Title DPP-2 SOLUTION.pdf ✅

CLASS : XITH SUBJECT : PHYSICS DATE : DPP NO. : 2 1 (d) Given equation , y = a sin(bt − cx) Comparing the given equation with general wave equation y = a sin ( 2πt T − 2πx λ ), We get b = 2π T , c = 2π λ Dimension of b c = 2π/T 2π/λ = [LT −1 ], and other three quantity is dimensionless 3 (b) Units of a and PV 2 are same and equal to dyne × cm4 4 (d) f = 1 2π√LC ∴ ( C L) does not represent the dimensions of frequency 5 (c) P1 = [ML 2T −1 ] D2 = [(2M)(2L) 2(2T) −1 ] P2 = 4[ML 2T −1 ] = 4P1 6 (a) Time period of a simple pendulum T = 2π√ L 8 Or g = 4π 2L T 2 ... . (i) Differentiating Eq. (i), we have Topic :-.UNITS AND MEASUREMENTS Solutions OM COACHING CLASSES

12 (a) Y = Stress Strain = Force/Area Dimensionless ⇒ Y ≡ Pressure 13 (c) Coefficient o friction= Applied force Normal reactiom = [MLT −2] [MLT−2] =no dimensions Unit= N N =no unit 14 (c) [kx] = Dimension of ωt = (dimensionless) Hence K = 1 X = 1 L = [L −1 ] ∴ [K] = [L −1 ] 15 (a) Magnetic field = Force Charge × velocity = [MLT−2 ] [AT][LT−1] = [MA−1T −2 ] 17 (c) Percentage error in measurement of a side = 0.01 1.23 × 100 Percentage error in measurement of area = 2 × 0.01 1.23 × 100 18 (a) Charge = current × time 19 (c) From the principle of dimensional homogenity [v] = [at] ⇒ [a] = [LT −2 ]. Similarly [b] = [L] and [c] = [T] 20 (d) Given, U = A√x x+B ... (i) Dimensions of U = dimensions of potential energy = [ML 2T −2 ] From Eq. (i), Dimensions of B = dimensions of x = [M0LT 0 ] ∴ Dimensions of A
= dimensions of U × dimensions of (x + B) dimension of √x = [ML 2T −2 ][M0LT 0 ] [M0L 1/2T 0] = [ML 5/2T −2 ] Hence, dimensions of AB = [ML 5/2T −2 ][M0LT 0 ] = [ML 7/2T −2 ] ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. D A B D C A D A B D Q. 11 12 13 14 15 16 17 18 19 20 A. D A C C A C C A C D