A metallic conductor at `10^(@)C` connected in the left gap of Meter Bridge given balancing length 40cm. When the conductor is at `60^(@)C`, the balancing point shifts by__cm, (temperature coefficient of resistance of the material of the wire is `(1//220)//^(@)C`)

Options

A. 4.8
B. 8.7
C. 15
D. 7

Answer

To solve the problem step by step, we will use the concepts of the meter bridge and the temperature dependence of resistance.

### Step 1: Understand the meter bridge principle
The meter bridge is based on the principle of Wheatstone bridge. It states that when the bridge is balanced, the ratio of the lengths of the bridge is equal to the ratio of the resistances.

### Step 2: Set up the initial conditions
Let:
- \( R \) be the resistance of the metallic conductor at \( 10^\circ C \).
- The balancing length \( l_1 = 40 \, \text{cm} \).
- The length on the other side of the meter bridge \( 100 - l_1 = 60 \, \text{cm} \).

Using the meter bridge formula:
\[ \frac{x}{R} = \frac{l_1}{100 - l_1} \]
where \( x \) is the resistance on the left side of the bridge.

### Step 3: Calculate the initial resistance
From the balancing condition:
\[ \frac{x}{R} = \frac{40}{60} \]
Cross-multiplying gives:
\[ x = R \cdot \frac{40}{60} = R \cdot \frac{2}{3} \]

### Step 4: Determine the resistance at \( 60^\circ C \)
The resistance of the conductor changes with temperature according to the formula:
\[ R_t = R_0 \left(1 + \alpha \Delta T\right) \]
where:
- \( R_0 \) is the resistance at \( 10^\circ C \).
- \( \alpha = \frac{1}{220} \, ^\circ C^{-1} \) (temperature coefficient).
- \( \Delta T = 60 - 10 = 50 \, ^\circ C \).

Substituting the values:
\[ R_{60} = R_0 \left(1 + \frac{1}{220} \cdot 50\right) = R_0 \left(1 + \frac{50}{220}\right) = R_0 \left(1 + \frac{5}{22}\right) = R_0 \left(\frac{27}{22}\right) \]

### Step 5: Set up the new balancing condition
Let the new balancing length at \( 60^\circ C \) be \( l_2 \). The new balancing condition is:
\[ \frac{x}{R_{60}} = \frac{l_2}{100 - l_2} \]

### Step 6: Substitute the new resistance
Using the expression for \( R_{60} \):
\[ \frac{x}{R_0 \cdot \frac{27}{22}} = \frac{l_2}{100 - l_2} \]
Substituting \( x = R_0 \cdot \frac{2}{3} \):
\[ \frac{R_0 \cdot \frac{2}{3}}{R_0 \cdot \frac{27}{22}} = \frac{l_2}{100 - l_2} \]
This simplifies to:
\[ \frac{2}{3} \cdot \frac{22}{27} = \frac{l_2}{100 - l_2} \]

### Step 7: Solve for \( l_2 \)
Cross-multiplying gives:
\[ 2(100 - l_2) = 3l_2 \cdot \frac{22}{27} \]
Expanding and simplifying:
\[ 200 - 2l_2 = \frac{66}{27} l_2 \]
Combining like terms:
\[ 200 = 2l_2 + \frac{66}{27} l_2 \]
Finding a common denominator (27):
\[ 200 = \left(54 + 66\right) \frac{l_2}{27} \]
\[ 200 = \frac{120l_2}{27} \]
Thus:
\[ l_2 = \frac{200 \cdot 27}{120} = \frac{5400}{120} = 45 \, \text{cm} \]

### Step 8: Calculate the shift in balancing point
The shift in the balancing point is:
\[ \Delta l = l_2 - l_1 = 45 \, \text{cm} - 40 \, \text{cm} = 5 \, \text{cm} \]

### Final Answer
The balancing point shifts by **5 cm**.