If plank's constent `(h)`, surface tension `(s)` and moment of inertia `(I)` are taken as fundamental quantities, then dimensions of linear momentum `(p)` will be given as

Options

A. `h s^(2) I^(1//2`
B. `h^(1//2) s^(1//2)I`
C. `h^(0)s^(1//2)I^(1//2)`
D. none of these

Answer

To find the dimensions of linear momentum \( p \) in terms of Planck's constant \( h \), surface tension \( s \), and moment of inertia \( I \), we will follow these steps:

### Step-by-Step Solution

1. **Identify the Dimensions of the Given Quantities**:
- Planck's constant \( h \): The dimension is given by \( [h] = M L^2 T^{-1} \).
- Surface tension \( s \): The dimension is given by \( [s] = M T^{-2} \).
- Moment of inertia \( I \): The dimension is given by \( [I] = M L^2 \).

2. **Express Linear Momentum in Terms of \( h \), \( s \), and \( I \)**:
We assume that the linear momentum \( p \) can be expressed as:
\[ p = h^x \cdot s^y \cdot I^z \]
where \( x \), \( y \), and \( z \) are the powers we need to determine.

3. **Write the Dimensions of Linear Momentum**:
The dimension of linear momentum \( p \) is:
\[ [p] = M L T^{-1} \]

4. **Substitute the Dimensions into the Equation**:
Substituting the dimensions of \( h \), \( s \), and \( I \) into our expression gives:
\[ [p] = (M L^2 T^{-1})^x \cdot (M T^{-2})^y \cdot (M L^2)^z \]

5. **Combine the Dimensions**:
This expands to:
\[ [p] = M^{x+y+z} \cdot L^{2x + 2z} \cdot T^{-x - 2y} \]

6. **Set Up the Equations by Comparing Dimensions**:
Now we can set up equations by comparing the coefficients of \( M \), \( L \), and \( T \):
- For \( M \): \( 1 = x + y + z \) (Equation 1)
- For \( L \): \( 1 = 2x + 2z \) (Equation 2)
- For \( T \): \( -1 = -x - 2y \) (Equation 3)

7. **Solve the Equations**:
- From Equation 2: \( 1 = 2x + 2z \) implies \( x + z = \frac{1}{2} \) (Equation 4).
- Substitute Equation 4 into Equation 1:
\[ 1 = \frac{1}{2} + y \implies y = \frac{1}{2} \]
- Substitute \( y = \frac{1}{2} \) into Equation 3:
\[ -1 = -x - 2 \cdot \frac{1}{2} \implies -1 = -x - 1 \implies x = 0 \]
- Substitute \( x = 0 \) into Equation 4:
\[ 0 + z = \frac{1}{2} \implies z = \frac{1}{2} \]

8. **Final Expression for Linear Momentum**:
Now we have:
- \( x = 0 \)
- \( y = \frac{1}{2} \)
- \( z = \frac{1}{2} \)

Therefore, we can express \( p \) as:
\[ p = h^0 \cdot s^{\frac{1}{2}} \cdot I^{\frac{1}{2}} = s^{\frac{1}{2}} \cdot I^{\frac{1}{2}} \]

### Final Result:
The dimensions of linear momentum \( p \) in terms of \( h \), \( s \), and \( I \) is:
\[ p = s^{\frac{1}{2}} I^{\frac{1}{2}} \]