Solution_jee Main Part Test-1

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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

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IITian’s TAPASYA...IITians creating IITians

JEE (MAIN & ADV.) / NTSE / KVPY / OLYMPIAD



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



IITian’s TAPASYA



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



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.



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



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



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



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



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



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



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



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



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



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



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



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



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



IITian’s TAPASYA

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



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, )



IITian’s TAPASYA

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



IITian’s TAPASYA

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

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Solution_jee Main Part Test-1