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AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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ANSWERS, HINTS & SOLUTIONS
CRT – III (Main)
PHYSICS CHEMISTRY MATHEMATICS 1. B B A 2. B B D 3. B B B 4. D D B 5. D D D 6. A A B 7. A B B 8. B C C 9. A A B 10. A B C 11. B B B 12. C B D 13. B B B 14. C A B 15. B B C 16. B B C 17. A B D 18. B B C 19. D D C 20. A B A 21. C B A 22. A D C 23. A D A 24. D C A 25. A D C 26. C A D 27. D D B 28. C C B 29. D D A 30. C B A
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AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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PPhhyyssiiccss PART – I
SECTION – A 1. Tdt Mv Mu and T d t Mv
uv2
2 2
A newv u 3MgT Mg M Mg Ml 4l 2
A oldT Mg
M
TA
Tdt Tdt
v
v u
2. Let equivalent resistance between A and B is R.
ViR V, i R3
ii3
2V 2iR3 3
R2
B
A R1 R1
R i i – i i
2R 2R
12ViR3
1
2
R 4R 3
3. Conserving angular momentum
2
23mRmvR 3mR2
2v 3R6R
4. Let at any time a charge q flows through the circuit. The
circuit is as shown
Time constant 2RC( )3
+ –
2C
4V
C
R + –
q
3t
2RC4CVq 1 e3
5. 1
MM 2
21
1
VI
1R LC
Similarly, 2
MM 2
22
2
VI
1R LC
Since, both 1MI and
2MI are n times less than resonance current amplitude,
AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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M1 M2I I
M M2 2
2 21 2
1 2
V V
1 1R L R LC C
or 1 21 2
1 1L LC C
or 2 21 2
1 2
LC 1 1 LCC C
or 1 2 1 2 1 2LC( ) ( ) 0 or 1 21 LC 0
21 2 0
1LC
0 1 2
6. 01 Tf2l
0
0
df 1 dTf 2 T
0Tf 2 15 100 15021
Hz.
7. W PdV and P V 9. The object distance for the mirror
2015 304 / 3
cm
For mirror,
1 1 1v u f
1 1 1v 30 20
v 60 cm
The distance from water surface equals 60 – 15 = 45 cm
12. Least count = PitchNumber of divisions on circular scale
= 1mm 0.01mm 0.001cm100
Zero error = – 0.03 mm Measurement = 3.76 – (– 0.03) = 3.79 mm
13. cma aˆ ˆr i j2 2
(considering A to be origin AD to x-axis and AB to be y-axis)
i cm cm 0a aˆ ˆ ˆ ˆL mr v m i j v (2i 4 j)2 2
= 0 0ˆ ˆ ˆmav (2k k) mav k
AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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2 2 2
2 2f
m a ma maL (a a ) m12 2 6 2
From conservation of angular momentum about point A f iL L
20
2 ma mav3
03v2a
0 0cm
3v 3va av2a2 2 2 2
14. If 0V be volume of the gas at C, 0 0 0P V 3RT W = 0 0 03R(2T )ln2 P V = 0 06RT ln2 3RT = 03RT [2 0.693 1] = 03RT [1.386 1] = 01.16RT 15. 0mv R = I + mvR
= 2 vmR mvRR
v = 0v2
iK = 20
1 mv2
fK = 2 2
2 2 20 02
v v1 1 1 1mv I m mR2 2 2 2 2 4R
= 20
1 mv4
So fW = 20
1 mv4
16. Initial current in 3R , i1 3
EiR R
Final current 2f
2 3 2 31
2 3
REiR R R RR
R R
= 2
1 2 3 2 3
R ER (R R ) R R
f1 3
1 32
EiR R R RR
As f ii i , so current in the resistor R3 will decrease.
AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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17. y
x
vdy 1dx v3
So x yv v 3
2 2 2x y 0v v v
2 2 2y y 03v v v
0y
vv
2
0x
3vv
2
So 0
0
3v223vy e
1y e 18. If the string is pulled down by distance x, it is equivalent that a mass of x is removed from the
upper end of the string and is connected to the lower end So g i fW U U U
gGM( x) GM xWR L R
= 1 1GM xR R L
19. amax = 2A = Km
A
If mass and amplitude both are doubled maximum acceleration will not change. 20. Resultant wave amplitude is given by A = 2 2
1 2 1 2 1 2A A 2A A cos( ) Value of cos(1 – 2) lies between 1 and – 1 Hence A1 – A2 A A, + A2 24. If three waves meet in same phase then the resultant amplitude will be algebraic addition of all
three amplitudes R = 3A A Amplitude of superimposed wave R Amplitude of resultant wave than I = 9I0 as [I R2] 25. If process starts and ends on same isotherm then product of P and V must be constant hence
process a, b, d will be on same isotherm.
26. 1 1 2 2mix
1 1 2 2
n Cp n Cpn Cv n Cv
=
7 52 R 1 R 192 25 3 132 R 1 R2 2
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27. 22 2
K
l 1 1R(Z 1)1 2
.....(i)
22 2
K
l 1 1R(Z 1)1 2
....(ii)
Solving (i) and (ii) Z = 9 28. = A( – 1) Slope of line will be A.
30. x = 2 (SS1 + S1O) – (SO) =
2
2 22 D d 2D2
2
2d2D 1 1
22D
d = D2
AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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CChheemmiissttrryy PART – II
SECTION – A 1. In gas phase inductive effect play a key role. 2. Geometrical isomers = 8 = 2n Optical isomer = 8 = 23 = 8 = 16 3. Due to bulky effect. 4.
O
O
H
O
OH
O
OH
O
OH
H
O
OH
Tautomerise
O
O
5. Birch reduction. 6. 3o carbocation is more stable. 7. 2: CCl intermediate. 8. 3 3 2
Red ppt.CH CHO 2CuO CH COOH Cu O
9.
D
H
D
H
D
H
D3D O
D
D
D
D
D
D
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10.
N OH
H
NC
H
O
Polymerisation NH (CH2)5 C
O
nCaprolactum Nylon-6
11. It is belong to f-block – Lu element. 12. IP4 – IP5 required more energy, So IVA group elements. 15.
Mg
H
H
H
H
Mg
16. 3NaOH + 2O3 2NaO3 + NaOH.H2O(s) + 21 O2
17. 2 2 2 2B.A 110 118 105 112
Cl O, ClO , F O, Br O
19. (II) and (IV) have unpaired electron. 20. 2 4 Yellow ppt
H S CdSO CdS
21. Let 1 mole of mixture has x mole N2O4 2 27.6 = x(92) + (1 – x) 46 x = 0.2
22. 2p 3p 3d
m 1 1 1s 1/ 2 1/ 2 1/ 2
23. % by weight 10 d 36.5 10 1.2M 12 MM.w 36.5
36.5 1000 1000m 15.7 m36.5 100 36.5 63.5
24. 3 4ClO ClO 2e x 2 oxidation x = 5, x = 7 3 2e ClO ClO x 1 reduction x = 5, x = 4
WME M2
25. Ideal gas has no force of attraction and has negligible volume. Hence it can’t be liquefied at any T
and P.
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26. H = nxCp T Cp = Cv + K
PVn 0.05RT
Cp = 20.794 JK–1
For reversible adiable r 1TV constant
r 1 r 12 2 1 1T V T V r = 1.66 for argon
T2 = 189.85 k T = – 110.15 Thus H = 0.05 20.794 –110.15) = – 114.52 J 27.
3 3 3 3CH COCH CH CH g CO g
At t 0 150 0 0At t t 150 x x x
x 1150 x 3
x = 75 mm. 28. 2
2 2ssMg OH Mg 2OH
4s3 = 8.9 10–12, s = 1.3 10–4 m [OH] = 2 1.305 10–4 pOH = 3.58832 pH = 10.4168 29. 1 mole of A will form 2 moles of B and C after completion of reaction when 75%. A converted into
B and C then total no. of moles 0.5 2 1.5 3.5 moles 30. Along one body diagonal 2X atoms from 2 corners, one Y particle (at the centre of cube) will be
removed.
