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AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
1
ANSWERS, HINTS & SOLUTIONS
FULL TEST V (Paper-1)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
1. A B B
2. A C A
3. D B C
4. B A D
5. A B C
6. B A C
7. A C B
8. C A D
9. A, C B, C A, C
10. A, D B, C A, D
11. B, C A, B, C, D A, C
12. A, B, D A, B, C C, D
13. C D B
14. C A A
15. B C D
16. D A D
17. A A C
18. C B B
1. 3 2 8
2. 2 4 4
3. 6 0 4
4. 6 1
6
5. 2 4
1
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AITS-FT-V-(Paper-1)PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
2
PPhhyyssiiccss PART I SECTION A
1. 1 1 2 2CM1 2
m v m vV 1 m / s
m m
At maximum extension both blocks will move in same direction with VCM now use energy y conservation.
2. 2MF N
2 12
F N = Ma N = Ma1
2
1MN
2 12
Acceleration of hinge O
= 1 1a a2 2
Nx = F/4 a = 5F/4M and = 9F/2M
oF a a i i
2 M
F
a1 1
N N a
3. 0y y sin kxcos t and 9 L4
4. 1 2d KA T Tdt L equation of heat flow through the box.
5. Just after collision.
a b v
eucos
2
22 1
Lm v ucos a b m a12
u sin
v
6. 2
21L constant2 2C
7. Using COM; vcos = u sin
and 22v sine 2tanucos
2tan2 < 1
tan < 1/2 1 1tan2
u
v v
8. Power factor = VRMS RMS cos
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AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
3
9. For maxima n pyd
for minima py 2n 12d
10. yy A sin t kx A cos t kxt
11. mediumCV
12. Isothermal PV cons tan t adiabatic rPV constant 13-14. 0h eV 15-16. 2E mc
17-18. 0 0S
C Vf f
C V
SECTION C
1. 3 2km/hr 2. The centre of mass of system will not move horizontally.
3. ReldMF Udt
4. In steady state the current in branch containing capacitor will become zero. 5. F q E V B
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AITS-FT-V-(Paper-1)PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
4
CChheemmiissttrryy PART II
SECTION A
1. 1 1 3
22 4 3NH OH H N O 2NH 2H 3H O
2.
P S
SP
SP
S
P
S
SS
S
S
Has 3 p - d P S bond Has 3 P = S bonds Has 3 tetrahedral units of P Has 6 P S P linkages. 3. 4 2 3 2KI 2KMnO H O KIO 2KOH MnO (nf = 6) (nf = 3) Moles of KI 6 = 3 0.2 10 10-3 Moles of KI = 10-3 = 1 milli mole = x
22 4 2S C N I H SO HCN I H O
To produce 10-3 moles I ,
Moles of 3SCN 6 10 1
Moles 3
410SCN 1.6 10 moles6
4. O CH3
2
HH O
OH
3CH OH
Tautomerises
O
6 5C H CHOOH
O
CH5C6 H
OH
2 4N H / OH A
CH5C6 H
OH
D 2 enantiomeric pairs possible for (D).
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AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
5
5. 2 2CoCl 2KCN Co CN 2KCl
Buff coloured
42 6Co CN 4KCN K Co CN 6. Half cell Reaction Eo L.H.S.
2H 2H 2e ooxE 0.00 V
R.H.S. 2Zn 2e Zn oredE 0.76 V Net 2 o2 cellzn H Zn 2H , E 0.76 V
2
2
H 0.0591K log KnZn
ocell cell0.0591E E log K
n
2
10log H0.05910.46 0.762 0.3
6H 4.60 10 M
2
3 3
6 3
HSO H SO
0.4 M 4.6 10 6.4 10
2 6 33 8
a3
H SO 4.6 10 6.4 10K 7.3 10HSO 0.4
pKa = 7.13 7.
MeOHOMe
OH
O
O
3PClOMe
Cl
O
O
2MeNHOMe
NHMe
O
O
NMe
O
O
C
O
C
O
O
8. Shortest of Lyman series 1 for H-atom H 1 H21
1 1 1R R1
9. (2) 2 23 4B OH NaOH NaBO Na B OH H O can be made to proceed forward by
adding cis-diol to stabilization by chelation. (3) The solution turning milky on passing H2S through group II solution indicates the presence of
oxidizing agent where H2S is oxidised to 8O2. 10. (A) The heat of formation for NO2 is calculated as 45 kJ mol-1. (B) For reaction 2 2 2N 2O 2NO , H 310 kJ / mole. i.e. it is exothermic.
