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THOMAStutorials

JEE (FINAL) Date : TEST NO: 18 Time : 03 HRS PCM MARKS: 360

1. The unit of surface tension in SI system is

a) 𝐷𝑦𝑛𝑒/𝑐𝑚2 b) 𝑁𝑒𝑤𝑡𝑜𝑛 /𝑚

c) 𝐷𝑦𝑛𝑒/𝑐𝑚 d) 𝑁𝑒𝑤𝑡𝑜𝑛/𝑚2

2. A car moves a distance of 200 m. it covers first

half of the distance at speed 60 kmh−1 and the

second half at speed 𝑣. If the average speed is

40 kmh−1, the value of 𝑣 is

a) 30 kmh−1 b) 13 kmh−1

c) 60 kmh−1 d) 40 kmh−1

3. In the above question, the percentage increase

in the time of flight of the projectile will be

a) 5% b) 10% c) 15% d) 20%

4. A motorcycle is travelling on a curved track of

radius 500 𝑚. If the coefficient of between

road and tyres is 0.5, the speed avoiding

skidding will be

a) 50 𝑚/𝑠 b) 75 𝑚/𝑠 c) 25 𝑚/𝑠 d) 35 𝑚/𝑠

5. A uniform chain of length 2 m is kept on a

table such that a length of 60 cm hangs freely

from the edge of the table. The total mass of

the chain is 4 kg. What is the work done in

pulling the entire chain on the table?

a) 7.2 J b) 3.6 J c) 120 J d) 1200 J

6. Three identical blocks 𝐴, 𝐵 and 𝐶 are placed on

horizontal frictionless surface. The blocks 𝐴

and 𝐶 are at rest. But 𝐴 is approaching towards

𝐵 with a speed 10 ms−1. The coefficient of

restitution for all collisions is 0.5. The speed of

the block 𝐶 just after collision is

a) 5.6 ms−1 b) 6 ms−1

c) 8 ms−1 d) 10 ms−1

7. A planet in a distant solar system is 10 times

more massive than the earth and its radius is

10 times smaller. Given that the escape

velocity from the earth is 11kms−1, the escape

velocity from the surface of the planet would

be

a) 1.1 kms−1 b) 11 kms−1

c) 110 kms−1 d) 0.11 kms−1

8. A steel wire of cross-sectional area

3 × 10−6m2 can withstand a maximum strain

of 10−3 . Young’s modulus of steel is

2 × 1011Nm−2. The maximum mass the wire

can hold is (take g = 10 ms−2 )

a) 40 kg b) 60 kg c) 80 kg d) 100 kg

9. Water is flowing through a pipe of constant

cross-section. At some point the pipe becomes

narrow and the cross-section is halved. The

speed of water is

a) reduced to zero

b) decreased by a factor of 2

c) increased by a factor of 2

d) unchanged

10. A body cools from 60℃ to 50℃ in 10 min.

if the room temperature is 25℃ and

assuming Newton’s law of cooling to hold

good, the temperature of the body at the

end of the next 10 min will be

a) 45℃ b) 42.85℃ c) 40℃ d) 38.5℃

11. A monoatomic gas is suddenly compressed to

(1/8)th of its innitial volume adiabatically. The

ratio of its final pressure to the initial pressure

is (Given the ratio of the specific heats of the

given gas to be 5/3)

a) 32 b) 40/3 c) 24/5 d) 8

12. The power radiated by a black body is 𝑃, and it

radiates maximum energy around the

wavelength 𝜆0. If the temperature of black

body is now changed so that it radiates

maximum energy around a wavelength 𝜆0/4,

the power radiated by it will increase by a

factor of

a) 4

3

b) 16

9

c) 64

27

d) 256

81

13. A block (B) is attached to two unstretched

springs 𝑆1 and 𝑆2 with spring constants k and

4 k, respectively (see figure I). The other ends

are attached to identical supports 𝑀1 and 𝑀2

not attached to the walls. The springs and

supports have negligible mass. There is no

friction anywhere. The block B is displaced

towards wall 1 by a small distance 𝑥 (figure II)

and released. The block returns and moves a

maximum distance 𝑦 towards wall 2.

