sri chaitanya iit academy., india
TRANSCRIPT
Sri Chaitanya IIT Academy., India A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI
A right Choice for the Real Aspirant ICON CENTRAL OFFICE, MADHAPUR - HYD
Sec: Sr. ICON ALL Jee-Main Date: 30-12-18 Time: 09:00 AM to12:00 Noon CODE-C Max.Marks:360
KEY SHEET PHYSICS
1 1 2 1 3 2 4 2 5 3 6 4 7 1 8 2 9 2 10 3 11 2 12 1 13 1 14 3 15 2 16 4 17 3 18 3 19 3 20 1 21 4 22 4 23 2 24 3 25 2
26 1 27 2 28 4 29 3 30 2
MATHS 31 2 32 1 33 2 34 1 35 1 36 1 37 2 38 3 39 4 40 3 41 2 42 3 43 4 44 2 45 1 46 3 47 4 48 4 49 2 50 2 51 3 52 4 53 3 54 2 55 3 56 3 57 2 58 1 59 3 60 2
CHEMISTRY
61 3 62 3 63 3 64 4 65 3 66 1 67 3 68 2 69 2 70 3 71 4 72 4 73 3 74 3 75 3 76 3 77 3 78 2 79 2 80 3 81 1 82 3 83 1 84 1 85 3 86 2 87 3 88 4 89 2 90 1
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
SOLUTIONS PHYSICS
1. Density,
Mass Md
Volume v
Here m = 2.42g, m 0.01g 3v 4.7cm V 0.1cc
Maximum error in density, d 100d
m V 0.01 0.1100 100 100 100 0.413% 2.127% 2.54%m V 2.42 4.7
2. A0v 0
1
2
l 3l 2
A1
A 2
V 3 3V 3 2 5
A1vv 3x 0.6v5
3. Conceptual 4. The acceleration of the horizontal movable pulley will be 6m/s2. So for the block B. 23 6, 2 /B Ba a m s 5. Free body diagram of the two bodies are as follows
Maximum friction between the two blocks is maxf mg , where m = 2kg maxf (0.5)(2)(10) 10N Let acceleration of both the blocks be a towards left. Then
f 2 20 fa
2 4
2f 4 20 f f 8N Now since, maxf f i.e., static region of 2kg on 4kg, hence friction force between the
two blocks is 8 N. 6. Conceptual 7. From work energy theorem W K
212
Pt mv
Or 2 ...(1)Ptvm
4kg2F 20 N
2kg1F 2N
f
f
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
Sec: Sr. ICON ALL Page 3
1/22ds Ptdt m
Or 1/2
0 0
2s tPds t dtm
Or 3/22 2 .... 23
Ps tm
From Eqs.(1) and (2) 23
s tv
8. From parallel axis theorem.
X CM
2
1 12MLI
22 1I I Mx
22 3
12 2 2ML LM
211
24ML
9. About point of contact I
23 0 8mgr mg mr
8gr
mg
10. 2 2 0 2cos 45 gO g Rw wR
11. If an bubble is formed, its radius is equal to that of capillary
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
Sec: Sr. ICON ALL Page 4
Required pressure = 02sghr
12. PP TT
,
oo
o
PT
3 32 2
oB o
o
PT
13. 3PT Constant
VPTConstant
2VT Constant 1/2TV Constant
11 1.52
14. Apparent Frequency of direct waves received by observer is
1s
Vv vV V
Apparent frequency of reflected waves received by the observer is 2s
Vv vV V
No.of beats per second 1 2v v v . 15. The wave pulse is travelling along positive x –a xis. Hence at and bx should have
opposite signs. Further wave speed
int
Coefficient of tvCoeffic of x
41
Coefficient of t
Coefficient of 14t s 16. S-Force of interaction
3 3
0
2 1 14 1
pqd d l
17. Conceptual 18. 0/ 2 ',E r where is the linear charged density of the inner cylinder.
And 0
ln2
b
a
bV Eda
Now, . .l J dA E dA
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
0
22
r drr
19.
The current through ammeter 4 3 2 1I I I I I
6 4 14 66 0 8 62 4 4 8
1 13 1 3 1 22 2
A
When only 4 resistance is shunted / 5gi i 20. / 5 4 4 / 5 16G i or G
21. For a circular wire
22. Conceptual 23. The emf induced in a conductor does not depend on its shape but only on its end
points. We replace the actual conductor by an imaginary straight conductor joining its two ends.
