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JEE-Mains – 2018 (Hints & Solutions) Narayana IIT/NEET Academy

Narayana Group of Educational Institutions -1-

This booklet contains 41 printed pages PAPER – 1 : PHYSICS, CHEMISTRY & MATHEMATICS Do not open this Test Booklet until you are asked to do so. Read carefully the instructions on the Back Cover of this Test Booklet. Important Instructions: 1. Immediately fill in the particulars on this page of the Test Booklet with only Black Ball Point Pen.

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take out the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and

Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

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A



JEE-MAIN-2018 SOLUTIONS PART A - PHYSICS

1. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is

(1) 2.5% (2) 3.5% (3) 4.5% (4) 6% Ans. (3) Sol. Density of the cube is

= 3

ML

Using error analysis method

= M 3 LM L

= [1.5 3(1)] = 4.5 % 2. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick

it up.

(1) (2)

(3)

(4)

Ans. (2) Sol. From the graphs it is clear that initial velocity is non-zero but in graph (2) initial velocity is zero. 3. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley,

are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is

(1) 18.3 kg (2) 27.3 kg (3) 43.3 kg (4) 10.3 kg Ans. (2) Sol. For the system to be at rest flim> m1g 2 1(m m )g m g 0.15(m + 10) > 5



m + 10 >50015

m + 10 >100

3

m >70 kg3

m > 23.33 kg Hence minimum mass will be 27.3 kg 4. A particle is moving in a circular path of radius a under the action of an attractive potential U =

2

k2r

. Its total energy is

(1) 2

k4a

(2) 2

k2a

(3) zero (4) 2

3 k2 a

Ans. (3)

Sol. U = 2

k2r

F = dUdr

= 3k ( 2) r2

= 3

kr

Now, F = 2mv

r

2mv

r = 3

kr

KE = 21 mv2

= 2

k2r

TE = PE + KE

= 2 2 2 2

k k k k2r 2r 2a 2a

( r = a)

= 0 5. In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same

mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is

(1) 0v4

(2) 02 v (3) 0v2

(4) 0v2

Ans. (2) Sol.

On conserving the momentum mv0 = mv1 + mv2 v0 = v1 + v2 …. (i) KEf> KEi it shows that explosion occurs during collision

2 21 2

1 1mv mv2 2

= 1.5 20

1 mv2

v0 v1 v2



2 21 2v v = 2

03 v2

… (ii)

On solving (i) and (ii),

v1 = 01 2 v

2

v2 = 01 2 v

2

Relative velocity = v2 – v1 = 02 v 6. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as

shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is

(1) 219 MR2

(2) 255 MR2

(3) 273 MR2

(4) 2181 MR2

Ans. (4) Sol. Centre of mass of system lies at O.

I0 = Idisc at centre + 6(Idisc surrounding it)

I0 = 2 2

2MR MR6 M(2R)2 2

[Using parallel axis theorem; I = Icm + md2;

d = OQ = 2R]

I0 = 255 MR2

IP = I0 + 7M(3R)2 [Using parallel axis theorem]

IP = 255 MR2

+ 63MR2 IP = 2181 MR2

7. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R3

is removed as

shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is

O

Q

P



(1) 4 MR2 (2) 240 MR

9 (3) 10 MR2 (4) 237 MR

9

Ans. (1) Sol. Mass of complete disc = 9 M Mass of removed portion = m

m = 2

9MR

× (area of removed portion)

m = 2

2

9M R.R 3

Solving we get, m = M IO = Icompletedisc – Iremoved portion

= 2 22(9M)R 1 R 2RM M

2 2 3 3

(Using parallel axis theorem)

IO = 2

29MR 1 MR2 2

= 4 MR2

8. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then

(1) 3/2T R for any n (2) n 12T R

(3) (n 1)/2T R (4) n /2T R Ans. (3)

Sol. nF R and F = 2mv

R

2

nmv RR

v2 R1 – n (1 n )/ 2v R

T Rv (1 n)/2

RTR 2 RT

v

n 12T R

9. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a

cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to

compress the liquid, the fractional decrement in the radius of the sphere, drr

, is

(1) Kamg

(2) Ka3mg

(3) mg3Ka

(4) mgKa

C

O

2R/3



Ans. (3)

Sol. P = 0mg Pa

p = mga

[ p = volumetric stress]

Volumetric deformation

V

V

= p

K

K Bulk modulus

V = 34 r3

VV

= 3 r

r

3 r

r

= mg / a

K

r

r

= mg3Ka

Decreament = drr

= r

r

drr

= mg3Ka

10. Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

(1) (a) 189 K; (b) 2.7 kJ (2) (a) 195 K; (b) –2.7 kJ (3) (a) 189 K; (b) –2.7 kJ (4) (a) 195 K; (b) 2.7 kJ Ans. (3)

Sol. 1TV = const. for adiabatic process and = 53

for monoatomic gas.

