Tiêu đề DPP-1 SOLUTION.pdf ✅

CLASS : XITH SUBJECT : PHYSICS DATE : DPP NO. : 1 1 (c) From h = ut + 1 2 gt 2 h = 0 + 1 2 × 9.8 × (2) 2 = 19.6 m ∆h h = ±2 ∆t t [∵ a = g = constant] = ±2 ( 0.1 2 ) = ± 1 10 ∴ ∆h = ± h 10 = ± 19.6 10 = ±1.96 m 2 (a) Given, W = 1 2 kx 2 Writing the dimensions on both sides [ML 2T −2 ] = k[M0L 2T 0 ] ∴ Dimensions of k = [MT −2 ] = [ML0T −2 ] 3 (a) Given, m = 3.513 kg and v = 5.00 ms−1 So, momentum, p = mv = 17.565 As the number of significant digits in m is 4 and v is 3, so, p must have 3 significant digits p = 17.6 kgms−1 4 (d) Modulas of rigidity = Shear stress Shear strain = [ML −1T −2 ] 5 (c) The unit of physical quantity obtained by the line intergral of electric field is JC−1 . Topic :-.UNITS AND MEASUREMENTS Solutions OM COACHING CLASSES
6 (b) F = Gm1m2 d 2 ⇒ G = Fd 2 m1m2 [G] = [MLT −2 ][L 2 ] [M2] = [M−1L 3T 2 ] Moment of inertia I = mK 2 = [ML2 ] 7 (c) Stress = Force Area = N m2 8 (a) n1u1 = n2u2 n2 = n1u1 u2 = 170.474L M3 = 170.474 × 10−3M3 M3 = 0.170474 9 (c) Intensity (I)= Energy Area×time 10 (d) By the principle of dimensions homogeneity F = at −1 [MLT−2 ] = a[T −1 ] a = [MLT−1 ] Similarly for b = [MLT−4 ] 11 (a) Let radius of gyration [k] ∝ [h] x [c] y[G] z By substituting the dimension of [k] = [L] [h] = [ML 2T −1 ] [c] = [LT −1 ] [G] = [M−1L 3T −2 ] And by comparing the power of both sides We can get x = 1/2, y = −3/2, z = 1/2

= [ 1 T ] = [T −1 ] 20 (a) Area of rectangle A = lb = 10.5 × 2.1 = 22.05 cm2 Minimum possible measurement of scale =0.1 cm So, area measured by scale = 22.0cm2