Tiêu đề DPP-4 SOLUTION.pdf ✅

On solving, we get a = − 5 6 , b = 1 2 , c = 1 3 5 (b) Average value = 2.63 + 2.56 + 2.42 + 2.71 + 2.80 5 = 2.62 sec Now |∆T1 | = 2.63 − 2.62 = 0.01 |∆T2 | = 2.62 − 2.56 = 0.06 |∆T3 | = 2.62 − 2.42 = 0.20 |∆T4 | = 2.71 − 2.62 = 0.09 |∆T5 | = 2.80 − 2.62 = 0.18 Mean absolute error ∆T = |∆T1 | + |∆T2 | + |∆T3 | + |∆T4 | + |∆T5 | 5 = 0.54 5 = 0.108 = 0.11 sec 6 (c) Y = 4MgL πD2I so maximum permissible error in Y = ∆Y Y × 100 = ( ∆M M + ∆g g + ∆L L + 2∆D D + ∆l l ) × 100 = ( 1 300 + 1 981 + 1 2820 + 2 × 1 41 + 1 87) × 100 = 0.065 × 100 = 6.5% 7 (d) τ = dL dt ⇒ dL = τ × dt = r × F × dt i. e., the unit of angular momentum is joule-second 8 (b) f = 1 2π√LC ⇒ LC = 1 f 2 = [M0L 0T 2 ] 9 (a) angular momentum linear momentum = [ML 2T −1 ] [MLT −1 ] = [M0LT 0 ] 10 (a) [e] = [AT], ∈0= [M−1L −3T 4A 2 ],[h] = [ML 2T −1 ] And [c] = [LT −1 ] ∴ [ e 2 4πε0hc ] = [ A 2T 2 M−1L −3T 4A2 × ML 2T −1 × LT −1 ] = [M0L 0T 0 ] 12 (a) The result has to be in one significant umber only. 13 (b)
v ∝ g ph q (given) By submitting the dimension of each quantity and comparing the powers on both sides we get [LT −1 ] = [LT −2 ] p[L] q ⇒ p + q = 1, −2p = −1, ∴ p = 1 2 , q = 1 2 14 (b) Force = Mass × acceleration = [M][LT −2 ] = [MLT −2 ] Torque = Force × distance = [MLT −2 ][L] = [ML 2T −2 ] Work = Force × distance = [MLT −2 ][L] = [ML 2T −2 ] Energy = [ML 2T −2 ] Power = Work Time = [ML 2T −2 ] [T] = [ML 2T −3 ] 16 (b) Positions x = ka mt n [M0LT 0 ] = [LT −2 ]m[T] n = [M0L mT −2m+n] On comparing both sides m = 1 −2m + n = 0 n = 2m n = 2 × 1 = 2 17 (a) ∵ R = PV T = [ ML −1T −2 × L 3 θ ] = [ML 2T −2θ −1 ] 18 (b) We know that Specific heat = Q m ∆t Unit of specific heat = unit of heat unit of mass × unit of temperature ∴ Unit of specific heat = J kg°C = Jkg−1 °C −1 19 (a) K = Y × r0 = [ML −1T −2 ] × [L] = [MT −2 ] Y=Young’s modulus and r0 = Interatomic distance