T1 = 300 k and V1 = V. Let T2 = T and V2 = 2V (given)

5 13300(V)

= 5 13T(2V)

T = 2/3

3002

T = 189 K

U = nCv T

U = 2 × 2 13 R[T T ]2

U = 3R[189 – 300] U = –333R U –2.7 kJ 11. The mass of a hydrogen molecule is 273.32 10 kg . If 1023 hydrogen molecules strike, per second, a

fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly :

(1) 3 22.35 10 N / m (2) 3 24.70 10 N / m (3) 2 22.35 10 N / m (4) 2 24.70 10 N / m Key (1)

Sol. odp dF (2N mvcos 45 )dt dt

d 1(2N mv )dt 2

m

r

area = amg + p a0

pa

45° 45°



dN dNF 2mv molecules strike / sec.dt dt

3 2FP so P 2.35 10 N / mA

12. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver = 108 and Avagadro number = 6.02 1023 gm mole-1)

(1) 6.4 N / m (2) 7.1 N / m (3) 2.2 N / m (4) 5.5 N / m Key (2) Sol. 2 2K m m(2 )

Where 3

23

108 10m6.02 10

K = 7.17 N/m 13. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations.

The density of granite is 2.7 103 kg/m3 and its Young’s modulus is 9.27 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?

(1) 5 kHz (2) 2.5 kHz (3) 10 kHz (4) 7.5 kHz Key (1) Sol. 3 3L 60cm 2.7 10 kg / m

10Y 9.27 10 Pa

L2

2L 120cm 1.2m

10

7 63

Y 9.27 10v 3.4333 10 34.3 102.7 10

35.86 10 m / s

3

3v 5.86 10f 4.88 10 Hz 5 kHz1.2

14. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities , and respectively.The potential of shell B is :

(1) 2 2

0

a b ca

(2) 2 2

0

a b cb

(3) 2 2

0

b c ab

(4) 2 2

0

b c ac

Key (2) Sol. 2

1Q 4 a

22Q 4 b

23Q 4 c

2 2 2

Bk(4 a ) k( 4 b ) k(4 c )V

b b c

2 2 2

2

a b ck(4 )b b c

a4 k b cb

2 2

0

a b cb

c

b

a

/ 2



15. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric

material of dielectric constant 5K3

is inserted between the plates, the magnitude of the induced

charge will be : (1) 1.2 n C (2) 0.3 n C (3) 2.4 n C (4) 0.9 n C Key (1) Sol. Q KQ Q (K 1)CV

35 1 90 10 203

32 1800 103

2 3.61.8 1.2 n C3 2

16. In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t

i 20sin 30t4

In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively :

(1) 50, 10 (2) 1000 ,102

(3) 50 ,02

(4) 50, 0

Key (2) Sol. av rms rmsP V I cos

100 20 12 2 2

av1000P

2

wattless rmsI I sin

20 1 10A2 2

17. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 . The internal resistances of the two batteries are 1 and 2 respectively. The voltage across the load lies between :

(1) 11.6 V and 11.7 V (2) 11.5 V and 11.6 V (3) 11.4 V and 11.5 V (4) 11.7 V and 11.8 V Key (2) Sol. 1 2 1 213 2i 12 1i 1 2i i

1 2 1 110(i i ) 10(i 1 2i )

1 1 110(i 1 2i ) 13 2i

1 130i 10 13 2i

132i 23

123i32

23V 13 2 11.5632

13V

12V

i1

i2

10

2

1(i +i )1 2


Narayana Group of Educational Institutions

-9-

18. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, r respectively in a uniform magnetic field B. The relation between re, rp, r is :

(1) re> rp = r (2) re< rp = r (3) re< rp< r (4) re< r < rp

Key (2)

Sol. mv 2mKrqB qB

ee e 0

2m Kr m x

eB

pp p 0

2m Kr m x

eB

p 0p

p 0

4m x2(4m )Kr m x

2eB 2

19. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic

field at the centre of the loop is B2. The ratio 1

2

BB

is :

(1) 2 (2) 3 (3) 2 (4) 12

Key (3)

Sol. 0 01 2

1 2

I IB . , B .2 R 2 R

But m = I A 2m I R

R m Hence 1R m

2R 2m

1 2

2 1

B RB R

Hence 1

2

B 2mB m

1

2

B 2B

20. For an RLC circuit driven with voltage of amplitude m and frequency 01LC

the current

exhibits resonance. The quality factor, Q is given by :

(1) 0LR

(2) 0RL

(3) 0

R( C)

(4) 0

CR

Key (1)

Sol. Quality factor for resonance is given by oL LXQR R

21. An EM wave from air enters a medium. The electric fields are 1 01zˆE E x cos 2 v tc



-10-

In air and 2 02 ˆE E x cos[k(2z ct)]

in medium, where the wave number k and frequency v refer to their values in air. The medium is non-magnetic. If

1r and

2r refer to relative permittivity of air and

medium respectively, which of the following options is correct ?

(1) 1

2

r

r

4

(2) 1

2

r

r

2

(3) 1

2

r

r

14

(4) 1

2

r

r

12

Key (3) In medium 1. speed = c In medium 2. Speed = c/2

10 0 r

1c

20 0 r

c 12

1

2

r

r

14

22. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is

placed behind A. The intensity of light beyond B is found to B I2

. Now another identical polarizer C

is placed between A and B. The intensity beyond B is not found to be I8

. The angle between

polarizer A and C is : (1) 0º (2) 30º (3) 45º (4) 60º Key. (3) Let after polarizer C is placed the intensity of the light beyond C is I1.

