 Ω0.2\,\Omega

Explanation

Solution

Given, length l=314ml=314m Specific resistance p=2.2×10−8Ω−mp=2.2\times {{10}^{-8}}\Omega -m External diameter D1=1.2×10−2m{{D}_{1}}=1.2\times {{10}^{-2}}m Internal diameter D2=1×10−2m{{D}_{2}}=1\times {{10}^{-2}}m As resistance of tube R=ρlAR=\frac{\rho l}{A} Or R=ρlπ[D12−D22]R=\frac{\rho l}{\pi [D_{1}^{2}-D_{2}^{2}]} [∵A=πr2][\because A=\pi {{r}^{2}}] Or R=4×2.2×10−8×314π[(1.2)2−(1.0)2]×10−4R=\frac{4\times 2.2\times {{10}^{-8}}\times 314}{\pi [{{(1.2)}^{2}}-{{(1.0)}^{2}}]\times {{10}^{-4}}} =0.2 Ω=0.2\,\Omega