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[1]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
Head Office : Pushpanjali Place, Boring Road, Patna, Tel. : 0612-3225103, Mob. : 7857966777Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna-14, Mob. : 7677222110
IITian’s TAPASYA
SOLUTION PAPER OF JEE MAIN 2016PART TEST - 1
DATE : 11th Oct. 2015
Ph.: 7857966777 / 8102926611/12/13Head Office : Pushpanjali Place, Boring Road, Patna-1
Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna-14, Mob. : 7677222110
www.iitianstapasya.com
IITian’s TAPASYA...IITians creating IITians
JEE (MAIN & ADV.) / NTSE / KVPY / OLYMPIAD
[2]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
Head Office : Pushpanjali Place, Boring Road, Patna, Tel. : 0612-3225103, Mob. : 7857966777Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna-14, Mob. : 7677222110
IITian’s TAPASYA
PHYSICSQ.1 (A)
Sol. Acceleration of blocks = 70 30
10
4 m/s2
Velocity at t = 2.5 is
v = u + at = 0 + 4(2.5) = 10 m/s2
After stopping B, A continues to move up with acceleration g downward.
B comes down with acceleration g.
String will tight again when their displacement become equal.
SA = SB
2 21 110 t gt gt t 1 sec2 2
AV 10 g(1) 0 m / s
Q.2 (D)
Sol. If Range is 6 m2
o ou sin26 18.5 ,71.5g
For R = 8 mo o26.5 ,63.5
For sucessful shot range of .o o o o18.5 26.5 and 63.5 71.5
Q.3 (B)
Sol. FBD of blocks in the frame of lift
2 4
T T
2g 4g2g 4g
a a
NLM equations.
16gT 4g 2a 8g T 4a T3
reading of spring balance 2T 32g 3
SOLUTION OF JEE MAIN PART TEST - 1
[3]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
Head Office : Pushpanjali Place, Boring Road, Patna, Tel. : 0612-3225103, Mob. : 7857966777Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna-14, Mob. : 7677222110
IITian’s TAPASYA
[4]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
Head Office : Pushpanjali Place, Boring Road, Patna, Tel. : 0612-3225103, Mob. : 7857966777Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna-14, Mob. : 7677222110
IITian’s TAPASYA
Q.4 (C)
Sol. By symmetry all will meet at centre. Comp. of velocity towards centre v 3vcos302
this is
also equal to avg. velocity.
v 30 v cos30o
A
B C
Q.5 (B)
Sol.
F
Fcosf
Fsin
N mg
N mg FsinFcos N (mg Fsin )
gFcos sin
For minimum Force dF 0 tand
min. 2
mgF1
contact force = 2 2N f
Q.6 (A)
Sol. FBD of block in frame of elevator
mg2 m
gsin
mgN
acceleration along the incline = 2gsin
2
2
1S ut at2
h 1o (2gsin )tsin 2
[5]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
Head Office : Pushpanjali Place, Boring Road, Patna, Tel. : 0612-3225103, Mob. : 7857966777Branch Office : Rukanpura Flyover, Near Sri Sai Yamaha, Bailey Road, Patna-14, Mob. : 7677222110
IITian’s TAPASYA
Q.7 (D)
Sol.
m1 m1
m2 m2
v
v
m g2
m g1 T
R B
R BT
1
2
R B T m g ....(i)R B T m g ....(ii)
1 2
2 1
T m g T m g1T (m m )g2
Q.8 (C)
Sol.Wsin
Wsin8
f k
f k
20
Net force will be zero.
k
k
W sin f 20 ...(i)8 W sin f ....(ii)
from (i) and (ii) fk = 14
Q.9 (D)
Sol. At x = 1 m
U = 3, K = 1 Total = 4J
When particle moves to the right of x = 1 its kinetic energy increase.
But beyond x = 2, KE decreases And force is along – ve x axis.
It will reverse its direction where K = 0, U = 4 that is at x = 6.
Q.10 (A)
Sol.A
(0,0,0) (15,0,0)
VAB B
VAB must be along AB.AB is along x axis.
[6]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
AB
A B B A
B
B
V V i
V V V i V V Vi
V 15 i 4j 3k V i
(15 V)i 4j 3k
| V | 13 m / sV 3
Q.11 (B)
Sol. Area under F-t graph = change in momentum
f i13 1 2 2 V V2
Q.12 (B)
Sol. FBD before cut
3 2
60o 60o
Kx1 Kx2
2gKx2
Kx = 2g2 Kx = 5g1 ,
Kx1
3g
After cut,
3 2
5g/2
5g2g
Net force
3g
5g
5g 32
5g 32
5g2
= 5g
Acceleration of 3 kg 25g 50 m / s3 3
Acceleration of 2 kg = 0
Q.13 (C)
Sol.
