Physics - Current Electricity Question with Solution | TestHub

Solution

Since the galvanometer shows null deflection, no current flows through the galvanometer branch. Therefore, the $2\Omega$ resistor and $r_0$ do not affect the null condition.

The potential difference between two points on the resistance wire is proportional to its resistance.

Given:

$$\rho (x)=\frac{8}{9\pi }{x}^2$$

For a wire of uniform cross-sectional area:

$$dR=\frac{\rho (x)}{A}dx$$

Hence:

$$R_{ab}\propto {\int }_a^b{x}^{2}dx$$

$$=\frac{b^3-a^3}{3}$$

From the figure:

$x_G=0.8m$

$x_{S_1}=1.0m$

$x_{S_2}=1.2m$

Therefore:

$$R_{G{S}_1}\propto \frac{1^3-(0.8{)}^3}{3}$$

$$=\frac{0.488}{3}$$

And:

$$R_{G{S}_2}\propto \frac{(1.2{)}^3-(0.8{)}^3}{3}$$

$$=\frac{1.216}{3}$$

Thus:

$$\frac{R_{G{S}_2}}{R_{G{S}_1}}=\frac{1.216}{0.488}$$

$$=\frac{152}{61}$$

First condition: S₁ closed

At null deflection:

$$V_{G{S}_1}=6.1V$$

Second condition: S₂ closed

Again, at null deflection:

$$V_{G{S}_2}=6.1+{\epsilon }_0$$

Since the current through the main resistance wire remains the same:

$$\frac{V_{G{S}_2}}{V_{G{S}_1}}=\frac{R_{G{S}_2}}{R_{G{S}_1}}$$

Therefore:

$$\frac{6.1+{\epsilon }_0}{6.1}=\frac{152}{61}$$

$$6.1+{\epsilon }_0=6.1\times \frac{152}{61}$$

$$6.1+{\epsilon }_0=15.2$$

Hence:

$$\boxed{{\epsilon }_0=15.2-6.1=9.1V}$$

So, with the values visible in the question, the exact result is:

$$\boxed{{\epsilon }_0=9.1V\approx 9V}$$

Thus, the given answer $\boxed{9V}$ is evidently using approximation/rounding.