24. 00 0 0
2
3
hc hc hceV
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
Sec: Sr. ICON ALL Page 6
0 0
1 2.4
Q hceR
25. Conceptual 26. If screen is perpendicular to the line joining the sources the fringes will be circular and
central fringe will be bright if 1 2S S n 27. When light passes through a medium of refractive index , the optical path it travels is
t Path difference t b b 1 For a small element ‘dx’, path difference, x 1 ax 1 dx axdx
For the whole length,
2
0
atx axdx2
For a minima to be at’O’,
x 2n 12
2at 2n 1
2 2
For minimum ‘a’, n = 0
2
2at a2 2
28. When a real object is placed 25 cm from a lens, a real image is formed. Mark the is
INCORRECT statement from the following. 1) The lens is a converging lens 2) The image may be magnified or diminished 3) The focal length of the lens is less than 25 cm 4) The focal length of the lens may be greater than 25 29. CONCEPTUAL
30. 8
8 19 10 1.5 106
v msk
Refractive index 8
8
3 10 21.5 10
cnv
Also r rn For a non – magnetic medium 1.r Therefore 4r rn
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
Sec: Sr. ICON ALL Page 7
MATHS
31. 2 2
2 15 5cosx y
2
221
1 bae = 1
225cos 1 cos ;
5ly encentricity of the elipse
2 2
2 125cos 25
x y is
22 22
25cos1 sin ;25e put 1 3e 2 2
2 1 23e e e
2 2 2 11 cos 3sin 2 4sin sin ]
2
32. conceptual
33. 2A=XY sin ; 4
2 2 2 2 24 sin ; ( ) ; '( ) 0 ]2 1 3
xA x y f x f x xx
34. Use sin x expansion, sin x 3 5
.......3! 5!x xx
35. 3 4 3 4 1 02 3 2 3 0 1
BC BC I
2( )
2 2r r rA At A t t
+……….
2
1 1( ) ( ) ......2 2r r rt A t A t A
( ) 2 ( ) 2(2 1) 6
1 (1 / 2)r
rt A t A
36. Conceptual 37. 2 3 2 2
0 1 2 1 2(1 2 5 10 )[ ......] 1 ......X X X C C X C X a x a x
1 2a n and 2( 1) 2 5
2n na n
Put 221
2aa
2( 2) ( 1) 4 10n n n n
2 24 4 5 10n n n n
6n 38. CONCEPTUAL 39. Lines are 1 0;x y 4 3 4 0x y and 0x y where 2 2 2
1 1 14 3 41
=0
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s 1 3 – 4 – 1 4 – 4 1 4 – 3 3 – 4 – 4 4 4 – 3
– 1 0 1 1 ] 40. Conceptual 41. Conceptual
42. f (x) = 23 ;
4t x x
x
f ' (x) =
2
2
( 4)(3 2 ) ( 3 )( 4)
x x t x xx
for maximum or minimum, f ' (x) = 0 – 2x2 + 11x – 12 – t – 3x + x2 = 0 – x2 + 8x – (12 + t) = 0 43. Shifting the origin at A equation is X2 = – 8Y Now (x – 2)2 = – 8(y – 2)
44. q2 - 4 p r = 0, p > 0, p > 0 f(x) = log (px3 + (p + q) x2 + (q + r) x + r) Let g(x) = px3 + (p + q) x2 + (q + r) x + r g(x) = (x + 1) (px2 + qx + r) Discreminant of px2 + qx + r = q2 – 4pr = 0 Domain (x + 1) (px2 + qx + r) > 0
p(x + 1) > 2 x – and x > – 1
x R – [(– , –1]
45. f is not differentiable at x = 12
g is not continuous in [0, 1] at x = 0 h is not continuous in [0, 1] at x = 1 & 0 k (x) = = (x + 3)p where 2 < p < 3 ] 46. x = cos , [0, ]
cos–1(cos ) + cos–1 =3
2
p2qx
p2q
5n2)3x( l
sin3
sincos3
cos
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
cos + cos–1cos 3
= 3 can hold only if 0
3 –
– 3 – – 2
3 x 1 ,1
2
0 03 ]
47. No of integral solutions 1 2 2 4 2 * 5 * 7 *11x x x x CASE-I ALL 0xi No of solutions = 4(4!) CASE-II All 0xi No of solutions = 4(4!)
CASE –III Two positive two negative 42C = 6 COMBINATIONS
No of solutions = 6* 4(4!)
Total no of integral solutions = 8* 4(4!) 48. Hint : find ‘c’ assuming the line is the tangent to the circle. And plot the inequalities
on x-y plane. 49. 22 + (x1 – 1)2 = x12
4 + x12 + 1 – 2x1 = x12 5 = 2x1 or x1 = 5/2 Equation of (1) from (2, 5/2) to the given base y – 5/2 = 2 (x – 2)
2y – 5 = 4 (x – 2) at y = 1 –3/4 = x – 2 or x = 5/4 ]
50. f '(x) =
Q f (0) = 0 Þ f ' (x) = f ' (x) = f ' (0) + | x | = | x | ] 51. Conceptual
52. [Hint :A = = =
f(a) = now f '(a) = = 0 Þ a3 = 1 Þ a = 1 53. Conceptual 54. Conceptual
hxhh|x|)h(Lim
h)x()hx(Lim
2
0h0h
fff
xh|x|h
)0()h(Lim0h
ff
dxx1
6xa2
a2
a2
a
2
x1
12x
a1
12a
a21
3a 22
a21
4a2
2a21
2a
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s 55. Conceptual 56. Conceptual
57. u = put x = tan q
= = = = ln 2 – u
u = ln 2 4u = ln 2 ....(1)
again v = (put 2x = t) ; v = = v = – ln 2 ....(2) (1) + (2) Þ 4u + v = 0 ] 58. Conceptual 59. x2y = c3
x2 + 2xy = 0 = = equation of tangent at (x,y)
Y – y =
Y = 0, gives , X = = a
and X = 0 , gives , Y = 3y = b Now a2b = = ] 60. Conceptual
CHEMISTRY 61. Meq of salt = Meq of 2 3Na SO 50 0.1 n 25 0.1 2 n 1 changein O.N.