21

II cos2

...(i)

21

I I cos8 …(ii)

By equation (i) & (ii) o45 23. The angular width of the central maximum in a single slit diffraction pattern is 60º. The width of the

slit is 1 m . The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance ?

(i.e. distance between the centres of each slit) (1) 25 m (2) 50 m (3) 75 m (4) 100 m Key. (1)

610 m6 a 6

2D 10d

6 62 2

D 0.5 50d 10 1010 10 6 6

24. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let n g, be the de Broglie wavelength of the electron in the nth state and the ground state



-11-

respectively. Let n be the wavelength of the emitted photon in the transition from the nth stage to the ground state. For large n, (A, B are constants)

(1) n 2n

BA

(2) n nA B (3) 2 2n nA B (4) 2

n

Key (1)

Sol. h 2 rmv n

gg

g

2 rhmv 1

…(i)

nn

n

2 rhmv n

…(ii)

m n

g 0

r nnr

…(iii)

Given 2

n2 2n

2

1 1 1 n 1 1R 11n R Rn 1 1n

2

1 11R n

(using binomial theorem)

2

g

n

1 1R R

[from (iii)]

n 2n

BA

25. If the series limit frequency of the Lyman series is vL, then the series limit frequency of the P fund series is :

(1) L25v (2) L16v (3) Lv /16 (4) Lv / 25 Key (4) Sol. For Lyman series limit

L 2

1n

2

L P

P L

nn

2

LP L

15 25

26. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is dp ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is cp . The value of dp and cp are respectively.

(1) ( 89, 28) (2) ( 28, 89) (3) (0, 0) (4) (0,1) Key. (1) Sol. a) for collision between neutron and deuterium Md = 2ma Using formula for 1- elastic collision

1 21 1

1 2

m mvm m



A d1 1 1

A d

m m 1vm m 3

Fractional loss in K.E. of neutron

2 2

n2 1

d21 1

n

1 m (u u )K (k k ) 2P 1k k m u2

d8 / 9P 8 / 9 0.89

1

b) for collision between neutron and carbon nucle cm 12m

1 111v

13

Fractional loss in K.E. C1

KPK

21 (11/13) cP 0.284 Hence

d

c

P 0.89P 0.284

27. The reading of the ammeter for a silicon diode in the given circuit is

(1) 0 (2) 15 mA (3) 11.5 mA (4) 13.5 Ma Key (3)

Sol. 3 0.7i 11.5 mA

200

28. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?

(1) 32 10 (2) 42 10 (3) 52 10 (4) 62 10 Key. (3)

Sol. available Band with 10 10GHz

100

631GH 10 KHz

Required Band with For a single channel = 5KHz No. of channels that

Can be transmitted = 610 KHz

5KHz = 52 10

29. In a potentiometer experiment, it sis found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a



resistance of , a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell

(1) 1 (2) 1.5 (3) 2 (4) 2.5 Key. (2)

Sol. 1

2

52 5 12r R 1 5 1 1.540 40

ll

30. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combinations is 1 k . How much was the resistance on the left slot before interchange the resistances?

(1) 990 (2) 990 (3) 550 (4) 910 Key. (3)

Sol. 1

2

R xR 1 x

Where x is in meter From the info in the question 1 2R R 1000 Let s assume 2 1R R Case – 1 Case – 2

1 1

2 1

R AJR J B

……(1) 2 2

1 2

R AJR J B

……..(2)

Here 1 2AJ J B x …… (3) 1 2J B AJ 1 x …… (4) Then 1 2AJ AJ 10cm 0.1m x (1 x) 2x 1 0.1 = x= 0.55 m Continued

then 1

2

R x 0.55 11R 1 x 0.45 9

1 29R 11R ……(5) 1 19R R 1000

11

1R 550

PART B - CHEMISTRY 31. The ratio of mass percent of C and H of an organic compound X Y ZC H O is 6 : 1. If one molecule of

the above compound X Y ZC H O contains half as much oxygen as required to burn one molecule of compound X YC H completely to CO2 and H2O. The empirical formula of compound X Y zC H O is:

(1) 3 6 3C H O (2) 2 4C H O (3) 3 4 2C H O (4) 2 4 3C H O Ans. (4)



-14-

In X Y ZC H O

Mass% of C = 12x 10012x y 16z

Mass% of H = y 10012x y 16z

mass% of C 6mass%of H 1

12x 6y

y = 2x

x y 2 2 2

y2xy2C H O xCO H O

2 2

y2x2z

2

4z 4x y Or 4z 4x 2x

3z x2

32. Which type of ‘defect’ has the presence of cations in the interstitial sites? (1) Schottky defect (2) Vacancy defect (3) Frenkel defect (4) Metal deficiency defect Ans. (3) Since cation is missing and present in interstitial site hence it will be Frenkel defect. 33. According to molecular orbital theory, which of the following will not be a viable molecule? (1) 2

2He (2) 2He (3) –2H (4) 2–

2H Ans. (4) (i) 2

2He 21s

B.O = 1 2 – 0 12

(ii) 2He 2 1

*1s 1s

B.O = 1 2 –1 0.52

(iii) –2H

2 1*

1s 1s

B.O. = 1 2 –1 0.52

(iv) 2–2H

2 2*

1s 1s

B.O.= 1 2 – 2 02

not viable.



