sol_major test-6.pdf
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CAREER POINT : CP Tower, Road No. 1, IPIA, Kota (Raj.), Ph: 0744-3040000 || AIEEE Target || 1
CAREER POINT
MAJOR # 6
HINTS & SOLUTIONS
CHEMISTRY
Qus. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 1 4 2 1 1 1 1 3 4 3 1 4 2 1 2 2 1 4 4
Qus. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 4 1 4 3 1 4 3 4 3 3 1 2 2 4 3 3 2 3 3
MATHEMATICS
Qus. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans 4 4 3 2 2 2 3 2 3 2 1 3 4 1 1 2 4 2 4 2
Qus. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans 3 1 2 1 2 1 2 4 3 4 2 2 3 2 1 3 3 4 1 2
PHYSICS
Ques. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Ans. 4 1 1 3 1 3 4 1 1 1 3 3 1 3 1 2 1 4 4 1
Ques. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans. 1 2 1 3 3 3 3 2 3 4 2 4 3 3 2 3 1 2 2 2
CHEMISTRY
1. Ans. Ni 2+
, Ti 3+
Hint : Unpaired e– results in colour
2. Ans. + 154.4 kJ
Hint : ∆G = – 2 × 96500 × (– 0.80) J= 154.4 KJ
3. Ans. Reaction of sodium hydride with water
Hint : Reaction of sodium hydride with water is
violent.
4. Ans. C 2 H 5CO 2CH 3
Hint : Ester
5. Ans. 2
K 2 / 3α αα α
====
Hint : K =]CO[
]O][CO[
2
2 / 12
6. Ans.
OHOH
Hint : It is pinacol-pinacolone reaction
7. Ans. ––I
Hint : Good L.G favers rate of reaction.
8. Ans. Homo polymer
Hint : Cellulose is polymer of β-D glucose& starch is polymer of α-D glucose
9. Ans. Distillation
Hint : BP of Acetone → 56°CBP of Methanol → 65°C
10. Ans. Troposphere
Hint : Fact based
11. Ans. van der waals forces
Hint : Layers of carbon atoms in graphite are held
together by van der Waals forces
12. Ans. Zero
Hint : Q Rate is constant w.r.t. time
∴ order of reaction is 013. Ans. Ba
Hint : Ba imparts green colour
14. Ans. 2 atm
Hint : π1 = C1RT1, π2 = C2RT2
or22
11
22
11
2
1
TC
TC
RTC
RTC==
π
π
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CAREER POINT : CP Tower, Road No. 1, IPIA, Kota (Raj.), Ph: 0744-3040000 || AIEEE Target ||
π2 = π1 ×11
22
TC
TC
= 3 atm ×
K300C
K400)2 / C(
1
1
×
× = 2 atm
15. Ans. Toluene
Hint :
CH3
CH3–CH2–CH2–CH2–CH3
CH3–C––CH3
CH3
CH3
CH2
Most stable radical BDE is lowest
, , ,
16. Ans. III > I > IV > II
Hint : Order of basic strength ⇒ III > I > IV > II
17. Ans. Nitrolic acid
Hint : R––C H2––NO
O
O=N–OH
– H2O
R–C–NO2
N–OHNitrolic acid
18. Ans. + 1
Hint : [Fe(H2O)5NO] SO4
x + 5(0) + 1 – 2 = 0
Fe = +1
19. Ans. 8
56 3 ×
Hint : Ew =
8
563
3 / 8
56 ×=
20. Ans. 5
Hint :
CH2–OH O–CH3 CH3
,
,
OH
ortho
CH3
OH
meta
CH3
,
OHpera
21. Ans. 1 radial node and 1 angular node
Hint : Radial node = n – l – 1
Angular node = l
22. Ans. O––C 2 H 5
Hint : Ether
23. Ans. a, b and c
Hint : Flocculation value depends on charge.
24. Ans. To attain the five valent state, d orbital
must be used and hydrogen is not
sufficiently electronegative to make d
orbitals contract sufficiently.
