Determine the dimensions of velocity (v) and pressure (P).

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] = ([F])/([A]) = [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 + (1)/(n)), the terms P and (1)/(n) must have the same dimensions. So, the dimension of (1)/(n) must be equal to the dimension of P: [(1)/(n)] = [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 + (1)/(n) is therefore also [M L^-1 T^-2]. Step 3: Determine the dimension of the numerator sqrt(P + (1)/(n)). Since [P + (1)/(n)] = [M L^-1 T^-2], the dimension of the square root is: [sqrt(P + (1)/(n))] = 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 = (sqrt(P + 1)/(n))x. We can write this in terms of dimensions: [v] = ([sqrt(P + 1)/(n))][x] Rearranging to solve for [x]: [x] = ([sqrt(P + 1)/(n))][v] Substitute the dimensions we found: [x] = [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: M^-1 L T^2 Dimension of x: M^1/2 L^-3/2