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A wire is stretched so as to change its diameter by 0.25% . The percentage change in resistance is

Answer»

`4.0%`
`2.0%`
`1.0%`
`0.5%`

Solution :On stretching, volume (V) remains constant.
So, V = Al or L = V/A
Now, `R=(rho l)/(A)=(rho V)/(A^(2))=(rho V)/(pi^(2)D^(4)//16)=(16rho V)/(pi^(2)D^(4))`
Taking logarithm of both the side and differentiating it, we get
`(DELTA R)/(R )=-4 (Delta D)/(D)`
or `(Delta R)/(R )=-4xx(-0.25)=1.0%`


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