Aiits 2016 Hct Vi Jeem Jeea Main Solutions Solutions
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
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JEE(Main)-2016 ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST-VI (Main)
Q. No. 1.
PHYSICS A
CHEMISTRY B
MATHEMATICS A
2.
B
A
A
3.
B
D
B
4.
B
D
A
5.
B
B
D
6.
C
B
A
7.
D
B
8.
A D
D
B
9.
C
B
B
10.
B
B
B
11.
D
B
B
12.
B
D
D
13.
B
B
B
14.
B
D
C
15.
A
B
A
16.
C
B
C
17.
C
B
A
18.
D
B
B
19.
D
D
D
20.
B
B
C
21.
D
C
B
22.
C
C
A
23.
C
C
B
24.
C
A
B
25.
A
A
A
26.
C
A
A
27.
C
C
B
28.
C
D
29.
C
A D
30.
B
C
A
B
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
Physics
PART – I SECTION – A
1.
Friction force on block A, f K = N = 10 N (opposite to v) 10N 2 acceleration of A = 2.5 m/s (opposite to v) 4kg Time taken by the plank to come at rest, 10 m / s t = 4 sec (using, v = u + at) 2.5 m / s2 2 2 In 4s distance moved by the plank SA = 20 m (using v = u + 2as). To reach the other end block will have to move by 20 m + 40 m = 60 m. As B is moving with constant speed 60 m / s T 6 sec 10 m / s
2.
Reference frame is at the edge of the plane with the leftward direction as the x-axis. Let x0 be the distance of the cart at t = 0 mg T = ma T = Ma mg a= = 2.8 m/s2 Mm Since, M has an initial velocity to the left, a is in fact retardation for M. V = 7 + (2.8) 5 = 7 m/s i.e. after 5 sec the cart will have the same speed but in opposite direction. 1 x = x0 + 7 5 + (2.8) 5 2 x = x0 + 35 35 = x0 Let the cart go by S to the left from its initial position 02 72 = 2(2.8) S S = 8.75 m Distance covered in up and down journey = 8.75 + 8.75 = 17.5 m
3.
Let the pressures be P1 and P2 of the spheres having initial temperatures 0C and 20C respectively then. n 1 293 Initially P1 = P2 n 2 273 P1 ' 283 293 1 P1 ' P2 ' P2 ' 303 273 thus the mercury pellet will be displaced towards the sphere at higher temp.
If P1 ',P2 ' be the pressures after rasing temperature then
4.
No eddy current form in case (i) & (ii) and hence motion is not opposed.
7.
(1.2 103 10 10 10 2 ) (103 10 4 ) F = Buoyant force – weight F = 20 N in down ward direction
8.
Kinetic energy is maximum at centre.
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
2
10.
1 mgx 1 dU A.dx 2 LA y l
U
U 12.
13.
14.
1 m2 g 2 A 2 x dx 2 L2 A 2Y 0
m2 g 2l . 6 AY
A node is a point where there is no vibration. There is vibration at other points where nodes are not formed. 2Rb b b dr W F.dr .d r | d r | = 2 bJ. | dr | R R R
Using A1v 1 = A2v2 and Bernoulli’s principle we get
……(i)
1 1 P1 v12 P2 v 22 2 2
…..(ii)
From (i) and (ii)
v1 A 2
2(P1 P2 ) (A12 A 22 )
Rate of flow = A1v1 A1 A 2
2(P1 P2 ) (A12 A 22 )
But P1 – P2 = P rate of flow of glycerine A1 A 2 15.
16.
2P (A12 A 22 )
In adiabatic compression, the temperature always increases and since PV = nRT, the quantity PV also increases. V = 0 + (0.4 × 10) t = 1 m/s. 1 t .25 second. 4 WfK WF K 0
N (u=0) fK
WF WfK (fK S cos180) fK S
10g 10g fK
= (0.4 × 10 × 10) × .25 × 1= 40 × .25 = 10 J. F v=1m/s
17.
