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Check The Following Equation For Calculating Displacement Is Dimensionally Correct Or Not (a) X = X0 + Ut + (1/2) At2 Where, X Is Displacement At Given Time T Xo Is The Displacement At T = 0 U Is The Velocity At T = 0 A Represents The Acceleration. (b) P = (ρgh)½ Where P Is The Pressure, ρ Is The Density G Is Gravitational Acceleration H Is The Height. |
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Answer» (a) x = x0 + ut + (1/2) at2 Applying PRINCIPLE of homogeneity, all the sub-expressions of the equation must have the same DIMENSION and be equal to [LHS] Dimension of x = [M0L1T0] Dimensions of sub-expressions of [RHS] must be [M0L1T0] ⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS] Dimension of ut = VELOCITY x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS] Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS] ∴ The equation is dimensionally correct. (b) P = (ρgh)½ Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2] Dimensions of ρ = mass/volume = [M1L-3T0] Dimensions of G (acceleration) = [M0L1T-2] Dimensions of h (height) = [M0L1T0] Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½ = [M½L-½T-1] ≠ [LHS] (a) x = x0 + ut + (1/2) at2 Applying principle of homogeneity, all the sub-expressions of the equation must have the same dimension and be equal to [LHS] Dimension of x = [M0L1T0] Dimensions of sub-expressions of [RHS] must be [M0L1T0] ⇒ Dimension of x0 (displacement) = [M0L1T0] = [LHS] Dimension of ut = velocity x time = [M0L1T-1][M0L0T1] = [M0L1T0] = [LHS] Dimension of at2 = acceleration x (time)2 = [M0L1T-2][M0L0T-2] = [M0L1T0] = [LHS] ∴ The equation is dimensionally correct. (b) P = (ρgh)½ Dimensions of LHS i.e. Pressure [P] = [M1L-1T-2] Dimensions of ρ = mass/volume = [M1L-3T0] Dimensions of g (acceleration) = [M0L1T-2] Dimensions of h (height) = [M0L1T0] Dimensions of RHS = [(ρgh)½] = ([M1L-3T0]. [M0L1T-2].[M0L1T0])½ = ([M1L-1T-2])½ = [M½L-½T-1] ≠ [LHS] |
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