Solution:

Case 1: Using the galvanometer as a voltmeter

Given Data:

• Coil resistance of galvanometer:
  $R_g = 100 \, \Omega$ 

  ,

• Full-scale deflection current of the galvanometer:
  $I_g = 50 \, \mu A = 50 \times 10^{-6} \, \text{A}$ 

  ,

• Range of the voltmeter to be designed:
  $V = 50 \, \text{V}$ 

  .

Total Resistance Required ( $R_v$

):

The total resistance

$R_v$

of the voltmeter is determined using Ohm's law:

$$R_v = \frac{V}{I_g} = \frac{50}{50 \times 10^{-6}} = 10^6 \, \Omega.$$

Since the galvanometer already has a resistance

$R_g = 100 \, \Omega$

, the additional series resistance

$R_s$

required is:

$$R_s = R_v - R_g = 10^6 - 100 \approx 10^6 \, \Omega.$$

Conclusion:

• To use the galvanometer as a voltmeter of range 50 V, the series resistance required is
  $\mathbf{10^6 \, \Omega}$ 

  (Option A is correct).

Case 2: Using the galvanometer as an ammeter

Given Data:

• Full-scale deflection current of the galvanometer:
  $I_g = 50 \, \mu A = 50 \times 10^{-6} \, \text{A}$ 

  ,

• Resistance of the galvanometer:
  $R_g = 100 \, \Omega$ 

  ,

• Range of the ammeter to be designed:
  $I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A}$ 

  .

Shunt Resistance ( $R_s$

):

The shunt resistance is connected in parallel with the galvanometer to allow the additional current (

$I - I_g$

) to pass through it. The voltage across the galvanometer and the shunt must be equal:

$$V_g = V_s \quad \Rightarrow \quad I_g R_g = I_s R_s,$$

where

$I_s = I - I_g = 10 \times 10^{-3} - 50 \times 10^{-6} = 9.95 \times 10^{-3} \, \text{A}$

.

Using the above relation, the shunt resistance is:

$$R_s = \frac{I_g R_g}{I_s} = \frac{(50 \times 10^{-6}) \cdot 100}{9.95 \times 10^{-3}} = 0.5 \, \Omega.$$

Conclusion:

• To use the galvanometer as an ammeter of range 10 mA, the shunt resistance required is
  $\mathbf{0.5 \, \Omega}$ 

  (Option C is correct).

Final Answer:

The correct options are:

$$\boxed{\text{A and C}}$$