2019 apr jee-main q p k s - sri chaitanya...15. a person of mass m is, sitting on a swing of length...

2019_Apr_Jee-Main Question Paper_Key & Solutions

2019_Jee-Main Question Paper_Key & Solutions

Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081

www. srichaitanya.net, [email protected]

PHYSICS 1. Shown in the figure is a shell made of a conductor. It has inner radius a and outer

radius b, and carries charge Q. At its centre is a dipole p

as shown. In this case:

1. Surface charge density on the inner surface is uniform and equal to 2

Q / 24 a

2. Electric field outside the shell is the same as that of a point charge at the centre of the shell

3. Surface charge density on the inner surface of the shell is zero every where 4. Surface charge density on the outer surface depends on p

Ans. 2 Sol. Electric field outside the shell is the same as that of a point charge at the centre of

the shell ,From Gauss law as dipole does not create any charge inside 2. In a double slit experiment, when a thin film of thickness t having refractive index

is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ( is the wavelength of the light used).

1. 2 1

2.

21

3.

1

4.

2 1

Ans. 3

Sol: 1u t n 1Here m 1

t

03. Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (R = 8.3 J/mol K)

1.21.6 J/mol K 2.15.7 J/mol K 3.19.7 J/mol K 4.17.4 J/mol K Ans. 4

Sol: 1 21 2

1 2

3 52 32 22 3

V VV mix

R Rn C n C

Cn n

6 152 5V mix

RC

21 21 8.314 17.45 /10 10

V mix

RC J molk




04. An excited He ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength , energy

1240 eVE

in nm

.

1.n = 4 2. N = 5 3.N = 7 4.N = 6 Ans. 2

Sol: 1240 11.43 .108.5

E e v

2

4 13.612 .nE e vn

If election Jumps from 5 2n to n 11.43 .E e v 5n 05. A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates

at a constant angular speed of 40 1rad s about its axis, perpendicular to its plane.

If the magnetic field at its centre is 93.8 10 T , then the charge carried by the ring is close to 7 2

0 4 10 N / A

1. 53 10 C 2. 62 10 C 3. 54 10 C 4. 67 10 C Ans: 1

Sol: 0

2qB

r

7

9 4 10 403.8 100.1 2

62 10 C 06. The stopping potential 0V (in volt) as a function of frequency (v) for a sodium

emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be. (Given. Planck’s constant (h) = 346.63 10 Js, electron charge e = 191.6 10 C )

1.1.82 eV 2.2.12 eV 3.1.95 eV 4.1.66 eV




Ans: 4

Sol: 0W h 34 14

19

6.625 10 4 101.6 10

W

1.66 .W eV

07. The transfer characteristic curve of a transistor, having input and output resistance 100 and 100 k respectively, is shown in the figure. The Voltage and Power gain, are respectively.

1. 4 65 10 ,2.5 10 2. 4 62.5 10 ,2.5 10 3. 4 55 10 ,5 10 4. 4 65 10 ,5 10 Ans: 1

Sol: 0 0Cg

I B I

V I RVV I R

3 3

6

5 10 100 10100100 10gV

45 10gV

2

0Cg

b I

I RP

I R

3 3

6

5 10 100 10100100 10

6

310 10400gp 62.5 10gp

08. To verify Ohm’slaw, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained.

If 0V is almost zero, identify the correct statement.




1. The potential difference across the battery is 1.5 V when it sends a current of 1000 mA

2. The value of the resistance R is 1.5 3. The emf of the battery is 1.5 V and the value of R is 1.5 4. The emf of the battery is 1.5 V and its internal resistance is 1.5 Ans: 4 Sol: V E Ir When 0, 1.5I V E V 1000 , 0When I mA V 1.5 1. 0r 1.5r 09. A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless

surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 1ms with respect to the man. The speed of the man with respect to the surface is.

1.0.28 1ms 2.0.47 1ms 3.0.14 1ms 4.0.20 1ms Ans: 4 Sol: 1 2 0.70V V

1 1 2 2m V m V

2 220 0.70 50V V

2 214 20 50V V

230 14V 2 0.47 /V m s 10. A shell is fired from a fixed artillery gun with an initial speed u such that it hits the

target on the ground at a distance R from it. If 1 2t and t are the values of the time taken by it to hit the target in two possible ways, the product 1 2t t is

1.R/2g 2.R/g 3.R/4g 4.2R/g Ans: 4

Sol: 12 sinu t

g

22 cosu t

g

1 22Rt tg

11. The truth table for the circuit given in the fig. is.




1.