Displacements 𝑥 and 𝑦 are measured with

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respect to the equilibrium position of the block

B. The ratio 𝑦

𝑥 is

a)

4 b)

2 c) 1

2

d) 1

4

14. It takes 2.0 s for a sound wave to travel

between two fixed points when the day

temperature is 10°C. if the temperature rises

to 30°C the sound wave travels between the

same fixed parts in

a) 1.9s b) 2.0s c) 2.1s d) 2.2s

15. A sphere of radius 1 𝑐𝑚 has potential of

8000 𝑉, then energy density near its surface

will be

a) 64 × 105 𝐽/𝑚3 b) 8 × 103 𝐽/𝑚3

c) 32 𝐽/𝑚3 d) 2.83 𝐽/𝑚3

16. The electrostatic potential energy between

proton and electron separated by a distance

1Å is

a) 13.6eV b) 27.2eV c) 14.4eV d) 1.44eV

17. For which of the following the resistance

decreases on increasing the temperature

a) Copper b) Tungsten

c) Germanium d) Aluminium

18. Arrange the order of power dissipated in the

given circuits, if the same current is passing

through all the circuits. The resistance of each

resistor is 𝑟.

a) 𝑃1 > 𝑃2 > 𝑃 3 > 𝑃4

b) 𝑃2 > 𝑃3 > 𝑃 4 > 𝑃1

c) 𝑃4 > 𝑃3 > 𝑃 2 > 𝑃1

d) 𝑃1 = 𝑃2 = P 3 = 𝑃4

19. An electron, moving in a uniform magnetic

field of induction of intensity �� , has its radius

directly proportional to

a) Its charge b) Magnetic field

c) Speed d) None of these

20. A steel wire of length 𝑙 has a magnetic moment

𝑀. It is bent at its middle point at an angle of

60°. Then the magnetic moment of new shape

of wire will be

a) 𝑀/√2 b) 𝑀/2 c) 𝑀 d) √2𝑀

21. For a large industrial city with much load

variations the DC generator should be

a) Series b) Shunt wound

c) Mixed wound d) Any

22. An alternating 𝑒.m.f. of angular frequency 𝜔 is

applied across an inductance. The

instantaneous power developed in the circuit

has an angular frequency

a) 𝜔/4 b) 𝜔/2 c) 𝜔 d) 2𝜔

23. Solar radiation is

a) Transverse Electromagnetic wave

b) Longitudinal Electromagnetic wave

c) Stationary wave

d) None of the above

24. The wavelength of a certain colour in air is 600

nm. What is the wavelength and speed of this

colour in glass of refractive index 1.5?

a) 500 nm and 2 × 1010cms−1

b) 400 nm and 2 × 108ms−1

c) 300 nm and 3 × 109cms−1

d) 700 nm and 1.5 × 109ms−1

25. If the eighth bright band due to light of

wavelength 𝜆1 coincides with ninth bright

band from light of wavelength 𝜆2 in Young’s

double slit experiment, then the possible

wavelength of visible light are

a) 400 𝑛𝑚 and 450 𝑛𝑚 b) 425 𝑛𝑚 and 400 𝑛𝑚

c) 400 𝑛𝑚 and 425 𝑛𝑚 d) 450 𝑛𝑚 and 400 𝑛𝑚

26. In Bainbridge mass spectrograph a potential

difference of 1000 𝑉 is applied between two

plates distant 1 𝑐𝑚 apart and magnetic field

𝐵 = 1𝑇. The velocity of undeflected positive

ions in 𝑚/𝑠 from the velocity selector is

a) 107 𝑚/𝑠 b) 104 𝑚/𝑠 c) 105 𝑚/𝑠 d) 102 𝑚/𝑠

27. Which of the following transition in Balmer

series for hydrogen will have longest

wavelength?