So effective no. of X particle in a unit cell = 1 154 28 4
And effective no. of Y particle in a unit cell = 4 – 1 = 3
15X : Y : 34
5 : 4
AITS-CRT-III-PCM(Sol)-JEE(Main)/15
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MMaatthheemmaattiiccss PART – III
SECTION – A 1. n + n = –pn ….. (1)
Again since
is a root of equation
xn + 1 + (x + 1)n = 0
n n
1 1 0
n + n + ( + )n = 0 n + n = –( + )n = –(–p)n ….. (2) From equation (1) and (2), we get –pn = –(–p)n pn = (–p)n pn = (–1)n pn (–1)n = 1 n is an even integer 2. From the figure, OM is bisector of angle BOP of triangle OBP
250 OB50 OP
2 2300 x5x
25x2 x2 = (300)2 24x2 = (300)2
2 300 300x 3250 25 25 624
x 25 6.
P
M
B A
300 m
250 m
50 m
O (observer)
x m
3. 2 31 11
1 2 2 2P 10 10 10 10 …..
2 31 1 11 .....2 2 2P 10
=
111 2210 10 100
log0.01P = log0.01(100) = –1
4. 2
2ax b ax b x ab2
Consequently, for all x > 0
2x x 1y
ax b 2x ab 2 ab
5. x = n n n n
1 2 3 nn n n
1 2 n
0.1 1 C 2 C 3 C n C1 C C C
= n
nr
r 0r C
=
n 1n
r 1r 1
nn
rr 0
nr Cr
C
=
nn 1
r 1r 1
nn
rr 0
n C
C
=
n 1
nn2
2
= n2
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6. Greatest coefficient is r 3m r
n! 15!3! 4!q! q 1 !
ln = 15 and m = 4, q = 3 and r = 37 15 = 4 3 – 13 7. Two triangles may have equal areas f is not one-one Since each positive real number can represent area of a triangle f is onto
8. Area of PAB = 32sin2
(PAB)min when 22
4
h = OC cosec = 4 2
C
P(0, h)
B A O
9. Equation of chord joining and is x ycos sin cosa 2 b 2 2
+ = 3
x ycos 0a 2 b
It passes through the centre (0, 0)
10. Ordinate of the point of intersection of the line x y ma b and the hyperbola, given by
x y x y 2y 1a b a b b
i.e. 2ym m 1b
i.e. 2b 1 m
y2m
Similarly ordinate of the point of intersection of the line x y ma b and the hyperbola is given by
2b m 1y
2m
Sum of the ordinates is zero 11. Let 1 1 2 2P(x ,y ) , Q(x , y ) be the given points on x 2y 2
1x 21y 2 and 2x 2
2y 2
ˆOP i
= projection of OP
on x-axis = x1 = 1 y1 = 2 and ˆOQ i
= projection of OQ
on x-axis = x2 = 2 y2 = 16
ˆ ˆ ˆ ˆOP i 2 j , OQ 2i 16j
ˆ ˆOQ 4OP 6i 8i
ˆ ˆOQ 4OP 6i 8 j 10
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12. The point B is (–q, –p) and the point C is (–p, –q) mid-point of AC is (0, 0) 13. The point Q is (2, 1)
Image of P(1, 2) in the line x – 2y + 1 = 0 is given by x 1 y 2 41 2 5
Ordinate of the point are 9 2,5 5
Since this point lies on QR Equation of QR is 3x – y – 5 = 0 + = 2 14. Let (, 3 – ) be any point on x + y = 3 Equation of chord of contact is x + (3 – )y = 9 i.e. (x – y) + 3y – 9 = 0 The chord passes through the point (3, 3) for all values of 15. Equation of the two circles be (x – r)2 + (y – r)2 = r2 i.e. x2 + y2 – 2rx – 2ry + r2 = 0 where r = r1 and
r2 condition of orthogonally gives 2r1r2 + 2r1r2 = 2 21 2r r
4r1r2 = 2 21 2r r , circle passes through (a, b)
a2 + b2 – 2ra – 2rb + r2 = 0 i.e. r2 – 2r(a + b) + a2 + b2 = 0 4(a2 + b2) = 4(a + b)2 – 2(a2 + b2) a2 – 4ab + b2 = 0
