JEE Physics 2023 Question with Solution

Given: The null point is at $200 \, \text{cm}$ when the cell is shunted by $5 \, \Omega$, and at $300 \, \text{cm}$ when it is shunted by $15 \, \Omega$.

Find: The internal resistance $r$ of the cell.

Using the potentiometer balance condition from the extracted working:

$$\frac{\epsilon}{r+5} \times 5 = 200x$$

So,

$$\frac{\epsilon}{r+5} = \frac{200x}{5} = 40x$$

Hence,

$$\epsilon = 40x(r+5)$$

Now use the second balance equation:

$$\frac{\epsilon \times 15}{r+15} = 300x$$

Substituting $\epsilon = 40x(r+5)$,

$$\frac{40x(r+5) \times 15}{r+15} = 300x$$

So,

$$\frac{600x(r+5)}{r+15} = 300x$$

Cancel the common factor $x$:

$$\frac{600(r+5)}{r+15} = 300$$

Therefore,

$$600(r+5) = 300(r+15)$$

Expanding both sides,

$$600r + 3000 = 300r + 4500$$

Rearranging,

$$600r - 300r = 4500 - 3000$$ $$300r = 1500$$

Thus,

$$r = \frac{1500}{300} = 5 \, \Omega$$

Therefore, the internal resistance of the cell is $5 \, \Omega$.

Given: Balancing lengths are proportional to the terminal potential difference of the cell for a fixed potential gradient $x$.

Find: The internal resistance $r$.

For a cell of emf $\epsilon$ and internal resistance $r$ shunted by external resistance $R$, the terminal potential difference is

$$V = \frac{\epsilon R}{r+R}$$

Since potentiometer balancing length is proportional to potential difference,

$$V = lx$$

For $R = 5 \, \Omega$ and $l = 200 \, \text{cm}$,

$$\frac{\epsilon \cdot 5}{r+5} = 200x$$

For $R = 15 \, \Omega$ and $l = 300 \, \text{cm}$,

$$\frac{\epsilon \cdot 15}{r+15} = 300x$$

Dividing the second equation by the first,

$$\frac{\frac{15}{r+15}}{\frac{5}{r+5}} = \frac{300}{200}$$ $$\frac{15(r+5)}{5(r+15)} = \frac{3}{2}$$ $$\frac{3(r+5)}{r+15} = \frac{3}{2}$$

Cancelling $3$,

$$\frac{r+5}{r+15} = \frac{1}{2}$$

So,

$$2(r+5) = r+15$$ $$r = 5 \, \Omega$$

Therefore, the internal resistance is $5 \, \Omega$.