If a shunt `1//10` of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomes

Options

A. `10` fold
B. `11` fold
C. `(1)/(10)` fold
D. `(1)/(11)` fold

Explanation

To solve the problem of determining the sensitivity of a moving coil galvanometer when a shunt of resistance \( \frac{1}{10} \) of the coil resistance is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the resistance of the galvanometer coil be \( R_g \).
- The shunt resistance \( R_s \) is given as \( \frac{1}{10} R_g \).

2. **Current Distribution**:
- Let the total current flowing through the circuit be \( I \).
- The current through the galvanometer is \( I_g \).
- The current through the shunt is \( I_s = I - I_g \).

3. **Applying the Current Division Rule**:
- The current through the galvanometer and the shunt can be expressed as:
\[ \frac{I_g}{I} = \frac{R_s}{R_s + R_g} \]
- Substituting \( R_s = \frac{1}{10} R_g \):
\[ \frac{I_g}{I} = \frac{\frac{1}{10} R_g}{\frac{1}{10} R_g + R_g} \]
- Simplifying the denominator:
\[ \frac{I_g}{I} = \frac{\frac{1}{10} R_g}{\frac{1}{10} R_g + \frac{10}{10} R_g} = \frac{\frac{1}{10} R_g}{\frac{11}{10} R_g} = \frac{1}{11} \]

4. **Sensitivity Relation**:
- The sensitivity of the galvanometer is defined as:
\[ \alpha = \frac{\theta}{I_g} \]
- Initially, the sensitivity is \( \alpha_i = \frac{\theta}{I_g} \) and after adding the shunt, it becomes \( \alpha_f = \frac{\theta}{I} \).

5. **Finding the Ratio of Sensitivities**:
- The ratio of the initial sensitivity to the final sensitivity is:
\[ \frac{\alpha_i}{\alpha_f} = \frac{I_g}{I} \]
- From the previous step, we found \( \frac{I_g}{I} = \frac{1}{11} \).

6. **Conclusion**:
- Therefore, the sensitivity of the galvanometer with the shunt applied becomes \( \frac{1}{11} \) of the original sensitivity.

### Final Answer:
The sensitivity of the moving coil galvanometer with the shunt applied is \( \frac{1}{11} \) times its original sensitivity.