Tiêu đề 9.MECHANICAL PROPERTIES OF SOLIDS - Explanations.pdf ✅

1 (d) Energy stored per unit volume = 1 2 × Stress × Strain = 1 2 × Young ′ s modulus × (Strain) 2 = 1 2 × Y × x 2 3 (d) YA YB = tan θA tan θB = tan 60 tan 30 = √3 1/√3 = 3 ⇒ YA = 3YB 4 (d) Let L be the length of each side of cube. Initial volume of cube = L 3 . When each side of cube decreases by 2%, the new length L ′ = L − 2L 100 = 98L 100 New volume = L ′3 = (98L/100) 3 ∴ Change in volume ∆V = L 3 − (98L/100) 3 = L 3 ⌈1 − (1 − 2 100) 3 ⌉ = L 3 ⌈1 − (1 − 6 100 + ... )⌉ (from binomial expension) = L 3 [ 6 100] = 6L 3 100 ∴ Bulk strain = ∆V V = 6L3 /100 L 3 = 0.06 5 (a) Let the original unstretched length be l. Y = Stress Strain = T/A ∆l/l = T A × l ∆l Now, Y = 4 A l (l1−l) = 6 A l (l2−l) = 9 A l (l3−l) ∴ 4(l3 − l) = 9(l1 − l) ⟹ 4l3 + 5l = 9l1 ... (i) Again, 6(l3 − l) = 9(l2 − l) ⟹ 2l3 + l = 3l2 ... (ii) Solving Eqs. (i) and (ii), we obtain l3 = (2.5l2 − 1.5l1) 6 (a) Let us consider the length of wire as L and cross- sectional area A, the material of wire has Young’s modulus as Y. Then for 1st case Y = W/A l/L For 2 nd case, Y = W A 2l ′ L ∴ l ′ = l 2 So, total elongation of both sides = 2l ′ = l 7 (b) l = FL AY ∴ l ∝ 1 A [F, L and Y are constant] A2 A1 = l1 l2 ⇒ A2 = A1 ( 0.1 0.05) = 2A1 = 2 × 4 = 8mm2 8 (b) Strain ∝ Stress ∝ F A Ratio of strain = A2 A1 = ( r2 r1 ) 2 = ( 4 1 ) 2 = 16 1 10 (c) Energy U = 1 2 × 4Al 2 L = 1 2 × 2×1011×3×10−6×(1×10−3) 2 4 = 0.075 J 11 (b) Longitudinal strain l L = stress Y = 106 1011 = 10−5 Percentage increase in length = 10−5 × 100 = 0.001% 12 (c) ∆l = 4Fl πD2Y = 4×30×2×7 22 ×(3×10−3) 2×1.1 ×1011 = 7.7 × 10−5m = 0.077mm 13 (b) T = YAl L Increase in length of one segment of wire l = (L + 1 2 d 2 L ) − L = 1 2 d 2 L So, T = Yπr 2.d 2 2L 2 14 (a) When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Young’s modulus (Y) of the material of the body. Y = stress strain = F/A l/L Where F is force, A the area, l the change in length and L the original length. ∴ Y = FL πr2l r being radius of the wire. Given r2 = 2r1, L2 = 2L1, F2 = 2F1 Since, Young’s modulus is a property of material, we have Y1 = Y2 ∴ F1L1 πr1 2 l1 = 2F1 × 2L1 π(2r1) 2l2 l2 = l1 = l Hence, extension produced is same as that in the other wire.
15 (d) Coefficient of elasticity in increasing order is given by RubberIf diameter is made four times then force required will be 16 times, i. e. 16 × 103N 38 (d) τ = Cθ = πɳr 4θ 2L = constant ⟹ πɳr 4 (θ−θ0) 2l = πɳ( r 2 ) 4 (θ0−θ ′′) 2(l/2) ⟹ (θ−θ0 ) 2 = θ0 16 ⟹ θ0 = 8 9 θ 39 (a) Isothermal elasticity = p, Adiabatic elasticity = γP ∴ Eθ Eφ = 1 Υ , Υ > 1 ∴ Eθ Eφ < 1 40 (c) Isothermal bulk modulus = Pressure of gas 41 (b) Y = F πR2 × l ∆l F, l and ∆l are constants. ∴ R 2 ∝ 1 Y RS 2 RB 2 = YB YS = 1011 2 × 1011 = 1 2 Or RS RB = 1 √2 or RS = RB √2 42 (b) Y = Fl A∆l or ∆l = Fl AY = Fl πr 2Y In the given problem, ∆l = 1 r 2 ; when both l and r are double, ∆l is halved. 