A B Y0 0 00 1 01 0 11 1 1

2.

A B Y0 0 10 1 11 0 01 1 0

3.

A B Y0 0 10 1 11 0 11 1 1

4.

A B Y0 0 10 0 11 1 11 1 0

Ans: 2 Sol: .Y A A B .Y A A B

A

B

.Y A A B

0 0 1

0 1 1 1 0 1 1 1 0

12. A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass

per unit area of the disc varies as 0r

then the radius of gyration of the disc about

its axis passing through the centre is.

1. a b3 2.

2 2a b ab2

3. a b2 4.

2 2a b ab3

Ans: 4

Sol: 2 202b

mk rdr rra

2 02b

a

k rdrr

202 rdr rr

2 3 302

2 .3

k b a b a

2 2

3a b abk




13. A progressive wave travelling along the positive x – direction is represented by y x, t Asin kx t . Its snapshot at t = 0 is given in the figure.

For this wave, the phase is

1. 2.2 3.

2

4.0

Ans: 1

Sol: cosdyslope AK kx tdx

0; 0,at x t slope is ve Slope negative 14. Two identical parallel plate capacitors, of capacitance C each, have plates of area A,

separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants

1 2 3K ,K and K .The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig II.

If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be ( 1E refers to capacitor (I) and 2E to capacitor (II)).

1. 1 2 3 2 3 3 1 1 21

2 1 2 3

K K K K K K K K KEE K K K

2.

1 2 31

2 1 2 3 2 3 3 1 1 2

K K KEE K K K K K K K K K

3. 1 2 3 2 3 3 1 1 21

2 1 2 3

K K K K K K K K KEE 9K K K

4.

1 2 31

2 1 2 3 2 3 3 1 1 2

9K K KEE K K K K K K K K K




Ans: 4

Sol: 1 2 3 21 0

12 3

K K KE C V

21 2 3

1 01 2 2 3 3 1

312

K K KE C VK K K K K K

1 2 3 1 2 2 3 3 11

2 1 2 39K K K K K K K K KE

E K K K

15. A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude 0 .If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l(l<<L),is close to.

1.Mgl 2.2

0Mgl 12

3. 2

0Mgl 1 4. 20Mgl 1

Ans: 3 Sol: Applying angular momentum conservation

'Mv L l mvL ' vLv

L l

Using Work-energy theorem

2

m

2m 0

1 1W Mgl mv' m2 2

W mgl 1

16. The resistive network shown below is connected to a D. C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is.

1.1 2.16 3.8 4. 6 Ans: 3 Sol:

16

2R 4R RR

2 2564

8P

R R

8R




17. A point dipole 0p p x

is kept at the origin. The potential and electric field due to this dipole on the y – axis at a distance d are, respectively. (Take V = 0 at indignity)

1. 2 30 0

p p,4 d 4 d

2. 2 3

0 0

p p,4 d 4 d

3. 3

0

p0,4 d

4. 3

0

p0,4 d

Ans: 3

Sol:

V=0 304

PEd

0p

18. An electromagnetic wave is represented by the electric field

0E E n sin t 6y 8z

. Taking unit vectors in x, y and z directions to be i , j,k

,

the direction of propogation s

is

1. 4k 3 js5

2. 4 j 3 js

5

3. 3 j 4ks

5

4. 3 j 4 js

5

Ans: 3

Sol: 0 5 cos 3 4B E A n i j

6 8 3 45100

i K j kS

19. The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1cm s-

1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 ,the current in the loop at that instant will be close to.

1.170 A 2.150 A 3.115 A 4. 60 A Ans: 1 Sol: 2 2 41 5 10 10 5 10E BlV volt

445 10 1.66 10

1.7 1.3eIR

170I A




20. A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine (B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency v. The value of v is close to. (Speed of sound in water = 1500 ms-1)

1. 504 Hz 2.499 Hz 3.507 Hz 4.502 Hz Ans: 4

Sol: 518 5 /18AV m s 1500 /V m s

527 7.5 /18BV m s 500f

B

V A

1495 5001492.5

AA

B

V V tV V

500.841 2A

BB

A

V V fV V

B

AV

1507.5 500.841505

502 Hz 21. A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that

has water filled up to 5 cm (see figure). It a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is close to. (Refractive index of water = 1.33)

