If G be the gravitational constant and u be the energy density then which of the following quantity have the dimension as that the √(uG):

The correct answer is Force per unit mass.

To solve this, we perform a dimensional analysis of $\sqrt{uG}$ and compare it with each option.

Step 1: Find the dimensions of energy density (u)

Energy density is energy per unit volume:

$u=\frac{\text{Energy}}{\text{Volume}}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

Step 2: Find the dimensions of Gravitational constant (G)

From Newton's law of gravitation $F=\frac{G{m}_1{m}_2}{r^2}$, we get:

$G=\frac{F{r}^2}{m_1{m}_2}=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M^2\right]}=\left[M^{-1}{L}^3{T}^{-2}\right]$

Step 3: Compute the dimensions of uG

$\left[uG\right]=\left[M{L}^{-1}{T}^{-2}\right]\times \left[M^{-1}{L}^3{T}^{-2}\right]=\left[M^{1-1}{L}^{-1+3}{T}^{-2-2}\right]=\left[M^0{L}^2{T}^{-4}\right]$

Step 4: Take the square root

$\left[\sqrt{uG}\right]=\sqrt{\left[L^2{T}^{-4}\right]}=\left[L{T}^{-2}\right]$

Step 5: Compare with each option

- Force per unit mass: $\frac{F}{m}=\frac{\left[ML{T}^{-2}\right]}{\left[M\right]}=\left[L{T}^{-2}\right]$ ✓
- Gravitational potential: $\left[L^2{T}^{-2}\right]$ ✗
- Energy per unit mass: $\left[L^2{T}^{-2}\right]$ ✗
- Pressure gradient per unit mass: $\left[L^{-2}{T}^{-2}\right]$ ✗

Conclusion: The dimension of $\sqrt{uG}$ is $\left[L{T}^{-2}\right]$, which is the same as the dimension of Force per unit mass (i.e., acceleration). This is also the same as gravitational field intensity, which physically makes sense — the gravitational constant G and energy density u together naturally produce a quantity with units of acceleration.