A meter bridge is set-up as shown in figure, to determine an unknown resistance X using a standard 1

Q: Why do we add end corrections instead of subtracting them?

End corrections represent the extra resistance of the metal strips and soldering at the ends of the wire. Since this resistance acts in series with the wire segments, it effectively increases their length. Therefore, we add the end corrections to the measured lengths.

Q: What causes these end corrections in a real meter bridge?

In a practical meter bridge, the wire is soldered to thick copper strips. The resistance of these strips, the contact resistance at the joints, and the fact that the zero mark of the scale may not perfectly align with the start of the wire all contribute to the end correction.

Q: Why is the total length of the wire taken as $100\text{ cm}$?

A standard meter bridge uses a wire that is exactly $1\text{ meter}$ (or $100\text{ cm}$) long, hence the name 'meter bridge'. This makes calculations straightforward.

Q: Can we ignore end corrections if they are not given?

Yes, if the problem does not mention end corrections, you should assume the meter bridge is ideal and the resistance of the connecting strips is negligible.

Q: What if the null point was exactly at $50\text{ cm}$?

If the null point is at $50\text{ cm}$ and there are no end corrections, it means the two resistances in the gaps are exactly equal. If end corrections exist, you would still add them to $50\text{ cm}$ on both sides before taking the ratio.

Question

A meter bridge is set-up as shown in figure, to determine an unknown resistance $X$ using a standard $10\ \Omega$ resistor. The galvanometer shows null point when tapping-key is at $52	ext{ cm}$ mark. The end-corrections are $1	ext{ cm}$ and $2	ext{ cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is

Options:

(A) $10.2\ \Omega$

(B) $10.6\ \Omega$

(C) $10.8\ \Omega$

(D) $11.1\ \Omega$

Sigma Percentile · Accepted Answer

The meter bridge is a classic application of the Wheatstone bridge principle, used to measure unknown resistances with high precision. In this problem, we are tasked with finding an unknown resistance $X$ using a standard $10\ \Omega$ resistor.

### Analyzing the Setup

When the galvanometer shows zero deflection, the bridge is balanced. This means the ratio of the resistances in the two gaps is equal to the ratio of the resistances of the two segments of the meter bridge wire.

The null point is obtained at the $52	ext{ cm}$ mark from end $A$. Since the total length of the wire is $100	ext{ cm}$, the apparent lengths of the two segments are $52	ext{ cm}$ and $100 - 52 = 48	ext{ cm}$.

### The Catch: End Corrections

In a real-world meter bridge, the thick copper strips and the soldered joints at the ends $A$ and $B$ offer some small but non-negligible resistance. To account for this, we use **end corrections**.

The problem states that the end corrections are $1	ext{ cm}$ for end $A$ and $2	ext{ cm}$ for end $B$. These corrections must be added to the respective apparent lengths to get the true effective lengths:

- Effective length on the left ($L_1$) = $52	ext{ cm} + 1	ext{ cm} = 53	ext{ cm}$
- Effective length on the right ($L_2$) = $48	ext{ cm} + 2	ext{ cm} = 50	ext{ cm}$

### The Master Equation

Now, we apply the balanced Wheatstone bridge condition:

$$ \frac{X}{L_1} = \frac{R}{L_2} $$

Substituting the effective lengths and the known resistance $R = 10\ \Omega$:

$$ \frac{X}{53} = \frac{10}{50} $$

### Final Calculation

Simplifying the right side of the equation, we get:

$$ \frac{X}{53} = \frac{1}{5} $$

Now, solving for $X$:

$$ X = \frac{53}{5} $$

$$ X = 10.6\ \Omega $$

**The determined value of the unknown resistance $X$ is $10.6\ \Omega$.**

Select Answer:

Visualized Solution

$Understanding the Meter Bridge$

$Identifying the Wire Lengths$

$Applying End Corrections$

$The Balanced Bridge Equation$

$Substituting the Values$

$Solving for X$

$Final Calculation$

$Conclusion$

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Catch

The Master Equation

Final Calculation

Similar Questions

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

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

$R_{1}, R_{2}, R_{3}$ are different values of $R$ . Further, $A, B$ and $C$ are the null points obtained corresponding to $R_{1}, R_{2}$ and $R_{3}$ respectively. For which resistor, the value of $X$ will be the most accurate and why?

In the shown arrangement of the experiment of the meter bridge if $A C$ corresponding to null deflection of galvanometer is $x$ , what would be its value if the radius of the wire $A B$ is doubled ?

Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying the Wire Lengths

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

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

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

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

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

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