1.11.7 cm 2.8.8 cm 3.6.7 cm 4.13.4 cm Ans: 2

Sol: 1 1 1f V u




1 1 1420 53

V

1 3 1 1020 20

V cmV

145 10 5 11.71.33

d cm

22. A magnetic compass needle oscillates 30 times per minute at a place where the dip is 045 , and 40 times per minute where the dip is 030 . If a

1.3.6 2.0.7 3.2.2 4.1.8 Ans: 2

Sol:

4B

7B

1 cos2

BnI

1 1 1

2 2 2

cos 30cos 40

n Bn B

1

2

192

1632

B

B

1

2

9 3 2 016 2 1

BB

.7

23. The value of numerical aperture of the objective lens of a microscope is 1.25. If light

of wavelength 5000 0A is used, the minimum separation between two points, to be

seen as distinct, will be. 1.0.38 m 2.0.48 m 3.0.24 m 4.0.12 m Ans: 3

Sol: 21.22

2.sinmxxx

101.22 5000 10

2 1.25

0.24 m 24. A sample of an ideal gas is taken through the cyclic process abca as shown in the

figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is.

1.120 J 2.140 J 3.100 J 4.130 J




Ans: 4 Sol: dQ dU dW 250 60 180 dW 130dW J 25. Which of the following combination has the dimension of electrical resistance ( 0 is

the permittivity of vacuum and 0 is the permeability of vacuum)?

1. 0

0

2. 0

0

3. 0

0

4. 0

0

Ans: 4 Sol: 2 3 2R ML T A

2 20 MLT A

1 3 4 20E M L T A

0

0

RE

26. The trajectory of a projectile near the surface of the earth is given as 2y 2x 9x .

If it were launched at an angle 0 with speed 0 then 2g 10 ms .

1. 1 10 0

1 5cos and ms35

2. 1 10 0

2 5cos and ms35

3. 1 10 0

2 5sin and ms35

4. 1 10 0

1 5sin and ms35

Ans: 1 Sol: 2Tan

5 2

1

100

0

1cos5

1

01cos5

22 9y x x

2 20

92 cos

gV

2 20

10cos18

V

59

05cos

3V

05 /3

V m s




27. At 040 C ,a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 040 C to

020 C it regains its original length of 0.2 m. The value of M is close to. (Coefficient of linear expansion and Young’s modulus of brass are

5 0 11 210 / C and 10 N / m respectively ; g = 10 ms-2) 1.9 kg 2.0.9 kg 3.0.5 kg 4.1.5 kg Ans: 1 Sol: YA t Mg

2Y r tMg

11 6 510 10 10 20

10M

6.28M ks 28. A galvanometer of resistance 100 has 50 divisions on its scale and has sensitivity

of 20 A / division. It is to be converted to a voltmeter with three ranges, of 0-2V, 0- 10 V and 0-20V. The appropriate circuit to do so is:

1. 2.

3. 4. Ans: 2 Sol: V I R G 32 10 1900 100 310 10 9900 100 320 10 19900 100

29. When 1M gram of ice at 010 C (specific heat = 0.5 cal 01 1g C is added to 2M gram

of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in 1calg is:

1. 1

1

5M 50M

2. 2

1

50MM

3. 2

1

5M 5M

4. 2

1

50M 5M

Ans: 4 Sol: 1 1 210 0.5 50M M L M




2 1

1

50 5M MLM

2

1

50 5MLM

30. A uniform rod of length l is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?

1. 2.

3. 4. Ans: 1

Sol:

L x

x

2dT dmx 2T dx x

L x

2

x

T xdx

2

2 2T L x x2

2

2T L 2Lx2

T

L




CHEMISTRY 31. In the following reaction: xA yB

10 10

d A d Blog log 0.3010

dt dt

‘A’ and ‘B’ respectively can be: 1. 2 4 4 8C H and C H 2. n – Butane and Iso – butane 3. 2 4 2N O and NO 4. 2 2 6 6C H and C H Ans: 1 Sol: d x d x x

dt dt y

2C H C H2 4 84 x y

32. The basic structural unit of feldspar, zeolites, mica, and asbestos is:

1. 2SiO 2. 23SiO 3. 44SiO 4.

|

|

R

Si O R Me

R

n

Ans: 3 Sol: These are 3.0 silicates and basic unit is 33. The major product of the following addition reaction is

Cl /H O3 2

2 2H C CH CH

1. 3H C

O

2. 3 2| |

H C CH CH

OH Cl

3. 3 2| |

CH CH CH

Cl OH

4.

