Tiêu đề DPP-5 SOLUTION.pdf ✅

Substituting dimensions for corresponding quantities in Eq. (i), we have [K] = [ML 2T −2 ][L] [L 2][θ][T] = [MLT −3θ −1 ] 7 (b) F − 32 9 = K − 273 5 ⇒ x − 32 9 = x − 273 5 ⇒ x = 574.25 8 (c) 1 fermi = 10−15 metre 9 (d) [Planck constant ] = [ML 2T −1 ] and [Energy] = [ML 2T −2 ] 10 (b) MeV-sec is not a unit of energy. While others are units of energy. 11 (b) F = 1 4πε0 q1q2 r 2 ⇒ ε0 = |q1 ||q2| [F][r 2] = [A 2T 2 ] [MLT −2][L 2] = [A 2T 4M−1L −3 ] 12 (d) R1 = (6 ± 0.3)kΩ, R2 = (10 ± 0.2)kΩ Rparallel = R1R2 (R1 + R2 ) Let (R1 + R2 ) = x ⇒ RP = R1R2 x Taking log of both sides ln RP = ln R1 + ln R2 − ln x Differentiating, ∆RP RP = ∆R1 R1 + ∆R2 R2 + (− ∆x x ) ∆xmean = 0.3 + 0.2 2 = 0.25 Ω