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A wire has resistance of `10Omega`. If it is stretched by 1/10th of its length, then its resistance is nearlyA. `9Omega`B. `10Omega`C. `11Omega`D. `12Omega`

Answer» Correct Answer - D
Resistance of wire `R = rho l//A =10 Omega,` New length,
`l_(1)=l+l//10=11//10`
`therefore` New are, `A_(1)=Al // l_(1)=(10A)/(11)`
`therefore` New resistance, `R_(1)=rho l_(1)//A_(1)=rho(11//10)//(10//11)A`
`= (121 rho l)/(100 A)=(121)/(100)xx10`
`= 12.1 Omega`


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