Determine the physical quantities that would be obtained from the products of each of the following dimensions: (i) [M ¹L ³T ¹] [M ²]L ²]; (ii) [M][L ³] [LT 2][L]; (iii) [M][L ²] [T 2]

Step 1: Simplify the dimensions for part (i). We are given the product of dimensions [M^-1L^3T^-1][M^2L^-2]. To simplify, we add the exponents for each base dimension (M, L, T): For M: -1 + 2 = 1 For L: 3 + (-2) = 1 For T: -1 The resulting dimension is [MLT^-1]. This is the dimension for momentum (mass × velocity). Step 2: Simplify the dimensions for part (ii). We are given the product of dimensions [M][L^-3][T^-2][L]. To simplify, we add the exponents for each base dimension (M, L, T): For M: 1 For L: -3 + 1 = -2 For T: -2 The resulting dimension is [ML^-2T^-2]. This is the dimension for force density (force per unit volume). Force has dimensions [MLT^-2] and volume has dimensions [L^3], so force density is [MLT^-2][L^3] = [ML^-2T^-2]. Step 3: Simplify the dimensions for part (iii). We are given the product of dimensions [M][L^2][T^-2]. To simplify, we add the exponents for each base dimension (M, L, T): For M: 1 For L: 2 For T: -2 The resulting dimension is [ML^2T^-2]. This is the dimension for energy or work (e.g., force × distance). Force has dimensions [MLT^-2] and distance has dimensions [L], so work is [MLT^-2] × [L] = [ML^2T^-2]. The physical quantities are: (i) Momentum (ii) Force density (iii) Energy or Work Send me the next one 📸