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If length of a conductor is doubled by keeping volume constant, then what is its new resistance if initial were `4Omega` ?A. `16Omega`B. `8Omega`C. `4Omega`D. `2Omega` |
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Answer» Correct Answer - A `R_(1)=4 Omega, R_(2)=? ` if `l_(2)=2l_(1)` by keeping volume constant. `(R_(2))/(R_(1))=((l_(2))/(l_(1)))^(2)` `R_(2)=((2l_(1))/(l_(1)))^(2) R_(1)=4xx4=16 Omega`. |
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