The given equation is: P=ρnv2 We need to find the value of n, and for this, we can perform dimensional analysis. Let's break down the dimensions of each quantity involved. The dimensions of pressure P are:
[P]=ML−1T−2 The dimensions of density ρ are: [ρ]=ML−3 The dimensions of velocity v are:
[v]=LT−1 Substitute these dimensions into the given equation: [P]=[ρn][v2] Let's plug in the dimensions: [ML−1T−2]=[(ML−3)n][(LT−1)2] Simplify the dimensions on the right-hand side: [ML−1T−2]=[MnL−3n][L2T−2] Combine the terms on the right-hand side: [ML−1T−2]=[MnL−3n+2T−2] Now, equate the exponents of corresponding dimensions from both sides: For mass (M): 1=n For length (L): −1=−3n+2 For time (T):−2=−2 We already get the value of n from the mass dimension equation: n=1 To verify, substitute n=1 in the length dimension equation:
−1=−3(1)+2 −1=−3+2 −1=−1 This is correct, so the value of n is confirmed as: n=1 Therefore, the correct option is: Option B: n=1