O

3H C 3CH

Ans: 2

Sol:

3 2CH CH CH

OH Cl

3 2 2 2CH CH CH cl / H o




34. The electrons are more likely to be found:

1. in the region a and b 2. Only in the region a 3. only in the region c 4. In the region a and c Ans: 4

Sol:

x

xa

b

c

x

2Pr obability chemistry a.c

35. The correct sequence of thermal stability of the following carbonates is: 1. 3 3 3 3BaCO SrCO CaCO MgCO 2. 3 3 3 3BaCO CaCO SrCO MgCO 3. 3 3 3 3MgCO CaCO SrCO BaCO 4. 3 3 3 3MgCO SrCO CaCO BaCO Ans: 3 Sol: 3 3 3 3MgCO CaCO SrCO BaCO

36. Enthalpy of sublimation of iodine is 124cal g at 200 C . If specific heat of

2 2I s and I vap are 0.055 and 1 10.031 cal g K respectively, then enthalpy of

sublimation of iodine at 250°C in 1cal g is: 1. 2.85 2. 11.4 3. 22.8 4. 5.7 Ans: 3

Sol: 2 1P

2 1

H H CT T

2H 24 0.024250 200

2H 24 1.2 22.8




37. 5 moles of 2AB weigh 3125 10 kg and 10 moles of 2 2A B weigh 3300 10 kg . The

molar mass of AA M and molar mass of 1BB M in kg mol are:

1. 3 3A BM 10 10 and M 5 10 2. 3 3

A BM 25 10 and M 50 10

3. 3 3A BM 50 10 and M 25 10 4. 3 3

A BM 5 10 and M 10 10 Ans: 4 Sol: Given the (1) weight of 5 mol of AB 3125 10 125 kg g or mass of 1 mol of 2 25AB g (2) weight of 10 mol of 3

2 2 300 10 300 A B kg g mass of 1 mol of 2 2 30A B g consider the molar mass as per molecular formula 2 25A BM M 2 2 30A BM M

on solving; 5AM g and 10BM g or 35 10AM kg and 310 10BM kg 38. An element has a face – centred cubic (fcc) structure with a cell edge of a. The

distance between the centres of two nearest tetrahedral voids in the lattice is:

1. 3 a2

2. 2a 3. a 4. a2

Ans: 4 Sol: In case of face-centred cubic structure, the distance between the centres of two nearest

tetrahedral voids is 2a which are parallel to edge of unit cell.

39. Given: 3 1 2Co e Co ; E 1.81V

4 2Pb 2e Pb ; E 1.67V

4 3Ce e Ce ;E 1.61V

3Bi 3e Bi;E 0.20V Oxidizing power of the species will increase in the order: 1. 3 4 4 3Bi Ce Pb Co 2. 3 4 3 4Co Ce Bi Pb 3. 4 4 3 3Ce Pb Bi Co 4. 3 4 4 3Co Pb Ce Bi Ans: 1 Sol: Greater SRP, greater oxidising power 40. The correct set of species responsible for the photochemical smog is: 1. 2 2 2CO , NO , SO and hydrocarbons 2. 2 2N , NO and hydrocarbons 3. 2 3NO, NO ,O and hydrocarbons 4. 2 2 3N , O ,O and hydrocarbons Ans: 3




Sol: Conceptual 41. The correct statement among the following is : 1. 3 3SiH N is pyramidal and less basic than 3 3CH N

2. 3 3SiH N is planar and less basic than 3 3CH N

3. 3 3SiH N is planar and more basic than 3 3CH N

4. 3 3SiH N is pyramidal and more basic than 3 3CH N Ans: 2 Sol: The loss pair ‘M’ is donated to 3d orbital of Si 42. Which of the following is a thermosetting polymer? 1. PVC 2. Bakelite 3. Nylon 6 4. Buna – N Ans: 2 Sol: Bakelite 43. An organic compound ‘A’ is oxidized with 2 2Na O followed by boiling with 3HNO .

The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is:

1. Fluorine 2. Sulphur 3. Nitrogen 4. Phosphorus Ans: 4

Sol: 4 42 2 3 2NH MoONa O HNO3 4 3 4 4 4 23P Na PO H PO NH PO .12MoO yellow PPt

44. An example of a disproportionation reaction is: 1. 4 2 4 2 22KMnO K MnO MnO O 2. 22CuBr CuBr Cu

3. 24 2 22MnO 10 I 16H 2Mn 5I 8H O

4. 2 22NaBr Cl 2NaCl Br Ans: 2 Sol: Cu under go both oxidation reduction 45. Complete removal of both the axial ligands (along the z – axis) from an octahedral

complex leads to which of the following splitting patterns? (relative orbital energies not on scale).