a) 𝑛=2 to 𝑛=1 b) 𝑛 =6 to 𝑛=1

c) 𝑛 =3 to 𝑛=2 d) 𝑛 =6 to 𝑛=2

28. The nucleus which has radius one-third of the

P1 B

B

B

B

i

i

i

i

P2

P3

P4

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radius of 𝐶𝑠189 is

a) 𝐵𝑒9 b) 𝐿𝑖7 c) 𝐹19 d) 𝐶12

29. When a p-n junction diode is connected in

forward bias its barrier potential

a) Decreases and less current flows in the

circuit

b) Decreases and more current flows in the

circuit

c) Increases and more current flows in the

circuit

d) Decreases and no current flows in the circuit

30. Range of frequencies allotted for commercial

UHF TV broadcast is

a) 470-960 kHz b) 47-960 MHz

c) 470-960 MHz d) 47-960 kHz

31. The mass of 112cm3 of CH4 gas at STP is

a) 0.16 g b) 0.8 g c) 0.08 g d) 1.6 g

32. What is the minimum energy that photons

must posses in order to produce photoelectric

effect with platinum metal? The threshold

frequency for platinum is 1.3× 1015s−1

a) 3.6 × 10−13erg b) 8.2× 10−13erg

c) 8.2× 10−14erg d) 8.6× 10−12erg

33. Which one of the following orders presents the

correct sequence of the increasing basic nature

of the given oxides?

a) Al2O3 < 𝑀𝑔𝑂 < 𝑁a2O < K2O

b) MgO < K2O < 𝐴l2O3 < 𝑁a2O

c) Na2O < K2O < 𝑀𝑔𝑂 < 𝐴l2O3

d) K2O < 𝑁a2O < 𝐴l2O3 < 𝑀𝑔𝑂

34. The percentage of p – character in the orbitals

forming P – P bonds in P4 is

a) 25 b) 33 c) 50 d) 75

35. Which of the following is a Boyle plot at very

low pressure?

a)

b)

c)

d)

36. In view of the signs of ∆𝑟𝐺° for the

following reactions

PbO2 + Pb → 2PbO, ∆𝑟𝐺° < 0,

SnO2 + Sn → 2SnO, ∆𝑟𝐺° > 0,

Which oxidation states are more

characteristic for lead and tin?

a) For lead+4, for tin+2

b) For lead+2, for tin+2

c) For lead+4,for tin+4

d) For lead+2, for tin+4

37. The 𝑝𝐾𝑎of acetylsalicylic acid (aspirin) is 3.5.

The pH of gastric juice in human stomach is

about 2 − 3 and the pH in the small intestine is

about 8. Aspirin will be

a) Unionised in the small intestine and in the

stomach

b) Completely ionised in the small intestine and

in the stomach

c) Ionised in the stomach and almost unionised

in the small intestine

d) Ionised in the small intestine and almost

unionised in the stomach

38. Which of the following reaction involves

oxidation and reduction?

a) NaBr + HCl ⟶ NaCl + HBr

b) HBr +AgNO3 ⟶ AgBr + HNO3

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c) H2 + Br2 ⟶ 2HBr

d) Na2O + H2SO4 ⟶Na2SO4 + H2O

39. Calgon used as water softner is

a) Na2[Na4(PO3)6] b) Na4[Na2(PO3)6]

c) Na2[Na4(PO4)5] d) None of these

40. Pick out the statement (s) which is (are) not

true about the diagonal relationship of Li and

Mg.

(i) Polarising powers of Li+ and Mg2+ are

almost same.

(ii) L like Li, Mg decomposes water very fast.

(iii) LiCl and MgCl2 are deliquescent.

(iv) Like Li, Mg does not form solid

bicarbonates.

a) (i) and (ii) b) (ii) and (iii)

c) Only (ii) d) Only (i)

41.

Silicate structure unit of

a) (Si4O11)𝑛−6𝑛 b) (Si2O11)𝑛

−2𝑛

c) (Si2O3) d) (SiO4)−4

42. IUPAC name of

CH2 = CH − CH(CH2CH3)C = CH2

|

Br

a) 4-bromo-3-ethyl-1,4-pentadiene

b) 2-bromo-3-ethyl-1,4pentadiene

c) 2-bromo-3-ethyl-1-5-pentadiene

d) None of the above

43. The following reaction is called

a) Michael addition reaction

b) Diels-alder reaction

c) Wolff-Kishner reaction

d) None of the above

44. Which of the following is a primary pollutant?

a) CO b) PAN

c) Aldehydes d) H2SO4

45. Copper crystallises in fcc with a unit cell length

of 361 pm. What is the radius of copper atom?