(C) o2 21 1N O NO, H 55 kJ / mol2 2
(D) o2 21NO O NO , H 100 kJ / mol,2
it is exothermic.
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AITS-FT-V-(Paper-1)PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
6
11. (A) No H-bonding can occur in chlorobenzene, so +I effect increases acidic strength. (B) Due to both steric and polar effect has no ortho effect is observed in case of o-MeOC6H4NH2.
MeOC6H5NH2. p-MeOC6H4NH2 > o-MeOC6H4NH2 > m-MeOC6H4NH2 pKa = 5.24 (pKa = 4.45) (pKa = 4.2)
(C) N > NH2 >
N
N
pKa = 5.25 pKa = 4.63 pKa = 1.3
(D)
NH>
Pyrrolidine
(CH3CH2)2NH
(pKa = 11.27) (pKa = 10.98)
12. (A) is correct, 2sys1
VS nR nV
, V2 > V1.
sysS ve (B) PV = constant, as is high, slope is higher. (D) At boiling point, the process is at equilibrium G 0 . 14. x = 6, y = 3, z = 2 Solution for the Q. No. 13 & 14. 53 2 6 2IO 6OH Cl IO 3H O 2Cl
5 H6 5 6IO H IO
o o
2
100 C 200 C5 6 4 2 54H O2H IO 2HIO I O .
17. V. D. = 28.75
D d 0.6n 1 d
2 4 2N O 2NO
At equilibrium : 1 0.6 1.20.4
2 42
N O
NO
P 0.4P 1.2
If moles of N2O4 escaping out is x, then for NO2 it is (1 x)
x 0.4 46 x 0.1881 x 1.2 92
2 4 2N O escaping out NO escaping outX 0.188, X 0.812
18. 22 4
2
NOP
N O
1.2P 1.6K 2.2
0.4P1.6
KP does not change.
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AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
7
SECTION C 1.
As
Cl
Cl
AsCl
ClCl
Cl
Cl
Cl
2.
3 5 5 2
2 2
CrO C H NCH Cl
Collins reagent
B
O
OH
O
O
O
O
H
ArOArO
3. copolymerisation2 2 2nCH CH CH CH nCH CH
CN
2 2 2CH CH CH CH CH CH
CN
n
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AITS-FT-V-(Paper-1)PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
8
MMaatthheemmaattiiccss PART III
SECTION A 1. If [x3 + x2 + x + 1] = [x3 + x2 + 1] + x, then x is integer log |[x]| = 2 |[x]| has same solutions as log |x| = 2 |x|, x is integer No integral solution 0 2. Each element has 5 choices, as it may be in any one of A1, A2, A3, A4 or may not be in any one The required number of ways = 57 3. (1 + x)n = C0 + C1x + C2x2 ..... Cnxn
n 1 2 3 n 11 2 n
01 x 1 C x C x C x
C x .....n 1 2 3 n 1
n 2 2 3 n 20 1 n1 x C x C x C xx 1 .....
n 1 n 2 n 1 n 1 n 2 2 1 3 2 n 1 n 2
Put x = 2, we get
2 3 n 2n 20 1 nC 2 C 2 C 23 1 2 .....
