A sensitive galvanometer, the maximum curent that its coil can measure is $$ 5 \mathrm{~m} / $$, and when it is used to measure an electne current of intensity $$ 2 imes 10^{\prime} 1 \mathrm{LA} $$ its pointer deflects by an angle $$ 30^{\circ} $$, so the maximum angle of deflection of the galvanometer pointer from zero position is .......... (A) $$ 25^{\circ} $$ (B) $$ 50^{\circ} $$ (C) $$ 75^{\circ} $$ (D) $$ 90^{\circ} $$

Question

A sensitive galvanometer, the maximum curent that its coil can measure is $$ 5 \mathrm{~m} / $$, and when it is used to measure an electne current of intensity $$ 2 	imes 10^{\prime} 1 \mathrm{LA} $$ its pointer deflects by an angle $$ 30^{\circ} $$, so the maximum angle of deflection of the galvanometer pointer from zero position is .......... (A) $$ 25^{\circ} $$ (B) $$ 50^{\circ} $$ (C) $$ 75^{\circ} $$ (D) $$ 90^{\circ} $$

UpStudy AI · Accepted Answer

To solve this problem, we need to determine the maximum angle of deflection of the galvanometer pointer based on the given information.

### Known Conditions:
1. Maximum current ($$ I_{max} $$) that the galvanometer can measure: $$ 5 \, \text{mA} $$ (which is $$ 5 \times 10^{-3} \, \text{A} $$).
2. Current being measured ($$ I $$): $$ 2 \times 10^{1} \, \text{A} $$ (which is $$ 20 \, \text{A} $$).
3. Angle of deflection for the current being measured: $$ 30^{\circ} $$.

### Step-by-Step Solution:

1. **Understanding the relationship between current and angle of deflection**:
   The angle of deflection ($$ \theta $$) is directly proportional to the current ($$ I $$) flowing through the galvanometer. This can be expressed as:
   $$
   \frac{\theta}{\theta_{max}} = \frac{I}{I_{max}}
   $$
   where $$ \theta_{max} $$ is the maximum angle of deflection.

2. **Substituting the known values**:
   We know that when $$ I = 20 \, \text{A} $$, $$ \theta = 30^{\circ} $$. We can rearrange the formula to find $$ \theta_{max} $$:
   $$
   \theta_{max} = \frac{I_{max}}{I} \cdot \theta
   $$

3. **Calculating $$ \theta_{max} $$**:
   Substituting the values:
   - $$ I_{max} = 5 \times 10^{-3} \, \text{A} $$
   - $$ I = 20 \, \text{A} $$
   - $$ \theta = 30^{\circ} $$

We can calculate $$ \theta_{max} $$:
   $$
   \theta_{max} = \frac{5 \times 10^{-3}}{20} \cdot 30^{\circ}
   $$

4. **Calculating the fraction**:
   $$
   \frac{5 \times 10^{-3}}{20} = \frac{5}{20} \times 10^{-3} = \frac{1}{4} \times 10^{-3} = 0.25 \times 10^{-3}
   $$

5. **Calculating $$ \theta_{max} $$**:
   $$
   \theta_{max} = 0.25 \times 10^{-3} \cdot 30^{\circ} = 0.25 \cdot 30^{\circ} = 7.5^{\circ}
   $$

6. **Finding the maximum angle of deflection**:
   Since the galvanometer can measure up to $$ 5 \, \text{mA} $$ and the current being measured is $$ 20 \, \text{A} $$, we need to find the maximum angle of deflection. The maximum angle of deflection can be calculated as:
   $$
   \theta_{max} = \frac{I_{max}}{I} \cdot 30^{\circ}
   $$

However, since the current exceeds the maximum limit, we can assume that the maximum angle of deflection is $$ 90^{\circ} $$ as the galvanometer cannot measure beyond its limit.

### Conclusion:
The maximum angle of deflection of the galvanometer pointer from the zero position is $$ 90^{\circ} $$.

Thus, the answer is:
(D) $$ 90^{\circ} $$
The maximum angle of deflection is $$ 90^{\circ} $$.

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