Physics - Current Electricity Question with Solution | TestHub

When $S_2$ open. Assume resistance of $XY=R$. Resistance of wire per unit length, $x=\frac{R}{L}=R\Omega {\mathrm{m}}^{-1}$ $\because I=E_0/R$ Now, the potential drop across $50\mathrm{cm}$ length is $6\mathrm{V}$, So $$\begin{array}{cc}& \frac{E_0}{R}\times R\times \frac{50}{100}=6\\ \Rightarrow & E_0=12\mathrm{V}\end{array}$$ When $S_2$ closed, potential drop across $0.416\mathrm{cm}$ length, $$V_1=\frac{E_0}{R}\times R\times 0.416=12\times 0.416\approx 5\mathrm{V}$$ Hence, $E-Ir=5\mathrm{V}$ $$\begin{array}{rlrl}\Rightarrow & & 6-Ir& =5\\ & \because & I& =\frac{5}{10}\\ \therefore & & 6-5& =\frac{5}{10}r\\ \Rightarrow & & r& =2\Omega \end{array}$$