Tiêu đề DPP-6 SOLUTION.pdf ✅

CLASS : XITH SUBJECT : PHYSICS DATE : DPP NO. : 6 1 (b) Bulk modulus K = normal stress volumetric strain = F/A −∆V/V = − FV A ∆V Now, F A = p ∴ K = pV ∆V As volumetric strain is dimensionless. ∴ Dimensions of K = dimensions of normal stress ⇒ [K] = [ML −1T −2 ] 2 (a) R = V I ⇒ ± ∆R R = ± ∆V V ± ∆I I = 3 + 3 = 6% 4 (d) n(xm) 2 = 1m2 or n = 1 x 2 5 (d) Given, v = at + bt 2 Applying the law of homogeneity [v] = [bt 2 ] Or [LT −1 ] = [bT 2 ] Or [b] = [LT −3 ] Topic :-.UNITS AND MEASUREMENTS Solutions OM COACHING CLASSES
6 (a) V = W Q = [ML 2T −2Q −1 ] 7 (c) Volume of sphere (V) = 4 3 πr 3 % error in volume = 3 × ∆r r × 100 = (3 × 0.1 5.3 ) × 100 8 (d) Given, v = at + b t+c Since, LHS is equal to velocity, so at and b t+c must have the dimensions of velocity. ∴ at = v Or a = v t = [LT −1] [T] = [LT −2 ] Now, c = time (∵ like quantities are added) ∴ c = t = [T] Now, b t + c = v ∴ b = v × time = [LT−1 ][T] = [L] 10 (a) Dimensions of E = [ML 2T −2 ] Dimensions of G = [M−1L 3T −2 ] Dimensions of I = [MLT −1 ] And dimension of M = [M] ∴ Dimensions of GIM2 E2 = [M−1L 3T −2 ][MLT −1 ][M2 ] [ML 2T −2] 2 = [T] = Dimensions of time 11 (a) Percentage error inside = 1 2 [ 0.2 100 × 100] = 0.1 Absolute error inside = 0.1 100 × 10 = 0.01
12 (d) The second is the duration of 9192631770 period of the radiation corresponding to the transition between the two hyperfine levels of the ground state of cesium-133 atom. Therefore, 1 ns is 10−9 s of Cs-clock of 9192631770 oscillations. 14 (a) Weight in air = (5.00 ± 0.05)N Weight in water =(4.00 ± 0.05)N Loss of weight in water =(1.00 ± 0.1)N Now relative density= weight in air weight loss in water i. e. R.D = 5.00 ± 0.05 1.00 ± 0.1 Now relative density with max permissible error = 5.00 1.00 ± ( 0.05 5.00 + 0.1 1.00) × 100 = 5.0 ± (1 + 10)% = 5.0 ± 11% 15 (c) Angular momentum = [ML 2T −1 ], Frequency = [T −1 ] 17 (a) By the principle of dimensional homogenity [P] = [ a V2 ] ⇒ [a] = [P] × [V 2 ] = [ML −1T −2 ][L 6 ] = [ML 5T −2 ] 18 (a) [E] = [ML 2T −2 ] [M] = [M] [L] = [ML 2T −1 ] [G] = [M−1L 3T −2 ] [ EL 2 M5G2 ] = [ML 2T −2 ][ML 2T −1 ] 2 [M] 5[M−1L 3T −2] 2 = [ML 2T −2 ][M2L 4T −2 ] [M5][M−2L 6T −4] = [M3L 6T −4 ] [M3L 6T −4] = [m0L 0T 0 ] = Angle 19 (c) [MT −3 ] = [ML 2T −2] [L 2][T] =energy /area ×time=dimensions of solar constant. 20 (b) We know that kinetic energy = 1 2 mv 2 Required percentage error is 2%+2× 3% ie, 8%