Joint Entrance Exam/JEE Mains 2015 Code –...

Vidyamandir Classes

VMC/JEE Mains 1 Solutions/JEE-2015

Joint Entrance Exam/JEE Mains 2015 Code – A

PART-A PHYSICS

1.(3) Δ 40 10 30ry u t t t (As long as in air) When one reaches ground graph will be a parabola opening downwards.

2.(2) 2 2 22 4Lt n g n Ltg

Δ Δ Δ2g L tg L t

0 1 1220 90.

0 005 0 022 . . 0 027. 2 7. % 3.(3) For equation of A 1 20f For equilibrium of B 1 2 2100 120f f f 4.(3) From Conservation of momentum 13 2 2mV mV

12 2

3VV

% energy loss

22 2

2 2

1 1 1 2 22 2 32 2 2 3

1001 12 22 2

m V mV m V

m V mV

2

25 3 100

3mV /

mV

5 100 569

%

5.(2) cm

x dmx

M

2

0

h

x r dx

M

34h

6.(3) For Max. volume

2 3R a

3

3

2 ,43 π3

R aa M MR

23

M

Vidyamandir Classes


2 2' 2 4

6 3 63cubeM a M RI

24

9 3MR

7.(2) 8c full cut cutMV V V M

= 2

23

3 832 2 22

GM R GM /R ·R /R

= GMR

8.(1) 2

0 22π4π

l gTT lg

Also, 2

2'4π

gTml

0

1 Δa lY Fl

2

2 202 2

4π4π

A g Tm TMg gT

2

2 1A TmMg T

9.(3) 4P T (Given) PV nRT (Ideal gas)

33

1 14 π3

TV R

1TR

10.(2) We have assumed temperature to be in Kelvin

Sequentially keeping in contact with 2 reservoirs

Δ dQ dTS msT T

150 200

100 150

2dT dT nT T

Sequentially keeping in contact with 8 reservoirs

1 2

1 7

200

100

Δs 2T T

T T

dT dT dT............ nT T T

Where 1 2 3 7T T T ...... T are temperature after each 1st, 2nd, 3rd ………. reservoirs have supplied heat to body.

11.(3) ( ) = mean free pathTime tmolecular speed

We know that: mean free path volume

Molecular speed T

Hence τ volumeT

So, QV VT

Vidyamandir Classes


1QV T const

2 2QV T const

We know that, for adiabatic

Process : γ 1TV const

γ 1 2 2QTV T ·V γ 1 2 2Q

γ 12

Q

12.(2) 212

PE K x

2 21 ( )2

KE K A x

Hence (2).

13.(2) 0 0320 320Δ

320 20 320 20f f f

= 01 1320

300 340f

0

Δ 40100 320 100300 340

ff

12%

14.(1) At P, the field due to upper part is E1 & due to lower point is E2 Hence net field will be in the direction of Enet Hence (1). 15.(3,4) For solid sphere.

2 2 2 203 2

03 3

8πε 2in

VQV R r R rR R

004πεout

Q RV Vr r

Given 004πε

QVR

0 0in outV V ,V V

R1 = 0

2 20 02 22

5 34 22V V RR R R

R

0 03

3

3 44 3V V R RR

R

0 04 4 3 2 4 3

4

84 Hence4 3

V V R RR R, R R R R RR

Also R4 > 2R

+ + + + +

+ – –

– – – – – –

E1 E2 Enet

Vidyamandir Classes


16.(2) 22

1 2 3Q QQ

3 33 3

C CEQ EC C

22 3 23 3 3

CE CEQC C

, whose graph will be as shown in (2)

at 2214 2E EC , Q

at 22 333 3

EC , Q E

17.(4) ρ

d

V lneAv A

28 13 45 ρ 0 1

8 10 1 6 10 2 5 10

5

55 5 10ρ

3 28 1 6 2 5 10 0 1

550 1032

51 6 10 Ωm 18.(3) from kirhhoffs law 1 16 3 0i i i . . . (i) 19 2 3 0i i i . . . . . (ii) 1 0 13 toi . A Q P 19.(1) As the solenoids are coaxial net force by the solenoid on other should be zero (by symmetry)

1 2 0F F

20.(2) 2

0μsinθ2π(2 sin θ)

I LTL

cosθ (λ )T L g

2

0μtanθ4π sinθ λ

I LL Lg

2

0

4π λ sinθ tanθμ

L gI

0

πλ2sinθμ cosθ

gLI

21.(3) For stable equilibrium magnetic moment should be in the direction of external field and for unstable equilibrium magnetic moment should be opposite to external field.

22.(4) 015 1 0 1

0 15 1000

I . A.

