DC pandey Mechanics part 1 soluti PDF
Preview DC pandey Mechanics part 1 soluti
Basic Mathematics 1 & Measurements Section I Single Correct Option 1 p 1, K = mv2 p = 1 2 2 9 ∴ ∆K = ∆m +2∆v ∴ Percentage increase in pressure = 100 K m v 9 max =11.1 ∴ Maximum error =2% +2(3%) Option (a) is correct. In the estimate of, p2 kinetic energy (K)=8% 5. K = 2m Option (c) is correct. m m ∴ Error in the measurement of kinetic 2. d= = energy (K) V l3 =2×100% ∴ ∆d ∆m ∆l =200% ×100 = ×100 +3 ×100 d m l Option (d) is correct. max = 4% +3(3%) 6. 3400 =3.400×103 =13% ∴ Number of significant figures =2 Option (d) is correct. Option (d) is correct. F F 3. p= = 7. A =3.124m ×3.002 m A L2 =9.378/248m2 ∴ Permissible error in pressure (p) =9.378m2 = 4% +2(2%) Option (a) is correct. =8% GM Option (a) is correct. 8. g = R2 4. pV = p V constant 1 1 2 2 = V R2 ⇒ p = p 1 2 1 V2 K = 1Iω2 V 2 = p1 V1 −10%1 of V1 = 12MR2ω 25 p = 90%1 = constant × R2 Basic Mathematics & Measurements | 3 tI ∴ Decrease in R (radius) by 2% world − increase g by 4% and decrease K 16. Q= Iα e (∆V)ε0β (given) (rotational kinetic energy) by 4%. We know that Option (b) is correct. Q= It 9. Heat (H) =22Rt − tI ∴ t=α e (∆V)ε0β ∴ Maximum error in measuring heat (H) ⇒ [α]=[t] =2(2%)+1% +1% tI =6% and [β]= Option (b) is correct. (∆V)ε0 β tI I 1 10. V = lbt ⇒ = = α (∆V)ε0t (∆V)ε0 =12×6×2.45 1 1 ==11.7766.44×102 cm3 = [Resistance]ε0 β 1 1 =2×102 cm2 or α = [ML2T–3A–2][M–1L–3T4A2] Option (b) is correct. 1 P = = [velocity] 11. I = [L–1T] 4πr2 1 i.e., Ir2 = constant = [µ ε ]1/2 0 0 i.e., if r is increased by 2% the intensity Option (a) is correct. will decrease by 4%. pq Option (d) is correct. 17. a= r2s3 12. Option (b) is correct. 13. V = 34πr3 ∆aa ×100max = 12∆pp×100 + 12∆qq×100 ∆V ∆r +2∆r ×100 +3∆s ×100 =3 r s V r 1 1 =3(1%) = (1%)+ (3%) +2(0.5%)+3(0.33%) 2 2 =3% =0.5% +1.5% +1% +1% Option (c) is correct. = 4% 14. a3 =6a2 (given) Option (c) is correct. ∴ a=6 18. Least count of main scale ⇒ V =63 =216m3 2 mm = =0.5 mm Option (b) is correct. 4 l least count of main scale 15. g = 4π2 Least count = T2 50 ∆g ∆l ∆T =0.1 mm ∴ ×100 = ×100+2 ×100 g l T Zero error = −30×0.01 mm max 1mm 0.15 = −0.3 mm = ×100 +2 ×10 100m 2003 (–ive sign, zero of circular scale is lying above observed reading of plate thick) =0.1% +0.1% =0.2% =2 MSR + 20 CSR Option (a) is correct. 