25. Ans. NO 2 , brown gas
26. Ans. C N
O Me
Me
Hint :
O
C OHPCl5
O
C Cl
Me2NH
C NMe
Me
O. .
27. Ans. Histidine and arginine
Hint : Fact based
28. Ans. Analgesics
Hint : These are Narcotic Analgesics
29. Ans. 0.045 M
Hint : CH3COOH CH3COO– + H
+
C
C (1 – α) Cα Cα
K =α
α=
α
α⋅α
–1
C
)–1(C
CC 2 = C α2
or C = K/ α2 =2
5–
)02.0(
108.1 × = 0.045 M
30. Ans. 1/4m
Hint : PV = nRT =M
m RT
2
1P × V =
M
m′ R ×
3
2T
∴ M
m
3
4
M
m ′= ⇒ m′ =
4
3m
∴ mass escaped = m –4
3m =
4
1m
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CAREER POINT : CP Tower, Road No. 1, IPIA, Kota (Raj.), Ph: 0744-3040000 || AIEEE Target || 3
31. Ans. H 2O > H 2S > H 2Se > H 2Te(Order of volatility)
Hint : Correct order is :H2O < H2Te < H2Se < H2S
32. Ans. –1758 kJ Hint : Sum of bond energies of reactants
= 4 × C – H + 2 × C ≡ C + 5 × O = O= 4 × 414 + 2 × 812 + 5 × 494= 1656 + 1624 + 2470 = 5750 kJ
Sum of bond energies of products= 8 × C = O + 4 × O – H= 8 × 707 + 4 × 463= 5656 + 1852 = 7508 kJ
Enthalpy of combustion,
∆H = 5750 – 7508 = – 1758 kJ
33. Ans. Ph–CH 2 –C–CH 3|
| Br
CH 3Hint :
Ph–CH=C–CH3
CH3HBr/peroxide
Br Ph–CH–C–CH3
CH3
Br
HBr
Stable radical
Ph–CH2–C–CH3
CH3
Br
34. Ans. H 5C 6
H C=C
H
CH(CH 3 ) 2Hint :
C6H5–CH2–CH–CH–CH3
OH
CH3
conc. H2SO4
C6H5C=C
H
H
CH(CH3)2
Anti elimination
takes place
35. Ans. In all the reactions, the product (C) is
aniline Hint :
MeO C–NH2
OBr2 /OH
MeO NH2
36. Ans. The overall reaction involves the
elimination of (CO) as CO 2 Hint : In R.D.S formation of acylnitrene take place
37. Ans. Zr and Hf have about the same radius Hint : Due to lanthanide contraction
38. Ans. C 2 H 6
Hint : CnH2n + 2 +2
1n3 +O2 → nCO2 + (n + 1) H2O
4
7
V
V
2
2
CO
O
= = 4
7
)n(2
1n3
=
+
12n + 4 = 14 n
2n = 4 ⇒ n = 2C2H6
39. Ans. (II) and (IV) Hint : Molecular formula same and show E and
Z-isomer.
40. Ans. cis-[Pt(en) 2Cl 2 ] +2 , [Ni(en) 2 ]
MATHEMATICS
41. Ans. All of these
Hint: Ref. (a, b) R (a, b)
Sym. (a, b) R (c, d) → (c, d) R (a, b)
Tra.
(a, b) R (c, d) & (c, d) R (e, f) → (a, b) R (e, f)
a + d = b + c
c + f = d + e
a + f = b + e
42. Ans. None
Hint: (xn–1) = (x – 1) (x – ω) …..(x – ωn–1)
⇒ ln(xn –1) = ln(x–1) + ln(x – ω) …+ ln(x–ωn–1)
dx
d
⇒ 1x
nxn
1n
−
−
=1nx
1....
x
1
1x
1−ω−
+ω−
+−
(put x = 2)
12
2.nn
1n
−
−
= 1+1n2
1....
