Question

When a metal conductor is connected to left gap of meter bridge is heated, the balancing point

• A. remains unchanged
• B. Shifts towards right
• C. Shift to the center
• D. Shifts towards left

Answer 1

The Correct Option is B

A meter bridge works on the principle of the Wheatstone bridge. Let the resistance in the left gap be \( R_L \) (the metal conductor) and the resistance in the right gap be \( R_R \). 

Let the balancing point be found at a distance \( l \) from the left end (usually denoted as A or 0 cm mark) on the uniform meter bridge wire of total length \( L \) (usually 100 cm). The length of the wire segment to the right of the balancing point is \( L - l \).

For a uniform wire, resistance is proportional to length. Therefore, the Wheatstone bridge condition at balance is:

\[ \frac{R_L}{R_R} = \frac{\text{Resistance of wire segment of length } l}{\text{Resistance of wire segment of length } L - l} = \frac{l}{L - l} \]

Now, consider the effect of heating the metal conductor connected in the left gap.

For a metal conductor, its resistance increases with an increase in temperature. This is because the increased thermal agitation of the atoms in the conductor hinders the flow of electrons.

So, when the metal conductor is heated, its resistance \( R_L \) increases.

The resistance in the right gap, \( R_R \), remains unchanged. The meter bridge wire properties also remain unchanged.

Let the initial balancing length be \( l_1 \) and the final balancing length (after heating) be \( l_2 \). Let the initial resistance be \( R_{L1} \) and the final resistance be \( R_{L2} \).

Initially: \( \frac{R_{L1}}{R_R} = \frac{l_1}{L - l_1} \)

After heating: \( \frac{R_{L2}}{R_R} = \frac{l_2}{L - l_2} \)

Since \( R_{L2} > R_{L1} \) (due to heating) and \( R_R \) is constant, the ratio on the left side has increased:

\[ \frac{R_{L2}}{R_R} > \frac{R_{L1}}{R_R} \]

This implies that the ratio on the right side must also increase:

\[ \frac{l_2}{L - l_2} > \frac{l_1}{L - l_1} \]

The function \( f(l) = \frac{l}{L - l} \) is an increasing function of \( l \) (for \( 0 < l < L \)). This means that if the value of the ratio increases, the value of \( l \) must also increase.

Therefore, \( l_2 > l_1 \).

An increase in the balancing length \( l \) means the balancing point moves further away from the left end (0 cm) and closer to the right end (100 cm).

Conclusion: The balancing point Shifts towards right.

Answer 2

Given the equation:

\( \ell_R = \frac{100 - \ell_S}{R} \) 

When temperature increases, resistance increases. As the resistance \( R \) increases, the balancing length \( \ell_R \) also increases.

Therefore, as the temperature increases, the balancing length will shift towards the right.

Answer 3

Answer: When the metal conductor is heated, the balancing point will shift towards the right.

Explanation:

Heating a metal conductor causes an increase in its resistance. This is because the resistance of most materials increases with temperature due to the increased vibration of atoms, which impedes the flow of electrons.

In the case of the meter bridge, the resistance in the left gap (where the heated conductor is connected) increases. Since the meter bridge operates on the principle of balancing resistances, an increase in the left gap's resistance causes the balancing point to shift towards the side with lower resistance, which is the right side.

Therefore, the balancing point shifts towards the right side of the meter bridge when the metal conductor is heated.