QJEEM24Ph495f: JEE Main 2024 Physics - Class 11 JEEM24Ph495f Solution: If be the gravitational constant and be the energy density then …

To determine the dimension of the quantity $\sqrt{uG}$ , we first need to understand the dimensions of both the gravitational constant (G) and the energy density (u).

The gravitational constant $G$ has dimensions given by:

$[G] = M^{-1}L^{3}T^{-2}$

where $M$ stands for mass, $L$ for length, and $T$ for time.

Energy density $u$ is defined as the energy per unit volume. Since energy has dimensions of $ML^{2}T^{-2}$ (from the dimension of work or energy, which is force times distance, and force itself has dimension $MLT^{-2}$ ), and volume has dimensions of $L^{3}$ , the dimensions of energy density would be:

$[u] = \frac{ML^{2}T^{-2}}{L^{3}} = M L^{-1} T^{-2}$

Now, we find the dimensions of $\sqrt{uG}$ by multiplying the dimensions of $u$ and $G$ , and then taking the square root:

$[\sqrt{uG}] = \sqrt{[u][G]} = \sqrt{(M L^{-1} T^{-2})(M^{-1}L^{3}T^{-2})} = \sqrt{L^{2}T^{-4}} = LT^{-2}$

So, the dimension of $\sqrt{uG}$ is $LT^{-2}$ , which corresponds to acceleration (length per square time).

Now, let's match this with the provided options:

• Option A (Gravitational potential) has dimensions of $[L^{2}T^{-2}]$ , not matching our target of $LT^{-2}$ .
• Option B (Pressure gradient per unit mass) would have dimensions of $[M^{-1}L^{-2}T^{-2}][L^{-1}]$ ( $Pressure\ Gradient = \frac{Pressure}{Length} = \frac{ML^{-1}T^{-2}}{L}$ , and then divided by mass, $M$ ), which simplifies to $L^{-3}T^{-2}M^{-1}$ , not matching.
• Option C (Energy per unit mass) has dimensions $ML^{2}T^{-2}M^{-1}$ which simplifies to $L^{2}T^{-2}$ , also not a match for the target dimension.
• Option D (Force per unit mass) has dimensions $MLT^{-2}M^{-1}$ which simplifies directly to $LT^{-2}$ , an exact match for our target dimension.

Thus, the correct answer is Option D (Force per unit mass), which has the same dimensions as that of $\sqrt{\mathrm{uG}}$ .