Column-1 gives three physical quantities. Select the appropriate units for the choices given in column-II. Some of the physical quantities may have more than one choice.
`{:("Column-I",,"Column-II"),("Capacitance",,"ohm-second"),("Inductance",,"Coulomb"^(2)//"Joule"),("Magnetic induction",,"coulomb"//"volt"),(,,"newton"//"ampere metre"),(,,"volt-second"//"ampere"):}`

Answer

To solve the problem of matching physical quantities with their appropriate units, we will analyze each quantity in Column-I and find the corresponding units in Column-II.

### Step-by-Step Solution:

1. **Capacitance**:
- The formula for capacitance (C) is given by:
\[ C = \frac{Q}{V} \]
where \(Q\) is charge (in coulombs) and \(V\) is voltage (in volts).
- Voltage can also be expressed as:
\[ V = \frac{W}{Q} \]
where \(W\) is work done (in joules).
- Therefore, substituting for voltage, we have:
\[ C = \frac{Q}{\frac{W}{Q}} = \frac{Q^2}{W} \]
- The units of capacitance can be expressed as:
- \(C = \frac{\text{Coulomb}}{\text{Volt}}\) (Option B)
- \(C = \frac{\text{Coulomb}^2}{\text{Joule}}\) (Option C)
- Thus, capacitance corresponds to options B and C.

2. **Inductance**:
- The formula for inductance (L) can be derived from the relationship of electromotive force (emf):
\[ \text{emf} = -L \frac{di}{dt} \]
- Rearranging gives:
\[ L = -\frac{\text{emf}}{\frac{di}{dt}} \]
- Here, emf is measured in volts, \(di\) in amperes, and \(dt\) in seconds. Thus:
\[ L = \frac{\text{Volt} \cdot \text{Second}}{\text{Ampere}} = \text{Volt-Second/Ampere} \]
- Since \( \text{Volt/Ampere} = \text{Ohm} \), we can also express:
\[ L = \text{Ohm} \cdot \text{Second} \quad \text{(Option A)} \]
- Therefore, inductance corresponds to options A and A.

3. **Magnetic Induction**:
- The magnetic induction (B) is defined in terms of force per unit charge:
\[ B = \frac{F}{q} \]
- In SI units, this can be expressed as:
\[ B = \frac{\text{Newton}}{\text{Coulomb}} \quad \text{(Option D)} \]
- Additionally, we can express it in terms of other units:
\[ B = \frac{\text{Newton}}{\text{Coulomb}} \cdot \frac{1}{\text{Second}} = \frac{\text{Newton}}{\text{Coulomb} \cdot \text{Second}} = \frac{\text{Newton}}{\text{Ampere} \cdot \text{Meter}} \quad \text{(Option D)} \]
- Therefore, magnetic induction corresponds to option D.

### Final Matching:
- **Capacitance**: B, C
- **Inductance**: A, A
- **Magnetic Induction**: D

### Summary of Answers:
- Capacitance: B, C
- Inductance: A
- Magnetic Induction: D