h d G has the dimensions of (h = height , d = density , G = Gravitational constant)

Options

A. Pressure
B. Power
C. Torque
D. Acceleration

Answer

To solve the problem of finding the equivalent dimensions of the product \( h \cdot d \cdot G \) where \( h \) is height, \( d \) is density, and \( G \) is the gravitational constant, we will follow these steps:

### Step 1: Identify the dimensions of height (h)
Height is a measure of length. The dimension of height is:
\[ [h] = L \]

### Step 2: Identify the dimensions of density (d)
Density is defined as mass per unit volume. The dimensions of density can be expressed as:
\[ [d] = \frac{M}{L^3} = M L^{-3} \]

### Step 3: Identify the dimensions of the gravitational constant (G)
The gravitational constant \( G \) has dimensions that can be derived from Newton's law of gravitation. The dimensions of \( G \) are:
\[ [G] = M^{-1} L^3 T^{-2} \]

### Step 4: Combine the dimensions of \( h \), \( d \), and \( G \)
Now we will multiply the dimensions together:
\[ [h \cdot d \cdot G] = [h] \cdot [d] \cdot [G] \]
Substituting the dimensions we found:
\[ [h \cdot d \cdot G] = (L) \cdot (M L^{-3}) \cdot (M^{-1} L^3 T^{-2}) \]

### Step 5: Simplify the expression
Now we can simplify the expression:
\[ = L \cdot M L^{-3} \cdot M^{-1} L^3 T^{-2} \]
Combining like terms:
\[ = (M^{1} M^{-1}) \cdot (L^{1 - 3 + 3}) \cdot (T^{-2}) = M^{0} L^{1} T^{-2} \]
This simplifies to:
\[ = L^{1} T^{-2} \]

### Step 6: Identify the physical quantity
The dimensions \( L^{1} T^{-2} \) correspond to acceleration, which is typically expressed in units of meters per second squared (m/s²).

### Conclusion
Thus, the product \( h \cdot d \cdot G \) has the dimensions of acceleration.

### Final Answer
The correct option is **acceleration**.

---