1.

E

x yd 2 2

xyd

zd 2

xz yzd , d 2.

E x yd 2 2

xyd

zd 2

xz yzd , d

3.

E

x yd 2 2

xydz

d 2

xz yzd , d 4.

E

x yd 2 2

xyd

zd 2

yz xzd , d




Ans: 1 Sol: For square plane complex splitting

dxydyzdzx

2 2 2dx y dz

2 2dx y

dxy

2dz

dyzdxz 46. The group number, number of valence electrons, and valencey of an element with

atomic number 15, respectively, are: 1. 16, 6 and 3 2. 16, 5 and 2 3. 15, 5 and 3 4. 15, 6 and 2 Ans: 3 Sol: 2 3ns np 47. The complex ion that will lose its crystal field stabilization energy upon oxidation of

its metal to +3 state is:

N N

(Phen =

andignore pairing energy)

1. 2

3Co phen

2. 2

3Zn phen

3. 2

3Fe phen

4. 2

3Ni phen

Ans: 3 Sol: 2CFSE for Fe 0.4 6 0.6 0

3CFSE for Fe 0.4 5 0.6 0

2.0 decreases CFSE 48. But – 2 – ene on reaction with alkaline 4KMnO at elevated temperature followed by

acidification will give: 1. 3 3

| |CH CH CH CH

OH OH

2. one molecule of 3CH CHO and one molecule of 3CH COOH 3. 2 molecules of 3CH CHO 4. 2 molecules of 3CH COOH




Ans: 4

Sol:

4KMno

H32CH COOH

4.2 ,moles of 3CH COOH

49. Glucose and Galactose are having identical configuration in all the positions except position.

1. C – 2 2. C – 5 3. C - 3 4. C – 4 Ans: 4 Sol: C – 4 configuration is different 50. The increasing order of the bpK of the following compound is:

A.

F

N

H

S

N

H B.

N

H

S

N

H

3CH O

C.

N

H

S

N

H

2O N

D.

N

H

S

N

H

3H C

1. (C) < (A) < (D) < (B) 2. (B) < (D) < (A) < (C) 3. (A) < (C) < (D) < (B) 4. (B) < (D) < (C) < (A) Ans: 1 Sol: B > D > A > C 51. What is the molar solubility of 3Al OH in 0.2 M NaOH solution? Given that,

solubility product of 243Al OH 2.4 10 :

1. 193 10 2. 2112 10 3. 223 10 4. 2312 10 Ans: 3 Sol:

33

3aq aqAl OH s aq Al OH

3 324 3 32.4 10 0.2spK Al OH Al

3 223 10Al

Thus solubility of 3Al OH is 223 10 .




52. The major product of the following reaction is:

HO

HO

1 CrO2 SOCl /3

32

1.

O

Cl 2.

O

HO

3.

O

Cl 4.

O

HO Ans: 4

Sol:

COOH

HO

2Socl /

2 2CH CH C Cl

0

HO

0

HO 53. The major product(s) obtained in the following reaction is/are:

i KO Buii O /Me S

t

3 2

Br

1. CHOCHO

2. CHO and OHC - CHO

OHC

3. CHOOHC

4. CHOOHC

tO Bu

Ans: 2




Sol:

KOtBu

Br3 2,O Me S

CH O

2CH

2CHCH O

CH O

CH O

54. The metal that gives hydrogen gas upon treatment with both acid as well as base is: 1. magnesium 2. zinc 3. mercury 4. Iron Ans: 2 Sol: 2 2 2Zn 2NaOH Na ZnO H

2 2Zn 2HCl ZnCl H 55. An ideal gas is allowed to expand from 1L to 10 L against a constant external

pressure of 1 bar. The work done is kJ is: 1. -0.9 2. -2.0 3. -9.0 4. +10.0 Ans: 1

SOL: W Pdv 1 9 9lt atm 9 101.325 KJ

1.01325 1000

0.9 KJ

56. Which of the following statements is not true about RNA?

1. It usually does not replicate

2. It has always double stranded helix structure

3. It is present in the nucleus of the cell

4. It controls the synthesis of protein

Ans: 2

Sol: RNA is single Stranded

57. Peptization is a:

1. process of converting soluble particles to form colloidal solution

2. process of bringing colloidal molecule into solution

3. process of converting a colloidal solution into precipitate

4. process of converting precipitate into colloidal solution

Ans: 4




Sol: Convergent of a freshly prepared precipitate into colloidal practical is known as petition

58. The major products of the following reaction are:

1 CHCl /aq.NaOH2 HCHO, NaOH conc. \

3 H O

3

3

OH

Cl

1.

