A resistance of 2 ohm is connected across one gap of a meter-bridge (the length of the wire is 100 c

Q: Why does the balance point shift to the right when interchanged?

Initially, the smaller resistance ($2\,\Omega$) is on the left, so the balance point is closer to the left ($x < 50$). When the larger resistance $R$ is moved to the left, the balance point must shift towards the right to maintain the ratio.

Q: Can I solve this without the symmetry trick?

Yes! You can set up the second equation as $R / 2 = (x + 20) / (80 - x)$ and substitute $R$ from the first equation. It leads to a quadratic equation which also gives $x = 40$ cm.

Q: What if $R$ was less than $2\,\Omega$?

If $R 50$ cm. Interchanging them would shift the balance point to the left, meaning the new balance point would be $x - 20$.

Q: Why is the total length 100 cm?

A standard meter bridge uses a wire that is exactly 1 meter long, which is 100 cm. This makes calculations involving percentages and ratios very straightforward.

Q: What are the 'corrections' mentioned in the question?

In real experiments, there are 'end corrections' due to the resistance of the copper strips and solder joints at the ends of the wire. Neglecting them means we assume the wire's resistance starts exactly at the 0 and 100 cm marks.

Question

A resistance of $2\,\Omega$ is connected across one gap of a meter-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than $2\,\Omega$, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is

Options:

(A) $3\,\Omega$

(B) $4\,\Omega$

(C) $5\,\Omega$

(D) $6\,\Omega$

Sigma Percentile · Accepted Answer

The meter bridge is a classic application of the Wheatstone bridge principle, and this problem is a beautiful example of how physical intuition can save you from pages of tedious algebra. Let's dive into the setup and uncover the elegant symmetry hidden within. ### Analyzing the Setup Imagine the meter bridge in front of you. It consists of a 100 cm long uniform wire. In the left gap, we place a known resistance of $2\,\Omega$. In the right gap, we place an unknown resistance $R$, which we are told is strictly greater than $2\,\Omega$. When we slide the jockey along the wire to find the null point (where the galvanometer shows zero deflection), let's say we find it at a distance $x$ from the left end. Because the wire is uniform, its resistance is proportional to its length. ### The Master Equation According to the balanced Wheatstone bridge principle, the ratio of the resistances in the gaps must equal the ratio of the balancing lengths. Therefore, we can write our first master equation: $$ \frac{2}{R} = \frac{x}{100 - x} $$ Before we move on, let's use a bit of logic. Since $R > 2\,\Omega$, the denominator on the left side is larger than the numerator. This means the right side must also have a larger denominator. Consequently, $100 - x > x$, which implies $x < 50 ext{ cm}$. The initial balance point is somewhere on the left half of the wire. ### The Symmetry Trick Now, the problem states that we interchange the two resistances. The unknown resistance $R$ is now in the left gap, and the $2\,\Omega$ resistance is in the right gap. We are told the balance point shifts by $20 ext{ cm}$. Since $R$ is now on the left, the balance point must shift towards the right half of the wire. Thus, the new balance point is $x + 20$. We could set up a second equation: $\frac{R}{2} = \frac{x + 20}{80 - x}$ and solve the resulting quadratic equation. But there is a much more elegant way! Think about the symmetry of the meter bridge. If you completely swap the left and right gaps, the entire physical setup is just mirrored. Therefore, the new balance point must simply be the mirror image of the old balance point across the center of the wire. If the original balance point was at $x$ from the left, the new balance point must be at $x$ from the right! Measuring from the left, this new position is exactly $100 - x$. ### Final Calculation We now have two different expressions for the new balance point. We can equate them to find $x$ instantly: $$ 100 - x = x + 20 $$ Solving this simple linear equation: $$ 2x = 80 \implies x = 40 ext{ cm} $$ This perfectly matches our earlier deduction that $x < 50 ext{ cm}$. Now that we have the initial balancing length, we can substitute it back into our first master equation to find $R$: $$ \frac{2}{R} = \frac{40}{100 - 40} $$ $$ \frac{2}{R} = \frac{40}{60} = \frac{2}{3} $$ Cross-multiplying gives us the final answer: $$ R = 3\,\Omega $$ By leveraging the physical symmetry of the system, we completely bypassed the quadratic algebra and arrived at the solution with pure logic!

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Visualized Solution

$I ni t ia l S e t u p$

$B a l an ce d B r i d g e C o n d i t i o n$

$F i r s tE q u a t i o n$

$I n t er c han g in g R es i s t an ces$

$S y mm e t r y o f M e t er B r i d g e$

$S hi f t in B a l an ce P o in t$

$S o l v in g f or x$

$C a l c u l a t in g U nk n o w n R es i s t an ce$

$C o n c l u s i o n$

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

The Symmetry Trick

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; } InitialSetup

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

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

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

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

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

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

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

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