If Surface tension (S) ,Moments of Inertia (I) and Plank's Constant(h), Where to be taken as the fundamental units, the dimentional formula for linear momentum would be:

Options

A. `S^((1)//2)I^((3)//2)h^(-1)`
B. `S^((1)//2)I^((1)//2)h^(-1)`
C. `S^((1)//2)I^((1)//2)h^(0)`
D. `S^((3)//2)I^((1)//2)h^(0)`

Answer

To find the dimensional formula for linear momentum (P) in terms of surface tension (S), moment of inertia (I), and Planck's constant (h), we will follow these steps:

### Step 1: Write the dimensional formulas for S, I, and h.
- Surface tension (S) has the dimensional formula:
\[ [S] = \frac{M}{T^2} \quad \text{(Force per unit length)} \]
- Moment of inertia (I) has the dimensional formula:
\[ [I] = ML^2 \]
- Planck's constant (h) has the dimensional formula:
\[ [h] = ML^2T^{-1} \]

### Step 2: Assume the dimensional formula for linear momentum (P).
The dimensional formula for linear momentum is given by:
\[ [P] = ML^1T^{-1} \]

### Step 3: Express P in terms of S, I, and h.
Assume that the dimensional formula for linear momentum can be expressed as:
\[ [P] = S^A I^B h^C \]
where A, B, and C are the powers we need to determine.

### Step 4: Substitute the dimensional formulas into the equation.
Substituting the dimensional formulas we have:
\[ ML^1T^{-1} = \left(\frac{M}{T^2}\right)^A (ML^2)^B (ML^2T^{-1})^C \]

### Step 5: Expand the right-hand side.
Expanding the right-hand side gives:
\[ ML^1T^{-1} = M^A T^{-2A} M^B L^{2B} M^C L^{2C} T^{-C} \]
Combining the terms, we have:
\[ ML^1T^{-1} = M^{A+B+C} L^{2B+2C} T^{-2A-C} \]

### Step 6: Equate the powers of M, L, and T.
Now we can equate the coefficients of M, L, and T from both sides:
1. For M:
\[ A + B + C = 1 \quad \text{(1)} \]
2. For L:
\[ 2B + 2C = 1 \quad \text{(2)} \]
3. For T:
\[ -2A - C = -1 \quad \text{(3)} \]

### Step 7: Solve the equations.
From equation (2):
\[ B + C = \frac{1}{2} \quad \text{(4)} \]
Substituting equation (4) into equation (1):
\[ A + \frac{1}{2} = 1 \implies A = \frac{1}{2} \]
Now substituting \(A = \frac{1}{2}\) into equation (3):
\[ -2\left(\frac{1}{2}\right) - C = -1 \implies -1 - C = -1 \implies C = 0 \]
Now substituting \(C = 0\) into equation (4):
\[ B + 0 = \frac{1}{2} \implies B = \frac{1}{2} \]

### Step 8: Write the final expression for P.
Now we have:
- \(A = \frac{1}{2}\)
- \(B = \frac{1}{2}\)
- \(C = 0\)

Thus, the dimensional formula for linear momentum in terms of S, I, and h is:
\[ P = S^{\frac{1}{2}} I^{\frac{1}{2}} h^{0} = S^{\frac{1}{2}} I^{\frac{1}{2}} \]

### Final Answer:
The dimensional formula for linear momentum (P) in terms of surface tension (S), moment of inertia (I), and Planck's constant (h) is:
\[ P = S^{\frac{1}{2}} I^{\frac{1}{2}} \]