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Meter Bridge MCQs list down important choice-based questions from the chapter Current Electricity. These MCQs are as per the latest CBSE Class 12 Physics Syllabus. The MCQs on Meter bridges cover questions on finding the unknown resistances in an electrical circuit and understanding their working principle.
Meter bridges are also called slide wire bridges that work on the principle of a Wheatstone Bridge. The unknown resistance in the conductor can be obtained by using the Meter bridge. The following figure shows how the unknown resistance can be evaluated:
Calculating Unknown Resistance
The above figure shows an electrical circuit wherein R is the Resistance, P is the Resistance across AB, S is the Unknown Resistance, and Q is the Resistance between the joints BD.
The wire Ac measures 1 m and is made from constantan and manganin that have an area such that the sum of the lengths is 100.
Therefore, L1 + L2 = 100
Let L1= L = L2, = 100-L
The unknown resistance X is given by:
X = RL2/L1 = R(100 – L)/L
The specific resistance of the wire material is = (3.14) r2X/l
Where,
r = the radius of the cable and also
l = length of the wire.
Check Out NCERT Solutions for Class 12 Physics Current Electricity
Class 12 Meter Bridge MCQs
Ques 1: What happens to the null deflection in the galvanometer when the wire radius is tripled?
- Halves
- No change
- Reduces by 1/3
- Thrice the original value
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Ans: The correct option is B. No Change
Explanation: In a blanched Meter bridge,
P/Q = x/(100-x)
Thus it is clear that the radius of the wire does not have any effect on the null deflection of the galvanometer.
Ques 2: When the resistance of 5 ohms is connected across the gap of a Meter Bridge, the unknown resistance connected has a value greater than 5 ohms connected across the other gap. On interchanging these resistances, there is a shift in the balance point by 50 cm. What will be the unknown resistance if the correction is neglected. Assume the length of the wire is 150 cm.
- 5 ohms
- 10 ohms
- 3 ohms
- 7 ohms
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Ans: The correct option is B. 10 Ohms
Explanation: If X is the unknown resistance, and ‘l’ is the length of the wire, then
Case 1: 5/x = l/(150-l)
750 - 5l = xl — (1)
Case 2: x/5= (l+50)/(100-l)
Therefore, on solving we get,
xl = 100x – 5l – 250 — (2)
Equating (1) with (2)
750 - 5l = 100x – 5l – 250
We get,
x= 10 ohms
Ques 3: The sensitivity of the meter bridge is minimum when all resistors have the same order.
- True
- False
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Ans: The correct option is B. False
Explanation: The sensitivity of the meter bridge increases when the current value in the galvanometer is high. This is achieved by reducing the resistor values that are used in the meter bridge. Hence, the sensitivity of the meter bridge is at the highest point when all the resistors are of the same order.
Ques 4: The error eliminated when the known and unknown resistances are interchanged in a meter bridge is ____________.
- End Error
- Measurement Error
- Index Error
- Percentage Error
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Ans: The correct option is A. End Error
Explanation: When the known and unknown resistances are interchanged in a Meter Bridge, the error eliminated is the End error. The end resistance comes from the additional length. So when the error arises, the mean of the resistances is taken to eliminate the end error.
Do Check Out:
Ques 5: On what principle does Meter bridge work on?
- Ohm’s Law
- Kirchoff’s law
- Wheatstone Bridge
- Potentiometer
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Ans: The correct option is C. Wheatstone Bridge
Explanation: Meter bridge works on the Wheatstone bridge principle and is used to find the unknown resistances.
Ques 6: What is the formula for unknown resistance, S, in a meter bridge if r is the unknown resistance and ‘l’ is the wire length?
- S= (100-l)R/l
- S= (100-R)/l
- S= R/l(100-R)
- S= R/l(100)
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Ans: The correct option is A.S = (100-l)R/l
Explanation: If R is the known resistance and l is the length of the wire, the meter bridge follows the Wheatstone bridge principle. Thus, the unknown resistance is given by:
\(S=\frac{(100-l)R}{l}\)
Ques 7: Why are constantan and manganin wires used in meter bridges?
- Low-temperature coefficient of resistance
- High-temperature coefficient of resistance
- No resistance to current
- Low conductivity.
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Ans: The correct option is A. Low-temperature coefficient of resistance
Explanation: Constantan and manganin are materials that just like nichrome offer a very low-temperature coefficient of resistance. This makes them the material of choice for wires that are employed in meter bridges.
Ques 8: Sensitivity and accuracy are maximum in a meter bridge when the null point is at
- Right Corner
- Left Corner
- Centre
- Any point on the wire
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Ans: The correct option is C. Centre
Explanation: In the middle of the wire, the resistance ratios are equal. Hence when the four arm resistances are of the same order, the balance or the null point can be obtained in the middle of the wire. The sensitivity is also at its peak at this point.
Ques 9: The balancing length is not 0.25 m for which of the resistor combinations?
- 25 Ohms, 75 Ohms
- 1 ohm, 3 Ohms
- 3 Ohms, 4/3 Ohms
- 2 Ohms, 3 Ohms
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Ans: The correct option is D. 2 Ohms, 3 Ohms
Explanation:
The resistance wire length l = 100 cm in a meter bridge.
Thus, the balanced position of the bridge is l1 and 100 – l1 respectively.4
The unknown resistance would be (100 – l1)R/l1,
For a balancing length of 0.25 m, 100-l1 = 100 – 25
=75 m
Thus the ratio would be = 25/75 =⅓
In the given option, Only the option with 2 Ohms and 3 Ohms does not have a ratio of ⅓.
Ques 10: Resistances 4 Ω and 6 Ω are connected across the left gap and right gap respectively of a meter bridge. When a 2 Ω resistance is connected in series with the 4 Ω resistance in the left gap, calculate the shift in the balance point?
- 15 Ohms
- 10 Ohms
- 25 Ohms
- 20 Ohms
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Ans: The correct option is B. 10 Ohms
Explanation: When the left and right have 4 and 6 Ω, the balanced condition would be:
4/6 = L/(100-L)
Solving we get L= 40 cm
When 2 Ω is connected with 4 Ω in a series, the balancing length is 50 cm. Thus,
the Shift in length is 50-40
=10 Ohms
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