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A metal wire has a resistance of `35Omega`. If its length is increased to double by drawing it, then its new resistance will beA. `70 Omega`B. `140 Omega`C. `105 Omega`D. `35 Omega` |
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Answer» Correct Answer - B Given, `R_(1)= 35 Omega, I_(2)=2I_(1)` On increasing the length, `therefore m_(1)=m_(2)` rhoA_(1)I_(1)=rhoA_(2)I_(2)` `pir_(1)^(2)I_(1) = pir_(2)^(2)I_(2)` `r_(1)^(2)=I_(2)/I_(1)` `r_(1)^(2)/r_(2)^(2)=(2I_(1))/I_(1)` `r_(1)^(2)/r_(2)^(2)=2` `R_(1)/R_(2) = ((rho.I_(1))/(pir_(1)^(2)))/(rho.(I_(2))/(pir_(2)^(2)))=I_(1)/I_(2).r_(2)^(2)/r_(1)^(2)=I_(1)/(2I_(1)).(1/2)` `R_(1)/R_(2)=1/4` `R_(2)=4R_(1)= 4 xx 35=140Omega`` |
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