A certain amount of current when flowing in a properly set tangent galvanoment, produces a deflection of `45^(@)`. If the current be reduced by a factor of `sqrt(3)`, the deflection would

Options

A. decrease by `30^(@)`.
B. decrease by `15^(@)`.
C. increase by `15^(@)`.
D. increase by `30^(@)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between current and deflection
In a tangent galvanometer, the current (I) flowing through it is directly proportional to the tangent of the angle of deflection (θ). Mathematically, this can be expressed as:
\[ I \propto \tan(\theta) \]

### Step 2: Define the initial conditions
Let the initial current be \( I_1 \) and the initial deflection be \( \theta_1 = 45^\circ \).

### Step 3: Determine the new current
The current is reduced by a factor of \( \sqrt{3} \):
\[ I_2 = \frac{I_1}{\sqrt{3}} \]

### Step 4: Set up the ratio of currents and tangents of angles
Using the proportionality established earlier, we can write:
\[ \frac{I_2}{I_1} = \frac{\tan(\theta_2)}{\tan(\theta_1)} \]

### Step 5: Substitute the known values
Substituting \( I_2 \) and \( \theta_1 \) into the equation:
\[ \frac{\frac{I_1}{\sqrt{3}}}{I_1} = \frac{\tan(\theta_2)}{\tan(45^\circ)} \]
Since \( \tan(45^\circ) = 1 \), we have:
\[ \frac{1}{\sqrt{3}} = \tan(\theta_2) \]

### Step 6: Solve for the new angle of deflection
To find \( \theta_2 \), we take the inverse tangent:
\[ \theta_2 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Step 7: Calculate \( \theta_2 \)
The angle whose tangent is \( \frac{1}{\sqrt{3}} \) is:
\[ \theta_2 = 30^\circ \]

### Step 8: Determine the change in deflection
Now we find the change in deflection:
\[ \text{Change} = \theta_1 - \theta_2 = 45^\circ - 30^\circ = 15^\circ \]

### Conclusion
The deflection decreases by \( 15^\circ \).

### Final Answer
The deflection would decrease by 15 degrees.

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