A quantity f is given by `f=sqrt((hc^(5))/(G))` where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of:

Options

A. Volume
B. energy
C. area
D. momentum

Answer

To find the dimensions of the quantity \( f \) given by

\[ f = \sqrt{\frac{hc^5}{G}} \]

we will break down the dimensions of each component: Planck's constant \( h \), the speed of light \( c \), and the universal gravitational constant \( G \).

### Step 1: Determine the dimensions of \( h \) (Planck's constant)

The unit of Planck's constant \( h \) is Joule-seconds (J·s).

1 Joule (J) can be expressed in terms of basic dimensions as:
\[ 1 \text{ J} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^2 = \text{ML}^2\text{T}^{-2} \]

Thus, the dimensions of \( h \) are:
\[ [h] = \text{ML}^2\text{T}^{-1} \]

### Step 2: Determine the dimensions of \( c \) (speed of light)

The speed of light \( c \) has the unit of meters per second (m/s). Therefore, its dimensions are:
\[ [c] = \text{LT}^{-1} \]

### Step 3: Determine the dimensions of \( G \) (universal gravitational constant)

The unit of \( G \) is Newton meter squared per kilogram squared (N·m²/kg²).

1 Newton (N) can be expressed as:
\[ 1 \text{ N} = 1 \text{ kg} \cdot \text{m/s}^2 = \text{MLT}^{-2} \]

Thus, the dimensions of \( G \) are:
\[ [G] = \frac{\text{MLT}^{-2} \cdot \text{L}^2}{\text{M}^2} = \text{M}^{-1}\text{L}^3\text{T}^{-2} \]

### Step 4: Substitute the dimensions into the expression for \( f \)

Now we substitute the dimensions of \( h \), \( c \), and \( G \) into the expression for \( f \):

\[ f = \sqrt{\frac{hc^5}{G}} = \sqrt{\frac{(\text{ML}^2\text{T}^{-1})(\text{LT}^{-1})^5}{\text{M}^{-1}\text{L}^3\text{T}^{-2}}} \]

### Step 5: Simplify the expression

Calculating \( c^5 \):
\[ c^5 = (\text{LT}^{-1})^5 = \text{L}^5\text{T}^{-5} \]

Now substituting back into \( f \):
\[ f = \sqrt{\frac{(\text{ML}^2\text{T}^{-1})(\text{L}^5\text{T}^{-5})}{\text{M}^{-1}\text{L}^3\text{T}^{-2}}} \]

This simplifies to:
\[ f = \sqrt{\frac{\text{M} \cdot \text{L}^{7} \cdot \text{T}^{-6}}{\text{M}^{-1} \cdot \text{L}^3 \cdot \text{T}^{-2}}} \]

### Step 6: Further simplification

Now simplifying the fraction:
\[ = \sqrt{\text{M}^{1 - (-1)} \cdot \text{L}^{7 - 3} \cdot \text{T}^{-6 - (-2)}} = \sqrt{\text{M}^{2} \cdot \text{L}^{4} \cdot \text{T}^{-4}} \]

### Step 7: Final dimensions of \( f \)

Taking the square root:
\[ f = \text{M}^{1} \cdot \text{L}^{2} \cdot \text{T}^{-2} \]

Thus, the dimensions of \( f \) are:
\[ f \sim \text{ML}^2\text{T}^{-2} \]

### Conclusion

The dimensions of \( f \) are that of energy.