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Q.A conductor of radius rr and length ll has resistance 10 Ω10\ \Omega. The conductor is stretched in such a way that its volume is maintained constant and its length is increased by 10% of its original length. The resistance of the stretched conductor is

(a) 8 Ω8\ \Omega
(b) 100 Ω100\ \Omega
(c) 12.10 Ω12.10\ \Omega
(d) 20 Ω20\ \Omega
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2020MCQ· 1mImportance★★★★★
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At constant volume, resistance scales as the square of the length, since stretching also shrinks the cross-section proportionally.

Step 1 — RR in terms of ll at constant volume

R=ρlA=ρl2Al=ρl2VR = \rho\frac{l}{A} = \rho\frac{l^2}{Al} = \rho\frac{l^2}{V}

(using V=AlV=Al, constant). So R∝l2R \propto l^2.

Step 2 — Apply the 10% increase

…

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