34. Which of the following lines correctly show the temperature dependence of equilibrium constant , K, for an exothermic reaction?

(1) A and B (2) B and C (3) C and D (4) A and D Ans. (1) CG – RT ln K

Cln K = G–RT

Cln K = H S–RT R

For exothermic reaction H – ve

Hence CH Sln K

RT R

Slope = HR

So A and B are correct answer. 35. The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene

at constant volume is – 3263.9 kJ mol– 1 at 25°C; heat of combustion (in kJ mol– 1) of benzene at constant pressure will be (R = 8.314 JK– 1 mol–1)

(1) 4152.6 (2) –452.46 (3) 3260 (4) – 3267.6 Ans. (4)

6 6 2 2 215C H l O g 6CO g 3H O l2

n 6 – 7.5 –1.5 H U nRT H – 3263900 –1.5 8.314 298 H – 3267.67 kJ 36. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing

point? (1) 2 36Co H O Cl (2) 2 2 25Co H O Cl Cl .H O

(3) 2 2 24Co H O Cl Cl.2H O (4) 2 3 23

Co H O Cl .3H O Ans. (4) f fT ik m

For 2 3 23Co H O Cl .3H O ; i = 1

37. An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 × 10–7 and that of S2– from HS– ions is 1.2 × 10–13 then the concentration of S2– ions in aqueous solution is

(1) –85 10 (2) – 203 10 (3) – 216 10 (4) –195 10 Ans. (2)

–2H S H HS ;

–– 7

c2

H HSk 1.0 10

H S

ln K A

B

C

D

(0,0) 1

T K



-16-

–7

– 1.0 10 0.1HS0.2

– – 8HS 5 10

– 2–HS H S ; 2–

–13c –

S Hk 1.2 10

HS

–13 –8

2– 1.2 10 5 10S0.2

2– –21 – 20S 30 10 3 10 38. An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of

Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10–10. What is the original concentration of Ba2+?

(1) –95 10 M (2) – 92 10 M (3) –91.1 10 M (4) –101.0 10 M Ans. (3) Mili moles of 2 4Na SO in 50 ml solution = 50 × 1 = 50

Concentration of 2 4Na SO = 50 0.1500

2 2–sp 4k Ba SO

–10

sp2 –92–4

k 1 10Ba 1.0 100.1SO

Moles of Ba2+ in 1000 ml = 1.0 × 10–9

Moles of Ba2+ in 500 ml = –91.0 10

2

Moles of Ba2+ in 450 ml = –91.0 10

2

[Ba2+] in original solution = –9

–91.0 10 1000 1.1 10 M2 450

39. At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reached and 0.5 Torr s–1 when 33% had reached. The order of the reaction is:

(1) 2 (2) 3 (3) 1 (4) 0 Ans. (1) xr k A x = order x

1 1r k A

x2 2r k A

x

1

2

r 344.85 1 2r 243.21 0.5

x1.4179 2

x log 1.4179 log2 0.1516x 0.3010

0.3010x 1.98 20.1516



40. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)

(1) 6.4 hours (2) 0.8 hours (3) 3.2 hours (4) 1.6 hours Ans. (3) Molar mass of 2 6B H = 2 × 10.8 + 6 = 27.6

Mole of 2 6B H = 27.6 127.6

2 6 2 2 3 2B H 3O B O 3H O 1 mole of 2 6B H reached with 3 mole of O2 Weight of O2 = 3 × 32 gm

w i tE 96500

3 32 100 t8 96500

t = 3.2 hours 41. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is

required to make teeth enamel harder by converting [3Ca3(PO4)2.Ca(OH)2] to : (1) [CaF2] (2) [3(CaF2).Ca(OH)2] (3) [3Ca3(PO4)2.CaF2] (4) [3{Ca(OH)2}.CaF2] Key (3) Sol. 3 4 3 4 22 2 2

Fluorapalite

3Ca PO .Ca OH F 3Ca PO .CaF

42. Which of the following compounds contain(s) no covalent bond(s)? KCl, PH3, O2, B2H6, H2SO4 (1) KCl, B2H6, PH3 (2) KCl, H2SO4 (3) KCl (4) KCl, B2H6 Key (3) Sol. KCl does not have covalent bond. 43. Which of the following are Lewis acids? (1) PH3 and BCl3 (2) AlCl3 and SiCl4 (3) PH3 and SiCl4 (4) BCl3 and AlCl3 Key (4) Sol. BCl3 and AlCl3 always act as a Lewis acid as they have vacant ‘p’ and ‘d’ orbitals respectively. 44. Total number of lone pair of electrons in 3I

ion is (1) 3 (2) 6 (3) 9 (4) 12 Key (3)

