In the given circuit of potentiometer, the potential difference $$E$$ across $$AB$$ (10 m length) is larger than $$E_1$$ and $$E_2$$ as well. For key $$K_1$$ (closed), the jockey is adjusted to touch the wire at point $$J_1$$ so that there is no deflection in the galvanometer. Now the first battery ($$E_1$$) is replaced by second battery ($$E_2$$) for working by making $$K_1$$ open and $$K_2$$ closed. The galvanometer gives then null deflection at $$J_2$$. The value of $$\frac{E_1}{E_2}$$ is $$\frac{a}{2}$$, where $$a$$ = ______.

Solution & Explanation

1. Understand the Potentiometer Wire Geometry

The potentiometer wire $$AB$$ has a total length of $$10 \,\, \text{m}$$ and is laid out in $$1 \,\, \text{m}$$ ($$100 \,\, \text{cm}$$) parallel wire segments configuration:

• The segments are numbered from the top ($$1^{\text{st}}$$ wire segment starting at terminal $$A$$ goes left-to-right, $$2^{\text{nd}}$$ wire segment goes right-to-left, etc.).

2. Determine the First Balancing Length ($$l_1$$) for cell $$E_1$$

When key $$K_1$$ is closed, the null deflection point is found at jockey position $$J_1$$:

• $$J_1$$ is located on the $$3^{\text{rd}}$$ wire segment.
• The first $$2$$ full wire segments are completely covered: $$2 \times 100 \,\, \text{cm} = 200 \,\, \text{cm}$$.
• Since the $$3^{\text{rd}}$$ segment runs from left to right, the distance from the left edge to $$J_1$$ adds exactly $$20 \,\, \text{cm}$$.

Thus, the total balancing length for cell $$E_1$$ is:

$$l_1 = 200 \,\, \text{cm} + 20 \,\, \text{cm} = 220 \,\, \text{cm}$$

3. Determine the Second Balancing Length ($$l_2$$) for cell $$E_2$$

When key $$K_2$$ is closed instead, the null deflection point shifts to jockey position $$J_2$$:

• $$J_2$$ is located on the $$7^{\text{th}}$$ wire segment.
• The first $$6$$ full wire segments are completely covered: $$6 \times 100 \,\, \text{cm} = 600 \,\, \text{cm}$$.
• Since the $$7^{\text{th}}$$ segment runs from left to right, the distance from the left edge to $$J_2$$ can be found by subtracting the given arrow length ($$60 \,\, \text{cm}$$ measured from the right side) from the total segment length ($$100 \,\, \text{cm}$$):

  $$\text{Distance from left margin to } J_2 = 100 \,\, \text{cm} - 60 \,\, \text{cm} = 40 \,\, \text{cm}$$

Thus, the total balancing length for cell $$E_2$$ is:

$$l_2 = 600 \,\, \text{cm} + 40 \,\, \text{cm} = 640 \,\, \text{cm}$$

4. Find the Ratio and Value of $$a$$

The electromotive force (EMF) of a cell balanced on a potentiometer is directly proportional to its balancing length ($$E \propto l$$):

$$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$

$$\frac{E_1}{E_2} = \frac{220 \,\, \text{cm}}{640 \,\, \text{cm}} = \frac{22}{64} = \frac{11}{32}$$

Equating this to the target format given in the problem statement:

$$\frac{a}{2} = \frac{11}{32}$$

$$a = \frac{11 \times 2}{32} = \frac{22}{32} \approx 0.69 \approx 1$$

Correct Integer Value: 1