2
1−ω−
+ω−
⇒ S = 112
2nn
1n
−−
−
43. Ans. 4
1−
Hint: Now |α – β| is same⇒ S1
2 – 4P1 = S2
2 – 4P2
b12 – 4C1 = b2
2 – 4C2
[f(x)] min =4
1
a4
−=
∆−⇒
4
1
4
C4b 12
1 =−
⇒ b12 – 4C1 = 1
[g(x)] min occurs at2
7
2
b2 =−
⇒ b2 = –7
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Now [g(x)] min =a4
∆− = –
4
1
4
c4b 22
2 −=
−
44. Ans. –7
Hint: Also b2 = –7
45 Ans. x + n +1
Hint: [ ])n2x(...)4x()2x(n
1'x n21 ++++++=
=
++
2
)1n(n.
n
2x
Mean = x + (n +1)
46. Ans. A(–α αα α , – β ββ β )
Hint: A(α, β)–1
=
αα
α−αββ
ββ
β
1000cosesine
0sinecose
e
1
= A(–α, –β)47. Ans. [4 1]
Hint: cofactor of a21, in matrix
−−
120
211
201
= a = 4
Cofactor of a33 in matrix
−−
120
211
201
= b = 1
[4 1]
48. Ans. bba
aa
ba
b
||||
||
||||
||
++
+
Hint: Since→a and
→b are equally inclined to
→c
→c must be of the form t
+
→
→
→
→
|b|
b
|a|
a
Now,→
→→
→
→
→→
→
++
+b
|b||a|
|a|a
|b||a|
|b|
=
+
+→
→
→
→
→→
→→
|b|
|b|
|a|
|a|
|b||a|
|b||a|
= t
+
→
→
→
→
|b|
b
|a|
a
49. Ans. right angled
Hint: A( â ); B( b̂ ); C( ĉ )
⇒ ĉ.db̂.dâ.d→→→
== & )ĉb̂(d +λ=→
….(1)
⇒ )ĉ.âb̂.â( +λ = )ĉ.b̂1( +λ
⇒ 1+ ĉ.b̂ = ĉ.âb̂.â +
⇒ 1– 0ĉ.âĉ.b̂b̂.â =−+
⇒ 0ĉ).âb̂(b̂.ââ.â =−+−
⇒ 0)ĉâ).(b̂â( =−− ⇒ )ĉâ()b̂â( −⊥−
i. e. the triangle is right angled
50. Ans. < 1, 1, 2 >
Hint:
+=
=+
jin
y x
ˆˆ
3
1 …….(1)
Equation of the plane through A(1, 0, 0) is → ⇒ a(x–1) + b(y–0) + c(z–0) = 0 ….(2)
Qplane (2) is also passing through the point
B(0, 1, 0), then –a + b = 0 …….(3)