N
x
yma
mg
FBD wrt wedge
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IITian’s TAPASYA
Work energy theorem
Wg + Wpseudo + WN = 0
– mgy + max = 0 2xy x x 4 y
4
Q.14 (B)
Sol. mgsin mgcos
5mgsin mgcos
23
ll
Q.15 (C)
Sol.2
4
10
6Let blocks move together.
210 6 2a m / s6 3
FBD of 2 kg 2 10f
max
2 4 2610 f 2 f 10 N3 3 3
f 8N f
so, blocks move separately.
2 108
10 8a 12
86
4 8 6a 0.54
Q.16 (C)
Sol.
Vmr
Vr
Resultant of Vmr and Vr is along 53o with Horizontal.
o omr
r mr
V cos 4tan53 0V V sin 3
mr
d 600t 30secV 20
[8]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.17 (C)
Sol. Real force are same in every frame.
Q.18 (D)
Sol.3 4L LK
T L
1K T
Q.19 (C)
Sol.
x
ucos tdisp.V ucost t
Q.20 (B)
Sol.
i i f f
2
2 2
U K U Km b L 1b sin g 0 m sin g mVL 2 2 2
gsinV L bL
Q.21 (B)
Sol. 2
2
2
x at ,y bty ayt x parabolab b
Q.22 (B)
Sol. Finally system will move with common velocity = v
From conservation of linear momentum
5 4 5V 15V V 1m / s
Heat = loss in Kinetic energy
i
2f
1K 5 16 40J21K (5 15) 1 10J2
Loss 30J
[9]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.23 (C)
Sol. 2dU d 1 dxkx Kxdt dt 2 dt
40 0.1 (3 1)16J / s
Q.24 (B)
Sol.A B
2T TT
1010
a 3a
2
2B
3T 10 a, 10 T 3a a 2m / s1S 3 2 1 3m2
Q.25 (C)
Sol. From diagram
o
22
22
d 1 tan45 1dt v
1 dv dv1 vv dt dt
dv 1 1m / sdt 33
Q.26 (A)
Sol.
60o 30oRV
2RV 25
2
R3V2
R3V2
VR
VRM
Now
o R
R
R
V 3 / 2 1tan30 V 3252
V 25m / s
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IITian’s TAPASYA
Q.27 (C)
Sol.
2 2
W K1 1m a 0 ma2 2
l l
Q.28 (C)
Sol.
(3,4) y= 4/3xy
x
3
0
4W ydx xdx 6J3
Q.29 (A)
Sol. P = kt
2 2
dvm V ktdt
kVdv tdtm,
kv tm
Q.30 (D)
Sol. 1 x y z
x 2 2 y z
L M E TM (ML T ) T
x y 0 solving1 12y 1 x , y , z 1
2 22y z 0
[11]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
CHEMISTRYQ.31 (B)
Sol.2 2
2 2
1 4 1 9 32 2725 3.5RZ 3 RZ 8 24
135 1281 9 1 16 7 1120RZ 8 RZ 15
Q.32 (A)
Sol.2 2
22 3
2
2 4
zz z v znK.E 13.6eV , ,
nh n 2 r nz
K(ze)e zF (I,II,IV)r n
Q.33 (A)
Sol.
0
h h hmv 2mKE hc hc2m
Q.34 (C)
Sol. 2 22n 2 3 18
Q.35 (D)
Sol. 3-radial nodes, yz-plane, origin (6 dyz) (n – l – 1) = 3
Q.36 (B)
Sol. 3 Px has 1 radial and 1 angular node.
Q.37 (B)
Sol. 34 66.663 1.5
3 100 256
Q.38 (D)
Sol.
740 100 1.146 0.0821 300100 x760 x
28 32x 72.42%
Q.39 (C)
Sol. 2 3 3
2 3 3
3
2.222.22 mg CaCI 100mg CaCO 2mg CaCO111
1.919mg MgCI 100mg CaCO 2mg CaCO95
4mg CaCO in 1L 4ppm
[12]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.40 (A)
Sol. 1 L H2O2 solution — 33.6 L of O2 at STP
500 mL of H2O2 solution — 16.8 L of O2 at STP
50 % loss 8.4 L of O2 at STP
moles of O2 =
3 0.0821 3008.4 3 8, L57022.4 8760
Q.41 (C) Q.42 (A) Q.43 (D) Q.44 (A)
Sol. 4.925 g (CuCI2 + Cu Br2) (134.5x + 223.5 y = 4.925)
x mole y mole
5.74g AgCl Br AgBr CI5.74AgCI 2y
143.55.74 5.742y 2y 2y 143.5 2y 188 6.63
143.5 143.5y ??, x ??
