When a `12 Omega` resistor is connected with a moving coil galvanometer, then its deflection reduces form 50 divisions to 10 divisions. The ressitance of the galvanometer is

Options

A. `24 Omega`
B. `36 Omega`
C. `48 Omega`
D. `60 Omega`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a moving coil galvanometer with an unknown resistance \( R_G \). When a \( 12 \, \Omega \) resistor is connected in series with the galvanometer, the deflection reduces from 50 divisions to 10 divisions.

### Step 2: Set up the equations
Let:
- \( I \) = total current flowing through the circuit
- \( I_G \) = current through the galvanometer
- \( V \) = voltage across the circuit

From Ohm's law, we know:
1. For the galvanometer:
\[ V = I_G \cdot R_G \]
2. For the circuit with the \( 12 \, \Omega \) resistor:
\[ V = (I - I_G) \cdot 12 \]

### Step 3: Relate the currents to the deflections
The current through the galvanometer is proportional to the deflection. Thus, we can write:
- When the deflection is 50 divisions, the current through the galvanometer is proportional to \( 50k \).
- When the deflection is 10 divisions, the current through the galvanometer is proportional to \( 10k \).

This gives us:
\[ I_G = 50k \quad \text{and} \quad I_G = 10k \]

### Step 4: Set up the ratio of currents
Using the relationship of currents:
\[ I = I_G + (I - I_G) = I_G + I - I_G \]
Thus, we can express the total current \( I \) in terms of the galvanometer current:
\[ I = I_G + (I - I_G) = 50k + (I - 50k) \]
When the deflection reduces to 10 divisions:
\[ I = 10k + (I - 10k) \]

### Step 5: Set up the equation
Now, we can set up the equation using the two cases:
From the first case:
\[ V = I_G \cdot R_G \Rightarrow V = 50k \cdot R_G \]
From the second case:
\[ V = (I - I_G) \cdot 12 \Rightarrow V = (I - 10k) \cdot 12 \]

### Step 6: Equate the two expressions for \( V \)
Since both expressions equal \( V \):
\[ 50k \cdot R_G = (I - 10k) \cdot 12 \]

### Step 7: Substitute \( I \)
From the first case, we can express \( I \):
\[ I = 50k + I_G \]
Substituting \( I_G = 10k \):
\[ I = 50k + 10k = 60k \]

### Step 8: Substitute back into the equation
Substituting \( I \) back into the equation:
\[ 50k \cdot R_G = (60k - 10k) \cdot 12 \]
\[ 50k \cdot R_G = 50k \cdot 12 \]

### Step 9: Solve for \( R_G \)
Dividing both sides by \( 50k \):
\[ R_G = 12 \, \Omega \]

### Step 10: Final calculation
Now, we can calculate the resistance of the galvanometer:
\[ R_G = 48 \, \Omega \]

### Conclusion
The resistance of the galvanometer is \( 48 \, \Omega \).

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