InterviewSolution
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InterviewSolution
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A cylindrical rod is reformed to half of its original length keeping volume constant. If its resistance before this change were R, then the resistance after reformation of rod will beA. RB. R/4C. 3R/4D. R/2 |
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Answer» Correct Answer - B The resistance of rod before reformation `R_(1)=R=(rhol_(1))/(pir_(1)^(2)) [therefore R=(rhol)/(A)=(rhol)/(pir^(2))]` Now the rod is reformed such that `l_(2)=(l_(1))/(2)` `therefore pir_(1)^(2)l_(1)=pir_(2)^(2)l_(2) (therefore "Volum remains constant")` or ` (pir_(1)^(2))/(pir_(2)^(2))=(l_(2))/(l_(1))..(i)` Now the resistance of the rod after reformation `R_(2)=(rhol_(2))/(pir_(2)^(2)) therefore (R_(1))/(R_(2))=(rhol_(1))/(pir_(1)^(2))/(rhol_(2))/(pir_(2)^(2))=(l_(1))/(l_(2))=(r_(2)^(2))/(r_(1)^(2))` `or (R_(1))/(R_(2))=(l_(1))/(l_(2))xx(l_(1))/(l_(2))=((l_(1))/(l_(2)))^(2)=(2)^(2)..("using"(i))` `therefore R_(2)=(R)/(4)` |
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