The dimensionla formula for the product of two physical quantities `P` and `Q` is `[ML^(2) T^(-2)]`. The dimensional formula of `(P)/(Q)` is `[MT^(-2)]`. Then `P` and `Q` respectively are

Options

A. Force and velocity
B. Momentum and displacement
C. Force and displacement
D. Work and velocity

Explanation

To find the dimensional formulas for the physical quantities \( P \) and \( Q \), we start with the information given in the problem:

1. The dimensional formula for the product of \( P \) and \( Q \) is:
\[ [P \cdot Q] = [ML^{2} T^{-2}] \]

2. The dimensional formula for the ratio \( \frac{P}{Q} \) is:
\[ \left[\frac{P}{Q}\right] = [MT^{-2}] \]

### Step 1: Set up the equations
From the information provided, we can express the dimensional formulas of \( P \) and \( Q \) as follows:
- Let the dimensional formula of \( P \) be \( [M^{a} L^{b} T^{c}] \).
- Let the dimensional formula of \( Q \) be \( [M^{d} L^{e} T^{f}] \).

### Step 2: Write the equations based on the product and ratio
From the product \( P \cdot Q \):
\[ [M^{a} L^{b} T^{c}] \cdot [M^{d} L^{e} T^{f}] = [M^{a+d} L^{b+e} T^{c+f}] = [ML^{2} T^{-2}] \]
This gives us the following equations:
1. \( a + d = 1 \) (1)
2. \( b + e = 2 \) (2)
3. \( c + f = -2 \) (3)

From the ratio \( \frac{P}{Q} \):
\[ \frac{[M^{a} L^{b} T^{c}]}{[M^{d} L^{e} T^{f}]} = [M^{a-d} L^{b-e} T^{c-f}] = [MT^{-2}] \]
This gives us the following equations:
4. \( a - d = 1 \) (4)
5. \( b - e = 0 \) (5)
6. \( c - f = -2 \) (6)

### Step 3: Solve the equations
Now we can solve these equations step by step.

**From equations (1) and (4):**
1. From (1): \( a + d = 1 \)
2. From (4): \( a - d = 1 \)

Adding these two equations:
\[ (a + d) + (a - d) = 1 + 1 \implies 2a = 2 \implies a = 1 \]
Substituting \( a = 1 \) into equation (1):
\[ 1 + d = 1 \implies d = 0 \]

**From equations (2) and (5):**
1. From (2): \( b + e = 2 \)
2. From (5): \( b - e = 0 \)

Adding these two equations:
\[ (b + e) + (b - e) = 2 + 0 \implies 2b = 2 \implies b = 1 \]
Substituting \( b = 1 \) into equation (2):
\[ 1 + e = 2 \implies e = 1 \]

**From equations (3) and (6):**
1. From (3): \( c + f = -2 \)
2. From (6): \( c - f = -2 \)

Adding these two equations:
\[ (c + f) + (c - f) = -2 - 2 \implies 2c = -4 \implies c = -2 \]
Substituting \( c = -2 \) into equation (3):
\[ -2 + f = -2 \implies f = 0 \]

### Step 4: Write the final dimensional formulas
Now we have:
- \( P: [M^{1} L^{1} T^{-2}] \)
- \( Q: [M^{0} L^{1} T^{0}] \)

Thus, the dimensional formulas for \( P \) and \( Q \) are:
\[ P = [ML^{1} T^{-2}], \quad Q = [L^{1}] \]

### Final Answer:
- \( P \) has the dimensional formula \( [ML^{1} T^{-2}] \)
- \( Q \) has the dimensional formula \( [L^{1}] \)