Sol. 45. Which of the following salts is the most basic aqueous solution? (1) Al(CN)3 (2) CH3COOK (3) FeCl3 (4) Pb(CH3COO)2 Key (2) Sol. 3 2 3CH COOK H O CH COOH K OH 46. Hydrogen peroxide oxidizes [Fe(CN)6]4– to [Fe(CN)6]3– in acidic medium but reduces [Fe(CN)6]3– in

acidic medium but reduces [Fe(CN)6]3– to [Fe(CN)6]4– in alkaline medium. The other products formed are, respectively

(1) (H2O + O2) and H2O (2) (H2O + O2) and (H2O + OH–) (3) H2O and (H2O + O2) (4) H2O and (H2O + OH–) Key (3) Sol. 4 3

2 2 26 6Fe CN H O 2H Fe CN 2H O

3 4

2 2 2 26 6Fe CN H O OH Fe CN H O O

I – I – I

–



47. The oxidation state of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(O2)(O)2(NH3)] respectively are

(1) +3, +4 and +6 (2) +3, +2 and +4 (3) +3, 0 and +6 (4) +3, 0 and +4 Key (3)

Sol. 3 0 1 6 1 2 1 0

2 3 6 6 2 36 22 2 2

Cr H O Cl , Cr C H ,K Cr C N O O N H

48. The compound that does not produce nitrogen gas by the thermal decomposition is (1) Ba(N3)2 (2) (NH4)2Cr2O7 (3) NH4NO2 (4) (NH4)2SO4 Key (4) Sol. 4 4 3 2 32

NH SO NH H O SO 49. When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is

soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is

(1) Zn (2) Ca (3) Al (4) Fe Key (3)

Sol.

2 3 23NaOH excess

4Solub le

Al NaOH aq Al OH Al O .xH O

Na Al OH

50. Consider the following reaction and statements : 3 2 3 3 34 3

Co NH Br Br Co NH Br NH

(I) Two isomers are produced if the reactant complex ion is cis-isomer. (II) Two isomers are produced if the reactant complex ion is a trans- isomer (III) Only one isomer is produced if the reactant complex is a trans-isomer. (IV) Only one isomer is produced if the reactant complex ion is a cis-isomer. The correct statements are (1) (I) and (II) (2) (I) and (III) (3) (III) and (IV) (4) (II) and (IV) Key (2)

Sol.

51. Glucose on prolonged heating with HI gives : (1) n-Hexane (2) 1-Hexane (3) Hexanoic acid (4) 6-iodohexanal Key (1)

Sol 52. The trans – alkenes are formed by the reduction of alkynes with : (1) 2 4H Pd / C,BaSO (2) 4NaBH (3) 3Na / liq.NH (4) Sn – HCl

CO

NH3

NH3 Br

BrH3N

H3NCO

Br

NH3 Br

BrH3N

H3N+ Br– CO

NH3

NH3 Br

BrBr

H3N+

facial mercis

CO

Br

BrNH3

NH3H3N

H3NCO

Br

BrBr

NH3H3N

H3N+ Br–

mer



Key (3)

Sol 53. Which of the following compound will be suitable for Kjeldahl’s method for nitrogen estimation ?

(1) (2) (3) (4) Key (2) Sol Compound containing nitrogen in cyclic compound, –NO2 group and N N group are not

suitable for kjeldahl’s method. 54. Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces

compound X as the major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces

(1) (2)

(3) (4) Key (1) Sol. 55. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a

correct combination ? Base Acid End point (1) Weak Strong Colourless to pink (2) Strong Strong Pinkish red to yellow (3) Weak Strong Yellow to pinkish red (4) Strong Strong Pink to colourless



Key (3) Sol. Color of methyl orange in weakly basic medium is yellow which turns pinkish red in acidic medium

56. The predominant form of histamine present in human blood is a(pK ,Histidine 60)

(1) (2)

(3) (4) Key (4) Sol. At PH = 6 histamine is monoprotonated form 57. Phenol reacts with methyl chloro formate in the presence of NaOH to form product A. A reacts with

2Br to form product B. A and B are respectively

(1) (2)

(3) (4) Key (3)

Sol. 58. The increasing order of basicity of the following compounds is

(a) (b)

(c) (d) (1) (a) < (b) < (c) < (d) (2) (b) < (a) < (c) < (d) (3) (b) < (a) < (d) < (c) (4) (d) < (b) < (a) < (c) Key (3)



Sol b < a < d < c 59. The major product formed in the following reaction is

(1) (2)

(3) (4) Key (4)

Sol. 60. The major product of the following reaction is

(1) (2)

(3) (4) Key (2)

Sol



PART C - MATHEMATICS

61. Two sets A and B are as under: A a, b : a – 5 1and b – 5 1 ; 2 2B a,b :4 a – 6 9 b – 5 36 . Then

(1) B A (2) A B (3) A B = (an empty set) (4) Neither nor Ans. (2)

Set A is an interior of square obtained by Cartesian product of intervals (4, 6) × (4, 6). B is interior of

an ellipse. 62. Let S x :x 0and 2 x – 3 x x – 6 6 0 . Then S:

(1) is an empty set (2) contains exactly one element (3) contains exactly two elements (4) contains exactly four elements Ans. (3) 2

