If young's modulus y, surface tension s and time T are the fundamental quantities then the dimensional formula of density is

Options

A. `s^(2)y^(3)t^(-2)`
B. `s^(3)y^(3)T^(-2)`
C. `s^(-2)y^(3)T^(2)`
D. `s^(-2) y^(2)T^(3)`

Answer

To find the dimensional formula of density using Young's modulus (Y), surface tension (S), and time (T) as fundamental quantities, we can follow these steps:

### Step 1: Understand the relationship
Density (D) is defined as mass (m) per unit volume (V). The dimensional formula of density is given by:
\[ D \propto \frac{m}{V} \]
Since volume (V) is \( L^3 \), the dimensional formula of density can be expressed as:
\[ [D] = [M][L]^{-3} \]

### Step 2: Write the dimensions of the given quantities
1. **Young's Modulus (Y)**:
- Young's modulus is defined as stress (force per unit area) divided by strain (dimensionless).
- The dimensional formula of Young's modulus is:
\[ [Y] = \frac{[F]}{[A]} = \frac{[M][L][T]^{-2}}{[L]^2} = [M][L]^{-1}[T]^{-2} \]

2. **Surface Tension (S)**:
- Surface tension is defined as force per unit length.
- The dimensional formula of surface tension is:
\[ [S] = \frac{[F]}{[L]} = \frac{[M][L][T]^{-2}}{[L]} = [M][T]^{-2} \]

3. **Time (T)**:
- The dimensional formula of time is:
\[ [T] = [T] \]

### Step 3: Express density in terms of Y, S, and T
Assume the dimensional formula of density can be expressed in terms of Y, S, and T as:
\[ [D] = [Y]^a [S]^b [T]^c \]

### Step 4: Substitute the known dimensions
Substituting the dimensional formulas we have:
\[ [M][L]^{-3} = ([M][L]^{-1}[T]^{-2})^a \cdot ([M][T]^{-2})^b \cdot ([T])^c \]

### Step 5: Expand the equation
This expands to:
\[ [M][L]^{-3} = [M]^a [L]^{-a} [T]^{-2a} \cdot [M]^b [T]^{-2b} \cdot [T]^c \]
Combining the terms gives:
\[ [M]^{a+b} [L]^{-a} [T]^{-2a - 2b + c} \]

### Step 6: Equate the dimensions
Now, we can equate the coefficients from both sides:
1. For mass (M):
\[ a + b = 1 \] (Equation 1)

2. For length (L):
\[ -a = -3 \]
\[ a = 3 \] (Equation 2)

3. For time (T):
\[ -2a - 2b + c = 0 \] (Equation 3)

### Step 7: Solve the equations
From Equation 2, we have \( a = 3 \).
Substituting \( a = 3 \) into Equation 1:
\[ 3 + b = 1 \]
\[ b = 1 - 3 = -2 \]

Now substituting \( a = 3 \) and \( b = -2 \) into Equation 3:
\[ -2(3) - 2(-2) + c = 0 \]
\[ -6 + 4 + c = 0 \]
\[ c = 2 \]

### Step 8: Write the final expression
Now substituting the values of \( a \), \( b \), and \( c \) back into the expression for density:
\[ [D] = [Y]^3 [S]^{-2} [T]^2 \]

### Conclusion
Thus, the dimensional formula of density in terms of Young's modulus, surface tension, and time is:
\[ [D] = S^{-2} Y^{3} T^{2} \]

### Final Answer
The correct option is:
\[ S^{-2} Y^{3} T^{2} \]

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