4 | Mechanics-1 =(2×0.5 mm) + (20 × 0.01 mm) = observed reading − zero error = 1 mm + 0.2 mm = 1.2 mm + 0.3 mm = 1.2 mm. = 1.5 mm Plate thickness (corrected reading) Option (d) is correct. More than One Correct Options 1. Maximum percentage error in x : =(%errorin a)+2(%error in b) +3 (% error in c) =15% Assertion and Reason 1. Least count of screw gauge Thus, assertion is true. Pitch From the above relation we conclude that = Number of divisions of circular scale least count of screw gauge is inversely Less the value of pitch, less will be least proportional to the number of divisions of count of screw gauge leading to len circular scale. uncertainty that is more accuracy in the Thus reason is false. measurement. Option (c) is correct. Match the Columns GM M 2 1. (a) F = r12 2 = qvBsinθ qB ∴ GM M = Fr2 1 2 =v2sin2θ i.e., [GM1M2]=[MLT–2][L2] F2 =[ML3T–2] ∴ q2B2 =[LT–1]2 =[L2T–2] ∴ (a)→(q) 3RT 3pV 3work ∴ (c) →(r) (b) = = GM M n nM (d) g = e R2 ∴ 3MRT = MLM2T−2 =[L2T−2] GRMe = gRee e ∴ (b) → (r) (c) qF2B22 = qFB2 ⇒ GRMee =[LT–2][L] =[L2T–2] ∴ (d) →(r) 2 Units & Dimensions Vectors Section I Single Correct Option Force 1. Pressure (p) = 4. φ= Li Area ∴ [φ ] =[L][i] [MLT−2] ∴ [p] = =[ML2 T−2 A−2][A] [L2] =[ML2 T−2 A−1] =[ML−1T−2] Option (a) is correct. Option (d) is correct. 5. Linear impulse (I)= F⋅∆t 2. W = I2Rt [I] =[MLT−2][T] [ML2T−2] [ML2T−2A−2] ∴ [R] = = …(i) =[MLT−1] [A2T] [T] Option (c) is correct. dI V = L dt and W = Vq 6. F =G m1m2 d2 W dt ∴ L= [F][d2] [MLT−2][L2] q dI [G] = = [m m ] [M2] [ML2T−2][T] 1 2 [L] = [A2T] =[ML2T−2A−2] =[M−1L3T−2] Option (c) is correct. Using Eq. (i) µ i ⋅i [L] 7. F = 0 1 2 [R] = 4π d [T] [MLT−2][L] i.e., [T]= L ∴ [ µ0] = [A2] =[ML2 T−2 A−2] R Option (c) is correct. Option (c) is correct. [L] 3. F =6πηav 8. [k]= [LT−1]=[T] [F] ∴ [η]= Option (c) is correct. [av] = [MLT−2] 9. [a]= [MLT−2]=[ML T−3] [LLT−1] [T ] [MLT−2] =[ML−1T−1] [b]= =[ML T−4] [T2] Option (d) is correct. Option (c) is correct. 6 | Mechanics-1 10. E= hν 17. q=CV ∴ [h]= [ML2 T−2]=[ML2 T−1] and V = iR [T−1] ∴ q= iCR nh Angular momentum (J)= it= iCR 2π ⇒ [CR] =[t]=[M0L0 TA0] [J]=[h]=[ML2 T−1] Option (a) is correct. Option (b) is correct. 1 q q 18. F = 1 2 11. [Energy] =[ML2 T−2] 4πε r2 0 =[M][LT−1]2 ∴ Unit of ε = Newton-metre2/coulomb2. 0 ∴ [Mass] =[Ev−2] Option (b) is correct. Option (c) is correct. nh 19. Angular momentum (J)= 12. 