Physics - Current Electricity Question with Solution | TestHub

Before heating, resistance of wire is $R_0=\left(\frac{\rho l_0}{A_0}\right)$

After heating, resistance is $R=\frac{\rho l}{A}$As volume = constant

Therefore,

$A_0{l}_0=1.2l_0\text{A}$

$\Rightarrow \text{A}=\left(\frac{A_0}{1.2}\right)$

So,

$$\begin{gathered} R=\frac{\rho l}{A} \\ =\frac{\rho \times \left(1.2l_0\right)}{\left(\frac{A_0}{1.2}\right)} \\ =\frac{\rho l_0}{A_0}\times 1.44 \\ =1.44R_0 \end{gathered}$$

Percentage increase in $R$ $=\frac{R-R_0}{R_0}\times 100=44\%$