0

RtLI I e

K1 closed for long time

Vidyamandir Classes


3 30 15 10 1 100 030 1

..eI .

50 1 0 1 1000

150

. .I mAe

0 67. mA

23.(2) 20 2

12 4π

PE Cr

22

0

24π

PECr

2 98

2 0.19 10 2.45 /3 10 1

E E V m

24.(1) For transmission at AC 2r C

12

1μ

r sin

Also, 1θ μsin sinr = 2μ sin A r

1 1 1sin sin A sin

25.(4) As μ is increasing in the upward direction, hence speed is decreasing as we go up. So the rays at the bottom will

be ahead of the rays at the top, hence the wavefront bends as shown. And the waves move to the wavefront. Hence the beam will bend upwards.

26.(2) 1 22Δθ .d

9

225 1 22 500 1025 Δθ) 30

100 0 5 10.x . ( m

.

27.(1) 2z nV r

n z

212

K mV

1 2Kq qVr

2

213 6 zE .n

28.(3) (A) – (ii) (B) – (i) (C) – (iii) 29.(3) Resultant frequencies are c m c, and c m .

A

B C

r1 r2

A θ

K1 open K2 closed

Vidyamandir Classes


30.(1) As 2

btm

maxA t Ae in Damped SHM

Similarly 20

RtLmaxQ t Q e

2 20

RtLmaxQ Q e

LR

A B

A BA B

L L L LR R

Then A correspond to 1L

And B correspond to 2L

Solutions JEE Mains-2015

PART-B CHEMISTRY

31.(4) Mole of 1g resin 1206

Mol of 2Ca uptake 1 1 1206 2 412

mol of 2Ca per g of resin

28 7 3 8 7 3 22 ( ) 2C H SO Na Ca C H SO Ca Na

32.(1) 4r 3 a

o o1.732r 4.29 A 1.86 A

4

33.(3) 2

n 2zE 13.6 eVn

z = Atomic Number For hydrogen (z = 1)

n 213.6E eVn

Energy for possible excited states is given by :

2 213.6n 2, E 3.4 eV2

34.(2) ion-dipole 31

Fr

35.(4) 2 22NO(g) O (g) 2NO (g) (1)

ofG NO 86.6 kJ / mol or 86600 J / mol

o2fG (O ) 0

Vidyamandir Classes


o2fG (NO ) ?

oG (for the above reaction ) 12RT ln k R(298) ln(1.6 10 )

Writing for the reaction:

reac no

tioG = o o2f f2 G (NO ) 2 G NO 0

12 o2fR 298 ln 1.6 10 2 G (NO ) 2 86600

o 122fG (NO ) 0.5[2 86600 R(298) ln(1.6 10 )]

36.(2) A BBo AA

| P | nx

nP

B1.2 / M2185 100 / 58

1BM 64 g mol

37.(2) oeqG 2494.2J RT ln K

eq1Ke

2

1 22Q 412

c eqQ K reaction goes backward

38.(2) 2Cu (aq) 2e Cu(s)

2 Faraday gives 1 mole of Cu. Weight of deposited Cu = 63.5 g 39.(1) Probability of collision of more than three particles simultaneously at certain orientation is very low. 40.(1) Initial millimoles of acetic acid

50 0.06 1 3 After filtration, millimoles of acetic acid left

50 0.042 2.1 Adsorbed millimoles of acetic acid 3 – 2.1 0.9 Wt. of acetic acid adsorbed 0.9 60 54 .mg

Wt. of acetic acid adsorbed per gm of charcoal 54 / 3 18 41.(3) Order of ionic radii in N3– > O2– > F– Hence correct order is 1.71, 1.40 and 1.36

Vidyamandir Classes


42.(4) Equation at cathode

3

2

22

(melt) 3 ( )

( ) ( ) ( ) 2

( ) 2 ( ) ( ) 4

Al e Al lAt Anode

C s O melt CO g e

C s O melt CO g e

In the metallurgy of aluminium, purified 2 3Al O is mixed with 3 6Na AlF or 2CaF which lowers the

melting point of the mix and brings conductivity.

The fused matrix is electrolyzed. 3 6Na AlF alone is not the electrolyte.

43. (1) 2 2 2 22H O 2H O O

22 2O O 2e

2 22O 2e 2O

So H2O2 can act as both reducing as well as oxidizing agent. 44.(2) For alkaline earth metal sulphates solubility decreases down the group. So only for the soluble BeSO4, hydration

energy is greater than its lattice enthalpy. 45.(4) For inter-halogen compounds (X X ) bond energy is lesser than general X – X bond {except for F - F},

and also inter-halogen bond is polar. So they are more reactive. 46.(2) TiCl3 Ziegler-Natta polymerization PdCl2 Wacker process CuCl2 Deacon’s process

2 5V O Contact process

47.(4) Down the group in inert gases inter-particle forces increase, so boiling point increases.

