Meter Bridge Experiment

The meter bridge experiment is a practical application of Wheatstone bridge principle to measure the unknown resistance of a conductor. In this experiment, a uniform wire of length one meter, known as the meter bridge, is connected in series with a resistance box and a galvanometer.

The principle of Wheatstone bridge states that if a bridge circuit is balanced, then the ratio of the two resistances in the bridge is equal to the ratio of the other two resistances in the bridge.

In the meter bridge experiment, the unknown resistance is connected in one of the gaps of the meter bridge, and a known resistance, R, is connected in the other gap. The jockey is moved along the wire until the galvanometer shows zero deflection, indicating a balanced bridge.

Formula for Meter Bridge Experiment

Using the principle of Wheatstone bridge, the unknown resistance can be calculated as follows:

Let the unknown resistance be x ohms and the known resistance be R ohms.

Then, at balance point, the ratio of the two resistances is given by:

x / R = (100 - l) / l

where l is the length of the wire from the left end to the balance point, in cm.

If the positions of the unknown resistance and the known resistance are interchanged, the new balance point is obtained by moving the jockey from the other end of the wire to the new balance point. The new balance point is at a distance of d cm from the point where the jockey was initially placed.

Formula for Interchanged Resistance

Using the principle of Wheatstone bridge, the unknown resistance can be calculated as follows:

Let the unknown resistance be x ohms and the known resistance be R ohms.

Then, at balance point, the ratio of the two resistances is given by:

x / R = d / (100 - d)

where d is the distance of the new balance point from the end where the jockey was initially placed, in cm.

Solution

Given that the resistances in the two gaps of the meter bridge are 10 ohm and 30 ohm respectively.

Let the unknown resistance be x ohms and the known resistance be R ohms.

At balance point, the ratio of the two resistances is given by:

x / R = (100 - l) / l

where l is the length of the wire from the left end to the balance point, in cm.

If the positions of the unknown resistance and the known resistance are interchanged, the new balance point is obtained by moving the jockey from the other end of the wire to the new balance point. Let the new balance point be at a distance of d cm from the point where the jockey was initially placed.

At the new balance point, the ratio of the two resistances is given by:

x / R = d / (100 - d)

Equating the two ratios, we get:

(100 - l) / l = d / (100 - d)

Solving for d, we get:

d = 100l / (l + 100 - 2l)

Substituting l = 50 cm, we get:

d = 100 x 50 / (50 + 100 - 2 x 50) = 50 cm

Therefore, the balance point shifts by 50 cm when the resistances are interchanged.

Hence, the correct answer is option 'D'.

Meter Bridge Experiment:

The Meter Bridge is an experimental setup that is used to measure an unknown resistance. It consists of a long wire of uniform cross-sectional area AB, which is divided into two parts by the help of a metallic strip (jockey) called a ‘galvanometer’. The two parts of the wire are of different resistances.

Working Principle:

The working principle of the meter bridge is based on the Wheatstone bridge concept. When a current is passed through the wire, a potential difference is developed across the wire. If a variable resistor is connected across the wire, the potential difference across the wire varies accordingly. By adjusting the variable resistor, the potential difference across the wire can be made zero. This is the balanced condition of the meter bridge.

Formula:

AB/AD = R2/R1

Where,
AB = Total length of the wire
AD = Distance between the jockey and R1
R1 = Resistance to be measured
R2 = Known resistance

Given:

R1 = 10 ohm
R2 = 30 ohm

Interchanging of Resistances:

When the resistances are interchanged, the resistance of 10 ohm will become R2 and 30 ohm will become R1. Therefore,

R1’ = 30 ohm
R2’ = 10 ohm

Let the initial balance point be at distance ‘x’ from A.

Balance Point:

AB/AD = R2/R1
AB/(x+AD) = 30/10
3(x+AD) = AB

Now, when the resistances are interchanged, the balance point shifts by ‘d’.

New balance point = x+d

AB/AD’ = R2’/R1’
AB/(x+d+AD’) = 10/30
(x+d+3AD’) = AB/3

Subtracting the above two equations, we get:

d + 3AD’ - 3AD = AB/3 - 3(x+AD)

d = (AB/3) - 3(x+AD) - 3AD’ + 3AD

Substituting the values of AD, AD’, R1, and R1’, we get:

d = (AB/3) - 3x

d = (100/3) - 3x

d = 33.33 - 3x

Therefore, the balance point shifts by 50 cm (i.e., 33.33 - 16.67). Hence, the correct answer is option D.