2 x – 3 x – 6 x 6 0

22 x – 3 x – 3 – 3 0

2

x – 3 2 x – 3 – 3 0

x – 3 3 x – 3 –1 0

x – 3 1

x – 3 1, –1 x 4,2 x = 16, 4 63. If , C are the distinct roots, of the equation 2x – x 1 0, then 101 107 is equal to (1) – 1 (2) 0 (3) 1 (4) 2 Ans. (3) – , 2– (where , 2 are the complex cube roots of unity)

107101101 107 2– –

2– (since 21 0 ) = – (– 1) = 1

64. If 2x – 4 2x 2x2x x – 4 2x A Bx x – A2x 2x x – 4

then the ordered pair (A, B) is equal to:

(1) (– 4, – 5) (2) (– 4, 3) (3) (– 4, 5) (4) ( 4, 5) Ans. (3) 1 2 3R R R

(4,6) (6,6)

(6,4)(4,4)

(6,5)

(6, –1)

(3, 5) (9, 5)

(6, 4)



Put x = 0 : 3

–4 0 00 – 4 0 A A – 40 0 – 4

Put x = 1 : 2–3 2 22 – 3 2 – 4 B 1 42 2 – 3

25 = (B – 4) . 25 B = 5 65. If the system of linear equations: x ky 3z 0 , 3x ky – 2z 0 and 2x 4y – 3z 0 has a non-zero

solution (x, y, z), then 2

xzy

is equal to:

(1) – 10 (2) 10 (3) – 30 (4) 30 Ans. (2) For non-trivial solution A 0

1 k 33 k –2 0 k 112 4 –3

x 11y 3z 0 ……(1) 3x 11y – 2z 0 ……(2) 2x 4y – 3z 0 …….(3) Solve (1) and (2) : 2y + z = 0 Put y = 1 then z = – 2 Put these in (3): x = – 5

22

–5 –2xz 10y 1

66. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is

(A) at least 1000 (B) less than 500 (C) at lest 500 but less than 750 (D) at least 750 but less than 1000 Ans. (1)

No. of ways of choosing Novels = 6C4 No. of ways of choosing 1 dictionary = 3C1 Total no. of arrangements = 6C4 × 3C1 × 4! = 1080

67. The sum of the co-efficients of all odd degree terms in the expansion of 53x x 1 +

53x x 1 , (x > 1) is

(A) –1 (B) 0 (C) 1 (D) 2 Ans. (4) Let 3x 1 = y (x + y)5 + (x – y)5 = 2(x5 + 10.x3.y2 + 5x.y4) = 2(x5 + 10.x3.(x3 – 1) + 5.x(x3 – 1)2) = 2(x5 + 10.x6 – 10.x3 – 5.x7 + 5x – 10x4) Sum of coefficients for odd terms 2 – 10 + 5 + 5 = 2

D



-24-

68. Let a1, a2, a3, ….., a49 be in A.P. such that 12

4k 1k 0

a = 416 a9 + a43 = 66. If 2 2 2

1 2 17a a .... a = 140 m,

then m is equal to (A) 66 (B) 68 (C) 34 (D) 33 Ans. (3) a1 + a5 + ….. + a49 = 416

Now, AM = 41613

S49 = 416 4913

Let a1 = a and a2 = a + d

49 [2a 48d]2

= 416 4913

a + 24d = 41613

a + 24d = 32 … (i) Also, a9 + a43 = 66 2a + 50d = 66 ... (ii) From (i) and (ii) a = 8 and d = 1

Now, 17

2

r 1

(a(r 1)d)

= 17

2

r 1

(8 (r 1))

= 17

2

r 1(7 r)

= 17

2

r 1(49 14r r )

= 49 × 17 + 14 17 18 17 18 332 6

= 4760 4760 = 140 m m = 34 69. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2.22 +

32 + 2.42 + 52 + 2.62 + …. If B – 2A = 100 , then is equal to (A) 232 (B) 248 (C) 464 (D) 496 Ans. (2) S2k = (12 + 22 + …. + (2k)2 + 22(12 + 22 + …. + k2)

= 2k(2k 1)(4k 1) 4k(k 1)(2k 1)6 6

= k(2k 1)2(8k 3)6

S2k = k(2k 1)(8k 3)3

n = 10, n = 20 B – 2A = S40 – S20 = 24800 70. For each t R , let [t] be the greatest integer less than or equal to t. Then

x 0

1 2 15lim x .....x x x



-25-

(A) is equal to 0 (B) is equal to 15 (C) is equal to 120 (D) does not exist (in R) Ans. (3)

x 0

klim xx

= k, k

x 0

1 2 15lim x ....x x x

= 1 + 2 + …. + 15 = 120

71. Let | |{ : | | . 1 sin | |xS t R f x x e x is not differentiable at t}. Then the set S is equal to: (1) (an empty set) (2) {0} (3){ } (4){0, } Key (1) Sol. | || | 1 sin | |xf x x e x

Possible points are 0 and

0lim 0h

f hh

lim 0

h

f hh

72. If the curves 2 2 26 ,9 16y x x by intersect each other at right angles, then the value of b is:

(1) 6 (2) 72

(3) 4 (4) 92

Key (4) Sol. 2 6y x

2 6y x

1 1P x y

29 6 16x bx

Slope of tangent of parabola at 1 1x y

m1 = 12 6dyydx

…..(i)

