Tiêu đề DPP-5 SOLUTION.pdf ✅

Rmean = 6 + 10 2 = 8 Ω ∴ x = 6 + 10 2 = 8 Ω ⇒ ∆x x = 0.25 8 ∴ Total error = 0.3 6 + 0.2 10 + 0.25 8 = 0.05 + 0.02 + 0.03125 = 0.10125 ∴ ∆RP RP = 10.125% 13 (d) Fm = Bqv or [B] = [ Fm qv] = [ Fm Itv] = [MLT −2 ] [A][T][LT −1] = [ML 0T −2A −1 ] Alternate F = BIl ⇒ [B] = [ F Il] = [ML 0T −2A −1 ] 14 (d) R=8.3 J/K-mol n1u1 = n2u2 ∴ n2 = n1u1 u2 = 8.3J/K − mol atm L/K − mol = 8.3J/K − mol (1.013 × 105N/m2)(10−3m3)/K − mol = 8.12 102 = 0.0812 ∴ 8.3 J/K-mol=0.0812 atm L/K-mol 15 (d) P = nu ∴ n ∝ 1 u 16 (b) From Coulomb’s law F = 1 4πε0 q1q2 r 2 Or ε0 = q1q2 4πFr 2
∴ Units of ε0 (permittivity) = C 2 N − m2 = C 2N −1m−2 17 (d) Work done W = ε∆q ∴ ε = W ∆q = [ML 2T −2 ] [AT] ∴ [ε] = [ML 2T −3A −1 ] 18 (a) Maximum absolute error is Δa + Δb. Now work out the relative error ad finally the percentage error. 19 (b) Potential energy = mgh = g ( cm sec2 ) cm = g ( cm sec ) 2 20 (c) Resistivity, ρ = m ne 2τ ∴ [ρ] = [M] [L −3][AT][T 2] = [ML3A −2T −3 ] So, electrical conductivity σ = 1 ρ ⇒ [σ] = 1 [ρ] = [M−1L −3A 2T 3 ]