Solution of Mechanics - Part 1. Lesson 1st to 8th

207 Pages·7.321 MB·English

by  DC Pandey

Basic Mathematics
1
& Measurements
Section I
Single Correct Option
1 p
1, K = 2mv2 p2 = 91
\ ÊËÁDKKˆ¯˜ = Dmm +2Dvv \ Percentage increase in pressure = 1900
max
\ Maximum error =2% +2(3%) =11.1
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%
ÊÁDd¥100ˆ˜ =ÊÁDm¥100ˆ˜ +3ÊÁDl¥100ˆ˜ =200%
Ë d ¯ Ë m ¯ Ë 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
\ Permissible error in pressure (p) =9.378/248m2
= 4% +2(2%) =9.378m2
Option (a) is correct.
=8%
GM
Option (a) is correct. 8. g =
R2
4. p1V1 = p2V2 = constant
V R2
fi p2= p1 V21 K = 1Iw2
V 2
= p1 V1 -10%1 of V1 = 1ÊÁ2MR2ˆ˜w
p 2Ë5 ¯
1
= 90% = constant ¥ R2
Basic Mathematics & Measurements | 3
\ Decrease in R (radius) by 2% world - tI
increase g by 4% and decrease K 16. Q= Ia e (DV)e0b (given)
(rotational kinetic energy) by 4%. We know that
Option (b) is correct. Q= It
9. Heat (H) =22Rt - tI
\ t=a e (DV)e0b
\ Maximum error in measuring heat (H)
=2(2%)+1% +1% fi [a]=[t]
tI
=6% and [b]= È ˘
Option (b) is correct. ÎÍ(DV)e0˚˙
10. V = lbt fi Èb˘ = È tI ˘ = È I ˘È1˘
ÎÍa˚˙ ÎÍ(DV)e0t˚˙ ÎÍ(DV)˚˙ÎÍe0˚˙
=12¥6¥2.45
1 1
==11.7766.44¥102 cm3 = [Resistance]ÈÎÍe0˘˚˙
=2¥102 cm2 or ÈÎÍab˘˚˙ = [ML2T1–3A–2][M–1L–13T4A2]
Option (b) is correct.
1
P = = [velocity]
11. I = [L–1T]
4pr2 1
i.e., Ir2 = constant = [µ0e0]1/2
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 = 34pr3 ÊËÁDaa ¥100ˆ¯˜max = 12ÊËÁDpp¥100ˆ¯˜ + 12ÊËÁDqq¥100ˆ¯˜
DV =3Dr +2ÊËÁDrr ¥100ˆ¯˜ +3ÊËÁDss ¥100ˆ¯˜
V r
1 1
=3(1%) = 2(1%)+ 2(3%) +2(0.5%)+3(0.33%)
=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
fi V =63 =216m3 = 2 mm =0.5 mm
Option (b) is correct. 4
l least count of main scale
15. g = 4p2 T2 Least count = 50
\ ÊÁDg ¥100ˆ˜ = Dl ¥100+2ÊÁDT ¥100ˆ˜ =0.1 mm
Ë g ¯ l Ë T ¯ Zero error = -30¥0.01 mm
max
= ÊËÁ110m0mm ¥100ˆ¯˜ +2ÊËÁ200.1053 ¥10ˆ¯˜ (–ive sign, z=e-r0o .o3f mcimrcular 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 = ÊÁqvBsinqˆ˜
qB
Ë ¯
\ GM1M2 = Fr2 =v2sin2q
i . e . , =[G[MML1M3T2–]2=] [MLT–2][L2] \ È F2 ˘ =[LT–1]2 =[L2T–2]
Íq2B2˙
Î ˚
\ (a)Æ(q)
3RT 3pV 3work \ (c) Æ(r)
(b) = = GM
M n nM (d) g = e
R2
\ È3RT˘ = ÈML2T-2˘ =[L2T-2] GM e
ÎÍ M ˚˙ ÎÍ M ˚˙ R e = gRe
e
\ (b) Æ (r) GM
F2 F 2 fi È e˘ =[LT–2][L]
(c) q2B2 = ÊÁqBˆ˜ ÎÍ Re ˚˙
Ë ¯
=[L2T–2]
\ (d) Æ(r)

2
Units & Dimensions
Vectors
Section I
Single Correct Option
Force
1. Pressure (p) = 4. f= Li
Area
\ [f ] =[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◊Dt
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)
[R] = [[TL]] 7. F = µ4p0 i1d◊i2
[MLT-2][L]
i.e., [T]= ÈL˘ \ [ µ0] = [A2] =[ML2 T-2 A-2]
ÍR˙
Î ˚
Option (c) is correct.