43 (b) U = 1 2 × (stress) 2 Y × volume = 1 2 × F 2 × A × L A2 × Y = 1 2 × F 2L AY = 1 2 × (50) 2 × 0.2 1 × 10−4 × 1 × 1011 = 2.5 × 10−5 J 44 (d) Y = F/A Breaking strain Or a = F Y×Breaking strain = 104×100 7×10×0.2 = 0.71× 10−3 = 7.1 × 10−4 45 (a) Y = Fl A∆l or ∆l ∝ F r 2 Or ∆l2 ∆l1 = F2 F1 × r1 2 r2 2 Or ∆l2 ∆l1 = 2 × 2 × 2 = 8 Or ∆l2 = 8∆l1 = 8 × 1 mm = 8 mm 46 (d) τx = πηr 4 2l θx and τy = πη(2r) 4 2l θy Since, τx = τy, ∴ θx = 16θy or θx θy = 16 47 (a) Because dimension of invar does not vary with temperature 48 (a) Area of hysterisis loop gives the energy loss in the process of stretching and unstretching of rubber band and this loss will appear in the form of heating 49 (c) Work done = 1 2 Fl = Mgl 2 50 (c) B = P ∆V/V ∆V V = P B = ρgh B = 1.36% 51 (d) Y = Mg A × L/2 ∆L (Length is taken as L 2 because weight acts as CG ) Now, M = ALρ (For the purpose of calculation of mass, the whole of geometrical length L is to be considered.) ∴ Y = ALpgL 2A∆L Or ∆L = pgL 2 2Y = 1.5×103×10×8×8 2 ×5 ×106 = 9.6 × 10−2m = 9.6 × 10−2 × 103 mm = 96 mms 52 (a) Y = F A ∆l l = Fl A∆l Or Y = Fl ×4 πD2×∆l or∆l ∝ 1 D2 or ∆L2 ∆L1 = D1 2 D2 2 = n 2 1 53 (b) W = 1 2 × F × l = 1 2 mgl = 1 2 × 10 × 10 × 1 × 10−3 = 0.05 J
54 (a) Y = FL Al = 1000 × 100 10−6 × 0.1 = 1012N/m2 55 (b) Work done = 1 2 F × ∆l = 1 2 Mgl 57 (c) l = FL AY = FL 2 (AL)Y = FL 2 VY If volume is fixed then l ∝ L 2 58 (a) ω = √ K m = √ YA lm = √ (n × 109) (4.9 × 10−7) 1 × 0.1 Given,ω = 140 rad s −1 in above equation, we get, n = 4 59 (b) 2π√ m k = 0.6 ...(i) and 2π√ m+m′ k = 0.7 ...(ii) Dividing (ii) by (i), we get ( 7 6 ) 2 = m+m′ m = 49 36 m + m′ m − 1 = 49 36 − 1 ⇒ m′ m = 13 36 ⇒ m′ = 13m 36 Also k m = 4π 2 (0.6)2 Desired extension = m′g k = 13 36 × mg k 13 36 × 10 × 0.36 4π 2 = 3.5 cm 60 (d) l = FL AY = FL 2 (AL)Y = FL 2 VY ∴ l ∝ L 2 if volume of the wire remains constant l2 l1 = ( L2 L1 ) 2 = ( 8 2 ) 2 = 16 ∴ l2 = 16 × l1 = 16 × 2 = 32mm = 3.2cm 61 (d) l = FL AY ∴ l ∝ 1 r 2 [F, L and Y are constant] l1 l2 = ( r2 r1 ) 2 = (2) 2 = 4 62 (b) ∆l l = stress Y = 1000 × 980)/(10−1) 2 1010 = 0.0098 % increase in length of wire = ∆l l × 100 = 0.0098 × 100 = 0.98% 63 (a) Per unit volume energy stored = 1 2 × Y × (strain) 2 = 1 2 × Y × ( l L ) 2 given l = L × 1% or l = L 100 × stored energy Y = 1 2 × 2 × 1010 × ( L 100L ) 2 = 106 Jm−3 64 (d) A1l1 = A2l2 ⟹ l2 = A2l1 A1 = A × l1 3A = l 3 ⟹ l1 l2 = 3 ∆x1 = F1 Aγ l1 . . . (i) ∆x2 = F2 3Aγ l2 .....(ii) Here ∆x1 = ∆x2 F2 3Aγ l2 = F1 Aγ l1 F2 = 3F1 × l1 l2 = 3F1 × 3 = 9F 65 (b) Y = F A × l ∆l or F = YA ∆l l = (5.0×108)×106) ×(2×10−2) (10×10−2) = 100 N 66 (a) To twist the wire through the angle dθ, it is necessary to do the work dW = τdθ And θ = 10′ = 10 60 × π 180 = π 1080 rad W = ∫ τ dθ = ∫ ηπr 4θdθ 2l = ηπr 4θ 4l θ 0 θ 0 W = 5.9 × 1011 × 10−5 × π(2 × 10−5) 4π 2 10−4 × 4 × 5 × 10−2 × (1080) 2 W = 1.253 × 10−12 J 67 (a) Energy stored in the wire U = 1 2 Y × (strain) 2 × volume or U = 1 2 Y × ( x l ) 2 × Al or U = 1 2 Yx2 l ×A or U = 1 2 YA l x 2 68 (c) K = 100 0.01/100 = 106atm = 1011N/m2 = 1012dyne/cm2 69 (d) From the definition of Bulk modulus, B = − dp (dV/V) Substituting the values we have, B = (1.165 − 1.01) × 105 ( 10 100) Pa = 1.55 × 105 Pa 71 (d)