1ε E2 = Energy density = Energy 2π 2 0 Volume I = Σmr2 ∴ 12ε0E2 = [ML[L23T]−2] ∴ hI = 2πΣmJr/2n = Σmmvrr2 Optio n ( b ) i s c o rr e c t . =[ML−1T−2] hI = [LTL−1] =[T−1] 13. [a]=[T2] = Frequency [T2] [b]= Option (a) is correct. [L][ML−1T−2] b 20. v= at+ ∴ a =[MT−2] t+ c b [c]=[T] Option (b) is correct. b =[v] 14. Velocity gradient = dv t+ c dx or [b]=[LT−1][T]=[L] [LT−1] [Velocity gradient] = [at]=[v]=[LT−1] [L] ⇒ [a] =[LT−2] =[T−1] Option (a) is correct. =[M0L0 T−1] 2π 21. y= Asin (ct− x) Option (a) is correct. λ 15. [Force] =[MLT−2] = Asin2π ct− 2πx [F] λ λ ∴ [Mass] = [L T−2] 2πx =θ (angle) λ =[FL−1T2] ∴ [x]=[λ]=[L] Option (a) is correct. Further, y= Asinθ 16. Coefficient of friction (µ) ∴ [A]=[y]=[L] Limittingfrictionalforce = Option (a) is correct. Normalforce 22. [X]=[M−1L−3T3A2] ∴ [µ] =[M0L0 T0] [TA2] = Option (b) is correct. [ML2 T−2] Units & Dimensions Vectors 7 [t][i2] 1 = =(3)(5) [Work] 2 ∴ X is resistance. [(cid:81) W = i2Rt] =7.5 23. F→ =2^i−3^j+ 4k^ Option (b) is correct. → → → 30. A +B=C →r =3^i+2^j+3k^ → → → → → → ∴ (A +B)⋅(A +B)=C⋅C ^i ^j k^ → → → → → → → → → → → or A⋅A +2A⋅B+B⋅B=C⋅C ∴ τ = r ×F =3 2 3 2 −3 4 or A2 +2→A⋅→B+ B2 =C2 → → or →τ =17^i−6^j−13k^ or A⋅B =0 → → 24. (0.5)2 +(0.8)2 +(c)2 =1 or |A||B|cosθ=0 or 0.25+0.64+ c2 =1 or cosθ=0 π or c2 =1−0.89 or θ= 2 c= 0.11 Option (d) is correct. Option (b) is correct. 31. Magnetic field intensity. → → → → 25. |A +B|=|A −B| Option (d) is correct. (A→ +B→)⋅(A→ +B→)=(A→ −B→)⋅(A→ −B→) 32. P→ +Q→ =R→ A2 + B2 +2→A⋅→B= A2 + B2 −2→A⋅→B (P→ +Q→)⋅(P→ +Q→)=R→⋅R→ i.e., A→⋅B→ =0 P2 +Q2 +2→P⋅Q→ = R2 ∴ Angle between A→ and B→ =90° 122 +52 +2→P⋅Q→ =132 26. (A→ +B→)⋅(A→ −B→)=0 P→⋅Q→ =0 A→⋅A→ +B→⋅A→ −B→⋅B→ − A→⋅B→ =0 ∴ Angle between P→ and Q→ = π 2 A2 − B2 =0 Option (b) is correct. A = ± B 33. Option (b) is correct. → → |A|=|B| 34. P→ +Q→ +R→ =0 Option (d) is correct. → → → ∴ P+Q= −R → → 27. Work (=F⋅s) is a scalar quantity. → → → → → → or (P+Q)⋅(P+Q)=(−R)⋅(−R) Option (d) is correct. → → → → → → → → → or P⋅P+Q⋅Q+2P⋅Q=R⋅R 28. Speed =|v| Option (d) is correct. or P2 +Q2 +2→P⋅Q→ = R2 …(i) 29. |A→|=3, |B→|=5 and angle between A→ and B→ Let Q2 = P2 and R= P 2 Thus, Eq. (i) takes the form is 60°. → → → → P2 + P2 +2PQcosθ=2P2 ∴ A⋅B=|A||B|cos60° or 2PQcosθ=0 8 | Mechanics-1 or cosθ=0 Qsinθ (−Qsinθ) − or θ=90° = P+Qcosθ P−Qcosθ → → Qsinθ (−Qsinθ) ∴ Angle between P and Q is 90° 1− ⋅ (P+Qcosθ) (P−Qcosθ) → → → P+Q+R=0 2PQ sinθ = → → → P2 +Q2cos2θ ∴ P+R= −Q → → → → → → → → This implies that angle between P+Q and or (P+R)⋅(P+R)=(−Q)⋅(−Q) → → or