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Q.
A wire is stretched so as to change its diameter by $0.25 \%$. The percentage change in resistance is
Current Electricity
Solution:
On stretching, volume $(V)$ remains constant.
So, $ V=A l$ or $l=V / A$
Now, $ R=\frac{\rho l}{A}=\frac{\rho V}{A^{2}}=\frac{\rho V}{\pi^{2} D^{4} / 16}=\frac{16 \rho V}{\pi^{2} D^{4}}$
Taking logarithm of both the side and differentiating it, we get
$\frac{\Delta R}{R}=-4 \frac{\Delta D}{D} \text { or } \frac{\Delta R}{R}=-4 \times(-0.25)=1.0 \%$
Resistance will increase.