a) 108 pm b) 127 pm c) 157 pm d) 181 pm

46. The modal elevation constant of water is

0.52℃. The boiling point of 1.0 modal aqueous

KCl solution (assuming complete dissociation

of KCl), therefore, should be

a) 98.96℃ b) 100.52℃

c) 101.04℃ d) 107.01℃

47. The equivalent conductances of two strong

electrolytes at infinite dilution in H2O (where

ions move freely through a solution) at 25℃

are given below

Λ°CH3COONa = 91.0 S cm2/equiv

Λ°HCl = 426.2 S cm2/equiv

What additional information/quantity one

needs to calculate Λ° of an aqueous solution of

acetic acid?

a) Λ° of NaCl

b) Λ° of CH3 COOK

c)

The limiting equivalent conductance of

H+(λ°H+)

d) Λ° of chloroacetic acid (ClCH2 COOH)

48. Observe the following reaction,

2𝐴 + 𝐵 → 𝐶

The rate of formation of C is 2.2 × 10−3𝑚𝑜𝑙

𝐿−1min−1.

What is the value of −𝑑[𝐴]

𝑑𝑡(𝑚𝑜𝑙 𝐿−1min−1) ?

a) 2.2 × 10−3 b) 1.1 × 10−3

c) 4.4 × 10−3 d) 5.5 × 10−3

49. Micelles have

a) Same colligative property as that of common

colloidal solution

b) Lower colligative property as that of

common colloidal solution

c) Higher colligative property as that of

common colloidal solution

d) None of the above

50. When a metal is to be extracted from its ore, if

the gangue associated with the ore is silica,

then

a) A basic flux is needed

b) An acidic flux is needed

c) Both basic and acidic flux are needed

d) Neither of them is needed

51. Which one is true peroxide?

a) NO2 b) MnO2 c) BaO2 d) SO2

52. Verdigris is

a) Basic lead b) Basic copper acetate

c) Basic lead acetate d) None of the above

53. Give the IUPAC name of the following

a) 5-ethyl-4, 4-dimethyloctane

b) 4-ethyl-5, 5-dimetyloctane

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c) 3-ethyl-2-methyl-2-propyl hexane

d) 4-ethyl-5-methyl, 5-propyl hexane

54. Identify A and B in the following reaction

C2H5Cl A → C2H5OH

B ← C2H5Cl

a) A= aqueous KOH; B= AgOH

b) A= alcoholic KOH/ ∆; B=aqueous NaOH

c) A= aqueous NaOH; B= AgNO2

d) 𝐴 = AgNO2; 𝐵 = KNO2

55. The reaction of ethanol with concentrated

H2SO4 at room temperature gives

a) CH3CH2OH2+HSO4

−

b) CH3CH2OSO2OH

c) CH3CH2OCH2CH3

d) H2C = CH2

56.

Product is

a)

b)

c)

d)

57. Ag2O → H2O

𝑋 400 K →

The products of above sequence of reactions

are

a) CH2 = CH2 and (CH3)2CHCN

b) CH3CH2CN and C2H5NH2

c) CH2 = CH2 and (CH3)3N

d) (CH3)2 C = CH2and NH3

58. Which of the following statement (s) is/are

true?

(i) All amino acids contain one chiral centre

(ii) Some amino acids contain one, while some

contain more chiral centre or even no chiral

centre

(iii) All amino acids in protein have L-

configuration

(iv) All amino acids found in proteins have 1°

amino group

a) (ii), (iii)and (iv) b) (ii) and (iii)

c) (i), (iii) and (iv) d) (i) and (iv)

59. Which one of the following statement is

wrong?

a)

The IUPAC name of [Co(NH3)6Cl3] is

hexamine cobalt III chloride.

b) Dibenzol peroxide is a catalyst in the

polymerization of PVC.

c) Borosilicate glass is heat resistant.

d)

Concentrated HNO3 can be safely

transported in aluminium containers.

60. Which of the following is an antipyretic?

a) Quinine b) Paracetamol

c) Luminal d) Poperazine

61. The shaded region in the figure represents

a)

𝐴 ∩ 𝐵

b)

𝐴 ∪ 𝐵

c)

𝐵 − 𝐴

d) (𝐴 − 𝐵)

∪ ( 𝐵

− 𝐴)

62. If 𝒬 denotes the set of all rational numbers and

𝑓 (𝑝

𝑞) = √𝑝2 − 𝑞2 for any

𝑝

𝑞∈ 𝒬, then observe

the following statements.

