During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pre

Q: Why do we add the errors $\Delta l/l$ and $\Delta l/(100-l)$ instead of subtracting them?

In error analysis, to find the maximum permissible error, we always add the absolute values of the relative errors, even if the terms are in the denominator or have a negative sign in the derivative.

Q: Why is $\Delta l$ taken as $0.1\text{ cm}$?

The problem states that the scale used in the meter bridge is $1\text{ mm}$. This implies the least count of the scale is $1\text{ mm}$, which is equal to $0.1\text{ cm}$. The least count is taken as the maximum absolute error in any single measurement.

Q: Does the known resistance $R = 90\ \Omega$ have any error?

The problem calls it a "standard resistance" and does not provide any error limits for it. Therefore, we assume it to be exact, meaning $\Delta R = 0$.

Q: What if the null point was measured from the other end?

The formula would simply swap $l$ and $100-l$. The nominal value of $X$ would change, but the error analysis method remains exactly the same.

Q: Why is the error in $(100-l)$ also $\Delta l$?

The length $(100-l)$ is not measured directly; it is calculated. The error in $l$ propagates to $(100-l)$. Since $d(100-l) = -dl$, the absolute error is $|-dl| = \Delta l$.

Question

During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at $40.0	ext{ cm}$ using a standard resistance of $90\ \Omega$, as shown in the scale used in the meter bridge is $1	ext{ mm}$. The unknown resistance is

Options:

(A) $60 \pm 0.15\ \Omega$

(B) $135 \pm 0.56\ \Omega$

(C) $60 \pm 0.25\ \Omega$

(D) $135 \pm 0.23\ \Omega$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a physics laboratory, looking at a classic meter bridge setup. The meter bridge is a practical application of the Wheatstone bridge principle, designed to find an unknown resistance with high precision.

In our specific experiment, we have an unknown resistance $X$ connected in the left gap and a standard, known resistance $R = 90\ \Omega$ in the right gap. The jockey, which slides along the $100	ext{ cm}$ long uniform wire, detects a null point (zero deflection in the galvanometer) at exactly $l = 40.0	ext{ cm}$ from the left end.

This means the wire is divided into two segments: the left segment of length $l = 40.0	ext{ cm}$ and the right segment of length $100 - l = 60.0	ext{ cm}$.

### The Master Equation

When the galvanometer shows zero deflection, the bridge is perfectly balanced. The core principle of a balanced Wheatstone bridge dictates that the ratio of the resistances in the upper arms equals the ratio of the resistances of the corresponding wire segments.

Mathematically, this is expressed as:

$$ \frac{X}{R} = \frac{l}{100 - l} $$

Let's substitute our known values into this elegant equation:

$$ \frac{X}{90} = \frac{40}{100 - 40} $$

$$ \frac{X}{90} = \frac{40}{60} $$

Solving for $X$, we get:

$$ X = 90 	imes \frac{2}{3} = 60\ \Omega $$

So, the nominal value of our unknown resistance is exactly $60\ \Omega$. But in the world of experimental physics, a measurement is never complete without its error limits.

### Error Analysis: The Catch

This is where many students make a silly mistake. The question explicitly mentions that the scale used in the meter bridge has a least count of $1	ext{ mm}$. This least count represents the maximum absolute error in our length measurement, meaning $\Delta l = 1	ext{ mm} = 0.1	ext{ cm}$.

To find the maximum permissible error in $X$, we need to look at our master equation again:

$$ X = R \left( \frac{l}{100 - l} ight) $$

To extract the relative error, we take the natural logarithm on both sides and differentiate. Remember, in error analysis, errors always add up to give the maximum possible deviation. We never subtract errors!

$$ \frac{\Delta X}{X} = \frac{\Delta l}{l} + \frac{\Delta l}{100 - l} $$

### Final Calculation

Now, let's plug in our calculated nominal value $X = 60\ \Omega$ and the absolute error $\Delta l = 0.1	ext{ cm}$:

$$ \frac{\Delta X}{60} = \frac{0.1}{40} + \frac{0.1}{60} $$

Let's isolate $\Delta X$ and solve the fractions by taking a common denominator of $120$:

$$ \Delta X = 60 \left( \frac{0.3 + 0.2}{120} ight) $$

$$ \Delta X = 60 \left( \frac{0.5}{120} ight) $$

$$ \Delta X = \frac{30}{120} = 0.25\ \Omega $$

Combining our nominal value with the calculated absolute error, the final, scientifically rigorous value of the unknown resistance is:

$$ X = (60 \pm 0.25)\ \Omega $$

This perfectly matches option (c). The beauty of this problem lies in seamlessly blending a fundamental circuit principle with the rigorous discipline of experimental error analysis.

Animated Solution for Physics - Current Electricity: During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at $40.0 cm$ using a standard resistance of $90 Ω$ , as shown in the scale used in the meter bridge is $1 mm$ . The unknown resistance is

Select Answer:

Visualized Solution

$Meter Bridge Setup$

$Balancing Condition$

$Substituting Values$

$Calculating X$

$Error Analysis Formula$

$Substituting Errors$

$Calculating Δ X$

$Final Answer$

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Error Analysis

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Meter Bridge Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Error Analysis Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Errors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔX

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer