Tiêu đề 2.UNITS AND MEASUREMENTS - Explanations.pdf ✅

1 (b) PV =[energy] Vander Waal’s equation is (P + a V2 ) (V − b) = nRT The dimensions of a V2 should be that of P and b is that of volume Work done (or energy) should have the dimensions of PV ∴ [ a V2 × b] = [Energy] [bP] = [Energy] [ a V2 ] = [P] is having dimensions different from energy 2 (d) [Calorie]=[ML 2T −2 ] Comparing with general dimensional formula [MaL bT c ], we get a = 1, b = 2, c = −2. n2 = 4.2 [ 1kg αkg] 1 [ 1m βm ] 2 [ 1s γs ] −2 =4.2α −1β −2γ 2 4 (a) k = [ R N ] = [ML 2T −2θ −1 ] 6 (d) Dimensions of β 3 = dimensions of density = [ML−3 ] β = [M1/3L −1 ] Also α = force × density = [MLT−2 ][ML−3 ] = [M2L −2T −2 ] 7 (d) Given, v = at + b t+c Since, LHS is equal to velocity, so at and b t+c must have the dimensions of velocity. ∴ at = v Or a = v t = [LT −1] [T] = [LT −2 ] Now, c = time (∵ like quantities are added) ∴ c = t = [T] Now, b t + c = v ∴ b = v × time = [LT−1 ][T] = [L] 8 (c) ∆l l × 100 = 0.01 15.12 × 100 = 0.07, ∆b b × 100 = 0.01 10.15 × 100 = 0.1, ∆t t × 100 = 0.01 5.28 × 100 = 0.2 Required percentage = 0.07 + 0.1 + 0.2 = 0.37% 9 (d) [C] = [M−1L −2T 4A 2 ],[R] = [ML 2T −3A −2 ] [L] = [ML 2T −2A −2 ] and [I] = [M0L 0T 0A] 1. [CR] = [M−1L −2T 4A 2 ] [ML2T −3A −2 ] = [M0L 0TA 0 ] 2. [L] [R] = [ML 2T −2A −2] [ML 2T−3A−2] = [M0L 0TA 0 ] 3. (√LC) = ([ML 2T −2A −2 ] × [M−1L −2A 2 ]) 1/2 = [M0L 0TA 0 ] 4. [LI 2 ] = [ML 2T −2A −2 ] [M0L 0T 0A 1 ] 2 = [ML 2T −2A 0 ] 10 (d) Poisson’s ratio is a unitless quantity 11 (d) Energy = Work done [Dimensionally] 12 (a) y = a sin (ωt + kx). Here, ωt should be dimensionless ∴ [ω] = [ 1 t ] [ω] = [M0L 0T −1 ] 13 (b) Time constant in an R − C circuit τ = R − C [τ] = [R][C] = [ML 2T −3A −2 ][M−1L −2T 4A 2 ] = [M0L 0T] 15 (b) 1 kWh = 1 × 103 × 3600 W × sec = 36 × 105 J 16 (c) Impulse = Force × time = (kg − m/s 2) × s = kg-m/s 18 (b) Time period of simple pendulum is T = 2π√ l g Or ∆T T = 1 2 ( ∆l l − ∆g g )
Or ∆g g = ∆l l − 2∆T T ∴ Maximum percentage error in equation ∆g g × 100 = ∆l l × 100 + 2∆T T × 100 = 1 × 100 + 2 × 2 × 100 = 5 × 100 = 5% 19 (a) Let T ∝ S x r yp z By substituting the dimensions of [T] = [T] [S] = [MT −2 ],[r] = [L],[ρ] = [ML −3 ] and by comparing the power of both the sides x = −1/2, y = 3/2, z = 1/2 so T ∝ √ρr 3/S ⇒ T = k√ ρr 3 S 21 (c) Given, voltage V = (100 ± 5) volt, Current I = (10 ± 0.2)A From Ohm’s law V = IR ∴ Resistance R = V I Maximum percentage error in resistance ( ∆R R × 100) = ( ∆V V × 100) + ( ∆I I × 100) = ( 5 100 × 100) + ( 0.2 10 × 100) = 5 + 2 = 7% 22 (d) v = √ T m = [ m′g M t ] 1/2 = [ m′lg M ] 1/2 It follows from here, ∆v v = 1 2 [ ∆m′ m + ∆l l + ∆M M ] = 1 2 [ 0.1 3.0 + 0.001 1.000 + 0.1 2.5 ] = 1 2 [0.03 + 0.001 + 0.04] = 0.036 Percentage error in the measurement=3.6 23 (d) e = Ldi dt ⇒ [e] = [ML 2T −2A −2 ][ A T ] [e] = [ML 2T −2Q −1 ] 24 (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 ] 25 (c) [planck ′ s constant] [linear momentum] = [ML 2T −1 ] [MLT −1] = [M0LT 0 ] 26 (a) According to Coulombs law F = 1 4πε0 q1q2 r 2 ∴ 1 4πε0 = Fr 2 q1q2 = (newton)(meter) 2 (coulomb)(coulomb) = Nm2 C 2 = C −2Nm2 27 (a) Let m = KF aL bT c Substituting the dimension of [F] = [MLT −2 ],[C] = [L] and [T] = [T] And comparing both sides, we get m = FL −1T 2 28 (a) Dimension of αt = [M0L 0T 0 ] ∴ [α] = [T −1 ] Again [ v0 α ] = [L] so [v0 ] = [LT −1 ] 29 (b) 1 yard = 36 inch = 36 × 2.54 cm = 0.9144m 30 (c) Force F⃗ = qv⃗ × B⃗ Or F = qvB sin θ ∴ [B] = [ F qv] = [MLT −2 ] [ATLT −1] = [MT −2A −1 ] 32 (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% 33 (d) F ∝ v ⇒ F = kv ⇒ [k] = [ F v ] = [ MLT 2 LT −1 ] = [MT −1 ] 34 (b) watt ampere = volt 35 (c) Electric displacement, D = εE Unit of D = C 2 Nm2 N C ∴ [D] = ( C m2 ) = [AT] [L 2] = [L −2TA]
37 (c) Torque = [ML 2T −2 ], Angular momentum = [ML 2T −1 ] So mass and length have the same dimensions 38 (d) [Pressure]=[Stress ]=[coefficient of elasticity] =[ML−1T −2 ] 39 (a) Kinetic energy =1 2 mv 2 = M[LT −1 ] 2 = [ML 2T −2 ] 40 (b) The height of a tree, building tower, hill etc, can be determined with the help of a sextant. 41 (a) Quantities having different dimensions can only be divided or multiplied but they cannot be added or subtracted 42 (c) [ 1 2 ∈0 E 2 ] = [Energy density] = ML 2T −2 L 3 = ML −1T −2 43 (c) 1 fermi = 10−15 metre 44 (a) 1 C.G.S. unit of density = 1000 M.K.S. unit of density ⇒ 0.5 gm/cc = 500 kg/m3 45 (b) In g = ln h − 2 ln t ( ∆g g × 100) max = ∆h h × 100 + 2 ∆t t × 100 = e1 + 2e2 49 (b) Capacitance C = Charge potential = q V Also potential = work charge (∵ V = W q ) ∴ C = q 2 J as well as C = J V2 . Thus, (a), (c), (d) are equivalent to farad but (b) is not equivalent to farad. 51 (a) Time defined in terms of the rotation of the earth is called universal time (UT). 52 (d) The number of significant figures in all of the given numbers is 4. 53 (b) RC = T ∵ [R] = ML 2T −3A −2 ] and [C] = [M−1L −2T 4A 2 ] 54 (b) Surface tension, T = F l ∴ [T] = [F] [l] = [MLT −2 ] [L] = [ML 0T −2 ] = [MT −2 ] 55 (a) ρ = RA l i. e. dimension of resistivity is [ML 3T −1Q −2 ] 56 (d) Y = X 3Z 2 = M−1L −2T 4A 2 [MT −2A−1] 2 = [M−3L −2T 8A 4 ] 57 (a) R = V I ⇒ ± ∆R R = ± ∆V V ± ∆I I = 3 + 3 = 6% 58 (b) Both force constant and surface tension represent force per unit length. 59 (a) Momentum p ∝ f av bρ c [MLT −1 ]=[T −1 ] a[LT −1 ] b [ML −3 ] c [MLT −1 ] = [McL b−3cT −a−b ] ⟹ c = 1. b − 3c = 1 ⟹ b = 4 −a − b = −1 a + b = 1, a = −3 ∴ [p] = [f −3v 4ρ] 60 (b) Required volume = 75×104×26 103×103×103 (km) 3 61 (a) The formula for fine structure constant is = e 2 4πε0 ( h 2π ) c 62 (a) Charge = current × time 63 (b) Both 1 2 LI 2 and 1 2 CV 2 represent energy. 64 (c) Snth represents the distance covered in nth sec. 65 (c) Intensity (I)= Energy Area×time 66 (d) Joule-sec is the unit of angular momentum where as other units are of energy 67 (d) Required error in density = 3% + 3 × 2% = 9%. 68 (a) Bxt is unitless. ∴ Unit of B is m−1 s −1 69 (c) % error in velocity = % error in L + % error in t = 0.2 13.8 × 100 + 0.3 4 × 100 = 1.44 + 7.5 = 8.94 % 70 (d) Let L = [h a c bG c ]
∴ [L ′ ] = [ML 2T −1 ] a[LT −1 ] b [M−1L 3T −2 ] c ⇒ a = 1 2 , b = − 3 2 , c = 1 2 Hence, L = [h 1/2 c −3/2G 1/2 ] 71 (b) We know that kinetic energy = 1 2 mv 2 Required percentage error is 2%+2× 3% ie, 8% 72 (b) [Pressure]=[stress]= [ML −1T −2 ] 73 (d) The number of significant figures in 4.8000 × 104 is 5 (zeros on right after decimal are counted while zeros in powers of 10 are not counted). The number of significant figures in 48000.50 is 7 (all the zeros between two non-zero digits are significant). 74 (a) [surface tension]=[ML 0T −2 ], [viscosity]=[ML −1T −1 ]. Clearly, mass has the same power in the two physical quantities. 75 (a) φ = BA = F I × L A = [MLT −2 ][L 2 ] [A][L] = [ML 2T −2A −1 ] 76 (b) Pyrometer is used the for measurement for temperature 77 (d) E = F/q = Newton/coulomb 78 (a) Couple = Force × Arm length = [MLT −2 ][L] = [ML 2T −2 ] 79 (a) T = 32 × 10−5 (10)−2 = 32 × 10−3Nm−1 = 0.032Nm−1 . 80 (a) Volume V = I 3 = (1.2 × 10−2m) 3 = 1.728 × 10−6m3 ∵ length lhas two significance figures. Therefore, the correct answer is V = 1.7 × 10−6m3 81 (a) The mass of electron = 9.1×10−31 1.67×10−27 ∴ E = 9.1×10−31 1.67×10−27 × 931 MeV = 0.5073 MeV 82 (b) Percentage error in length = 1 50 × 100 = 2 Percentage error in breadth = 0.1 2.0 × 100 = 5 Percentage error in thickness= 0.1 1.00 × 100 = 1 Percentage error in volume= 2 + 5 + 1 = 8 83 (b) Frequency f = cmxk y , k = Force/Length [M0L 0T −1 ] = [M] x [ML 0T −2 ] y = [M] x+y [L] 0 [T] −2y Comparing the powers on M, L and T −2y = −1 ⇒ y = 1 2 And x + y = 0 ∴ x = −y = − 1 2 84 (d) [ε0L] = [C] ∴ X = ε0LV t = C × V t = Q t = current 85 (b) Time constant = L R ∴ [ L R ] = [T] ∴ [ R L ] = [T −1 ] 86 (a) Impulse = force × time = [MLT −2 ][T] = [MLT −1 ] 87 (d) Time period of a simple pendulum is T = 2π√ L g ⇒ g = 4π 2L T 2 ∴ ∆g g × 100 = ( ∆L L + 2 ∆T T ) × 100 = 1% + 2 × 2% = 5% 88 (c) Angular momentum = [ML 2T −1 ], Frequency = [T −1 ] 89 (c) We know, f = 1 2π√LC Or √LC = 1 2πf = time Thus, √LC has the dimension of time. 90 (b)