Question

The resistance per unit length of potentiometer wire of uniform cross section is $\left(\frac{3R_0x}{\ell^2}\right)$, where x is measured from end A. Balance length is also measured from A. Select correct option (s) :

• A. When S is open balance length is at $\frac{\ell}{2}$
• B. When S is open balance length is at $\ell \sqrt{\frac{5}{6}}$
• C. When S is closed balance length is at $\ell \sqrt{\frac{3}{4}}$
• D. When S is closed balance length is at $\ell \sqrt{\frac{7}{8}}$

Answer 1

Answer: (B), (D)

When S is open balance length is at $\ell \sqrt{\frac{5}{6}}$ , When S is closed balance length is at $\ell \sqrt{\frac{7}{8}}$

Answer 2

The total resistance of the potentiometer wire of length $\ell$ is $R_{AB} = \int_0^\ell \frac{3R_0x}{\ell^2} dx = \frac{3R_0}{\ell^2} \left[\frac{x^2}{2}\right]_0^\ell = \frac{3R_0}{2}$.
The current in the primary circuit is $I_p = \frac{E_0}{R_0 + R_{AB}} = \frac{E_0}{R_0 + \frac{3R_0}{2}} = \frac{2E_0}{5R_0}$.
The potential drop across the wire from A to a point at distance $x$ is $V_{Ax} = \int_0^x I_p dR = \int_0^x \frac{2E_0}{5R_0} \frac{3R_0x'}{\ell^2} dx' = \frac{6E_0}{5\ell^2} \int_0^x x' dx' = \frac{6E_0}{5\ell^2} \frac{x^2}{2} = \frac{3E_0x^2}{5\ell^2}$.
The potential at a point at distance $x$ from A is $V_x = V_A - V_{Ax} = V_A - \frac{3E_0x^2}{5\ell^2}$, assuming current flows from A to B and A is at higher potential.

Case 1: Switch S is open.

The galvanometer is connected between the jockey at distance $x$ from A and the negative terminal of the battery $E_0/2$. The positive terminal of $E_0/2$ is connected to A.
Potential at A is $V_A$. Potential at the positive terminal of $E_0/2$ is $V_A$. Potential at the negative terminal of $E_0/2$ is $V_A - E_0/2$.
For balance, potential at the jockey is equal to the potential at the negative terminal of $E_0/2$.
$V_x = V_A - E_0/2$.
$V_A - \frac{3E_0x^2}{5\ell^2} = V_A - E_0/2$.
$\frac{3E_0x^2}{5\ell^2} = E_0/2$.
$\frac{3x^2}{5\ell^2} = \frac{1}{2}$.
$x^2 = \frac{5\ell^2}{6}$.
$x = \ell \sqrt{\frac{5}{6}}$.
Since $\sqrt{5/6} < 1$, this balance length is on the wire.

Case 2: Switch S is closed.

The galvanometer is connected between the jockey at distance $x$ from A and the negative terminal of the battery $E_0$. The positive terminal of $E_0$ is connected to A.
Potential at A is $V_A$. Potential at the positive terminal of $E_0$ is $V_A$. Potential at the negative terminal of $E_0$ is $V_A - E_0$.
For balance, potential at the jockey is equal to the potential at the negative terminal of $E_0$.
$V_x = V_A - E_0$.
$V_A - \frac{3E_0x^2}{5\ell^2} = V_A - E_0$.
$\frac{3E_0x^2}{5\ell^2} = E_0$.
$\frac{3x^2}{5\ell^2} = 1$.
$x^2 = \frac{5\ell^2}{3}$.
$x = \ell \sqrt{\frac{5}{3}}$.
Since $\sqrt{5/3} > 1$, this balance length is beyond the wire AB. Balance cannot be achieved on the wire AB in this configuration.

Let's assume the problem intended the connection to be positive terminal to A, negative terminal through galvanometer to jockey, with the resistors being internal resistances. However, the diagram shows external resistors.

Let's look at the options again.
When S is open balance length is at $\ell \sqrt{\frac{5}{6}}$. This option is consistent with our calculation assuming the galvanometer is connected to the negative terminal of $E_0/2$ and the positive terminal is at A.
When S is closed balance length is at $\ell \sqrt{\frac{7}{8}}$. Let's see if we can get this result.
If the secondary EMF is $E_{sec}$, then $\frac{3E_0x^2}{5\ell^2} = E_{sec}$.
For $x = \ell \sqrt{\frac{7}{8}}$, we have $\frac{3E_0(\ell \sqrt{7/8})^2}{5\ell^2} = \frac{3E_0 \ell^2 (7/8)}{5\ell^2} = \frac{21E_0}{40}$.
So, if the secondary EMF when S is closed is $E_{sec} = \frac{21E_0}{40}$, then balance length is $\ell \sqrt{7/8}$.