At the farthest point X, mv0R = mv x 4R 1 GMm 1 GMm m v2x = mv 02 2 4R 2 R v0 = 1.6 gR .
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
18.
19.
20.
T = period 2
m 10 2 s. K1 K 2 360 3
The maximum velocity is always at equilibrium position since at any other point there will be a restoring force attempting to slow the mass. impulse 50 Vmax 5ms 1 mass 10 V 5 5 A amplittude= max m. (2 ) 6 3 3 Their average translational K.E. is kB T, kB Boltzmann's constant 2 1 1 3 2 2 mH2 CH2 mHe CHe = kBT 2 2 2 CH2 : CHe = mHe : mH2 = 4 : 2 = 2 : 2 7 5 R R 2 2 3 For the mixture, = (1) (2) 5 3 2 CV CV R R 2 2 Applying TV - 1 = constant, the temperature increases by a factor of (2) C(1) P CP
8.
21.
Impulse = Force time; net momentum of the system is conserved as the net force on it is zero.
22.
Since velocity of man w.r.t trolley is greater than velocity of trolley w.r.t. earth, after the man turns back displacement of the man will decrease, so maximum displacement will be at the moment when man turns back. L t 1.5V L 5 Displacement (v 1.5v)t 2.5v L 1.5v 3
23.
pV1/4 = constant RT 1/4 v = constant V 3 -3/4 dT V - dV.V 7/4 .T = 0 4 dQ dU dV 4V 5 4 C= P Cv p. R R dT dT dT 3T 2 3
24.
or TV-3/4 = constant or
dV 4 V dt 3T
23 R. 6
(for first harmonic) 4l1 3 v= (for third harmonic) 2l3 l 1 1 = 6 l3 v=
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5
25.
velocity of upper block when lower block hits a obstacle, F F v 2a. 2. 2M M now after collision retardation of upper block w.r.t. earth, Mg a g M v 2 u2 2as 0
26.
AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
F F 2g. . M 2 Mg
1 2 I 2 Let x is the distance of CM from A I M.x2 4M( x)2 and if I is minimum, W will be minimum w
dI 2Mx 4M 2( x) ( 1) = 2 Mx – 8M ( x) dx
dI 10Mx 8M dx 10Mx 8M 0 x
4 . 5
27.
Total internal energy of system will remain conserve nCV1 T1 nCV2 T2 nCV1 nCV2 T 3 5R 3R 5R 2 R 300 2 600 2 2 T 3900 9 30 T 100 8 = 487K (approx) 8 3 3 E R 487 300 187 R = 280.5 R 2 2
28.
Let initial velocity of the shot be u and P denote the force of resistance of penetration. In the 1st case plate is fixed 1 0 mu2 PS(COE) …(i) 2 nd In the 2 case plate is free to move. The shot will penetrate up to both acquire a common velocity V (say), there will be no further penetration thus P seases to act 1 1 (M m)V 2 mu2 PS'(COE) …(ii) 2 2 and (M m)V mu(COLM) …(iii) From equation (iii) and (ii) 1 m 2 u2 1 mu2 PS ' 2 (M m) 2 1 M mu2 PS' …(iv) 2 (M m) equation (i) and (ii)
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
S' M S (M m) SM S' . (M m) (COE = conservation of energy) (COLM = conservation of linear momentum). 29.
Stress =
T 1 E= Stress × Strain × Volume A 2 2
Stress 1 Stress = Y, E = × Volume Strain 2 Y
T = (L –x)
30.
F x
2
L
F = ma, E =
dx
1 L–x F 1 2 L A Y A dx o F F 2L L – x = = F M 6YA L
0 at 0 0 t 0
v a r 0 2
v 1 0r 2
at 0 t
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Chemistry
AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
PART – II SECTION – A
4.
Na 1s2 2s2 2p6 3s1 3s1 orbitals has n = 3, 0 Number of radial nodes = n – 1 = 3 – 0 – 1 = 2
6.
