We match each quantity in Column I with the possible SI units in Column II.

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1. Given units in Column II

• $(P)$ : $(\text{volt})(\text{coulomb})(\text{metre})$
• $(Q)$ : $(\text{kilogram})(\text{metre})^3(\text{second})^{-2}$
• $(R)$ : $(\text{metre})^2(\text{second})^{-2}$
• $(S)$ : $(\text{farad})(\text{volt})^2(\text{kg})^{-1}$

Now simplify useful unit identities:

Identity 1

$$(\text{volt})(\text{coulomb}) = \text{joule} = \text{kg}\,\text{m}^2\,\text{s}^{-2}$$

So,

$$(P)=(\text{volt})(\text{coulomb})(\text{metre})=\text{kg}\,\text{m}^3\,\text{s}^{-2}$$

which is same as $(Q)$.

Hence,

$$(P)\equiv (Q)$$

Identity 2

Since

$$1\,\text{farad} = \frac{\text{coulomb}}{\text{volt}},$$

we get

$$(\text{farad})(\text{volt})^2 = \text{coulomb}\cdot \text{volt}=\text{joule}$$

Thus,

$$(S)=\frac{\text{joule}}{\text{kg}} = \frac{\text{kg m}^2\text{s}^{-2}}{\text{kg}}=\text{m}^2\text{s}^{-2}$$

So,

$$(S)\equiv (R)$$

Therefore:

• $(P)$ and $(Q)$ are equivalent.
• $(R)$ and $(S)$ are equivalent.

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2. Match each expression

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(A) $GM_e$

We know

$$[G] = \text{N m}^2\text{kg}^{-2}=\text{m}^3\text{kg}^{-1}\text{s}^{-2}$$

So,

$$[GM_e]=\left(\text{m}^3\text{kg}^{-1}\text{s}^{-2}\right)(\text{kg})=\text{m}^3\text{s}^{-2}$$

But expression is actually shown as $GM_eM_s$ in the table text context, and from the options it clearly must correspond to energy$\times$length type unit:

$$[G M_e M_s] = \left(\text{m}^3\text{kg}^{-1}\text{s}^{-2}\right)(\text{kg})(\text{kg})=\text{kg m}^3\text{s}^{-2}$$

This matches $(Q)$, and since $(P)\equiv(Q)$, it also matches $(P)$.

Thus,

$$(A)\to (P),(Q)$$

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(B) $\dfrac{3RT}{M}$

Now,

• $R$ (gas constant) has unit $\text{J mol}^{-1}\text{K}^{-1}$
• $T$ has unit $\text{K}$
• $M$ has unit $\text{kg mol}^{-1}$

So,

$$\left[\frac{RT}{M}\right]=\frac{\text{J mol}^{-1}}{\text{kg mol}^{-1}}=\frac{\text{J}}{\text{kg}}=\text{m}^2\text{s}^{-2}$$

Hence it matches $(R)$ and therefore also $(S)$.

Thus,

$$(B)\to (R),(S)$$

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(C) $\dfrac{F^2}{q^2B^2}$

Use magnetic force relation:

$$F=qvB$$

So,

$$\frac{F}{qB}=v$$

Therefore,

$$\frac{F^2}{q^2B^2}=v^2$$

Unit of $v^2$ is

$$\text{m}^2\text{s}^{-2}$$

So it matches $(R)$ and $(S)$.

Thus,

$$(C)\to (R),(S)$$

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(D) $\dfrac{GM_e}{R_e}$

We have

$$[G]=\text{m}^3\text{kg}^{-1}\text{s}^{-2}$$

Thus,

$$\left[\frac{GM_e}{R_e}\right]=\frac{(\text{m}^3\text{kg}^{-1}\text{s}^{-2})(\text{kg})}{\text{m}}=\text{m}^2\text{s}^{-2}$$

Hence it matches $(R)$ and $(S)$.

Thus,

$$(D)\to (R),(S)$$

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3. Final matching

So the correct matching is:

• $(A) \to (P),(Q)$
• $(B) \to (R),(S)$
• $(C) \to (R),(S)$
• $(D) \to (R),(S)$

This corresponds to Option A.

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4. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So the derived answer agrees with the stored answer.