Key Fact
Wheatstone bridge balance
Bridge is balanced when P/Q = R/S (no current through galvanometer). Used to measure unknown resistance with high precision.
-- NCERT Class 12 Physics, Ch. 3, p. 101Metre Bridge
8 MCQs1 revision card9-step worked example
Source: NCERT Current ElectricityPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 21 Sep 2026
Metre Bridge, explained for NEET
The trap first: in a metre-bridge problem, many students write the balance ratio upside down — putting the right-gap length over the left-gap length, or the known resistor over the unknown. The answer then comes out as the reciprocal ratio, and that wrong value is usually sitting among the options.
The principle. NCERT Class 12 Physics Part I, Chapter 3, Section 3.13 (pages 100–101) treats the Wheatstone bridge: when the galvanometer current is zero, the four arms satisfy R₂/R₁ = R₄/R₃ (Eq. 3.64a). NCERT then adds one sentence on page 101: a practical device using this principle is called the meter bridge. The current reprint has no separate metre-bridge section, derivation or example; the NTA NEET (UG) 2026 syllabus, Physics Unit 12, still lists "Metre Bridge", so it remains examinable. What follows is the standard application of the balance condition.
The device. A uniform wire of length 100 cm is stretched between two ends. One resistor (say X) sits in the left gap, another (S) in the right gap. A jockey slides along the wire until the galvanometer shows zero deflection at a length l measured from the left end. Because the wire is uniform, the resistance of each wire segment is proportional to its length, so the two segments act as the other two bridge arms. Balance gives:
X / S = l / (100 − l)
How to avoid the inversion. Pair each resistor with the wire segment on its own side: the left-gap resistor goes with l (left segment), the right-gap resistor with (100 − l). Write both ratios in the same order — left over right — every time.
NEET bridge. Typical questions ask for an unknown resistance from a balance length, the new balance point after the resistors are interchanged, or the shift when a resistor is added in series or parallel with one gap.
Watch out: if the resistors are swapped between gaps, the balance length becomes (100 − l), not l. And l is measured from the end on the side of the resistor it pairs with.
Can you answer these Metre Bridge MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A metre bridge works on the principle of which of the following?
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Physics Part I, Chapter 3, page 101 states that a practical device using the Wheatstone bridge principle is called the meter bridge; the balance condition is Eq. 3.64(a).
Why B is wrong: B is wrong because the metre bridge measures resistance by finding a null (zero galvanometer current), not by measuring heat produced in the wire.
Why C is wrong: C is wrong because terminal voltage versus emf describes a cell's internal resistance; the metre bridge balance does not depend on the cell's emf at all.
Why D is wrong: D is wrong because temperature dependence of resistance is a property the bridge could be used to study, not the principle the bridge operates on.
MCQ 2Easy RecallPractice
When a metre bridge is balanced, the current through the galvanometer is:
Show answer and why every option is right or wrong
Answer: C. Balance is defined by zero galvanometer current (I_g = 0), which is the condition used to derive Eq. 3.64(a) in NCERT Class 12 Physics Part I, Chapter 3, pages 100–101.
Why A is wrong: A is wrong because the jockey is moved to find the point of zero deflection, not maximum deflection; maximum current means the bridge is far from balance.
Why B is wrong: B is wrong because the cell current divides between the resistor branch and the wire; at balance none of it passes through the galvanometer.
Why D is wrong: D is wrong because there is no halving rule; at balance the two junctions joined by the galvanometer are at the same potential, so its current is exactly zero.
MCQ 3Easy RecallPYQ Pattern
In a metre bridge, resistor X is in the left gap and resistor S is in the right gap. The balance point is at length l cm from the left end of the 100 cm wire. The balance condition is:
Show answer and why every option is right or wrong
Answer: C. The left-gap resistor pairs with the left wire segment l, and the right-gap resistor with (100 − l). The Wheatstone balance condition (NCERT Class 12 Physics Part I, Chapter 3, Eq. 3.64a) then gives X/S = l/(100 − l).
Why A is wrong: A is wrong because it inverts the ratio — it pairs the left-gap resistor with the right wire segment (trap: ratio inversion).
Why B is wrong: B is wrong because it divides by the whole wire length; the other bridge arm is only the right segment (100 − l), not the full 100 cm.