48.(2) Complex is a Mabcd type square planar complex, so it has 3 possible geometrical isomers.

49.(3) KMnO4 is coloured due to charge transfer from oxide to Mn+7 i.e from ligand to metal (L M).

50.(1) Nitrogen and oxygen do not react at ambient temperatures. In order to produce various oxides of nitrogen, they have to undergo endothermic reaction at high temperatures. Generally the range of temperature to react N2 and

O2 to form oxides is o o300 C to 800 C .

51.(1) Mass of organic compound = 250 mg Mass of AgBr = 141 mg.

Mole of Br = mole of AgBr 141188

Mass of Br 141 80188

% of Bromine in organic compound

80 141188 100

250

= 24%.

Vidyamandir Classes


52.(1) 2 3Ph CH CH = CH CH

1-phenyl-2-butene exhibits geometrical isomerism 53.(2) 5-keto-2-methyl hexanal. 54.(4) Alkyl fluorides are generally formed by swartz Reaction.

2 2 2AgF/ Hg F / CoFR X R F ppt.

X Cl / Br 55.(4) 56.(3)

57.(2) Glyptal is a polyester of glycerol and pthalic acid used in paints and lacquers. 58.(1) Vitamin C is soluble in water 59.(3) Phenelzine is used as an anti-depressant and is not an antacid.

60.(1) 2Zn on reaction with 4 6K [Fe(CN) ] produce bluish white ppt. due to formation of 2 6Zn [Fe(CN) ]. All other

substances are yellow coloured.

1. O3 2. H2O/Zn

Vidyamandir Classes


Solutions JEE Mains-2015

PART-C MATHEMATICS 61.(1) 4n A

2n B

8n A B

Number of subsets having atleast three elements

8 8 8 80 1 22 C C C

82 1 8 28 256 37 219

62.(3) 1 22

1 2

21 1

2

z z, z

z z

2 21 2 1 22 2z z z z

1 2 1 2 1 2 1 22 2 2 2z z z z z z z z

2 2 21 2 14 4z z z

2 21 24 1 0z z

1 2z

63.(3) 2 6 2 0

2 6 2 0

1 26 2 0n n n

1 26 2 0n n n

Subtract above two equations we get: 1 26 2 0n n na a a 2

1

23

2n n

n

a a

a

64.(4) 1 2 22 1 2

2A

a b

9TA A I

1 2 2 1 2 9 0 02 1 2 2 1 2 0 9 0

2 2 2 0 0 9

a

a b b

4 2 0a b

2 4a b ….. (i)

2 2 2 0a b

1a b ….. (ii) 1 2b , a

Vidyamandir Classes


65.(3) 2 2 1

2 3 2 01 2

3 2 5 3 0

21 2 3 0

21 3 0

1 1 3, ,

66.(2) 3 5 6 7 8, , , ,

4 digit numbers 3 4 3 2 72 ( 3 & 5 can’t come at 1st place)

5 digit numbers 5 4 3 1 120

192 numbers

67.(1)

50 2 3 450 50 50 50 500 1 2 3 4

505050

1 2 2 2 2 2

2

x C C x C x C x C x

........... C x

50 2 5050 50 50 500 1 2 501 2 2 2 2x C C x C x .......... C x

On adding 50 50 50 50 50 500 2 501 2 1 2 2 4 2x x C C . .x ..... C

Putting x = 1 on both sides

50

501 3 1 3 12 2

68.(2) 2

nm

1 2 3, G , G , G , n in G.P.

14nr

3 1

4 41G n ,

1 12 2

2G n , 31

4 43G n

4 4 4 3 2 2 3 21 3 32 2 4G G G n n n m n

69.(2)

2

23 3 3

2

12 11 2

1 3 1 4n

n nn....... nt

...... n n

9

2 2 2

1

1 1 10 11 212 3 10 1 964 4 6n

n

. .t ...

70.(3) 1 2 340

cos x cos xlim

x tan xx

22 4

40 44

sin xlimtan xx x . x

x

1 4 22

Vidyamandir Classes


71.(1) 3 3 3g g g 2 3 2k m

3 34kg g m 4k m

8 3 2m m 5 2m

2 85 5

m , k

2k m

72.(4) 2 22 3 0x xy y ……..(i)

Differentiating w.r.t. x

2 2 2 6 0dy dyx y x ydx dx

3 0dyx y x ydx

3

x ydydx x y

3dx x y

dy x y

At (1, 1) 1 3 2 11 1 2

dxdy

Equation of normal at (1, 1)

1 1 1y x

Or 2y x 2y x ….(ii)

Substituting value of y from (ii) in (i)

We get 2 22 2 3 4 4 0x x x x x

2 2 22 3 4 12 12 0x x x x x 24 16 12 0x x

2 4 3 0 1 3 0x x x x

3 1x , y which is in 4th quadrant.

