Meter Bridge MCQ Quiz - Objective Question with Answer for Meter Bridge - Download Free PDF

Explanation:

• Initially, the null point is at 1 cm from the left end.
• When the cross-sectional area is doubled, the resistance of the wire decreases by half, which causes the null point to shift towards the left end.
• The new position of the null point will be at 1 cm from the left end.

Additional Information:

• The resistance of a conductor is inversely proportional to its cross-sectional area. Doubling the area reduces the resistance by half, which impacts the position of the null point in the meter bridge experiment.

The correct option is: 2

Wheatstone meter bridge formula:

(R₁ / R₂) = (l₁ / l₂)

Therefore, (R₁ / 90) = (40 / 60)

R₁ = 60 Ω

Now,

We also have:

(dR / R) = (L₁ + L₂) / (L₁ * L₂) * dL

(dR / 60) = (60 + 40) / (60 * 40) * 0.1

Therefore, dR = 0.25 Ω

\(\dfrac{X}{Y}=\dfrac{20}{100-20}\)

\(Y=4X\)

so when \(4X\) balanced against \(Y\): \(\dfrac{4X}{Y}=\dfrac{l}{100-l}\)

\(\dfrac{4X}{4X}=\dfrac{l}{100-l}\)

\(100-l=l\)

\(l=50 \, \text{cm}\)

\(R_{AB} = \rho \dfrac{x}{A}\) where A is the radius of the wire and x is the length of wire in cm between A and C.

Resistance of wire between B and C \(R_{BC} = \rho \dfrac{100-x}{A}\)

Using \(\dfrac{R_1}{R_{AB}} = \dfrac{R_2}{R_{BC}}\)

\(\therefore \dfrac{R_1}{\rho\dfrac{x}{A}} = \dfrac{R_2}{\rho\dfrac{100-x}{A}} \implies \dfrac{R_1}{x}=\dfrac{R_2}{100-x}\)

Thus position of balance point does not depend on the radius of wire.

Concept:

Meter Bridge Experiment:

• In the meter bridge experiment, the null point is the point where there is no current flowing through the galvanometer.
• At the null point, the ratio of the resistances on either side of the bridge is equal: R₁ / R₂ = l / (100 - l), where l is the distance of the null point from the left end, and R₁ and R₂ are the resistances in the left and right gaps respectively.
• When the resistances in both gaps are doubled, and then the resistances are interchanged, the equation changes, and the new position of the null point can be derived.

Calculation:

Let the initial resistances in the left and right gaps be R₁ and R₂, and the null point is at a distance l from the left end.

Initially, R₁ / R₂ = l / (100 - l)

Now, the resistances are doubled, so the new resistances are 2R₁ and 2R₂, and they are interchanged.

The new equation becomes: 2R₂ / 2R₁ = x / (100 - x), where x is the new position of the null point.

Simplifying, R₂ / R₁ = x / (100 - x)

From the initial condition, we have R₁ / R₂ = l / (100 - l), so x = 100 - l.

∴ The new position of the null point is 100 - l.
Hence, the correct option is 1)

CONCEPT:

• Meter Bridge: It is an electrical instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100\; - \;l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100\; - \;l} \right)}}{l}R\) Here S is unknown resistance that can be measured with the help of the meter bridge.

EXPLANATION:

• Thus the meter bridge is used to measure the resistance of a resistor. So option 3 is correct.

CONCEPT:

• Meter bridge: Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

EXPLANATION:

• From above it clear that the meter bridge works on the principle of Wheatstone bridge. Therefore, option 1 is correct.
• The Meter Bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance. Therefore option 2 is correct.
• The Meter bridge can't measure the internal resistance of a battery.
• A potentiometer is a device used to measure the internal resistance of the battery. Therefore option 3 is incorrect.

• Wheatstone Bridge: In the below diagram, BD is Wheatstone Bridge when there is no current across it. This happens when

\(\frac{R_1}{R_3} = \frac{R_2}{R_4}\)

• Meter Bridge: It is an electrical instrument used to measure an unknown resistance.
• It works on two principles, one is Wheatstone Bridge, and the other is 'direct proportionality relationship between length and resistance of a wire.'

• Working of meter Bridge: AC is long resistance wire 100 cm long.. Varying the position of tapping point B, the bridge is balanced.
• Bridge Balanced means the galvanometer is showing zero resistance and the condition satisfying wheatstone Bridge.
• If B is the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100\; - \;l} \right)}}{l}\)

Also, By condition of Wheatstone Bridge.

\(\frac{P}{Q} = \frac{R}{S}\)

⇒ \(S = \frac{{\left( {100\; - \;l} \right)}}{l}R\) 

Here S is unknown resistance that can be measured with the help of the meter bridge. 

Hence, we can say that the meter bridge works on the principle of Wheatstone Bridge and direct proportionality relationship between length and resistance of a wire.'

