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The resistance of wire is 10 Omega IF the length of wire is increase by n% the new resistance is 10.2 Omega then n….. |
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Answer» Solution :Here `l_2=l_1+(n% of l_1)` `=l_1+(l_1 times n/100)` `therefore l_2=l_1(1+ n/100)` When a wire is STRETCHED UNIFORMLY, its electrical resistance `R propl^2` `therefore R_1/R_2=l_1^2/l_2^2` `therefore R_1/R_2=l_1^2/(l_1^2(1+n/100)^2)` `therefore R_1/R_2=1/(1+n/100)^2` `therefore 10/10.2=1/(1+n/100)^2` `therefore(1+n/100)^2=10.2/10=1.02` `therefore 1+n/100=sqrt1.02=1.00995` `therefore n/100=0.00995` `therefore n=0.995 =1` |
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