To find the dimensions of $n$ and $x$, we first need to establish the dimensions of the known quantities: velocity ($v$) and pressure ($P$). Step 1: Determine the dimensions of velocity ($v$) and pressure ($P$). The dimension of velocity is length divided by time: $$[v] = [L T^{-1}]$$ Pressure is defined as force per unit area. Force ($F$) has dimensions of mass times acceleration ($F=ma$), so $[F] = [M L T^{-2}]$. Area ($A$) has dimensions of length squared, $[A] = [L^2]$. Therefore, the dimension of pressure is: $$[P] = \frac{[F]}{[A]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$$ Step 2: Apply the principle of dimensional homogeneity to the terms inside the square root. For terms to be added or subtracted, they must have the same dimensions. In the expression $\sqrt{P + \frac{1}{n}}$, the terms $P$ and $\frac{1}{n}$ must have the same dimensions. So, the dimension of $\frac{1}{n}$ must be equal to the dimension of $P$: $$\left[\frac{1}{n}\right] = [P]$$ $$[n]^{-1} = [M L^{-1} T^{-2}]$$ To find the dimension of $n$, we take the reciprocal: $$[n] = [M^{-1} L T^2]$$ The dimension of $P + \frac{1}{n}$ is therefore also $[M L^{-1} T^{-2}]$. Step 3: Determine the dimension of the numerator $\sqrt{P + \frac{1}{n}}$. Since $[P + \frac{1}{n}] = [M L^{-1} T^{-2}]$, the dimension of the square root is: $$\left[\sqrt{P + \frac{1}{n}}\right] = \sqrt{[M L^{-1} T^{-2}]} = [M^{1/2} L^{-1/2} T^{-1}]$$ Step 4: Use the full equation to find the dimension of $x$. The given equation is $v = \frac{\sqrt{P + \frac{1}{n}}}{x}$. We can write this in terms of dimensions: $$[v] = \frac{\left[\sqrt{P + \frac{1}{n}}\right]}{[x]}$$ Rearranging to solve for $[x]$: $$[x] = \frac{\left[\sqrt{P + \frac{1}{n}}\right]}{[v]}$$ Substitute the dimensions we found: $$[x] = \frac{[M^{1/2} L^{-1/2} T^{-1}]}{[L T^{-1}]}$$ $$[x] = [M^{1/2} L^{-1/2 - 1} T^{-1 - (-1)}]$$ $$[x] = [M^{1/2} L^{-3/2} T^0]$$ $$[x] = [M^{1/2} L^{-3/2}]$$ The dimensions are: Dimension of $n$: $\boxed{M^{-1} L T^2}$ Dimension of $x$: $\boxed{M^{1/2} L^{-3/2}}$