16. 2 3y 6 x2
Equation of directrix x = 0
Let coordinate P be 23 3 t , 3t2 2
Coordinate of M are (0, 3t) 2MS 9 9t
23 3MP t2 2
9 + 9t2 = 2
22 23 3 9t 1 t2 2 4
4 = 1 + t2 Length of side = 6 17. yy1 = 2a(x + x1)
x2 = 4by = 11
2a4b x xy
y1x2 – 8ab x – 8ab x1 = 0 D = 0 gives xy = –2ab
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18. Equation of the tangent to the ellipse at P(5 cos , 4 sin ) is xcos y sin 15 4
It meets the line x = 0 at Q(0, 4 cosec ) Image of in the line y = x is R(4 cosec , 0) Equation of the circle is x(x – 4 cosec ) + y(y – 4 cosec ) = 0 i.e. x2 + y2 – 4(x + y) cosec = 0 Each member of the family passes through the intersection of x2 + y2 = 0 and x + y = 0 i.e. The point (0, 0)
19. Equation of normal at (13 cos , 5 sin ) of 2 2x y 1
169 25
13siny 5 sin x 13cos5cos
it passes through (0, 6) then cos (15 + 72 sin ) = 0
cos = 0, 5sin24
1 5sin24
21. Given, f(x) = (x – 1)(x – 2)(x – 3)(x – 4) Consider the interval [1, 2] Since, f(x) is a polynomial function, it is continuous in [1, 2] and differentiable in (1, 2) Also, f(1) = 0 = f(2) By Rolle’s theorem these exists a point x = c in (1, 2) such that f(c) = 0 i.e. these exists a root of
f(x) = 0 in the interval (1, 2) which is positive Similarly, two positive roots of f(x) = 0 exist in the open interval (2, 3) and (3, 4) Hence, the three roots of f(x) are positive 22. f(x) g(y) + f(x) g(y) = g(y) (f(x) + f(x)) g(y) = g(y)
f(x) + f(x) =
g' yg y
= constant ….. (1)
Put y = 0
g' 0 1g 0 1
Now from equation (1), we get f(x) + f(x) = 1 and
g' y1
g y
g(y) = ey 23. 10 10 10 10
1 2 3 4C C C C 10 45 120 210 385 24. Total number of numbers without restriction two numbers have all the digits equal Hence, the required number of numbers = 25 – 2 = 30
25. Given 7a – 9b = 0 b = 7 a9
So, the number of pairs (a, b) can be (9, 7), (18, 14) (27, 21) and (36, 28)
Required probability = 392
4 4741C
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26. Two numbers are selected from {1, 2, 3, 4, 5, 6} n(S) = 6 5 (as one by one with out replacement) favourable events = minimum of the two
numbers is less than 4 n(E) = 6 4 As for the minimum of the two is less then 4, we can select one from {1, 2, 3, 4} and other from
{1, 2, 3, 4, 5, 6}
Required probability =
n E 24 4n S 30 5
27. 4
dx
sin xcot x = 2
dxsin x cot x =
2cosec x dxcot x
Put cot x = t2 cosec2 x dx = –2t dt
= 2t dtt
= –2t + c = 2 cot x c = 2 ctanx
28. Let a = 2x, b = –3x – 1, c = –1 a3 + b3 + c3 = 3abc
Now, a3 + b3 + c3 – 3abc = 21 a b c a b2
factorizing and get a + b + c = 0 as a b c
Hence, the given equation is; 2x – 3x – 1 – 1 = 0 By trial x = 1 and 2 are solution of this Now, 2x = 3x – 1 + 1 has no other solution Number of real solution exactly two 29. 2 sin2 – 5 sin + 2 > 0
1sin2
( –1 sin 1)
50, , 26 6
30. Let h(x) = f2(x) + (f(x))2 h(x) = –2x(f(x))2 g(x) x = 0 is the point of maximum