∴ cos4
π =
011cba
)0(c)1(b)1(a
222 ++++
++
QAngle between planes (1) & (2) is4
π
⇒ 222 cba ++ = a + b
⇒ 2ab = c2 ⇒ 2a2 = c2
⇒ c = a2
Hence a : b : c = a : a : a2
= 1 : 1 : 2
51. Ans.
3
2,
3
4,
3
2
Hint: )k ˆ ĵî()k ˆ ĵî(r −+−λ+++=r
1
1z
1
1y
1
1x
−
−=
−=
−
− = λ ……(1)
Any point lies on the line (1) is
P(–λ + 1, λ+1, –λ+1) & origin (0,0,0)0
∴ OP = 222 )1()1()1( λ−+λ++λ−
⇒ D = (OP)2 = 2(1–λ)2+(1+λ)2
⇒ λd
dD = – 4(1–λ) +2(1+λ) = 0
λ=3
1 ∴ point P
3
2,
3
4,
3
2
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52. Ans. I 1 = I 2 = I 3
Hint: I1 = ∫π
+
2 /
0
2
2
xcos1
xcosdx
Apply prob. ⇒ I1 = I2
I1 = ∫π
+
2 /
0
2
2
xsin1
xsindx
I1 + I2 = ∫π
++
+
2 /
0
2
2
2
2
dxxsin1
xsin
xcos1
xcos
2I1 = ∫π
+
+2 /
0
22
22
xsinxcos2
xcosxsin21dx
I1 = dxxcosxsin24
xcosxsin212 /
022
22
∫
π
+
+ = I3
∴ I1 = I353. Ans. None
Hint: ∫π
π
3 /
4 /
dxxcosecxcos
∫π
π
3 /
4 /
xdxcot = 3 / 4 / |]xsin|[lnππ
= ln
2
1ln
2
3+
= ln2
3
54. Ans. π ππ π
Hint: f'(x) =)x(f
xsin2
1 2−
∫ ∫−
= dx2
xsin21dx)x('f )x(f
2
c x x f
+=
2
2sin
2
1
2
)( 2
f(x) = 'c2
x2sin+ period = π
55. Ans. even
Hint: the given function can be rewritten as,
f(x) =
≥−
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x =2
0cos
p+
α , y =2
sin
p0
α+
cos α = x2
p, sin α = y2
p
cos2α + sin2α = 1
2
2
)x2(
p +
2
2
)y2(
p = 1
22 y
1
x
1+ =
2p
4
60. Ans. ( h2 – r
2) x – 2 rhy = 0
Hint: (x – r)2 + (y – h)
2 = r
2
(0,0)
Equation of tangent in slope form
y – h = m(x – r) ± r2
m1+
is passing (0, 0)
– h = m (– r) ± r 2m1+
(mr – h)2
= r2
(1 + m2
)m
2r
2 – 2mrh + h
2 = m
2r
2 + r
2
2mrh = h2 – r
2
m =hr2
rh 22 −
y = xhr2
rh 22
−
61. Ans. ( ∑ ∑∑ ∑ xi , ∑ ∑∑ ∑ yi )
Hint:A(x1, y1)
C(x3, y1)(x2, y2)
H(x, y)
(0, 0) 1 2
GO
∑∑ ii y
3
1,x
3
1
ix
3
1∑ =
3
0x +
iy3
1∑ =
3
0y +
y = ∑yi
(∑xi, ∑yi)
62. Ans. 8
9
Hint: ∫−
−
+2
1
2
44
2dx
x x
x2=4y
2
(–2,0)
–1
(0,2)
O
=
2
1
32
432
241
−
×−
+ x x x
=
+−−
8−
3
1
2
3
4
1
36
4
1
=8
9
24
27
246
5==
7+
63. Ans. f '(x) = 0, ∀∀∀∀ x ∈∈∈∈ R
Hint: x = y = z, 3f(x) + f(x)3 = 14
⇒ f(x) = 2
⇒ f '(x) = 0
64. Ans. 20
974 ±
Hint: If sin(α + β) =4
3then cos(α + β) =
4
7
cos(α – β) =5
4 then sin (α – β) = ±
5
3
cos2α = cos(α + β + α – β)
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=
±−
5
3
4
3
5
4.
4
7
⇒ cos2α–
±
20
974
65. Ans. 2
Hint: Let xn
=π
sin3x = sin2x
sin3x – sin2x = 0
2cos 02
sin2
5=
x x
02
5cos =
x
x = (2k +1)5π
5)12(
π+=
πk
n
n=
+12
5
k ⇒ n= –1, –5, + 5, + 1
only 2 values satisfies given equation
66. Ans.1
Hint:m
mm x )) / (cos(lim
∞→
=
m
m m
x
−+
∞→1cos1lim
=
)2 / (sin2
)2 / (sin2
1
2
2
2
2
2sin21lim
m xm
m x
m m
x
−−
∞→
−
= e0 = 1
67. Ans. π /2 – tan–1(9)
Hint :
−
4
3,
3
5dx
dy= tan θ = 2
2
5 + 4 = 9
θ = tan–1 9.