(a) 2y 223.5% CuBr 100%
4.925
(b) x y 63.55% Cu 100%
4.925
(b) 2y 100%
5.74143.5
(d) (2x + 2y)
Q.45 (C)
Q.46 (C)Sol. AI – O & O – H bonds are not equal in energy.
However, AI(OH)3 behaves Amphoteric.
Q.47 (C)
Sol. (i) 2 3Na ,Mg ,AI size decreasing2 3I ,S ,N decreasing size in aqueous state.
2O,O ,O decreasing order of electron Affinity..
2IE N,F,O increasing order of (IE)2
[13]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.48 (C)
Sol.
CHOCHO
OHCCHO
O O
O
(1)
(1)
(2)
(2)
(3)
(3)
(4)
Q.49 (C)
Sol. (a) (b) NOH
:
(c) N
OH
:
(d) NOH
:
Q.50 (C)
Sol.
Q.51 (D)
Sol.
Q.52 (B)
Sol. Angle of rotation mass concentration.
For 1 L solution angle of rotation = o o200mL 20g30 620g 1000mL
lc
= specific angle of rotation at fixed temperature ( ) and fixed wave length ( ) = angle through which PPL is rotated.c = mass concentration (g/mL)l = length (dm)
Substituiting values
o
o 1 16 30 mLg dm20g 10dm
1000mL
[14]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.53 (B)
Sol.CH3
CH3
CH3
CH3
H
H
II
Bulky groups are on anti position.
Q.54 (C)
Q.55 (C)
Sol. gauche conformation
Q.56 (D)
Sol. C CHH
CH3CH3
OHOHC C
HH
** If the stereochemistry about the double bond in X is cis, the number
of enantiomers possible for X is 2
Q.57 (C) 1 2 3 4
Sol. CH3---CH
OH
---CH
CH3
---CH3 3-Methyl-2-butanol
— OH is functional group.
Q.58 (C)
Sol.
O
CCI Benzene carbonyl chloride
Q.59 (A)
Sol. ---SO3H, ---COOH, ---CONH2, ---CHO
Q.60 (B) OH
CNBr
1
23
45
— CN is most prior functional group.Hence, 2-Bromo-5-hydroxybenzonitrile
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IITian’s TAPASYA
MATHEMATICSQ.61 (D)
Sol. We can write sin x sinx lnsin x(sin x) e
now maximize & minimize sinx ln sin x
Q.62 (A)
Sol. From given condition we can obtain x ytan 72
now 2
x y2tan72sin(x y) x y 251 tan
2
Q.63 (C)Sol. Graphically
0 23
45
6x
hence upto 12 total no. of intersection is 10 so no. of solution is 10
Q.64 (A)
Sol. 7 4 7cos 1 sin 0 butcos 1
so cos 1; sin 0
0, 2
Q.65 (C)Sol.
Above expression can be written aso o o o
o o(cos 1 sin 1 ) (cos2 sin2 ) ................
cos 1 cos 2
o o
ocos 45 sin 45
cos 45
now convert numrator in term of complete sin
[16]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.66 (C)
Sol.2012 1006 1006
2 4 20102
1 (1 x ) (1 x ) (1 x )1 x x .......x1 x (1 x) (1 x)
503 503
1006 (1 x ) (1 x )(1 x )(1 x) (1 x)
1006 2 502 2 2 502(1 x ) (1 x x .......x ) (1 x x x .....x )
this is divisible by 2 n 11 x x .....x if n 1 502
n 503
Q.67. (C)
Sol. above expression can we writen as logb loga clog a log b
now use A.M G.M
Q.68 (C)Sol. Check by option
Q.69 (A)
Sol. Let P denote the desired product, and letQ sina sin2a sin3a.....sin999a .
Then9992 PQ (2 sina cos a) (2 sin2a cos2a).....(2 sin999a cos999a)
sin2a sin4a.....sin1998a
(sin2a sin 4a....sin998a) [ sin(2 1000a)]
[ sin(2 1002 a)].....[ sin(2 1998a)]
sin2a sin4a.....sin998a sin999a sin997a....sina Q
It is easy to see that Q 0 . Hence the desired product is 9991P
2 .
Q.70 (A)Sol. o o o2 sin2 4 sin4 ..... 178sin178 ,
which is equivalent too o o o o o
o2 sin2 sin1 2(2 sin4 sin1 ) ..... 89(2 sin178 sin1 )
sin 1
.Note that
o o o o2 sin 2k sin1 cos(2k 1) cos(2k 1) .We have
o o o o o o
o2 sin2 sin1 2 (2 sin4 sin1 ) ..... 89(2 sin178 sin1 )
sin 1
[17]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
o o o o o o
o(cos1 cos3 ) 2 (cos3 cos5 ) ..... 89(cos177 cos179 )
sin 1
o o o o
ocos1 cos3 .... cos177 89 cos179
sin 1
o o o o o o
ocos 1 (cos3 cos177 ) ..... (cos 89 cos91 ) 89 cos1
sin 1
o o
ocos 1 89 cos 1
sin 1
= o90cot 1 hence average is ocot 1 .