7.
rO2 rCH4
PO2
MCH4
PCH4
MO2
nO2 nCH4
16 3 16 16 3 = 32 32 2 32 4 2
Half life t1/2 = 140 days 280 140 3 140
Number of half lives in total time = R = R0(1/2)n 3000 = R0(1/2)3, R0 = 24000 dps 9.
11.
58 1.63 0.082 300 Mass of Al(s) 0.05 101 20 20 61.13 13.56 gm nCl2
So mass of Al(s) dissociated = 0.5 moles So eq is established 3 Al(s) Cl2(g) AlCl3(s) 2 3 1.63 0.5 0.5 2 3
3 0.88 0.082 300 2 K p PCl2 2.7 2 8 (Volume of the vessel is taken 8 lit approx even after eq is established).
13.
HO
3 2
OH 2
B HO 14.
OH
HO
CH2
HO
CH2
H2C
O
O
CH2 4H2 O
B H2C
O
(i)
Na2 S2 O3 2HCl 2NaH SO2 H2 O S
(ii)
KI3 2Na2S2 O3 KI 2NaI Na 2S4 O6
(iii)
2AgNO3 Na2 S 2O3 Ag2 S2 O3 2NaNO3
O
CH2
white
H2O Ag2 S H2 SO4
Black
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
15. 16.
Stability of intermediate carbocation increases rate of reaction. CH3
CH3
CH3
CH3
NH2
NH2 NH3
H
Cl
NH2
NH2
18. O
O
(Aromatic)
O
19. O
C OH
KMnO4 H
O
O OH
21.
Mg2 C3 2H2 O MgOH2 CH3 C CH
24.
pOH pKb log10
NH4 Cl 0.1 4.7 log 5.7 NH OH 0.01 4
pH = 14 – 5.7 = 8.3
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Mathematics
AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
PART – III SECTION – A
1.
The equation of the line joining A(1, 0) and B(3, 4) is y 2x 2 . This
8 6 cuts the circle x 2 y 2 4 at Q(0, 2) and P , . 5 5 7 3 We have BQ 3 5, QA 5, BP and PA 5 5
BP 7 / 5 7 BQ 3 5 and 3 PA 3 / 5 3 QA 5
Y
B(3,4) P(8/5,6/5)
X
O
A(1,0)
X
Q(0,–2) Y
, are roots of the equation x 2 x( ) 0 7 7 i.e., x 2 x 3 (3) 0 or 3x 2 2x 21 0 . 3 3 2.
ax2 + by2 + 2fx + 2gy + c = 0 2fx 2 2gy a x2 b y c 0 a b 2 2 f 2 g2 f g a x b y c a b b a 2 2 (x f / a) (y g / b) 2 2 1 f f g2 g2 c / a c / b b b a a
f 2 g2 f 2 g2 c c a b a b (1 e2 ) i.e. e is eccentricity, then b a
1 e2
3.
a ba ba e2 e b b b
(1 3x)1/2 (1 x)5/3 x 2 1 4
1/2
1 1 1 1 5 521 2 2 1 2 (3x) 2 2 2 ( 3x) .... 1 3 ( x) 3 3 2 ( x) .... 2 1 x 1 1 1 x 2 1 .... 2 4 2 2 2 4 53 2 19 1 12 x 144 x .... 1 35 x .... x 1 24 2 1 2 8 x .... Neglecting higher powers of x , then 35 35 a bx 1 x a 1,b . 24 24
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
4.
A(0,0)
y mx . It makes an angle of 45o with 2x 3y 6 .
m ( 2 / 3) 1 1 m( 2 / 3) 1 or 3m 2 (3 2m) m , 5 5 Hence sides are x 5y 0, 5x y 0 and 2x 3y 6 . Solving in pairs, vertices are (0,0) ,
90°
tan( 45o )
B
45°
45° (2x+3y=6)
C
6 30 30 6 13 , 13 , 13 , 13 . 5.