P2 + R2 +2PRcosφ=Q2 P−Q will vary from 0 to π. or 2PRcosφ=Q2 − P2 − R2 Option (b) is correct. or 2PRcosφ= − R2 36. R2 = P2 +Q2 +2PQcosθ or 2Pcosφ= − R for R= P=Q or 2Pcosφ= − P 2 P2 = P2 + P2 +2PPcosθ 1 1 or cosφ= − or cosθ= − 2 2 ∴ φ=135° or θ=120° → → Option (b) is correct. ∴ Angle between P and R is 135°. → → 37. W =F⋅s 135° =(3^i+ 4^j)⋅(3^i+ 4^j) → R → P =25 J 90° Option (b) is correct. → 135° Q 38. P→⋅Q→ =(a^i+ a^j+3k^)⋅(a^i−2^j− k^) Option (a) is correct. = a2 −2a−3 35. Angle (φ) between P→ +Q→ and P→ −Q→ For P→ ⊥Q→, P→⋅Q→ =0 Q→ →P +→Q i.e., a2 −2a−3=0 θ or (a−3)(a+1)=0 φ → P ⇒ a=3 → φ' – Q P→ –Q→ Other value is − ive. Option (d) is correct. tanφ= Qsinθ 39. If a vector makes angles α, β and γ with P+Qcosθ the co-ordinate axes, then → → → cos2α +cos2β +cos2 γ =1 Angle φ′ between P−Q and P Qsin(π +θ) Now,32 = 9 , 62 = 36, 22 = 4 tanφ′= 7 49 7 49 7 49 P+Qcos(π +θ) 9 36 4 −Q sinθ and + + =1 = 49 49 49 P−Qcosθ ∴ Option (a) is correct. tanφ− tanφ′ tan[φ+(−φ′)]= 1+ tanφtanφ′ 40. A→ = 4^i−3j and B→ =8^i+8(cid:36)j ∴ A→ +B→ =C→ =12^i+5^j Units & Dimensions Vectors 9 ^ C→ 12^i+5^j ∴ B=2 N C= = |C| 122 +52 Option (c) is correct. 12 5 46. Angle between A→ =2^i+3^j = ^i+ ^j 13 13 and B→ = ^i+ ^j Option (b) is correct. → → 41. A→ =2^i+3(cid:36)j−2k^, B→ =5^i+ n(cid:36)j+ k^, θ= A⋅B = 2+3 |→A||→B| 22 +32 ⋅ 2 C→ = − ^i+2(cid:36)j+3k^ 5 = ∴ Vectors A→,B→,andC→ will be coplanar if 13⋅ 2 5 their scalar triple product is zero i.e., = 26 → → → (A ×C)⋅B=0 Component of A→ along ^i+ ^j (cid:36)i (cid:36)j k(cid:36) 2 3 −2⋅(5^i+ n^j+ k^)=0 C→ = 5 (2^i+3^j) 26 −1 2 3 |C→|= 5 2 or (13^i− 4^j+7k^)⋅(5^i+ n^j+ k^)=0 Option (a) is correct. or 65− 4n+7=0 47. R2 =(3P)2 +(2P)2 +2×3P×2Pcosθ or n=18 or R2 =13P2 +12P2cosθ …(i) Option (a) is correct. Further 42. Option (a) is correct. (2R)2 =(6P)2 +(2P)2 +2×6P×2Pcosθ 43. (→a +→b)×(→a −→b) or 4R2 = 40P2 +24P2cosθ …(ii) → → → → → → → → Dividing Eq. (ii) by Eq. (i), =a ×a + b×a −a × b− b× b 10P2 +6P2cosθ=13P2 +12P2cosθ → → → → =0−a × b−a × b−0 or 6cosθ= − P → → → → or θ=120° = −2(a × b) =2(b×a) Option (b) is correct. Option (a) is correct. Qsinα 44. A→ =3^i+ 4^j+5k^ 48. tanθ= P+Qcosα B→ =3^i+ 4^j−5k^ → → (A⋅B) cosθ= R Q = 2P → → |A||B| θ = 90° 9+16−25 = P 32 + 42 +52 As θ=90°, tanα = ∞ =0 ∴ P+Qcosα =0 ⇒ θ=90° P i.e., cosα = − Option (c) is correct. Q 45. A + B=7 P = − A − B=3 2P 10 | Mechanics-1 1 = − or − P×16= −96 2 P= +6 N ∴ α =120° ∴ Q=10 N Option (a) is correct. Option (a) is correct. → → 49. A⋅B=0 → → → → 51. |A ×B|= 3(A⋅B) → A → → → → → → ⇒ |A||B|sinθ= 3|A||B|cosθ B × C ⇒ tanθ= 3 ⇒ θ=60° ∴ |A→ ×B→|2 =|A→|2 ×|B→|2 +2|A→||B→|cosθ → → B C = A2 + B2 +2ABcos60° → → ⇒ A ⊥B …(i) = A2 + B2 + AB A→⋅C→ =0 |A→ ×B→|=[A2 + B2 + AB]1/2 → → → → → ⇒ A ⊥C …(ii) 52. C is perpendicular to both A and B → → → From Eq. (i) and Eq. (ii), we conclude that C =A × B → A is perpendicular to the plane containing → → B and C. → This implies that A is perpendicular to → → → → B⋅C. A B Option (c) is correct. ∴ C→⋅A→ =0 50. P2 +Q2 +2PQcosα = R2 → → → or (A ×B) ⋅A =0 or P2 +Q2 +2PQcosα =82 Option (d) is correct. or P2 +Q2 +2PQ+2PQcosα −2PQ=64 53. (2^i+3^j+8k^)⋅(−4^i+ 4^j+αk^)=0 or (P+Q)2 +2PQ(cosα −1)=64 ⇒ −8+12+8α =0 or (16)2 +2PQ(cosα −1)=64 1 ∴ α = − or 2PQ(cosα −1)= −192 2 Option (c) is correct. or PQcosα − PQ= −96…(i) → → → → Q sinα 54. P+Q+R= 0 tanθ= = ∞ (as P+Qcosα → P θ=90°) γ β → α ∴ P+Qcosα =0 → R Q Qcosα = − P …(ii) → → → |P| |Q| |R| Using Eq. (ii) and Eq. (i), If, = = sinα sinβ sinγ P(−P)− PQ= −96 or −P(P+Q)= −96 Units & Dimensions Vectors 11 55. O→A + A→B=O→B 56. Using answer to questions no. 35, as angle → → → → B (–2,6,4) between A +B and A −B is 90° A2 + B2cos2θ=0 Z or A2 = − B2cos2θ or A2 = − B2cos2π (cid:81)θ= π O 2 2 Y or A2 = − B2cosπ A (0, 3, –1) X or A2 = B2 → → → ∴ AB=OB−OA ⇒ A = B =(−2^i+6^j+ 4k^)−(0^i+3^j− k^) Option (a) is correct. = −2^i+3^j+5k^ Option (c) is correct. Match the Columns → → → → → → → → → → → → 1. (a) |A ×B|=|A⋅B| ⇒ (A +B)⋅(A +B)=(A −B)⋅(A −B) → → → → → → → → → → → → or |A||B|sinθ= ±|A||B|cosθ or A⋅A +B⋅A +B⋅A +B⋅B or tanθ= ±1 = →A⋅→A −→B⋅→A − →A⋅→B+→B⋅→B π 3π ⇒ θ= , → → 2 4 or 4A⋅B=0 Thus, (a) → (r) (s). ⇒ A→ ⊥B→ → → → → (b) A ×B=B× A (given) Thus, (c) → (q). → → → → → → → or A ×B= −(B× A) (d) A +B=C or |A→||B→|sinθ= −|A→||B→|sinθ or (A→ +B→)⋅(A→ +B→)=C→⋅C→ or sinθ= −sinθ or A→⋅A→ +B→⋅A→ +B→⋅A→ +B→⋅B→ =C→⋅C→ or 2sinθ=0 or A2 +2→A⋅→B+ B2 =C2 ⇒ θ=0 rad Thus, (b) → (p). or 2A→⋅B→ =0 ((cid:81) A2 + B2 =C2) → → → → (c) |A ×B|=|A ×B| → → ⇒ A ⊥B or |A→ +B→|2=|A→ −B→|2 Thus, (d) → (q). Section II Subjective Questions 1. 2×1011N/m2 = (2×1011)(105 dyne) [ωt]=[M0L0T0] ∴ [ω]=[T−1] (104 cm2) [θ]=[M0L0T0] =2×1012 dyne/cm2 4. h= E = J = J-s (72)(10−5N) ν per sec 2. 72dyne/cm = =0.072N/m (10−2m) [h]=[ML2T–2][T]=[ML2T–1] 3. [a]=[y]=[L] 5. [b]=[x2]=[L2]
The list of books you might like
