A moving coil galvanometer experience torque `=ki`, where i is current. If N coils of area A each and moment of inertia I is kept in magnetic field B.
(i) If for current `i` deflection is `(pi)/2`, the torsional constant of spring is `(x BiNA)/(pi)`. Find `x^2`
(ii) If a charge Q is passed suddenly through the galvanometer, the maximum angle of deflection is `Qsqrt((BNpiA)/(xI))`. Find `x^2`

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (i)

1. **Understanding Torque in a Galvanometer**:
The torque \( \tau \) experienced by the coil in a magnetic field is given by:
\[ \tau = k \cdot i \]
where \( k \) is a constant and \( i \) is the current.

2. **Torque from Magnetic Field**:
The torque can also be expressed as:
\[ \tau = N \cdot A \cdot B \cdot i \]
where \( N \) is the number of coils, \( A \) is the area of the coils, and \( B \) is the magnetic field strength.

3. **Equating the Two Expressions for Torque**:
From the above two equations, we can equate:
\[ k \cdot i = N \cdot A \cdot B \cdot i \]
Dividing both sides by \( i \) (assuming \( i \neq 0 \)):
\[ k = N \cdot A \cdot B \]
This is our **Equation 1**.

4. **Finding Torsional Constant**:
The torsional constant \( C \) is defined as:
\[ C = \frac{\tau}{\theta} \]
where \( \theta \) is the angular deflection. Given that for current \( i \), the deflection \( \theta = \frac{\pi}{2} \):
\[ C = \frac{N \cdot A \cdot B \cdot i}{\frac{\pi}{2}} = \frac{2N \cdot A \cdot B \cdot i}{\pi} \]

5. **Setting Equal to Given Torsional Constant**:
We are given that:
\[ C = \frac{x \cdot B \cdot i \cdot N \cdot A}{\pi} \]
Setting the two expressions for \( C \) equal:
\[ \frac{2N \cdot A \cdot B \cdot i}{\pi} = \frac{x \cdot B \cdot i \cdot N \cdot A}{\pi} \]
Canceling \( B \), \( i \), \( N \), and \( A \) (assuming they are non-zero):
\[ 2 = x \]

6. **Finding \( x^2 \)**:
Thus, we find:
\[ x^2 = 2^2 = 4 \]

### Part (ii)

1. **Understanding Charge Passage**:
When a charge \( Q \) is passed through the galvanometer, it causes a change in angular momentum. The change in angular momentum \( \Delta L \) is given by:
\[ \Delta L = \tau \cdot dt \]

2. **Expressing Angular Momentum**:
The angular momentum can also be expressed as:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Setting Up the Equation**:
From the torque we calculated earlier:
\[ \tau = N \cdot A \cdot B \cdot i \]
Thus:
\[ I \cdot \omega = N \cdot A \cdot B \cdot i \cdot dt \]

4. **Integrating**:
The integral of \( i \cdot dt \) gives the charge \( Q \):
\[ I \cdot \omega = N \cdot A \cdot B \cdot Q \]
Therefore:
\[ \omega = \frac{N \cdot A \cdot B \cdot Q}{I} \]

5. **Kinetic Energy and Potential Energy**:
The loss in kinetic energy is:
\[ KE = \frac{1}{2} I \omega^2 \]
The gain in potential energy is:
\[ PE = \frac{1}{2} C \theta_{max}^2 \]
Setting these equal gives:
\[ \frac{1}{2} I \left(\frac{N \cdot A \cdot B \cdot Q}{I}\right)^2 = \frac{1}{2} C \theta_{max}^2 \]

6. **Substituting for C**:
Substitute \( C = \frac{2NAB}{\pi} \):
\[ \frac{1}{2} I \left(\frac{N \cdot A \cdot B \cdot Q}{I}\right)^2 = \frac{1}{2} \cdot \frac{2NAB}{\pi} \cdot \theta_{max}^2 \]

7. **Solving for \( \theta_{max} \)**:
Rearranging gives:
\[ \theta_{max}^2 = \frac{N^2 A^2 B^2 Q^2}{\pi I} \]
Setting this equal to the given expression:
\[ \theta_{max} = Q \sqrt{\frac{BN \pi A}{xI}} \]

8. **Comparing Expressions**:
By comparing the two expressions for \( \theta_{max} \):
\[ \frac{N^2 A^2 B^2 Q^2}{\pi I} = Q^2 \frac{BN \pi A}{xI} \]
Cancelling \( Q^2 \), \( B \), \( N \), \( A \), and \( I \) gives:
\[ x = 2 \]

9. **Finding \( x^2 \)**:
Thus, we find:
\[ x^2 = 4 \]

### Final Answers
(i) \( x^2 = 4 \)
(ii) \( x^2 = 4 \)