Surface tension has the same dimensions as that of

Options

A. coefficient of viscosity
B. impulse
C. momentum
D. spring constant

Answer

To solve the problem of determining which quantity has the same dimensions as surface tension, we will follow these steps:

### Step 1: Determine the dimensions of surface tension.
- **Formula for surface tension (σ)**: Surface tension is defined as force per unit length.
- **Dimensions of force (F)**: The dimension of force is given by the formula \( F = m \cdot a \), where \( a \) (acceleration) has dimensions \( L T^{-2} \). Thus, the dimension of force is:
\[ [F] = M L T^{-2} \]
- **Dimensions of length (L)**: The dimension of length is simply:
\[ [L] = L \]
- **Calculating dimensions of surface tension**:
\[ [\sigma] = \frac{[F]}{[L]} = \frac{M L T^{-2}}{L} = M L^{0} T^{-2} \]
Thus, the dimensions of surface tension are \( M L^{0} T^{-2} \).

### Step 2: Check the dimensions of the coefficient of viscosity.
- **Formula for coefficient of viscosity (η)**: The coefficient of viscosity is defined as:
\[ \eta = \frac{F}{A \cdot \frac{dv}{dz}} \]
where \( A \) is area and \( \frac{dv}{dz} \) is the velocity gradient.
- **Dimensions of area (A)**: The dimension of area is:
\[ [A] = L^{2} \]
- **Dimensions of velocity gradient (\( \frac{dv}{dz} \))**: The dimension of velocity is \( L T^{-1} \) and since \( dz \) is length:
\[ \frac{dv}{dz} \text{ has dimensions } = \frac{L T^{-1}}{L} = T^{-1} \]
- **Calculating dimensions of viscosity**:
\[ [\eta] = \frac{[F]}{[A] \cdot [\frac{dv}{dz}]} = \frac{M L T^{-2}}{L^{2} \cdot T^{-1}} = M L^{-1} T^{-1} \]
This is not the same as the dimensions of surface tension.

### Step 3: Check the dimensions of impulse.
- **Impulse (J)**: Impulse is defined as the change in momentum.
- **Dimensions of momentum (p)**: The dimension of momentum is:
\[ [p] = [m] \cdot [v] = M \cdot (L T^{-1}) = M L T^{-1} \]
Thus, the dimensions of impulse are also \( M L T^{-1} \), which do not match the dimensions of surface tension.

### Step 4: Check the dimensions of momentum.
- As established in the previous step, the dimensions of momentum are \( M L T^{-1} \), which again do not match the dimensions of surface tension.

### Step 5: Check the dimensions of the spring constant.
- **Formula for spring constant (k)**: The spring constant is defined as:
\[ k = \frac{F}{x} \]
where \( x \) is the elongation (length).
- **Calculating dimensions of spring constant**:
\[ [k] = \frac{[F]}{[x]} = \frac{M L T^{-2}}{L} = M L^{0} T^{-2} \]
This matches the dimensions of surface tension.

### Conclusion:
The quantity that has the same dimensions as surface tension is the spring constant.

### Final Answer:
**The correct option is D: Spring constant.**