DC pandey Mechanics part 1 soluti

202 Pages·2014·4.23 MB·English

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 = 12MR2ω
25 
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 ×100max = 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 = qFB2 ⇒ 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] = Asin2π 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,32 = 9 , 62 = 36, 22 = 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 = − B2cos2π (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]