73.(3) 4 3 2f x ax bx cx dx e

20

1 3x

f xlim

x

20

2x

f xlim

x

4 3 2

202

x

ax bx cx dx elimx

0 0 2d , e , c

4 3 22f x ax bx x

3 24 3 4f x ax bx x

Vidyamandir Classes


4 3 4a b …… (i)

2 32 12 8 0f a b

8 3 2a b …… (ii)

Solving (i) and (ii) 4 2a

1 22

a , b

4 3 21 2 22

f x x x x

2 8 16 8 0f

74.(4)

3 3

2 4 2 44 41 1

dx dx

x x x x

Let 41 x t

54x dx dt

1 1

4 414 1 4 443 4

4

1 111444

dt t xI C t C x C Cxt

75.(3) 4

2

2 1 2 1I dx I

76.(4) 2 24 1 2 16 10 1 0x x x x

10 100 6432

x

1 12 8

,

112

y ,

1 11 2 2 3

1 2 1 21 2

1 1 1 94 2 4 2 2 3 32

/ //

y y y ydy y

77.(3) 2dy ydx x log x

I.F. 1 dx

x log xe log x

Required solution is

2y log x log x dx

2 1y log x x log x C

As x = 1 is in domain, replace x = 1 0 2(0 1) 0 2 2y C C C

2 1 2y log x x log x 2y e

Vidyamandir Classes


78.(4)

No. of points on L2

(1, 39)…………(39,1) = 39

Similarly points on other lines would be

38, 37,…….., 1

total points = 1 + 2 + ……….39 39 40 7802

79.(3)

Let 2 3P ,

Let point of intersection of 2 3 4 0x y and 2 3 0x y be Q

1 2Q ,

2 22 1 3 2 2PQ

Since PAR QAR

2AQ Point Q is always lies at distance of 2 unit from point A.

80.(3) Equation of circles are:

2 2 22 3 5x y ……..(i)

1 12 3 5C , ,r And 2 2 23 9 8x y …..(ii)

2 23 9 8C , , r

2 21 2 2 3 3 9C C = 13 = 1 2r r No. of common tangents = 3

Vidyamandir Classes


81.(4) 2 2x y 1

9 5 [Equation of tangent 2 1

9 3x y ]

Area of quadrilateral 1 9 6 272

82.(4) 44th t

2 224 2t tk

Eliminating t

2

2 22

hk x y

83.(4) 2 1 23 4 12

x y z k

3 2x k 4 1y k satisfies 16x y z 12 2z k 3 2 4 1 12 2 16k k k Point of intersection 11 11k 5 3 14, , 1k

Distance is 2 2 24 3 12 13

84.(3) Equation of family of planes containing line

2 5 3x y z and 4 5x y z is 2 5 3 4 5 0x y z k x y z

2 5 4 1 3 5 0x k y k z k k …..(i)

(i) is parallel to 3 6 1x y z …….(ii)

2 5 4 11 3 6

k k k 113 6 5

2k k k

Substituting value of k in (i), we get

11 11 112 5 22 1 3 5 02 2 2

x y z

7 21 42 49 0x y z 3 6 7 0x y z

85.(1) 13

a b c b c a 13

a.c b b .c a b c a

If 0a.c , then 13

b .c b c

Vidyamandir Classes


13

b c cos b c 13

cos 2 23

sin

86.

87.(4) Total = 16×16 = 256

New after removing one ‘16’

Total = 240 & no. of entries = 15

New total after adding 3,4, 5 = 240 + 12 = 252

No. of entries = 15 + 3 = 18

Mean = 252 14 018

.

88.(1) Let height of tower be PQ = h.

In PAQ,

30 htanAQ

3AQ h

In PBQ, BQ h

AB 3 1 h

In PCQ, 60 htanCQ

3

hCQ

BC = 113

h

AB:BC = 13 1 13

h : h

3 1:

89.(1) 1 1 12

21

xtan y tan x tanx

, 1

3x

12

21

xtan zx

x tan 2 2

,

1 2tan tan z

12 2z tan x 1 1x ,

For 13

x , 1 12

2 21

xtan tan xx

1 1 12tan y tan x tan x

1 13tan y tan x 3

231 3

x xyx

90.(4) ~ ~ s ( ~ r s

s ~ ~ r s s r ~ s s r s ~ s s r F s r

Joint Entrance Exam/JEE Mains 2015 Code –...

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