1

3dydx y

Slope of tangent of ellipse at 1 1x y

1 118 2 0dyx bydx

m2 = 1

1

9xdydx by

…..(ii)

1

1 1

93 1xy by

1 2 1m m

21 127x by 2

1 16y x

1 127 6x x b



-26-

276

b

92

b

73. Let 22

1f x xx

and 1g x xx

, { 1,0,1}x R . If

f xh x

g x , then the local minimum value

of h x is:

(1) 3 (2) – 3 (3) 2 2 (4) 2 2 Key (4)

Sol. 22

1 1,f x x g x xx x

22

1

1

xf x xh xg x x

x

21 2

1

xx

xx

=

1 21x

x xx

Let 1x t xx

, 2h x t x

t x

2

2' ' 'h x t x t xt x

2

't x

t x 2

2t x

2

2

11x

t x

21 2xx

= 22

1 4xx

= 4 24 1x x

= 22 2 3x

= 2 22 3 2 3x x

2 3 2 3 2 3 2 3x x x x

i) 2 3x

22

12 3, 2 3xx

= 2

22

1 1 2 2x xx x



-27-

1 22, 2 2 22

x h xx

(ii) 2 3x

2 2 3x

2

1 2 3x

1 2xx

2 2h x

74. The integral

2 2

25 3 2 3 2 5

sin cos

sin cos sin sin cos cos

x x dxx x x x x x

is equal to:

(1) 3

13 1 tan

Cx

(2) 3

13 1 tan

Cx

(3) 3

11 cot

Cx

(4) 3

11 cot

Cx

(Where C is a constant of integration) Key (2)

Sol.

2 2

25 3 2 3 5

sin cos

sin cos sin sin cos

x x dxx x x x x

= 2 2

210 5 2 3

sin .cos

cos tan tan tan 1

x x dxx x x x

=

2 6

22 3 5

tan sec

1 tan tan tan

x x dx

x x x

=

22 2 2

22 3 5

tan sec sec

1 tan tan tan

x x xdx

x x Let tant x

=

22 2

22 3 5

t 1

1

t dt

t t t

=

22 2

22 3 2

1

1 1

t t dt

t t t

=

2

23

dt

1

t

t

Let 3t z 23t dt dz

2

1 13 3 11

dzzz

3

13 1 tan

Cx



-28-

75. The value of 22

2

sin1 2x

x dx

is:

(1) 8 (2)

2 (3) 4 (4)

4

Key (4)

Sol. 2 22

0

sin 2 sin1 2 1 2

x

x x

x xI dx

22

0

sin 1 21 2

x

x

xI dx

2 2

2

0 0

1 cos 2sin2

xI xdx dx

0

0

1 sin 22 4

xI x

0 04 4

I

76. Let 2cos , ,g x x f x x and , be the roots of the quadratic equation 2 218 9 0x x .Then the area (in sq. units) bounded by the curve y = gof x and the lines

,x x and 0y , is:

(1) 1 3 12

(2) 1 3 12

(3) 1 3 22

(4) 1 2 12

Key (1) Sol. 2cos ,g x x f x x

218 9 0, ,3 6

x x x

cosy g f x x

3

6

3 1 area = cos2 2

xdx

77. Let y y x be the solution of the differential equation sin cos 4 , 0,dyx y x x xdx

. If 02

y

,

then 6

y

is equal to:

(1) 249 3

(2) 289 3

(3) 28

9 (4) 24

9

Key (3)

Sol. sin cos 4dyx y x xdx

siny x t

cos sin dy dty x xdx dx



-29-

4dt xdx

2 t = 2x C

2sin 2y x x C 2

02 2

y C

2 21 2

2 36 2y

2 28 8

2 18 9y y

78. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct point P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:

(1) 3 2 6x y (2) 2 3x y xy (3)3 2x y xy (4)3 2 6x y xy Key (3) Sol. R = (a, b) 3 2y m x

(0, b) Q

(a, 0)

(2, 3)

O P

3 2b m

3 2am

32

ma

32

bba

3 2 6b a

2 3 0ab b a 2 3 0xy y x 79. Let the orthocentre and centroid of a triangle be 3,5A and 3,3B respectively. If C is the

circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is:

(1) 10 (2) 2 10 (3) 532

(4) 3 52

Key (3) Sol. 3,5H

2 :1

G OH

A B C (–3, 5) (3, 3)

2

ACr



-30-

3 3

2 2 4AB ABr

2 23 402 6 34 4

r

= 2 10 3 103

4 2

= 532

80. If the tangent at (1, 7) to the curve 2 6x y touches the circle 2 2 16 12 0x y x y c then the value of c is

(1) 195 (2) 185 (3) 85 (4) 95 Key (4) Sol. 2 6x y Equation of tangent 2 4 5x Centre : 8, 6

Distance = 5 55

2 28 6 5x y

2 216 12 95 0x x y y

95c 2 2 26 ,9 16y x x by any point on parabola 2 , 2at at

Equation of tangent to parabola 2yt x at

Equation of normal to ellipse at 2 , 2at at = 2yt x at

Slop = 1dx

dy t

18 2 0dyx bydx

2

2 2 218 18

dx by b atdy x at

1 4

18b

t t

92

b

81. Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the

parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and CPB , then a value of tan is :