L12 x 3y 1 0 L 23 2x y 12 0 L13 3x 2y 4 0 Therefore, the required distances are 4 3 4 1 7 D3 10 10 D1 D2
6.
4 1 12 5
7 5
3 5 2 2 4 94
7 13
.
The equation of any line passing through the given point P(3, 4) and making an angle
with 6
x3 y4 r (say) ......(i) cos 30o sin30o Where ‘r’ represents the distance of any point Q on this line from the given point P (3, 4). The coordinates (x, y) of any point Q on line (i) are r 3 r (3 r cos30o ,4 r sin30o ) i.e., 3 ,4 2 2 If the point lies on the line 12x 5y 10 0 , then
x-axis is
r 3 r 132 12 3 . 5 4 10 0 r 2 2 12 3 5 7.
Suppose a point on circle is B(x1, y1 ) and that which divides A and B in 3 : 2 is P given by h
2 3x1 2 3y1 5h 2 5k 2 ,k or x1 , y1 5 5 3 3
As (x1, y1 ) lies on circle x 2 y 2 4 , we get on substituting, 25(x 2 y 2 ) 20(x y) 28 0 .
8.
(y - 3)2 = -4 (x - 3) (given equation of parabola) equation of latus rectum is x = 2.
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9.
AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
x t 2 t 1and y t 2 t 1 x y 2 2t 2 and x y 2t 2(x y 2) (x y)2 x 2 y 2 2xy 2x 2y 4 0 Comparing with the equation ax 2 2hxy by2 2gx 2fy c 0, abc 2fgh af 2 bg2 ch2 0 and h2 ab
10.
Since the number of students giving wrong answers to at least i question (i 1, 2,........,n) 2ni . The number of students answering exactly i (1 i 1) questions wrongly = {the number of students answering at least i questions wrongly, i 1, 2, .........,n)} – {the number of students answering at least (i 1) questions wrongly (2 i 1 n)} 2ni 2n(i1) (1 i n 1) . Now, the number of students answering all the n questions wrongly 2nn 20 . Thus the total number of wrong answers 1(2n 1 2n 2 2(2n 2 2n3 ) 3(2n3 2n 4 )
11.
We have, y z a(cos2 x sin2 x) c(sin2 x cos2 x) a c ( solution is (b)}
y z a(cos2 x sin2 x) 4b sin x cos x c(cos2 x sin2 x) (a c)cos2x 2b sin2x 1 tan2 x 2 tan x (a c). 2b. 2 2 1 tan x 1 tan x 2 2 2.2b / (a c) 1 4b / (a c) (a c). 2b. 2 2 2 2 1 4b / (a c) 1 4b / (a c)
Since tan x yz
2b , (a c)
(a c).{(a c)2 4b 2 } 8b 2 (a c) (a c)2 4b2
(a c)(a c)2 4b2
{(a c)2 4b2 } y z,( a c) .
12.
(a c)
Given that z2 (p iq)z r is 0 ......(i) Let z (where is real) be a root of (i), then
2 (p iq) r is =0 or 2 p r i(q s) =0 Equating real and imaginary parts, we have 2 p r 0 and q s 0 2
s s Eliminating , we get p r 0 q q
or s2 pqs q2r 0 or pqs s2 q2r
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
13.
1 t2 x 3 2 1 t
2t , y 1 t2
2
1 t 2 4t 2 x2 2 y 1 Which is an ellipse 2 9 1 t2
1 8 1 9 1 e e 1 9 9 2
2
3 5 Hence foci of the ellipse are ( 3, 0) Hence PF1 + PF2 = 10.
14.
Here a = 5, b = 4 and e =
15.