I. 𝑓 (𝑝

𝑞) is real for each

𝑝

𝑞∈ 𝒬.

II. 𝑓 (𝑝

𝑞) is a complex number for each

𝑝

𝑞∈ 𝒬.

Which of the following is correct?

a) Both I and II are true

b) I is true, II is false

c) I is false, II is true

d) Both I and II are false

63. If tan𝐴 = 2 tan𝐵 + cot𝐵, then 2 tan(𝐴 − 𝐵) is

equal to

a) tan𝐵 b) 2 tan𝐵 c) cot𝐵 d) 2 cot𝐵

64. If 𝑥2𝑛−1 + 𝑦2𝑛−1 is divisible by 𝑥 + 𝑦, if 𝑛 is

a) A positive integer

b) An even positive integer

c) An odd positive integer

d) None of the above

65. If 𝑧−1

𝑧+1 is purely imaginary number (𝑧 ≠ −1),

then |𝑧| is equal to

a) 1 b) 2 c) 3 d) 4

66. Let 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 and 𝑓(−1) <

1, 𝑓(1) > −1, 𝑓(3) < −4 and 𝑎 ≠ 0,then

a) 𝑎 > 0

b) 𝑎 < 0

c) Sign of a cannot be determined

d) None of the above

67. There are 6 letters and 3 post-boxes. The

number of ways in which these letters can be

posted, is

a) 63 b) 36 c) 6𝐶3 d) 6𝑃3

68. If 𝑎, 𝑏, 𝑐 are in AP, then the sum of

the coefficients of {1 + (𝑎𝑥2 − 2𝑏𝑥 +

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𝑐)2}1973 is a) −2 b) −1 c) 0 d) 1 69. The solution of the equation

(𝑥 + 1) + (𝑥 + 4) + (𝑥 + 7)+. . . +(𝑥 + 28) =

155 is

a) 1 b) 2 c) 3 d) 4

70. The equation of line through the point (1, 2)

whose distance from the point (3, 1) has the

greatest value, is

a) 𝑦 = 2𝑥 b) 𝑦 = 𝑥 + 1

c) 𝑥 + 2𝑦 = 5 d) 𝑦 = 3𝑥 − 1

71. The equation to the line touching

both the parabolas 𝑦2 = 4𝑥 and

𝑥2 = −32𝑦, is a) 𝑥 + 2𝑦 + 4 = 0 b) 2𝑥 + 𝑦 − 4 = 0 c) 𝑥 − 2𝑦 − 4 = 0 d) 𝑥 − 2𝑦 + 4 = 0 72. lim𝑥→0 [

2𝑥−1

√1+𝑥−1] is equal to

a) log𝑒 2 b) log𝑒 √2 c) log𝑒 4 d) 2

73. If 𝑆(𝑝, 𝑞, 𝑟) = (~𝑃) ∨ [~(𝑞 ∧ 𝑟)] is a compound

statement, then 𝑆(~𝑝,~𝑞, ~𝑟) is

a) ~𝑆(𝑝, 𝑞, 𝑟) b) 𝑆(𝑝, 𝑞, 𝑟)

c) 𝑝 ∨ (𝑞 ∧ 𝑟) d) 𝑝 ∨ (𝑞 ∨ 𝑟)

74. The variance of first 𝑛 numbers is

a) 𝑛2+1

12 b)

𝑛2−1

12

c)

(𝑛+1 )(2𝑛+1)

6 d) [

𝑛(𝑛+1)

2]2

75. If 𝑛 positive integers are taken at random and

multiplied together, the probability that the

last digit of the product is 2, 4, 6 or 8, is

a) 4𝑛 + 2𝑛

5𝑛 b)

4𝑛 × 2𝑛

5𝑛

c) 4𝑛 − 2𝑛

5𝑛

d) None of these

76. The incentre of the triangle formed by (0, 0),

(5, 12), (16, 12) is

a) (7, 9) b) (9, 7) c) (−9, 7) d) (−7, 9)

