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InterviewSolution
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The resistance of a 10 m long wire is 10 `Omega`. Its length is increased by 25% by stretching the wire uniformly . The resistance of wire will change toA. `12.5Omega`B. `14.5Omega`C. `15.6Omega`D. `16.6Omega` |
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Answer» Correct Answer - C Given, `l_(1)=l+(25)/(100)l=(5l)/(4)`. Since, volume of wire remains unchanged on increasing length, hence `A_(1)l_(1)=Al` `A_(1)xx(5l)/(4)=Al` or `A_(1)=4A//5` Given, `R= rho l//A= 10 Omega` and `R_(1)=(rho l_(1))/(A_(1))=(rho5l //4)/(4A//5)=(25 rho l)/(16 A)` `therefore " " R_(1)=(25)/(16)xx10=(250)/(16)=15.6 Omega` |
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