A beaker contains a fluid of density `rhokg//m^(3)`, specific heat `SJ//kg^(@)C` and viscosity `eta`. The beaker is filled up to height h. To estimate the rate of heat transfer per unit area (Q/A) by convection when beaker is put on a hot plate, a student proposes that it should depend on `eta,((SDeltatheta)/(h))` and `((1)/(rhog))` when `Deltatheta("in"^(@)C)` is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for (Q/A) is

Options

A. `eta(SDeltatheta)/(h)`
B. `eta((SDeltatheta)/(h))((1)/(rhog))`
C. `(SDeltatheta)/(etah)`
D. `((SDeltatheta)/(etah))((1)/(rhog))`

Explanation

To solve the problem of estimating the rate of heat transfer per unit area (Q/A) by convection in a fluid contained in a beaker, we will use dimensional analysis based on the parameters provided: viscosity (η), specific heat (S), temperature difference (Δθ), height (h), density (ρ), and gravitational acceleration (g).

### Step-by-Step Solution:

1. **Identify the Variables and Their Dimensions**:
- Viscosity (η): The dimension of viscosity is \( [η] = M L^{-1} T^{-1} \).
- Specific heat (S): The dimension of specific heat is \( [S] = L^2 T^{-2} \) (since it is in Joules per kg°C).
- Temperature difference (Δθ): The dimension of temperature is \( [Δθ] = Θ \) (where Θ is the dimension of temperature).
- Height (h): The dimension of height is \( [h] = L \).
- Density (ρ): The dimension of density is \( [ρ] = M L^{-3} \).
- Gravitational acceleration (g): The dimension of acceleration is \( [g] = L T^{-2} \).

2. **Formulate the Expression for Q/A**:
We propose that \( Q/A \) depends on the variables as follows:
\[ \frac{Q}{A} \propto η^A \left(\frac{S Δθ}{h}\right)^B \left(\frac{1}{ρg}\right)^C \]

3. **Write the Dimensions of Q/A**:
The dimension of heat transfer per unit area \( Q/A \) is:
\[ \left[\frac{Q}{A}\right] = M T^{-3} \]

4. **Substituting Dimensions into the Expression**:
Substitute the dimensions into the expression:
\[ M T^{-3} = (M L^{-1} T^{-1})^A \left(L^2 T^{-2}\right)^B \left(M^{-1} L^2 T^2\right)^C \]

5. **Expand and Combine Dimensions**:
Expanding gives:
\[ M T^{-3} = M^A L^{-A} T^{-A} \cdot L^{2B} T^{-2B} \cdot M^{-C} L^{2C} T^{2C} \]
Combining the dimensions:
\[ M^{A-C} L^{-A + 2B + 2C} T^{-A - 2B + 2C} \]

6. **Set Up the Equations**:
From the dimensions, we can set up the following equations:
- For mass (M): \( A - C = 1 \)
- For length (L): \( -A + 2B + 2C = 0 \)
- For time (T): \( -A - 2B + 2C = -3 \)

7. **Solve the System of Equations**:
From \( A - C = 1 \), we can express \( C = A - 1 \).
Substitute \( C \) into the other equations:
- \( -A + 2B + 2(A - 1) = 0 \) simplifies to \( 2B + A - 2 = 0 \) → \( A + 2B = 2 \).
- \( -A - 2B + 2(A - 1) = -3 \) simplifies to \( A - 2B = -1 \).

Now we have:
1. \( A + 2B = 2 \)
2. \( A - 2B = -1 \)

Adding these two equations:
\[ 2A = 1 \implies A = \frac{1}{2} \]
Substituting \( A \) back into \( A + 2B = 2 \):
\[ \frac{1}{2} + 2B = 2 \implies 2B = \frac{3}{2} \implies B = \frac{3}{4} \]
And substituting \( A \) into \( C = A - 1 \):
\[ C = \frac{1}{2} - 1 = -\frac{1}{2} \]

8. **Final Expression for Q/A**:
Now substituting back into the original expression:
\[ \frac{Q}{A} = η^{\frac{1}{2}} \left(\frac{S Δθ}{h}\right)^{\frac{3}{4}} \left(\frac{1}{ρg}\right)^{-\frac{1}{2}} \]

### Conclusion:
The correct option for \( \frac{Q}{A} \) is:
\[ \frac{Q}{A} \propto η^{\frac{1}{2}} \left(\frac{S Δθ}{h}\right)^{\frac{3}{4}} \left(ρg\right)^{\frac{1}{2}} \]