(1) 12

(2) 2 (3) 3 (4) 43

Key. (2)



Equation of tangent at ‘p’

2 16y x 16 16

.162

xy

12 16, 82

y x y x …. (1)

Equation of Normal at ‘p’ 16 2 16y x 16 2 32y x 2 48y x … (2)

Now 320 400 400cos2 20 320

1cos tan 25

82. Tangents are drawn to the hyperbola 2 24 36x y at the points P and Q. If these tangents intersect at the point T(0,3) then the area (in sq.units) of PTQ is

(1) 45 5 (2) 54 3 (3) 60 3 (4) 36 5 Key. (1)

Equation of hyperbola 2 2

19 36x y

Equation of tangent 2 2 2 y mx a m b

29 36 y mx m Passing through (0, 3) 29 9 36 5m m Equation of tangent PT is 5 3y x Equation of tangent QT is 5 3y x Now equation of chord of contact PQ

.0 .3 1 129 36

x y y

Now area of PQT = 1 15 6 52 45 5



83. If L1 is the line of intersection of the plane 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 an L2 is the line of intersection of the plane x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the plane containing the lines L1 and L2 is :

(1) 14 2

(2) 13 2

(3) 12 2

(4) 12

Key. (2) 2 2 3 2 0x y z 1 0x y z Both are passing through (0, 5, 4) Let , ,m n are direction ratios of normal to plane is 2 2 3 0m n 0m n

, , 1,1,02 3 3 2 2 2

m n m n

So, equation of L1 will be

0 5 4

1 1 0x y z

2 3 0x y z 3 2 1 0x y z 2 0m n 3 2 0m n

4 1 3 2 1 6

m n

, , 3,5,7m n

Let , ,m n be the direction ratios of normal to plane 0. 0m n 3 5 7 0m n

7 7 5 3

m n

Equation of plane containing the lines will be 7 0 7 8 8 4 0x y z 7 7 8 3 0x y z

Distance from origin (0, 0,0) will be = 3 3 3 149 49 64 162 9 2 3 2

84. The length of the projection of the line segment joining the points (5, –1,4) and (4, –1, 3) on the

plane, x + y + z = 7 is

(1) 23

(2) 23

(3) 13

(4) 23

Key. (4)

ˆˆ ˆ 2ˆˆˆ

33 3 3i j kAB n i k



-33-

85. Let u be a vector coplanar with the vectors ˆˆ ˆ2 3a i j k and ˆˆb j k

. If u is perpendicular to a

and . 24u b , then 2u is equal to :

(1) 336 (2) 315 (3) 256 (4) 84 Key. (1) ˆˆ ˆ. 0 . 24 2u a u b a i j k

u a b ˆ ˆˆ ˆ ˆ2 3i j k j k

ˆˆ ˆ2 3i j k Now given

u.b 24

3 24 2 2 24 12 … (1) . 0u a 4 3 3 0 7 0 … (2) Solving eq. (1) & (2) we get 14& 2

2 336u

86. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

(1) 3

10 (2)

25

(3) 15

(4) 34

Key. (2) 4 Red 6 Black

Let 1E 1st drawn ball is red

2E 1st drawn ball is black A IInd drawn ball is red P(A) = P(E1) P(A/E1) + P(E2) P(A/ E2)

4 6 6 4 1 1 2

10 12 10 12 5 5 5

87. If 9

1

5 9ii

x

and 9

2

1

5 45ii

x

, then the standard deviation of the 9 items x1, x2, ..,x9 is :

(1) 9 (2) 4 (3) 2 (4) 3 Key. (3)

9

1

5 9ii

x

9

1

45 9 54ii

x

9

1 69

ii

x

Standard deviation of 1 2 9, ,....,x x x will be =

9 92 2

1 16 5 1 2 5

9 9

i i ii i

x x x



45 9 2 9 4 29

88. If sum of all the solutions of the equation 18cos . cos .cos 16 6 2

x x x

in 0, is k ,

then k is equal to :

(1) 23

(2) 139

(3) 89

(4) 209

Key. (2)

18cos cos .cos 16 6 2

x x x

14cos cos 2 12

x x

2 34cos 2cos 12

x x

38cos 6cos 1x x

3 14cos 3cos2

x x

cos3x = 12

3 2 ,3

x m m I

2 ,3 9

mx m I

5 7, , , [0, ]

9 9 9 x as x

So, sum of solution 13

9

So 139

k

89. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45o, 30o and 30o, then the height of the tower (in m) is :

(1) 100 (2) 50 (3) 100 3 (4) 50 2 Key. (1) Let height of tower be ‘h’

11,3

h hPD QD

PD h 33

xh x h

Now in PQD

2 2 2PQ PD QD



-35-

2 240000 3h h 2 10000 100h h m 90. The Boolean expression ~ ~ p q p q is equivalent to : (1) ~p (2) p (3) q (4) ~q

p q ~p p q ~ p q ~ p q ~ p q (~ )p q

T T F T F F F

T F F T F F F

F T T T T F T

F F T F F T T

So it is equivalent to ~p

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