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Q.A metallic wire of resistance 40Ω40\Omega is stretched to twice its length. Its new resistance would be

(a) 20Ω20\Omega
(b) 80Ω80\Omega
(c) 160Ω160\Omega
(d) 120Ω120\Omega
Nagaland NbseNagaland Board of School Education 2016MCQ· 1mImportance★★★★★
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Resistance varies as the square of length when volume is conserved on stretching: R∝L2R \propto L^2, giving R′=4R=160 ΩR' = 4R = 160\,\Omega.

Given: Original resistance R=40 ΩR = 40\,\Omega, wire stretched to twice its length (L′=2LL' = 2L).

Step 1 -- Volume conservation.

Stretching does not change the volume of the wire: V=AL=A′L′=constantV = A L = A' L' = \text{constant}.

So A′=ALL′=AL2L=A2A' = \dfrac{AL}{L'} = \dfrac{AL}{2L} = \dfrac{A}{2}.

Step 2 -- Apply resistance formula.

R=ρLA,R′=ρL′A′=ρ2LA/2=4(ρLA)=4RR = \rho \dfrac{L}{A}, \qquad R' = \rho \dfrac{L'}{A'} = \rho \dfrac{2L}{A/2} = 4\left(\rho \dfrac{L}{A}\right) = 4R

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