Tiêu đề DPP-3 SOLUTION.pdf ✅

CLASS : XITH SUBJECT : PHYSICS DATE : DPP NO. : 3 1 (b) Given, p = a−t 2 bx or pbx = a − t 2 By the law of homogeneity of dimensional equation. Dimensions of a = dimensions of t 2 = [T 2 ] Dimensions of b =dimensions of t 2 px = [M−1T 4 ] So, dimensions of a b is [MT−2 ]. 2 (d) f = uv u + v , ∆f f = ∆u u + ∆v v + (u + v) u + v 4 (b) L = ∅ I = Wb A = Henry 6 (b) r1 = 10−15m, r2 = 1026m Log r = 1 2 [log10−15 + log1026] = 1 2 [−15 + 26] = 5.5 ≈ 6 ⇒ r = 106m 7 (d) The dimensions of x = dimensions of v0 A Therefore, out of the given options v0 has dimensions equal to [M0LT −1 ] and A has dimensions equal to [M0L 0T −1 ] So, that [v0 ] [A] = [M0LT −1] [M0L 0T−1] = [L] = dimension of x Topic :-.UNITS AND MEASUREMENTS Solutions OM COACHING CLASSES