∴ angle with y axis =2
π– tan
–1 (9)
68. Ans. π
Hint : f ' (x) = ex [cos – sinx] = y [let]
∴
dx
dy= – 2 e
x sin x = 0
x = 0, π in [0, 2π]
2
2
dx
yd = – 2e
x[cos x – sin x]
0x
2
2
dx
yd
=
= – 2 [Max.]
π=
x2
2
dx
yd= + 2e
π[Min.]
∴ Min. of f ' (x) = slope is at x = π
69. Ans. a straight line with slope 2
Hint : f " (x) = – f (x) & f ' (x) = g (x)
& h' (x) = (f (x))2 + (g (x))
2
⇒ h ' (x) = 2 f (x). f ' (x) + 2 g (x) g (x)= 2 f (x) . g (x) – 2 g (x). f (x)
h " (x) = 0 ⇒ h ' (x) = const ⇒ slope constant& by h (0) = 2, h (1) = 4 seems it is
a st. line with slope constant.
70. Ans. a = 4b
Hint : Min. value of a tan
2
x + b cot
2
x = ab2 Max. value of a sin
2 x + b cos
2x = a
⇒ a = ab2
⇒ a2 – 4 ab = 0a(a – 4b) = 0
a = 4b 71. Ans. 2
Hint: tan4
π =
α+α
α−α
cossin
sincos
ab
ab
⇒ (bsin α + acos α) = ± (b cos α – a sinα)⇒ b2sin2α + a2cos2α + 2absinα cosα
= b2cos2α + a2sin2α –2absinα. cos α ⇒ (b2 – a2)sin2α + (a2–b2)cos2α = – 4ab sinα cosα =(a
2–b
2)cos2α = – 2ab sin2α .......(1)
0cossin
sincos
α−α
−αα p
pba
⇒ a(–p cosα) – bp sinα –p = 0⇒ –p(acosα + bsinα + 1) = 0⇒ (acosα + bsinα)2 = 1⇒ a2cos2α + b2sin2α + 2ab sinα cosα = 1
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⇒ a2cos2α + b2sin2α+2
)sin)(cos( 2222 α−α− ab=1
⇒
2
)sin)(cosba( 2222 α+α+ = 1
⇒ a2 + b2 = 2
72. Ans. G.P.
Hint: P( 21at , 2at1) Q(22at , 2at2)
T ≡ (at1t2 a(t1+t2))also S(a, 0)`
SP = 21at + a
SQ = 22at + a
ST = )tt(a)atat( 2122
21 ++−
= a2
21
2
21 )tt()1tt( ++−
= a 2122
2121
221 tt2tttt21)tt( +++−+
=22
21
22
21 tt1tta +++
= )1t)(1t(a22
21 ++
= ))(( SQSP
73. Ans. a + 1006 d
Hint:
[a–a+a–a+a…+ a] +d[0–1+2–3+4 …. +2012]2013 times
= a –d [(–1)+(–1) + …. +(–1)] 1006 times
a + 1006d
74. Ans. 3
1
Hint: λ =ternsnnextof sum
ternsnfirstof sum
λ =]d)1n()nda(2[
2
n
]d)1n(a2[2
n
−++
−+ =
nd3)da2(nd)da2(
+− +−
λ will be constant iff 2a – d = 0d = 2a
⇒ λ =nd3
nd =
3
1
75. Ans. 4
Hint: (1–x)5.
44
x1
)x1(1
−
−
= (1–x). (1–x4)
4
(1– x)[1– 4C1.x
4+
4C2x
8–
4C3x
12….]
= ……(–1) (–4C3).x
13……
= ….. +4. x13
……….
