To increase the current sensitivity of a moving coil galvanometer by `50%` its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

Explanation

To solve the problem, we need to analyze the relationship between current sensitivity and voltage sensitivity in a moving coil galvanometer. Let's break it down step by step.

### Step 1: Understand the initial conditions
Let:
- \( I_s \) = initial current sensitivity
- \( R \) = initial resistance
- \( V_s \) = initial voltage sensitivity

The formulas for current sensitivity and voltage sensitivity are:
- Current sensitivity: \( I_s = \frac{\theta}{I} \)
- Voltage sensitivity: \( V_s = \frac{\theta}{V} = \frac{I_s}{R} \)

### Step 2: Calculate the new current sensitivity
The problem states that the current sensitivity is increased by 50%. Therefore, the new current sensitivity \( I_s' \) can be calculated as:
\[ I_s' = I_s + 0.5 I_s = 1.5 I_s = \frac{3}{2} I_s \]

### Step 3: Determine the new resistance
The resistance is doubled, so the new resistance \( R' \) is:
\[ R' = 2R \]

### Step 4: Calculate the new voltage sensitivity
Using the new current sensitivity and the new resistance, we can find the new voltage sensitivity \( V_s' \):
\[ V_s' = \frac{I_s'}{R'} = \frac{\frac{3}{2} I_s}{2R} = \frac{3}{4} \cdot \frac{I_s}{R} = \frac{3}{4} V_s \]

### Step 5: Determine the factor by which voltage sensitivity changes
The factor by which the voltage sensitivity changes can be calculated as:
\[ \text{Factor} = \frac{V_s'}{V_s} = \frac{\frac{3}{4} V_s}{V_s} = \frac{3}{4} \]

### Conclusion
The voltage sensitivity changes by a factor of \( \frac{3}{4} \).