Q.71 (D)Sol. | sin x| is periodic with period , while {x} is periodic with period 1
no common period is possibleQ.72 (C)
Sol. f(x) is odd if 2p0 1
x 1
p (0, 25)Q.73 (C)
Sol.2
2
1x 1 x5x2
22
1 15x 1 x x3
x
y
O 53
53
(0, 1)
2
1x 1 x
hence in positive region two curve are non intersecting
[18]TAITS / JM / PT-1 / 2016IITian’s TAPASYA...IITians creating IITians
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IITian’s TAPASYA
Q.74 (A)Sol. sin(A B) sin A cosB cos A sinB
3 31 1( sin B sinB) cosB (cos B cosB)sinB2 2
1 sin2B2 2
....(3) (Verify)
By 2 2(1) (2) , we have
6 6 2 2 4 42 cos B sin B cos B sin B 2 (cos B sin B)
2 2 3 2 2 2 2 2 2(cos B sin B) 3 cos B sin B(cos B sin B) 1 2 cos B sin B
2 21 3 cos B sin B 1 2cos2B
230 sin 2B 2cos2B
4 2 28 cos 2B 3sin 2B 3(1 cos 2B)
1cos 2B (Verify)3
2 2sin 2B3
1sin (A B) from (3)3
Q.75 (C)
Sol. above experssion can be written as
1 1 1 1f(x) ...........x x 1 (x 1) (x 2)
1 1(x n 1) (x n)
on simplifying we get
nf(x)x(x n)
1 x(x n)f(x) n
x( x n) n
n
2 2x nx n 0
| | 5 n
Q.76 (B)
Soln. f(x) can be written as 5f(x)
13 sin (x )
whre 1 12tan5
5f(x)
13 cosec (x )
5| f(x) |
13
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Q.77 (D)
Soln. 20 x 1 | x | 1
But 2x 1 | x | 1
if x 0 but x 0 is not domain of cot(sinx)
20 x 1 | x | 1
2x 1 | x | 0
Hence 2
1
x 1 | x |
is not defined
hence x
Q.78 (D)
Soln. f(x) 2 sin x4
114
2
y
x/ 4
5 / 4
hence f(x) is one - one but not onto so 1f (x) not existQ.79 (C)
Sol. 31log kx
, k
31log kx
kx 3
possible values of k are 1, 0, 1, 2, 3,.......
2 31 1 1S (3 1) ......3 3 3
(1/ 3) 1 94 4
1 (1/ 3) 2 2
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IITian’s TAPASYA
Q.80 (A)Sol. Given tan x cos x
or 2 2sin x cos x 1 sin x
now, 21 sin xcosec x sinx 1
sinx
(from (1)) ]
Q.81 (A)
Sol. Let 2f(x) sin x b now compare the coefficient and eliminate b. divide by 2cos
to get (tan 1) (tan 2) 0 4
or 1tan (2) ]
Q.82 (A)
Sol. We know that 1 cos Acot A / 2
sinA
o o
o1 1 cos165cot 82 6 2 3 22 sin165
now 6, k 2
as be know G.C.D (a, b) L.C.M (a, b) (a b)
hence answer is (A)
Q.83 (D)Sol. Make sign scheme of above expression
0 1 2 3 4
+++
x [1, 3] (4, )
Q.84 (A)
Sol. Make graph of [x] { x }
0 1 2 [x] {x}
0 1 2
hence range is [0, )
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Q.85 (B)
Sol. On solving given equation be get 2sin x3
all these values are repeated four times in [0, 2 )hence number of principle solution is 4
Q.86 (A)Sol. both equation are satisfied in fourth quadrants
/ 4
hence general solution is 2n , n z4
Q.87 (D)
Sol. If any one of A B C is obtuse angle then that sum may be negativehence answere is (D)
Q.88 (C)Sol. Use A. M H.M
Q.89 (C)Sol. Graphically
8
17 2 3
2
hence number of solution is 1
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Q.90 (C)Sol. On simplyfying 3sin x sin 3x
3sinx sin3x sin 3x
4
23 1sinx sin 3x sin 3x4 4
3 [cos 2x cos 4x]8
18
(1 cos6x)
1 3 3 1cos 2x cos 4x cos 6x8 8 8 8
on comparing L.H.S & R.H.S we get 01c
8
, 61c8
, 23c8
, 43c8
0 6 2 4| c c | |c c | 0