Option (c) is correct.
[L]
3. F =6phav 8. [k]= [LT-1]=[T]
[F]
\ [h]= Option (c) is correct.
[av]
= [[MLLLTT--12]] 9. [a]= [M[LTT ]-2]=[ML T-3]
[MLT-2]
=[ML-1T-1] [b]= =[ML T-4]
[T2]
Option (d) is correct.
Option (c) is correct.
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10. E= hn 17. q=CV
\ [h]= [ML2 T-2]=[ML2 T-1] and V = iR
[T-1] \ q= iCR
nh
Angular momentum (J)= it= iCR
2p fi [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] 4pe0 r2
=[M][LT-1]2 \ Unit of e0 = Newton-metre2/coulomb2.
\ [Mass] =[Ev-2] Option (b) is correct.
Option (c) is correct. nh
19. Angular momentum (J)=
12. 12e0E2 = Energy density = VEonleurmgye I = S2mpr2
\ ÈÎÍ12e0E2˘˚˙ = [ML[L23T]-2] \ hI = 2pSmJr/2n = Smmvrr2
Optio n ( b ) i s c o rr e c t . =[ML-1T-2] ÈÍhI˘˙ = ÈÍ[LTL-1]˘˙ =[T-1]
Î ˚ Î ˚
13. [a]=[T2] = Frequency
[T2]
[b]= Option (a) is correct.
[L][ML-1T-2] b
20. v= at+
\ Èa˘ =[MT-2] t+ c
ÎÍb˚˙ [c]=[T]
Option (b) is correct. È b ˘
14. Velocity gradient = dv ÎÍt+ c˚˙ =[v]
dx
[LT-1] or [b]=[LT-1][T]=[L]
[Velocity gradient] = [at]=[v]=[LT-1]
[L]
fi [a] =[LT-2]
=[T-1] Option (a) is correct.
=[M0L0 T-1] 21. y= AsinÈ2p(ct- x)˘
Option (a) is correct. ÎÍl ˚˙
15. [Force] =[MLT-2] = AsinÈ2p ct- 2px˘
[F] ÎÍl l ˚˙
\ [Mass] = [L T-2] 2px =q (angle)
=[FL-1T2] l
\ [x]=[l]=[L]
Option (a) is correct.
Further, y= Asinq
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
= [Work] =(3)(5)ÊËÁ2ˆ¯˜
\ X is resistance. [Q W = i2Rt] =7.5
Option (b) is correct.
23. FÆ =2^i-3^j+ 4k^
30. AÆ +BÆ =CÆ
Ær =3^i+2^j+3k^
\ (AÆ +BÆ)◊(AÆ +BÆ)=CÆ◊CÆ
Ω^i ^j k^Ω
\ Æt =Ær ¥ÆF =ΩΩ3 2 3ΩΩ or AÆ◊AÆ +2AÆ◊BÆ +BÆ◊BÆ =CÆ◊CÆ
ΩΩ2 -3 4ΩΩ or A2 +2ÆA◊ÆB+ B2 =C2
Ω Ω
or Æt =17^i-6^j-13k^ or AÆ◊BÆ =0
24. (0.5)2 +(0.8)2 +(c)2 =1 or |AÆ||BÆ|cosq=0
or 0.25+0.64+ c2 =1 or cosq=0
p
or c2 =1-0.89 or q= 2
c= 0.11 Option (d) is correct.
Option (b) is correct.
31. Magnetic field intensity.
25. |AÆ +BÆ|=|AÆ -BÆ| Option (d) is correct.