n 1 n 2 n 1 2 1 2 3 n 1 n 2
2 n 1 n 20 1 nC 2 C 2 C 2 1 1 3.....2 2 3 n 1 n 2 n 1 2 n 1 n 2 2 n 1 n 2
n 2
n3S
2 n 1 n 2
9
78
6
S 3 2 7 8 7S 32 8 9 3
4. It is given that |z| = 1 z = cos + i sin (let), (0, 2)
Now, z z 1z z gives 2 |cos 2| = 1
1 1cos2 ,2 2
5 7 11 2 4 5, , , , , , ,6 6 6 6 3 3 3 3
i.e. 8 values
5. As cot tan = 2 cot 2 tan = cot 2 cot 2
1 1 1tan tan cot cot2 2 4 4 4 4
6. Latus rectum, 4a 2 2
1a2
Vertex = 1 1,2 2
Axis = x y = 0 As origin lies outside the parabola Focus = (1, 1), Directrix = x + y = 0
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AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
9
Equation of parabola is 2
2 2x y x 1 y 12
x2 + y2 2xy 4x 4y + 4 = 0 7. Tangent at (0, 0) will be same (3 + sin B)x + (2 cos )y = 0 and 2 cos x + 2cy = 0 are same
22cosc
3 sin
cmax = 1 where sin = 1 and = 0
8. Consider g(x) = xf(x) g is continuous on [0, 1] and differentiable on (0, 1) f(1) = 0 g(0) = 0 = g(1) By Rolles Theorem g(c) = 0 some for c (0, 1) cf(c) + f(c) = 0 9. At x = 2, RHD = 5 + |1 x| = 5 + 1 = 6 whereas LHD = 5 11. Let the coordinates of point A are (ct, c/t) So, the slope of normal at A will be t2. And normal will be parallel to BC. So, t will be 2 c = 2. 12. a must be negative and x2 a = x should have no solution
Now, D < 0 1 + 4a < 0 1a4
0
y=x
13. ABC bPQR a
2b 3 3 3 3ABC a aba 4 4
14. Slope of OP and OQ will be 1amb
and 2amb
respectively
If POQ = than tan = 1 2 1 2
2 2 21 2
1 22
a m m ab m mba b a m m1 m mb
Now, area of sector OPQ of circle x2 + y2 = a2 = 2
2a 1a2 2
Area of sector OAB = 2b 1 1a aba 2 2
where = 1 21
2 21 2
ab m mtan
b a m m
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AITS-FT-V-(Paper-1)PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
10
15.-16. LCM = 90 and CL = CM
CLM = CML = 4
Let ACB = 2
LAC = 4 and LCB = 3
4
By sine Rule in ABC
a2 + b2 = 2 2 2 34R sin sin4 4
A
D 4
L
4
C
4
M
B
a2 + b2 = 2 2 24R sin cos4 4
= 4R2
Let ADC = ABC = CBM =
4
CAD = 2 4
BAC = CAD BC = CD
17.-18. We want to have an = k if k k 1 k k 1
n2 2
n is an integer, this is equivalent to k k 1 k k 11 1n2 8 2 8
2 21 1k k 2n k k4 4
1 1k 2n k2 2
1k 2n k 12
Hence, n1a 2n2
= 2 =
17. Now, a = 2, b = 3, c = 5 Let A = number of numbers which are divisible by 2 B = number of numbers which are divisible by 3 C = number of numbers which are divisible by 5 Required number = A + B + C A B B C C A + A B C
= 1000 1000 1000 1000 1000 1000 1000 7342 3 5 6 15 10 30
18. a = 2, b = 3, c = 5, d = 7 Hence, the given number is 25 35 53 73 4n + 1 is odd number therefore the factor 2 will not occur in divisor. 3 and 7 are of 4n + 3 form,
odd powers of 3 and 7 will be of 4n + 3 form and even powers will be 4n + 1 form 5 is 4n + 1 form and any power of 5 will be of 4n + 1 form Number of divisors of 4n + 1 type = number of terms in the product (1 + 32 + 34)(1 + 5 + 52 + 53)(1 + 72) + number of terms in the product (3 + 32 + 35)(1 + 5 + 52 + 53)(7 + 73) = 48
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AITS-FT-V-(Paper-1) PCM(Sol)-JEE(Advanced)/15
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
11
SECTION C 1. As g(x) is inverse of f(x) Domain of g(x) is range of f(x) a = 3, b = 11 2. Required number of ways = co-efficient of x30 in (x2 + x3 + x4 + x5 + x6 + x7 + x8)5
= co-efficient of x30 in
52 7x 1 x
1 x
= co-efficient of x20 in (1 x7)5(1 x)5 = 24C20 517C13 + 1010C6 = 826 A = 8, B = 2, C = 6 A + B C = 4
3. 2 2
3 3x 1 1 x 1 3 = 3 3x 1 1 x 1 3
For x [0, 2] 1 3x 1 3 f(x) = 2
2
0
f x dx 4
4. Orthocentre is foot of perpendicular drawn from origin on the plane