In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that the variation dR

Q: Why can't we just use the ratio of lengths $l/(1-l)$?

Because the wire is non-uniform, its resistance is not directly proportional to its length. You must integrate the given resistance gradient to find the actual resistance of each segment.

Q: Why did we integrate from $l$ to $1$ for the second part?

The total length of the meter bridge wire is given as $1\text{ m}$. So the second segment spans from the balance point $x=l$ to the end of the wire at $x=1$.

Q: What happens to the constant $k$?

It acts as a proportionality constant and cancels out when we equate the resistances of the two segments, so its exact value doesn't affect the balance point.

Q: Why are the upper resistances $R'$ equal?

It's given in the problem diagram. This simplifies the balance condition from a ratio to a direct equality: $R_{AP} = R_{PB}$.

Q: What if the total length was $L$ instead of $1\text{ m}$?

The upper limit of the second integral would be $L$, and the final equation would be $\sqrt{l} = \sqrt{L} - \sqrt{l}$, leading to $l = L/4$.

Question

In a meter bridge, the wire of length $1	ext{ m}$ has a non-uniform cross-section such that the variation $\frac{dR}{dl}$ of its resistance $R$ with length $l$ is $\frac{dR}{dl} \propto \frac{1}{\sqrt{l}}$. Two equal resistance are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point $P$. What is the length $AP$?

Options:

(A) $0.3	ext{ m}$

(B) $0.25	ext{ m}$

(C) $0.2	ext{ m}$

(D) $0.35	ext{ m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Look closely at the circuit. The galvanometer shows zero deflection. This means our meter bridge is in a perfectly balanced state. By the principle of a balanced Wheatstone bridge, the ratio of resistances in the adjacent arms must be equal.

Since the two upper resistors are identical (both are $R'$), the resistances of the two wire segments must also be equal. That is, the resistance of segment $AP$ must equal the resistance of segment $PB$:

$$R_{AP} = R_{PB}$$

### The Master Equation

Now, there is a catch here. The wire's cross-section is not uniform. The question states that the variation of resistance with length is proportional to $\frac{1}{\sqrt{x}}$. So, the resistance of a small element $dx$ will be:

$$dR = \frac{k}{\sqrt{x}} dx$$

To find the resistance of the first segment $AP$, we will integrate this expression from $0$ to $l$. The integral of $\frac{1}{\sqrt{x}}$ is $2\sqrt{x}$. Applying the limits, we get:

$$R_{AP} = \int_0^l \frac{k}{\sqrt{x}} dx = k [2\sqrt{x}]_0^l = 2k\sqrt{l}$$

Similarly, to find the resistance of the second segment $PB$, we will integrate from $l$ to $1$. Because the total length is $1	ext{ m}$. Applying the limits, $R_{PB}$ comes out to be:

$$R_{PB} = \int_l^1 \frac{k}{\sqrt{x}} dx = k [2\sqrt{x}]_l^1 = 2k(1 - \sqrt{l})$$

### Final Calculation

Now let's equate these two resistances.

$$2k\sqrt{l} = 2k(1 - \sqrt{l})$$

The constant $2k$ cancels out from both sides. Bringing $\sqrt{l}$ to one side, we get:

$$2\sqrt{l} = 1$$

Let's calculate the final value. $\sqrt{l}$ becomes $\frac{1}{2}$. Squaring both sides, the value of $l$ will be $\frac{1}{4}$, which is **$0.25	ext{ m}$**. This is our final answer.

Did you get the feel of it? Imagine if the wire was uniform, we wouldn't even need integration. In a uniform wire, resistance is directly proportional to length, making calculations very simple. That is why a wire of uniform cross-section is always used in a standard meter bridge.

Select Answer:

Visualized Solution

Visual Anchor

Balanced Wheatstone Bridge Condition

Resistance Variation

Resistance of Segment $A P$

Resistance of Segment $P B$

Equating Resistances

Final Calculation

The Way Forward

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Visual Anchor

Balanced Wheatstone Bridge Condition

Resistance Variation

Resistance of Segment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP

Resistance of Segment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PB

Equating Resistances

Final Calculation

The Way Forward

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Shown in the figure adjacent is a meter-bridge set up with null deflection in the galvanometer. The value of the unknown resistor R is

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