The given ellipse is
3 4 1 e2
x2 y 2 1 a 2,b 3 4 3 1 e 2
1 1 2 Hence, the eccentricity e1, of the hyperbola is given by
ae 2
1 e1 sin e1 cosec b2 sin2 cosec 2 1 cos2 2
Hence, equation of hyperbola is
2
x y 1 or x 2 cosec 2 y 2 sec 2 1 2 sin cos2
16.
Since (n+2)th term is the middle term in the expansion of (1 x)2n 2 , therefore p 2n 2Cn1 . Since (n+1)th and (n+2)th terms are middle terms in the expansion of (1+x)2n+1, therefore q 2n1Cn and r n1Cn1 But n1Cn 2n1Cn1 2n 2Cn1 qr p
17.
For the first set number of ways
52
C17 . Now out of 35 cards left 17 cards can be put for second in
C17 ways similarly for 3 in 18 C17 . One card for the last set can be put in only one way. Therefore the required number of ways for the proper distribution 52! 35! 18! 52! 1! . 35!17! 18!17! 17!1! (17!)3 35
18.
19.
rd
tan(cot x) cot(tan x) tan(cot x) tan tan x 2 (70)2 20h h2 (6)(70)(20) cot x n tan x cot x tan x n 2 2 2 2 4 2 2 (70) 20h h (6)(70)(20) . n sin2x (2n 1) sin 2x 2 n 2
1 sin x sin2 x .... 4 2 3
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
1 1 4 2 3 sin x 1 1 sin x 42 3
(4 2 3) 2 3 3 4 4 2 2 x or . 3 3
sin x 1
20.
Adding and subtracting the given relation, we get (m n) a cos3 3acos sin2 3a cos 2 . sin a sin 3
a(cos sin )3 and similarly (m n) a (cos sin )3 Thus, (m n)2/3 (m n)2/3
a2/3 {cos sin )2 (cos sin )2 } a2/3 {2(cos2 sin2 )} 2a2/3 . n sin A 1 sinB A B A B 2cos sin n 1 sin A sinB 2 2 A B A B n 1 sin A sinB 2sin cos 2 2 A B A B tan cot 2 2 n 1 A B A B tan tan 2 . n1 2
21.
We have sin A nsinB
22.
We have equal roots, therefore 2 4 . Now second equation x 2 x 12 0 has a root 2, so put x 2 4 2 12 0 4 16 Hence from 2 4 , we have 4 4 (, ) (4,4) .
23.
Let the roots are and n b b Sum of roots, n a a(n 1) c c and product, .n 2 a na From (i) and (ii), we get
.....(i) ....(ii)
2
b c b2 c 2 2 a(n 1) na na a (n 1)
nb2 ac(n 1)2 . Note: Students should remember this question as a fact.
24.
We have
3 i (a ib)(c id)
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AIITS-HCT-VI-PCM(Sol)-JEE(Main)/16
ac bd 3 and ad bc 1 b d -1 tan1 Now tan a c b d bc ad 1 1 tan1 a c tan1 n ,n I tan b d 6 ac bd 3 1 . a c
25.
We know that loga m loga n m n or m n , according as z x iy log(1/3) | z 1| log(1/3) | z 1|| z 1| | z 1|
a 1 or 0 a 1 . Hence for
1 0 1 3
| x iy 1|| x iy 1| (x 1)2 y 2 (x 1)2 y 2 4x 0 x 0 Re(z) 0 26.
x 2n1 y2n1 is always contain equal odd power. So it is always divisible by x y .
27.
n 1 Check through option, the condition n! is true for n 1 . 2
28.
H d tan and H h d tan
n
60 tan 60 tan 60 tan h 60 h tan tan 60 sin( ) h x cos sin . sin cos cos cos
29.
h T2 A h T2 A Hence 120 h 30 h h 45 m.
H=60m
h
d
T2
tan 45o 1
T1
h
h 45° T2
30.
45° A 30 B 120 m
T1
x 1 1 1 ..... to
We have x 1 x x2 1 x x2 x 1 0
1 1 4 1 5 2 2 1 5 As x 0 , we get x 2 x
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