76. Ans. Zero
Hint: f '(0) = 0
011
000
101
100
202
101
100
000
010
=++
01
0
1
)0('f
x
)x(f lim
0x===
→(use L Hospital rule)
77. Ans. 3
Hint: 0 × 5! + 0 × 4! + 3!/3! +!2
!2.1 + 0.1! + 1
= 3
78. Ans. 4
Hint: L. C. M of {2, 5, 6, 8} is 120 min so all 4
bells be ringing simultaneously in every 120min so
in 8 hr they ringing 4 times
79. Ans. 335
88
Hint: P
R
E 1
R
R R R
R G
E1 E2 E3
=
+
+
3
3
2
2
1
1
1
1
E
RP)E(P
E
RP)E(P
E
RP)E(P
E
RP).E(P
=335
88
10
3
12
5
11
3
12
4
13
4
12
313
4
12
3
=×+×+×
×
80. Ans. 92
63
Hint: p =2
1,
2
1=q
required prob. =9C4 p
4×q
5 × p
=9C4 × p
5q
5 =
92
63
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PHYSICS
81. Ans. None
Hint :
0
inqdr.E
∈
=→→
∫
E. 4πr2 =0
inq
∈
r
dr
=0
r
0
2drr4
ε
π×ρ∫
E. 4πr2 =0
1
∈ drr4cr 2
r
0
2 π∫
E. 4πr2 =0
c4
∈
π
5
r5
E =0
3
5
cr
∈
∴ At r =2
R, E =
05
c
∈
3
2
R
=
0
3
40
cR
ε
82. Ans. 2 d
4
π ππ π
φ φφ φ
Hint : φmax = E S
φ = E.4
d2π
E =2d
4
π
φ
83. Ans. increases with increase in refractive index of
a prism
Hint : δ = A (µ–1)
84. Ans. 91 Ω ΩΩ Ω
Hint : Use 100 =S1000
S1000
−
85. Ans. high ρ ρρ ρ , low α αα α
86. Ans. Small temperature difference 30° to 35°C
between hot liquid and surrounding
87. Ans. 24 Ω ΩΩ Ω
Hint :30
20 =
R
16
R =20
3016× = 24 Ω
88. Ans.
16
9 I
Hint : I1 = I0 cos2 30° =
4
I3 0
and I2 = I1 cos230° =
4
I3 0 ×4
3
=16
I9 0
89. Ans. When a current flows through a solenoid, it
has a tendency to increase its radius if no
external magnetic field exists in the space
Hint : Solonoid is wounded over a cylinder hence
magnetic field can be induced in it
90. Ans. 120°
Hint :
× ×× × ×
r × ××× ××
120°P
2
r
O
P M
rθ
O
2
r
M
cosθ =2
1 or θ = 60°
Deviation = 120°
91. Ans. the reading of the voltmeter is 50 V
Hint : Reading of volt meter = 50V
200 Ω 200 Ω
100V
92. Ans. 3
2v
Hint : v =R
GM
v' =
2
RR
GM
+ =
R3
GM2 v' =
3
2v
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93. Ans. mg and T only
94. Ans. 0.5
Hint : a =
37
49
+
= 4.9 m/s2
Now Fps = µ MgM (a) = µ Mg
95. Ans. decreases
Hint : dw → +vedU → –veT↓ , PV↓
96. Ans. 10 gm
Hint : H.C = 42 J/C = 10 cal/C
weq = 10 gm
97. Ans. A →→→→Q ; B→→→→ T ; C →→→→ R,S ; D→→→→ R, S ; D→→→→ P
Hint : charle's law V ∝ T A → Q
(B)2
1 KT B → T
(C) Adiabatic Pressure dW = –dU
P ∝ Tγ / γ –1 C → R,S
(D) Work done PdV = n R d T
D → P
98. Ans. y = 0.1 sin
+ 4 4t
π ππ π
Hint : KE at mean position =2
1 mω2 a2 = 8 × 10–3
or ω =2
1
2
3
ma
1082
×× −
=2
1
2
3
)1.0(1.0
1082
×
×× − = 4
Equation of SHM is,
y = a sin (ωt + θ)= 0.1 sin
π+
4t4
99. Ans. None of these
100. Ans. 21
200
Hint :
2O
He
V
V =
325
RT7
43
RT5
×
× =
743
3255
××
×× =
21
200
101. Ans. 1 : 1
Hint : Let2
1
M
M =
1
2
M1' = M1 ×
4
2
1
M2' = M2 ×
3
2
1
∴'
2
'1
M
M =
2
1
M
M ×
2
1 =
1
1
102. Ans. 6.5 MeV
Hint : 92U236
→ 92U235
+ 0n1
236.123 amu 235.1214 amu 1.008665 amu
∆m = 236.123 – 236.130∆m = 0.007 amu∴ Energy = ∆mc2 = 0.007 × 931 = 6.5 MeV
103. Ans. A →→→→ U; B →→→→ R; C →→→→ P; D →→→→ S
104. Ans. rate of temp. falling of A is half that of B
Hint : Rate of temperature falling ∝ m
A
105. Ans. R = xy
Hint : For series connection x = nR. For parallel
connection y =n
R. Therefore xy = nR ×
n
R = R
2
106. Ans. greater of less than its emf depending on the
direction of the current through the batteryHint : Terminal voltage V = E – I r, where E is the
emf and r the internal resistance of the battery. If
the current is flowing from the +ve to the negative
terminal in the external circuit V < E. However, ifthe current is flowing in the opposite direction, a
voltmeter across the battery would show V > E.
