Question

A potentiometer wire of length 10 m is connected in series with a battery. The e.m.f. of a cell balances against 250 cm length of wire. If length of potentiometer wire is increased by 1 m, the new balancing length of wire will be

• A. 2.00 m
• B. 2.25 m
• C. 2.50 m
• D. 2.75 m

Answer 1

Answer: (D)

2.75 m

Answer 2

Here's how to solve this problem:

1. Given:

   • Initial potentiometer wire length, $L_1 = 10 \, \text{m}$.
   • Balancing length for the cell, $l_1 = 250 \, \text{cm} = 2.50 \, \text{m}$.
   • New total length $L_2 = 10 + 1 = 11 \, \text{m}$.

2. Concept: The potential gradient $k$ is given by $k = \dfrac{V}{L}$ where $V$ is the battery voltage. For the cell, the balancing condition gives:

   $$\text{EMF of cell} = V_{\text{cell}} = k \times (\text{balancing length})$$

3. Step-by-Step Calculation:

   • Initially: $$V_{\text{cell}} = \left(\frac{V}{10}\right) \times 2.50$$
   • With the new wire: New potential gradient $k' = \dfrac{V}{11}$. Let the new balancing length be $l_2$. Balancing condition gives: $$V_{\text{cell}} = \left(\frac{V}{11}\right) \times l_2$$
   • Equate the expressions for $V_{\text{cell}}$: $$\left(\frac{V}{10}\right) \times 2.50 = \left(\frac{V}{11}\right) \times l_2$$
   • Simplify (cancel $V$): $$\frac{2.50}{10} = \frac{l_2}{11} \quad \Longrightarrow \quad l_2 = \frac{2.50 \times 11}{10} = 2.75 \, \text{m}$$

Therefore, the new balancing length is 2.75 m.