(AÆ +BÆ)◊(AÆ +BÆ)=(AÆ -BÆ)◊(AÆ -BÆ) 32. PÆ +QÆ =RÆ
A2 + B2 +2ÆA◊ÆB= A2 + B2 -2ÆA◊ÆB (PÆ +QÆ)◊(PÆ +QÆ)=RÆ◊RÆ
i.e., AÆ◊BÆ =0 P2 +Q2 +2ÆP◊QÆ = R2
\ Angle between AÆ and BÆ =90° 122 +52 +2ÆP◊QÆ =132
26. (AÆ +BÆ)◊(AÆ -BÆ)=0 PÆ◊QÆ =0
AÆ◊AÆ +BÆ◊AÆ -BÆ◊BÆ - AÆ◊BÆ =0 \ Angle between PÆ and QÆ = p2
A2 - B2 =0 Option (b) is correct.
A = ± B 33. Option (b) is correct.
|AÆ|=|BÆ| 34. PÆ +QÆ +RÆ =0
Option (d) is correct.
\ PÆ +QÆ = -RÆ
27. Work (=ÆF◊Æs) is a scalar quantity.
or (PÆ +QÆ)◊(PÆ +QÆ)=(-RÆ)◊(-RÆ)
Option (d) is correct.
28. Speed =|Æv| or PÆ◊PÆ +QÆ◊QÆ +2PÆ◊QÆ =RÆ◊RÆ
Option (d) is correct. or P2 +Q2 +2ÆP◊QÆ = R2 …(i)
29. |AÆ|=3, |BÆ|=5 and angle between AÆ and BÆ Let Q2 = P2 and R= P 2
Thus, Eq. (i) takes the form
is 60°.
\ AÆ◊BÆ =|AÆ||BÆ|cos60° P2 + P2 +2PQcosq=2P2
or 2PQcosq=0
8 | Mechanics-1
or cosq=0 Qsinq (-Qsinq)
-
or q=90° = P+Qcosq P-Qcosq
\ Angle between PÆ and QÆ is 90° 1- Qsinq ◊ (-Qsinq)
(P+Qcosq) (P-Qcosq)
PÆ +QÆ +RÆ =0 2PQ sinq
=
\ PÆ +RÆ = -QÆ P2 +Q2cos2q
or (PÆ +RÆ)◊(PÆ +RÆ)=(-QÆ)◊(-QÆ) This implies that angle between PÆ +QÆ and
or P2 + R2 +2PRcosf=Q2 PÆ -QÆ will vary from 0 to p.
or 2PRcosf=Q2 - P2 - R2 Option (b) is correct.
or 2PRcosf= - R2 36. R2 = P2 +Q2 +2PQcosq
or 2Pcosf= - R for R= P=Q
or 2Pcosf= - P 2 P2 = P2 + P2 +2PPcosq
1 1
or cosf= - or cosq= -
2 2
\ f=135° or q=120°
\ Angle between PÆ and RÆ is 135°. Option (b) is correct.
37. W =ÆF◊Æs
135°
=(3^i+ 4^j)◊(3^i+ 4^j)
Æ
R ÆP =25 J
90° Option (b) is correct.
Æ
135° Q 38. PÆ◊QÆ =(a^i+ a^j+3k^)◊(a^i-2^j- k^)
Option (a) is correct. = a2 -2a-3
35. Angle (f) between PÆ +QÆ and PÆ -QÆ For PÆ ^QÆ, PÆ◊QÆ =0
QÆ ÆP +ÆQ i.e., a2 -2a-3=0
q or (a-3)(a+1)=0
f Æ
P fi a=3
– QÆ PÆ –f' Other value is - ive.
QÆ Option (d) is correct.
tanf= Qsinq 39. If a vector makes angles a, b and g with
P+Qcosq the co-ordinate axes, then
Angle f¢ between PÆ -QÆ and PÆ cos2a +cos2b +cos2 g =1
3 2 9 6 2 36 2 2 4
tanf¢= PQ+sQinco(sp(+pq+)q) Now,ÊËÁ7ˆ¯˜ = 49, ÊËÁ7ˆ¯˜ = 49, ÊËÁ7ˆ¯˜ = 49
9 36 4
-Q sinq and + + =1
= 49 49 49
P-Qcosq
\ Option (a) is correct.
tanf- tanf¢
tan[f+(-f¢)]= 1+ tanftanf¢ 40. AÆ = 4^i-3j and BÆ =8^i+8$j
\ AÆ +BÆ =CÆ =12^i+5^j
Units & Dimensions Vectors 9
C^ =|CÆC|= 1122^i2++55^2j \O p t i o n ( c ) is cBor=re2c Nt.
