InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A wire with 15 Q resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will beA. `15.18Omega`B. `81.15Omega`C. `51.18Omega`D. `18.15Omega` |
|
Answer» Correct Answer - D If the wire is stretched by `(1//10)^(th)` of its original length then the new length of wire become `l_(2)=l+(l)/(10)=(11l)/(10)....(i)` As the volume of wire remains constant then `pir_(1)^(2)l=pir_(2)^(2)l_(1)=pir_(2)^(2)l((11l)/(10)) ("using" (i))` `Rightarrow r_(2)^(2)=(10)/(11)r_(1)^(2)....(ii)` Now the resistance of stretched wire. `R_(2)=(p((11)/(10)l))/(pir_(2)^(2))=(((11)/(10)rhol))/(pixx(10)/(11)r_(1)^(2))xx(rhol)/(pir_(1)^(2))` `(therefore R_(1)=(rhol)/(pir_(1)^(2))=15Omega)` `therefore R_(2)=((11)/(10))^(2)xx15=18.15Omega` |
|