Why D is wrong: D is wrong because the balance condition is a ratio of arms, not a product; multiplying the arms has no basis in the Wheatstone condition.
MCQ 4Direct ApplicationPYQ Pattern
In a metre bridge, an unknown resistor X is in the left gap and a 10.0 Ω resistor is in the right gap. The null point is at 40.0 cm from the left end. What is X?
Show answer and why every option is right or wrong
Answer: D. X/10.0 = 40.0/(100 − 40.0) = 40.0/60.0, so X = 10.0 × 2/3 = 6.7 Ω (Wheatstone balance, NCERT Class 12 Physics Part I, Chapter 3, Eq. 3.64a).
Why A is wrong: A is wrong because 4.0 Ω comes from X = 10.0 × 40.0/100, using the whole wire length as the denominator instead of the right segment 60.0 cm.
Why B is wrong: B is wrong because 25 Ω comes from X = 10.0 × 100/40.0, which both inverts the ratio and uses the full wire length.
Why C is wrong: C is wrong because 15 Ω comes from X = 10.0 × 60.0/40.0, which inverts the ratio by pairing X with the right segment (trap: ratio inversion).
MCQ 5Direct ApplicationPYQ Pattern
A 3.0 Ω resistor is in the left gap and a 2.0 Ω resistor is in the right gap of a metre bridge. How far from the left end is the balance point?
Show answer and why every option is right or wrong
Answer: B. l/(100 − l) = 3.0/2.0, so 2l = 300 − 3l, giving 5l = 300 and l = 60 cm. Check: 60/40 = 3/2.
Why A is wrong: A is wrong because 40 cm gives l/(100 − l) = 40/60 = 2/3, which is S/X — the ratio has been inverted (trap: ratio inversion).
Why C is wrong: C is wrong because the balance point is at the middle only when the two gap resistors are equal; here 3.0 Ω ≠ 2.0 Ω.
Why D is wrong: D is wrong because 67 cm comes from l/100 = 2/3, which both inverts the ratio and divides by the whole wire length.
MCQ 6Direct ApplicationPYQ Pattern
In a metre bridge the balance point is at 30.0 cm from the left end. The resistors in the left and right gaps are now interchanged. Where is the new balance point, measured from the left end?
Show answer and why every option is right or wrong
Answer: A. Initially X/S = 30.0/70.0. After interchange, the left gap holds S, so S/X = l′/(100 − l′) = 70.0/30.0, giving l′ = 70.0 cm — that is, (100 − l).
Why B is wrong: B is wrong because 40.0 cm comes from 100 − 2 × 30.0, which has no basis in the balance condition.
Why C is wrong: C is wrong because 60.0 cm comes from doubling 30.0 cm; interchanging the resistors inverts the ratio, it does not double the length.
Why D is wrong: D is wrong because the ratio of the left-gap to right-gap resistor has flipped, so the balance point cannot stay where it was.
MCQ 7CalculationPYQ Pattern
An unknown resistor X is in the left gap and a 30.0 Ω resistor is in the right gap of a metre bridge; the balance point is at 40.0 cm from the left end. An identical 30.0 Ω resistor is then connected in parallel with the one in the right gap. What is the new balance length from the left end?
Show answer and why every option is right or wrong
Answer: C. Step 1: X/30.0 = 40.0/60.0, so X = 20.0 Ω. Step 2: the right gap becomes 30.0 ∥ 30.0 = 15.0 Ω. Step 3: l/(100 − l) = 20.0/15.0 = 4/3, so 3l = 400 − 4l and l = 400/7 = 57.1 cm.
Why A is wrong: A is wrong because 25.0 cm uses 60.0 Ω for the right gap, treating the added resistor as in series instead of in parallel (20.0/60.0 = 1/3 gives l = 25.0 cm).
Why B is wrong: B is wrong because 42.9 cm comes from l/(100 − l) = 15.0/20.0, which inverts the ratio (trap: ratio inversion).
Why D is wrong: D is wrong because 80.0 cm assumes the balance length doubles when the right-gap resistance halves; l is not proportional to 1/S — the ratio l/(100 − l) is.