12 5 46. Angle between AÆ =2^i+3^j
= ^i+ ^j
13 13 and BÆ = ^i+ ^j
Option (b) is correct.
41. AÆ =2^i+3$j-2k^, BÆ =5^i+ n$j+ k^, q= ÆA◊ÆB = 2+3
|ÆA||ÆB| 22 +32 ◊ 2
CÆ = - ^i+2$j+3k^ 5
=
\ Vectors AÆ,BÆ,andCÆ will be coplanar if 13◊ 2
5
their scalar triple product is zero i.e., =
26
(AÆ ¥CÆ)◊BÆ =0
Ω $i $j k$ Ω Component of AÆ along ^i+ ^j
5
Ω 2 3 -2Ω◊(5^i+ n^j+ k^)=0 CÆ = 26 (2^i+3^j)
ΩΩΩ-1 2 3 ΩΩΩ |CÆ|= 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¥2Pcosq
or n=18 or R2 =13P2 +12P2cosq …(i)
Option (a) is correct.
Further
42. Option (a) is correct. (2R)2 =(6P)2 +(2P)2 +2¥6P¥2Pcosq
43. (Æa +Æb)¥(Æa -Æb) or 4R2 = 40P2 +24P2cosq …(ii)
Dividing Eq. (ii) by Eq. (i),
=Æa ¥Æa +Æb¥Æa -Æa ¥Æb-Æb¥Æb
10P2 +6P2cosq=13P2 +12P2cosq
=0-Æa ¥Æb-Æa ¥Æb-0 or 6cosq= - P
= -2(Æa ¥Æb) =2(Æb¥Æa) or q=120°
Option (b) is correct.
Option (a) is correct.
Qsina
44. AÆ =3^i+ 4^j+5k^ 48. tanq= P+Qcosa
BÆ =3^i+ 4^j-5k^
cosq= (ÆA◊ÆB) Q = 2P R
|ÆA||ÆB|
q = 90°
9+16-25
= P
32 + 42 +52
As q=90°, tana = •
=0 \ P+Qcosa =0
fi q=90° P
i.e., cosa = -
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
\ a =120°
\ Q=10 N
Option (a) is correct.
Option (a) is correct.
49. AÆ◊BÆ =0 51. |AÆ ¥BÆ|= 3(AÆ◊BÆ)
Æ
A
Æ Æ fi |AÆ||BÆ|sinq= 3|AÆ||BÆ|cosq
B ◊ C
fi tanq= 3
fi q=60°
\ |AÆ ¥BÆ|2 =|AÆ|2 ¥|BÆ|2 +2|AÆ||BÆ|cosq
Æ Æ
B C
= A2 + B2 +2ABcos60°
fi AÆ ^BÆ …(i) = A2 + B2 + AB
AÆ◊CÆ =0 |AÆ ¥BÆ|=[A2 + B2 + AB]1/2
fi AÆ ^CÆ …(ii) 52. CÆ is perpendicular to both AÆ and BÆ
From Eq. (i) and Eq. (ii), we conclude that ÆC =ÆA ◊ ÆB
AÆ is perpendicular to the plane containing
BÆ and CÆ.
This implies that AÆ is perpendicular to
BÆ◊CÆ. AÆ ÆB
Option (c) is correct. \ CÆ◊AÆ =0
50. P2 +Q2 +2PQcosa = R2 or (AÆ ¥BÆ) ◊AÆ =0
or P2 +Q2 +2PQcosa =82
Option (d) is correct.
or P2 +Q2 +2PQ+2PQcosa -2PQ=64
53. (2^i+3^j+8k^)◊(-4^i+ 4^j+ak^)=0
or (P+Q)2 +2PQ(cosa -1)=64
fi -8+12+8a =0
or (16)2 +2PQ(cosa -1)=64 1
\ a = -
or 2PQ(cosa -1)= -192 2
Option (c) is correct.
or PQcosa - PQ= -96…(i)
Q sina 54. PÆ+QÆ +RÆ =Æ0
tanq= = • (as
P+Qcosa Æ
P
q=90°) g b
Æ
\ P+Qcosa =0 Æ a R
Q
Qcosa = - P …(ii)
|PÆ| |QÆ| |RÆ|
Using Eq. (ii) and Eq. (i), If, = =
sina sinb sing
P(-P)- PQ= -96
or -P(P+Q)= -96