Dimensional formula of the product of the two physical quantities P and Q is `ML^(2)T^(-2)` , the dimesional formula of `P//Q` is `MT^(-2)` . P and Q respectively are

Options

A. Force, velocity
B. Momemtum , displacement
C. Force ,displacement
D. Work, velocity

Answer

To solve the problem, we need to find the physical quantities \( P \) and \( Q \) based on their dimensional formulas. Let's break it down step by step.

### Step 1: Understand the given information
We are given:
1. The dimensional formula of the product \( P \cdot Q \) is \( [P \cdot Q] = ML^2T^{-2} \).
2. The dimensional formula of the ratio \( \frac{P}{Q} \) is \( \left[\frac{P}{Q}\right] = MT^{-2} \).

### Step 2: Write the dimensional formulas
Let’s denote the dimensional formulas of \( P \) and \( Q \) as:
- \( [P] = M^{a}L^{b}T^{c} \)
- \( [Q] = M^{d}L^{e}T^{f} \)

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

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

### Step 4: Solve the equations
Now we have a system of equations:
1. \( a + d = 1 \)
2. \( b + e = 2 \)
3. \( c + f = -2 \)
4. \( a - d = 1 \)
5. \( b - e = 0 \)
6. \( c - f = -2 \)

From equations 4 and 1:
- Adding gives \( 2a = 2 \) → \( a = 1 \)
- Substituting \( a = 1 \) into equation 1 gives \( 1 + d = 1 \) → \( d = 0 \).

From equations 5 and 2:
- Adding gives \( b + e = 2 \) and \( b - e = 0 \) → \( 2b = 2 \) → \( b = 1 \).
- Substituting \( b = 1 \) into equation 2 gives \( 1 + e = 2 \) → \( e = 1 \).

From equations 6 and 3:
- Adding gives \( c + f = -2 \) and \( c - f = -2 \) → \( 2c = -4 \) → \( c = -2 \).
- Substituting \( c = -2 \) into equation 3 gives \( -2 + f = -2 \) → \( f = 0 \).

### Step 5: Write the dimensional formulas for \( P \) and \( Q \)
Now we have:
- \( P: M^1L^1T^{-2} \) (which corresponds to force)
- \( Q: M^0L^1T^0 \) (which corresponds to length or displacement)

### Conclusion
Thus, we conclude that:
- \( P \) is force (dimension \( MLT^{-2} \))
- \( Q \) is displacement (dimension \( L \))