Answer:

Given, full scale value of the moving iron voltmeter = 100 V

Current drawn at full scale deflection = 1 mA

Current drawn at the meter reading of 50 V = 0.5 mA

We can use the principle of the moving iron voltmeter to find the meter reading at 0.5 mA.

Principle of Moving Iron Voltmeter:

When current flows through a moving coil, it produces a magnetic field. When this magnetic field interacts with the magnetic field of a fixed permanent magnet, a torque is produced, which tends to deflect the moving coil. In a moving iron voltmeter, the moving coil is replaced by a soft iron vane or piece. When current flows through the coil, the magnetic field produced causes the iron piece to move, and this movement is proportional to the current flowing through the coil.

Therefore, the deflection of the iron piece is proportional to the current flowing through the coil, and hence, the voltage being measured.

Calculation:

Let's use the principle of the moving iron voltmeter to find the meter reading at 0.5 mA.

The current drawn by the voltmeter is proportional to the deflection of the iron piece. Therefore, we can write:

Current drawn by the voltmeter ∝ Deflection of the iron piece

Let's assume that the full-scale deflection of the iron piece occurs when a current of I_F is flowing through the voltmeter.

We know that the full scale value of the voltmeter is 100 V when a current of 1 mA is flowing through it.

Therefore, we can write:

I_F ∝ 100 V

I_F = K × 100 V (where K is a constant of proportionality)

Let's assume that the deflection of the iron piece is proportional to the square of the current flowing through it.

Therefore, we can write:

Deflection of the iron piece ∝ (Current drawn by the voltmeter)^2

Let's substitute the values of current drawn by the voltmeter in the above equation:

Deflection of the iron piece ∝ (0.5 mA)^2

Deflection of the iron piece ∝ 0.25

Let's assume that the deflection of the iron piece is maximum when the full-scale value is measured.

Therefore, we can write:

Deflection of the iron piece ∝ 100 V

Now, let's equate the two expressions of the deflection of the iron piece:

0.25 ∝ 100 V

0.25 = K × 100 V

K = 0.0025

Therefore, the deflection of the iron piece is proportional to the current flowing through the voltmeter, and hence, the voltage being measured.

Let's find the meter reading at 0.5 mA:

Deflection of the iron piece at 0.5 mA = K × (0.5 mA)^2

Deflection of the iron piece at 0.5 mA = 0.0025 × 0.25

Deflection of the iron piece at 0.5 mA = 0.000625

Let's find the voltage corresponding to this deflection:

Deflection of the iron piece ∝ Voltage being measured

0.000625 ∝ Voltage being measured

Voltage being measured = (0.000625 / K)

Voltage being measured = (0.000625 /