So, the correct option is Both (1) and (2)

CONCEPT:

• Meter bridge: Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.

• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

EXPLANATION:

• The balance point of the meter bridge is independent of the radius of the wire.
• Though the radius of the wire is doubled, it will not affect the ratio of the length of the 2 parts of the wire. Hence the balancing length will also not get affected. So option 3 is correct.

Concept:

Meter Bridge:

• A meter bridge also called a slide wire bridge is an instrument that works on the principle of a Wheatstone bridge.
  • A meter bridge is used in finding the unknown resistance of a conductor as that of in a Wheatstone bridge.
• In a meter bridge, a wire having a length of a uniform cross-section of about 1m is used.
• One known resistance and an unknown resistance are connected as shown in the figure.
• The one part of the galvanometer is connected in between the known and unknown resistances, whereas the other part of the wire is used to find the null point where the galvanometer is not showing any deflection.
• At this point, the bridge is said to be balanced.

​

• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

Calculation:

Given: P = R₁ = 1Ω, Q = R₂ = 4Ω

\(\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{l}{L}= \frac{1}{4}\)

So, \(\frac{l}{L} = \frac{1}{4}\)

CONCEPT:

• Meter bridge: Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

EXPLANATION:

• A potentiometer is a device mainly used to measure emf of a given cell and to compare emf’s of cells. Therefore option 1 is incorrect.
• The Meter Bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance. Therefore option 2 is correct.
• Kirchoff's Laws are used to find the current and emf in the complicated circuits. Therefore option 3 is incorrect.
• Ohm’s law gives the relation between the voltage and current. Therefore option 4 is incorrect.

CONCEPT:

• Meter bridge: Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

EXPLANATION:

• When the balance point is in the middle of the bridge wire, the resistance in the four arms P, Q, R, and S become of the same order and consequently the bridge becomes more sensitive.
• Thus, the balance point is obtained in the middle of the meter bridge wire. Therefore option 4 is correct.

CONCEPT: 

Meter bridge:

• The Meter Bridge is instruments based on the principle of the Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} ⇒ S = \frac{{\left( {100 - l} \right)}}{l}R\)

EXPLANATION:

Given - Null point in 1^st case (l₁) = 33.7 cm, Null point in 2^nd case (l1) = 51.9 cm

• The null point in 1​st case (l1)

\(⇒ \frac{R}{X} =\frac{l}{100-l}=\frac{33.7}{66.3}\)

• When X is connected in parallel to R = 12 Ω, the resistance across the gap is R_eq 

\(⇒ \frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{X}=\frac{12+X}{12X}\)

\(⇒ R_{eq}=\frac{12X}{12+X}\)

• The null point in 2^nd case (l1)

\(⇒ \frac{R}{R_{eq}} =\frac{l}{100-l}=\frac{51.9}{48.1}\)

\(⇒ \frac{R}{\frac{12X}{12+X}}=\frac{R(12+X)}{12X}=\frac{51.9}{48.1}\)

Substitute \( \frac{R}{X} =\frac{33.7}{66.3}\) in the above equation, we get

\(⇒ \frac{33.7}{66.3}[\frac{12+X}{12}]=\frac{51.9}{48.1}=1.08\)

\(⇒ [\frac{12+X}{12}]=2.126\)

⇒ X = 13.5 Ω 

CONCEPT:

Meter Bridge:

• A meter bridge also called a slide wire bridge is an instrument that works on the principle of a Wheatstone bridge.
  • A meter bridge is used in finding the unknown resistance of a conductor as that of in a Wheatstone bridge.
• In a meter bridge, a wire having a length of a uniform cross-section of about 1m is used.
• One known resistance and an unknown resistance are connected as shown in the figure.
• The one part of the galvanometer is connected in between the known and unknown resistances, whereas the other part of the wire is used to find the null point where the galvanometer is not showing any deflection.
• At this point, the bridge is said to be balanced.

​

EXPLANATION:

• The Meter bridge is most sensitive when all four arms have resistances of the same order. It is so only for moderate resistances.
• Meter bridges become insensitive if the resistances are either too low or too high.
• While we are measuring low resistances, the resistance of copper strips and connecting wire becomes comparable to that of unknown low resistance and there cannot be ignored. Hence option 2) is correct.

CONCEPT:

• Meter bridge: Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance.
• In the case of a meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced.
• If in the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)

Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)

EXPLANATION:

• From above it clear that the meter bridge works on the principle of Wheatstone bridge. Therefore, option 1 is correct.
• The Meter Bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance. Therefore option 2 is correct.
• A potentiometer is a device which used to measure the internal resistance of the battery. Therefore option 3 is incorrect.
• The meter bridge is sensitive when the resistance of all four arms of the bridge are off the same order.
• Two parts of the meter bridge wire form two arms with small values of resistance. So other resistances to be compared should also be comparable and hence small. Therefore, option 4 is correct.