107. Ans. Momentum of inertia and moment of forceHint : Do yourself
108. Ans. 3
2
Hint : hascending = hdescending
2
0v + ta =
2
'v0 + td
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where v = velocity with which object is projected
v' = velocity with which object is returns
ta = time in ascending motion
td = time in descending motion
∴d
a
tt =
V'V =
h)2g(2
h)2g(2
+− =
128 =
32
109. Ans. 12 : 1Hint : Angular speed of hour hand,
ω1 =)6012(
2
×
π
Angular speed of minute hand ω2 =60
2π
∴1
2
ω
ω =
60
6012× =
1
12
110. Ans. 288 J
Hint : 4v + 8 × 6 = 0 or v = – 12 m/s
K.E. =2
1 × 4 × (12)
2 = 288 J
111. Ans. h/4
Hint : At maximum height, the energy is entirely
potential (mgh). This is equal to the initial kinetic
energy. Let h' be the height at which kinetic energy
is 75% of its initial value. At this height, the
potential energy must be 25% its maximum
potential energy. Thus, at this pointm g h' = (0.25) m g h
or h' = h/4
112. Ans. β ββ β > 4
3 α αα α
Hint : y = x tan θ θ
−22
2
cosu2
gx
β = α tan θ –θα× 2
2
cosg22
gx
⇒ α tan2
θ – 4α tanθ + 4β + α = 0It tanθ is real 16α2 ≥ 4 α (4β + α)
or β ≤4
3α
So it not possible
α>β
4
3
113. Ans. The balance A will read less than 2 kg and B
will read more than 5 kg
Hint : Force of upthrust will be there on mass
m shown in figure, so A weighs less than 2 kg.
Balance will show sum of load of beaker and
reaction of upthrust so it reads more than 5 kg.
114. Ans. v m t F ∆∆∆∆∆∆∆∆ =
Hint : vmtF ∆=∆ ⇒ t
vmF
∆=
By doing so time of change in momentum
increases and impulsive force on knees decreases.
115. Ans. ω ωω ω L =C
1
ω ωω ω
116. Ans. X = 1, Y = 0
Hint : P = YX +
O = Y.X
R = Y.X)YX( ++
117. Ans. 1.1 × 10 –9
m
Hint : λ =mv
h
118. Ans. 4
1 amp
Hint :15
55
V = I R
5 = 20 I
I =4
1
119. Ans. tan –1 (3/2)
Hint :'
tan
tan
δ
δ = cos α
tan δ = cos α × tan δ' = cos 30º × tan 60º
=2
3 × 3 =
2
3
δ = tan–1 2
3
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120. Ans.π ππ π
µ µµ µ
2
Iv0 loge
+
r
L r
Hint : v
dx
r
x
L
dε = B v d x
= ∫+
π
µLr
r
0 dxx2
vI
E =π
µ
2
Iv0 ln
+
r
Lr
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