3 2M e M 62. Polarity in a molecule gives rise to an increase in forces of attraction among molecules
and thus, the boiling point increases.
63. Let w g of each be taken, then initial mole of w wP ;molof Q10 20
Final mole of 1wP5 10
Final mole of 14wQ20 5
For 0 1N t
N
PP e
P For 0 2N t
N
QQ e
Q
For 1 20
2
w 5 10P e10 w
For 2 20
1
w 20 5Q e20 w 4
By Eqs. (i) and (ii) 1 2 204 e 1 2 e20 log 4
1
02 dx
1x)1x(nl
4
0
d)tan1(nl
4
0
dtan1tan11nl
4
0
dtan12nl
4
8
2
2
0
dx)x2(sinnl
0
dt)t(sinn21 l
2
0
dx)t(sinnl
dxdy
dxdy
xy2
)xX(xy2
2x3
y3.4x9 2
32 c4
27yx4
27
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s
Sec: Sr. ICON ALL Page 11
or e1/ 2
0.693 0.69320 log 410 t
t
64. Higher vapour pressure of 2H O in atmosphere will derive 2H O vapours to the solute particles
65. phenomenon of conversion of freshly precipitated mass into colloidal state by the action of solute or solvent is called peptization
66. 2 5 2 2d N O d NO 2d O1dt 2 dt dt
2
1 2 5 2 5 3 2 5kk N O N O 2k N O2
67.
22 32 5
c2 4
2 10 / 2NOK 1 10
0.2N O2
68. For NaX, 21 h h h
X H O HX OH
14
wh5
a
KK 10hC K C 10 0.1
8 410 10 4 2%H 10 100 10 0.01 69. T 300K, V 10 1 9L H E pV E 2 RT 0 2 8.314 300 4.98kJ E=0 for isothermal process 70. This gives rise to higher e nuclear charge in Na and the size of Na becomes smaller
due to more effective pull of valence shells towards nucleus. 71. Allotropes of an element have the same chemical properties but have different
arrangement of atoms and physical properties 72. N a is soluble in NaOH whereas 3cFe OH is insoluble. 73. 4 2 4 2FeSO 7H O FeSO 7H O; 4 2 3 2 32FeSO Fe O SO SO 74. 2 5 2 3HO SO OH PCl Cl SO Cl POCl 2HCl 75. The solubility of nobel gases increases with increase in molecular weight due to
increase in van der Waals’ forces. However, these are sparingly soluble. 76. According to Werner’s theory, only those ions are precipitated which are attached to
the metal atoms with ionic bonds and are present outside the coordination sphere. 77. PbS is black and 2S reacts with 2 2 7K Cr O to give 2 4 3Cr SO solution which is green.
78. cellE for the given change = 2
2
H I2
H II 1
H p0.059 log2 p H
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s Thus, for positive
H 2II2 Icell P HE P 79.
80. In D-(+)-tartaric acid, the (+) is due to positive optical rotation and is derived from
D glyceraldehydes.
81. It undergoes dehydration easily as the product obtained is conjugated, and is more
stable.
82. 2 2 2
2H O H O H
2 2 2 3 3 2NiHg catalystCaC C H CH CHO CH CH OH
83.
2 2
2H O Br
3 3 3 3 3NaOHH .HgCH C CH CH COCH CHBr CH COONa
Since, B and D are different, thus B is 3 2CH CH CHO and so A is 3 2 2CH CH CHCl .
84. Must be a tertiary alcohol as it gives alkene on treatment with Cu. Thus, 4 8C H O is a
ketone.
85. Follow applications of inductive effect. The negative charge on carboxylate ion is
dispersed more due to IE of F atom.
The carboxylate ion thus becomes more stable and the acid becomes more reactive.
86.
Sri Chaitanya IIT Academy 30-12-18_Sr.ICON ALL_JEE-Main_GTM-7_Code-C(12th Jan)Key & Sol’s 87.
88.
89. DNA has deoxyribose sugar; RNA has ribose sugar with three bases common as
adenine, guanine and cytosine, DNA has fourth base thymine; RNA has uracil
90. Bakelite is a step-growth polymer i.e., the condensation involving the reaction of
functional group e.g. terylene, Bakelite etc.