OH

Cl

COOH

and Methanol

2. OH

and Formic acid

OH

OH

3. OH

COOH

and Methanol

OH

4. Cl

and Formic acid

OH

OH

Ans: 4

Sol:

OH

Cl

3

3

1 CHCl2 HCHO/NaOH3 H O

OH

Cl

2CH OH

HCOOH

59. The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality

1in mol kg of the aqueous solution is:

1. 13.88 2. 313.88 10 3. 213.88 10 4. 113.88 10 Ans: 1

Sol: x 1000m1 x mwty solved

where 2x mole fraction of solved 0.2 1000 13.880.8 18

60. The ideal of froth floatation method came from a person X and this method is related to the process Y of ores. X and Y, respectively, are:

1. washer man and reduction 2. Fisher women and concentration 3. washer women and concentration 4. Fisher man and reduction Ans: 3 Sol: Washer woman and concentration




MATHEMATICS

61. The value of 1 112 3sin sin13 5

is equal to:

1. 1 33cos65

2. 1 9cos2 65

3. 1 56sin2 65

4. 1 63sin65

Ans: 3

Sol: 1 12 4 3 5sin . .13 5 5 13

1 56sin2 65

62. If the normal to the ellipse 2 23x 4y 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q (4, 4) then PQ is equal to:

1. 612

2. 2212

3. 1572

4. 5 52

Ans: 4

Sol: 2

2a y a tan 2

bb x

3 1a 2,b 2 , tan 3, sin ,cos2 2

P 2cos , 3 sin 1,3 / 2

5 5Q 4,4 PQ2

63. If the data 1 2 10x , x ,.....x is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000; then the standard deviation of this data is:

1. 2 2. 2 2 3. 4 4. 2 Ans: 1 Sol: ix 11 4 16 6 = 44+96 = 140 2

ix 2,000

S.D =22

i ix x10 10

= 2




64. The equation 2y sin x sin x 2 sin x 1 represents a straight line lying in: 1. third and fourth quadrants only 2. Second and third quadrants only 3. first, second and fourth quadrants 4. First, third and fourth quadrants Ans: 1 Sol: 2y sin x sin(x 2) sin x 1

1 cos 2x 21 cos 2 cos 2x 22 2

2cos 2 1 sin 12

65. The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is:

1. 202 2. 202 1 3. 212 4. 202 1 Ans: 1 Sol: No. of ways C C C10 9 021 21 ........ 21 S

21S S 2 20S 2 66. For x R , let [x] denote the greatest integer x , then the sum of the series

1 1 1 1 2 1 99.... is :3 3 100 3 100 3 100

1. -133 2. -153 3. -135 4. -131 Ans: 1 Sol: -1-1-.....-1-2-2....-2 67 times 33 times =-67-66=-133 67. Let f : R R be a continuously differentiable function such that f(2) = 6 and

1f ' 248

. If f x 36

4t dt x 2 g x , then x 2lim g x

is equal to:

1. 24 2. 18 3. 36 4. 12 Ans: 2

Sol: f x

2

64t d x 2 g x




f x2

64t dt

g xx 2

3

x 2 x 2

4 f x .f xlt g x lt

1

3 14.6 .48

= 18

68. Consider the differential equation, 2 1y dx x dy 0y

. If value of y is 1 when x = 1,

then the value of x for which y = 2, is:

1. 3 12 e 2. 5 1

2 e 3. 3 e

2 4. 1 1

2 e

Ans: 1

Sol: 2 1y dx x dy 0y

2 3dx x 1dy y y

1 1dy2y yIF e e

Sol is 1 1

y y3

1xe e . dy cy

1 1 1

y y y1xe e e cy

1y1x 1 c e

y

x 1, y 1 1 1 1 ce

1ce

1y1 1x 1 e

y e




121 1y 2 x 1 e

2 e

3 12 e

69. For 3x 0, , let f x x , g x tan x2

and 2

21 xh x1 x

. If x hof og x then

3

is equal to:

1. 5tan12 2. 11tan

12 3. tan

12 4. 7tan

12

Ans: 2

Sol: 1 tan xx tan x1 tan x 4

tan3 12

11tan12

70. If the line x 2 y 1 z 13 2 1

intersects the plane 2x + 3y – z + 13 = 0 at a point P and

the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to: 1. 14 2. 2 7 3. 14 4. 2 14 Ans: 4 Sol: P 3t 2,2t 1, t 1 lieson 2x 3y z 13 0 6t 4 6t 3 t 1 13 0 13t 13 t 1 P 1, 3,2 Q 3t 2,2t 1, t 1 lies on 3x y 4z 16 9t 6 2t 1 4t 4 16 7t 7 t 1 Q 5,}0

PQ 36 16 4 56 2 14

71. The coefficient of 18x in the product 910 21 x 1 x 1 x x is:

1. 126 2. -126 3. -84 4. 84




Ans: 4

Sol: 910 21 x 1 x 1 x x

931 x 1 x 1 x

2 3 6 91 2 31 x 1 c x c x c x ......

18c6coef x 9 84

72. The integral 3

42x 1dxx x

is equal to: (Here C is a constant of integration)

1. 23

e 3

x 11 log C2 x

2.

3

e 2

x 1log C

x

3. 3

e 2

x 11 log C2 x

4.

3e

x 1log Cx

Ans: 4

Sol: 3

42x 1dxx x

3

3

2 x 1 3dx

x x 1

2

3 32 3x dxx x x 1

33 3

2 1 1dx dxx x x 1

3 32log x log x log x 1 c

3x 1log c

x

73. If the volume of parallelopiped formed by the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k, j k and i k is minimum, then is equal to:

1. 3 2. 3 3. 13

4. 13




Ans: 3

Sol: 1 1

V 0 10 1

3V 1

2V 3 1 let V 0

13

V 6

1 6V 03 3

13

74. A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25cm/sec., then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1m above the ground is:

1. 53 2. 25 3. 25 3 4. 25

3 Ans: 4 Sol: 2 2x y 4

dx dyzx 2y 0dt dt

dx dyx y

dt dt

dx3 1 25dt

dx 25 m / secdt 3

2y 1 x 3 x 3




75. The equation z i z 1 , i 1 , represents: 1. a circle of radius 1

2. a circle of radius 12

3. the line through the origin with slope – 1 4. the line through the origin with slope 1. Ans: 4 Sol: z i z 1

2 22 2x y 1 x 1 y

2 2 2 2x y 2y 1 x 2x 1 y 2y 2x x y 0 y x

76. If 20

cot x dx m ncot x cosecx

, then m.n is equal to:

1. -1 2. 12

3. 1 4. 12

Ans: 1

Sol: 2 2

0I cosecx cot x cot x

2 2

0I cos ecx cot x cosec x 1 dx

20cosecx cot x x

1

2

1 2

2

1m n 22

1

1m , n 22




77. If ye xy e , the ordered pair 2

2dy d y,dx dx

at x = 0 is equal to:

1. 21 1,e e

2. 21 1,e e

3. 2

1 1,e e

4. 21 1,e e

Ans: 3

Sol: yyy

e x

1y 0e

4 3

24

e x y y e y 1y

e x

21y

e

21 1y , y ,

e e

78. If 5 2 1

B 0 2 13 1

is the inverse of a 3 3 matrix A, then the sum of all values of

for which det (A) + 1 = 0, is: 1. -1 2. 2 3. 1 4. 0 Ans: 3

Sol: 1det AB

1 1 0 B 1B

22 2 25 1 04, 3

79. Let a random variable X have a binomial distribution with mean 8 and variance 4.

If 16kP X 2

2 , then k is equal to:

1. 1 2. 121 3. 17 4. 137 Ans: 4

Sol: np = 8, npq = 4 1 3q ,p n 162 2




P x 2 P x 0 P x 1 P x 2

= 0 1 2C C C 16116 16 16

2

= 16 161 16 120 137

2 2

80. Let ˆ ˆ ˆa 3i 2 j 2k and ˆ ˆ ˆb i 2 j 2k

be two vectors. If a vector perpendicular to both

vectors a b and a b has the magnitude 12 then one such vector is:

