A metallic conductor at `10^@ C` connected in the left gap of meter bridge gives balancing length 40 cm. 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. 10
C. 15
D. 7

Answer

To solve the problem step by step, we will follow the outlined process to find the change in balancing length when the temperature of the metallic conductor changes from 10°C to 60°C.

### Step 1: Understand the given data
- Initial temperature (T1) = 10°C = 283 K
- Final temperature (T2) = 60°C = 333 K
- Initial balancing length (L1) = 40 cm
- Temperature coefficient of resistance (α) = 1/220 °C

### Step 2: Set up the equations for the meter bridge
Using the meter bridge principle, we have:
1. For the first case (at 10°C):
\[ \frac{R_1}{R_2} = \frac{L_1}{100 - L_1} \]
Let \( R_2 = q \) (resistance in the right gap). Thus:
\[ \frac{R_1}{q} = \frac{40}{60} = \frac{2}{3} \]
Therefore, we can express \( R_1 \) as:
\[ R_1 = \frac{2}{3}q \]

### Step 3: Express the resistance at different temperatures
The resistance of the conductor at temperature \( T \) can be expressed as:
\[ R = R_0(1 + \alpha T) \]
Where \( R_0 \) is the resistance at 0°C. Thus:
- At \( T1 = 10°C \):
\[ R_1 = R_0(1 + \alpha \cdot 10) \]
- At \( T2 = 60°C \):
\[ R_1' = R_0(1 + \alpha \cdot 60) \]

### Step 4: Set up the equation for the second case
For the second case (at 60°C), let the new balancing length be \( L \):
\[ \frac{R_1'}{q} = \frac{L}{100 - L} \]
Substituting \( R_1' \):
\[ \frac{R_0(1 + \alpha \cdot 60)}{q} = \frac{L}{100 - L} \]

### Step 5: Formulate the ratio of resistances
Now we can set up the ratio of the resistances from both cases:
\[ \frac{R_0(1 + \alpha \cdot 10)}{R_0(1 + \alpha \cdot 60)} = \frac{L_1}{L} \]
This simplifies to:
\[ \frac{1 + \frac{1}{220} \cdot 10}{1 + \frac{1}{220} \cdot 60} = \frac{40}{L} \]

### Step 6: Substitute values and solve for L
Substituting the values:
\[ \frac{1 + \frac{10}{220}}{1 + \frac{60}{220}} = \frac{40}{L} \]
Calculating the left side:
\[ \frac{1 + \frac{1}{22}}{1 + \frac{3}{11}} = \frac{22/22 + 1/22}{11/11 + 3/11} = \frac{23/22}{14/11} = \frac{23 \cdot 11}{22 \cdot 14} \]
Calculating this gives:
\[ \frac{253}{308} = \frac{40}{L} \]
Cross-multiplying gives:
\[ 253L = 40 \cdot 308 \]
Calculating \( 40 \cdot 308 = 12320 \):
\[ L = \frac{12320}{253} \approx 48.7 \text{ cm} \]

### Step 7: Calculate the change in balancing length
The change in balancing length is:
\[ \Delta L = L - L_1 = 48.7 - 40 = 8.7 \text{ cm} \]

### Final Answer
The balancing point shifts by approximately **8.7 cm**.