MCQ 8CalculationPYQ Pattern
In a metre bridge, X is in the left gap and S in the right gap, with the balance point at 20.0 cm from the left end. When a 15.0 Ω resistor is connected in series with X, the balance point moves to 40.0 cm from the left end. What is X?
Show answer and why every option is right or wrong
Answer: B. First balance: X/S = 20.0/80.0, so S = 4X. Second balance: (X + 15.0)/S = 40.0/60.0 = 2/3, so X + 15.0 = 8X/3, giving 5X/3 = 15.0 and X = 9.0 Ω (S = 36 Ω). Check: 24/36 = 2/3.
Why A is wrong: A is wrong because 6.4 Ω comes from using X/S = 20.0/100 for the first balance (whole wire length) but 40.0/60.0 for the second — mixing two different denominators.
Why C is wrong: C is wrong because 15 Ω comes from dividing by the whole wire length in both balances (X/S = 0.20 and (X + 15.0)/S = 0.40), which is not the balance condition.
Why D is wrong: D is wrong because 36 Ω is S, the right-gap resistor, not X; the question asks for the left-gap resistor.
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Metre Bridge: quick recall before you leave
How do you solve a Metre Bridge question? A worked example
- 1
Given
A metre bridge has an unknown resistor X in the left gap and a 5.0 Ω resistor in the right gap. The galvanometer shows zero deflection when the jockey is at 60.0 cm from the left end of the 100 cm wire.
- 2
Required
Find X.
- 3
Concept
At balance, the metre bridge is a balanced Wheatstone bridge (NCERT Class 12 Physics Part I, Chapter 3, Eq. 3.64a, pages 100–101). The uniform wire's two segments form the other two arms, with resistance proportional to length.
- 4
Formula
X / S = l / (100 − l)
- 5
Substitution
X / 5.0 = 60.0 / (100 − 60.0) = 60.0 / 40.0
- 6
Calculation
X = 5.0 × 60.0 / 40.0 = 5.0 × 1.5 = 7.5 Ω. The 100 cm wire length is exact (it defines the device) and does not limit the significant figures; the answer carries two significant figures from the 5.0 Ω resistor.
- 7
Final answer
X = 7.5 Ω
- 8
Common trap
Writing X/S = (100 − l)/l pairs the left-gap resistor with the right wire segment and gives X = 5.0 × 40.0/60.0 = 3.3 Ω — the inverted answer. Check: a larger resistor in the left gap must push the balance point past the middle toward the right, and 60.0 cm is indeed past the middle, so X must be larger than 5.0 Ω.
- 9
Similar NEET-style question
"In the same bridge, the 7.5 Ω and 5.0 Ω resistors are interchanged. Find the new balance point from the left end."
Strategy: l/(100 − l) = 5.0/7.5 = 2/3 → 3l = 200 − 2l → l = 40.0 cm, which is (100 − 60.0), as expected after an interchange.
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What to remember before solving Metre Bridge questions
Which Metre Bridge formulas do you need for NEET?
1 formula — click to collapse
Wheatstone bridge balance
Balance condition: no current through galvanometer when P/Q = R/S. Used to measure unknown resistance.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P,Q,R,S | four arm resistances | Ω |
Valid when
- Bridge balanced (galvanometer reads zero)
- DC, steady state
Where do students lose marks on Metre Bridge?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Category: Similar Terms
Student writes Q/P = R/S or S/R = P/Q instead of P/Q = R/S. Order of arms in the proportion matters.
When it triggers
Wheatstone bridge problem with 4 named arms.
How to avoid
Standard convention: P, Q in one branch (P top, Q bottom from input node); R, S in other branch. Balance: P/Q = R/S.
More in Current Electricity: 2 exam traps and mistakes · 6 formulas · 5 question patterns from its other lessons.
Metre Bridge questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys. — click to collapse
NEET 2020
11.5×10⁻² m
21.0×10⁻² m
31.0×10⁻¹ m
41.5×10⁻¹ m
NTA Answer: Option 3(final)
How does NEET ask about Metre Bridge?
1 recurring pattern from past papers — click to collapse
Apply P/Q = R/S to find unknown resistance. Common in metre-bridge problems.
Direct ApplicationMedium
Common distractors
inverts ratio
Uses Q/P or S/R instead
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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