1. ˆ ˆ ˆ4 2i 2j k 2. ˆ ˆ ˆ4 2i 2 j k 3. ˆ ˆ ˆ4 2i 2j k 4. ˆ ˆ ˆ4 2i 2j k

Ans: 4 Sol: a b 4,4,0 ,a b 2,0,4

a b a b 2 i 2 j 8u

3a b a b

12u 4 2 i 2 j k

81. If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equailateral is:

1. 320

2. 15

3. 110

4. 310

Ans: 3

Sol:

8A 3A

6A

1A 2A

3A

1 3 5A ,A ,A

2 4 6A ,A ,A Form Equilateral triangle

3C

2 2 16 20 10

82. If m is the minimum value of k for which the function 2f x x kx x is increasing in the interval [0, 3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to:

1. 4,3 3 2. 3,3 3 3. 4,3 2 4. 5,3 6




Ans: 1

Sol: 2kf x x kx x

kf 3 3 3k 9

1kf x 0 and for function to be , the minimum value of k = 4 where x = 3

m 4, m 3 12 9 3 3

m,m 4,3 3

83. If the area (in sq.units) of the region 2x, y :y 4x,x y 1, x 0, y 0 is a 2 b ,

then a – b is equal to

1. 103

2. 23

3. 6 4. 83

Ans: 3 Sol: 2y 4x x y 4

21 x 4x

2x 2x 1 4x 0 2x 6x 1 0

6 36 4x2

3 2 2

3 2 2 2

0

12 x 1 3 2 22

3 2 2 2

0

2 12 x x. 2 2 23 2

3 24 2 1 2 2 13

2 42 1 2 1 23

4 2 23 2 23

12 2 4 2 2

3




8 2 1023 8 10a , b

3 3

a b 6 84. Let nS denote the sum of the first n terms of an A.P.. If 4 6 10S 16 and S 48, then S

is equal to: 1. -260 2. -320 3. -380 4. -410 Ans: 2

Sol: 4 2a 3d 16,2

6 2a 5d 48,2

d = -12 , a = 44

10S 10 44 108 3202

85. The number of solutions of the equation

4 2 5 51 sin x cos 3x, x ,2 2

is:

1. 7 2. 5 3. 3 4. 4 Ans: 2 Sol: 4 21 sin x cos 3x 2 4sin 3x sin x 0 sin 3x 0 and sin x 0 x 2 , ,0, ,2 86. If the angle of intersection at a point where the two circles with radii 5cm and 12cm

intersect is 90°, then the length (in cm) of their common chord is:

1. 12013

2. 132

3. 135

4. 6013

Ans: 1

Sol:

P

BA

Area 1 21 1PAB dx r r2 2

2 21 1. 5 12 .x 5 122 2




5 12x13

1202x13

87. If A is a symmetric matrix and B is a skew symmetric matrix such that 2 3

A B5 1

, then AB is equal to:

1. 4 21 4

2. 4 21 4

3. 4 21 4

4. 4 2

1 4

Ans: 3 Sol: T TA A, B B

2 3

A B5 1

2 3 2 81 1A5 1 3 12 2

4 8 2 418 2 4 12

0 2 0 11B2 0 1 02

2 4 0 1AB

4 1 1 0

0 4 2 0 4 2

AB0 1 4 0 1 4

88. If and are the roots of the equation 2375 x 25x 2 0 , then

n nr r

n nr 1 r 1lim lim

is equal to:

1. 29358

2. 21346

3. 112

4. 7116

Ans: 3

Sol: 1 1

21 1




25 222 375 375

25 21 2 1375 375

25 4375 25 2

29 1

348 12

89. If the truth value of the statement p ~ q r is false (F), then the truth values of the statements p, q, r are respectively:

1. T, F, T 2. T, F, F 3. T, T, F 4. F, T, T Ans: 3 Sol: P qVr

T TVT T FVF isF

90. Let P be the point of intersection of the common tangents to the parabola 2y 12x

and the hyperbola 2 28x y 8 . If S and S' denote the foci of the hyperbola where S lies on the positive x – axis then P divides SS ' in a ratio:

1. 13:11 2. 14:13 3. 2:1 4. 5 : 4 Ans: 4

Sol: 3y mxM

2 2x y 11 8 2 8 1e 9

1

22

9 1m 8m

4 29 m 8m m 3

Common tangents y 3x 1

y 3x 12y 0

y 0 0 3x 1

1x3

1point P ,03

S 3,0 , S 3,0

13SP 103 5 / 41PS 833

... Prepared by Sri Chaitanya Faculty

2019 apr jee-main q p k s - sri chaitanya...15. a person of mass m is, sitting on a swing of length...

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