77. If 𝐴(−𝑎, 0) and 𝐵(𝑎, 0) are two fixed points,

then the locus of the point at which 𝐴𝐵

subtends a right angle, is

a) 𝑥2 + 𝑦2 = 2𝑎2 b) 𝑥2 − 𝑦2 = 𝑎2

c) 𝑥2 + 𝑦2 + 𝑎2 = 0 d) 𝑥2 + 𝑦2 = 𝑎2

78. If 𝑓: 𝑅 → 𝑅 is an even function having

derivatives of all orders, then an odd function

among the following is

a) 𝑓′′ b) 𝑓′′′

c) 𝑓′ + 𝑓′′ d) 𝑓′′ + 𝑓′′′

79. The relation tan−1 (1+𝑥

1−𝑥) =

𝜋

4+ tan−1 𝑥 holds

true for all

a) 𝑥 ∈ 𝑅 b) 𝑥 ∈ (−∞,1)

c) 𝑥 ∈ (−1,∞) d) 𝑥 ∈ (−∞,−1)

80. If 𝑥 [−34] + 𝑦 [

43] = [

10−5

] , then

a) 𝑥 = −2, 𝑦 = 1 b) 𝑥 = −9, 𝑦 = 10

c) 𝑥 = 22, 𝑦 = 1 d) 𝑥 = 2, 𝑦 = −1

81. If 𝐴 and 𝐵 are square matrices of order 3 such

that |𝐴| = −1, |𝐵| = 3 then |3𝐴𝐵| is equal to

a) −9 b) −81 c) −27 d) 81

82.

If 𝑓(𝑥) = {

sin5𝑥

𝑥2+2𝑥, 𝑥 ≠ 0

𝑘 +1

2, 𝑥 = 0

is continuous at

𝑥 = 0, then the value of 𝑘 is

a) 1

b) -2

c) 2

d) 1

2

83. If 𝜃 is the angle between the curves 𝑥 𝑦 = 2

and 𝑥2 + 4𝑦 = 0 , then tan 𝜃 is equal to

a) 1 b) -1 c) 2 d) 3

84. ∫𝑑𝑥

𝑥(𝑥+1) equals

Where 𝑐 is an arbitrary constant

a) log |

𝑥 + 1

𝑥| + 𝑐 b) log |

𝑥

𝑥 + 1| + 𝑐

c) log |

𝑥 − 1

𝑥| + 𝑐 d) log |

𝑥 − 1

𝑥 + 1| + 𝑐

85. ∫ |𝑥3 + 𝑥2 + 3𝑥|𝑑𝑥3

0 is equal to

a) 171

2

b) 171

4

c) 170

4

d) 170

3

86. The area bounded by the curves 𝑦2 = 𝑥 and

𝑦 = 𝑥2 is

a) 2

3 sq unit b) 1 sq unit

c) 1

2 sq unit d) None of these

87. The differential equation of all circles

which passes through the origin and whose

centre lies on 𝑦-axis, is

a) (𝑥2 − 𝑦2)

𝑑𝑦

𝑑𝑥− 2𝑥𝑦 = 0

b) (𝑥2 − 𝑦2)

𝑑𝑦

𝑑𝑥+ 2𝑥𝑦 = 0

c) (𝑥2 − 𝑦2)

𝑑𝑦

𝑑𝑥− 𝑥𝑦 = 0

d) (𝑥2 − 𝑦2)

𝑑𝑦

𝑑𝑥+ 𝑥𝑦 = 0

88. If �� , 𝐛 , 𝐜 be three non-coplanar vectors and

�� , �� , 𝐫 constitute the corresponding reciprocal

system of vectors then for any arbitrary vector

P a g e | 7

α

a) α = ( α ∙ �� )�� + ( α ∙ 𝐛 )𝐛 + (α ∙ 𝐜 )𝐜

b) α = ( α ∙ �� )�� + ( α ∙ �� )�� + (α ∙ 𝐫 )𝐫

c) α = ( α ∙ �� )�� + ( α ∙ �� )𝐛 + (α ∙ 𝐫 )𝐜

d) None of the above

89. If the distance of the point 𝑃(1,−2, 1) from the

plane 𝑥 + 2𝑦 − 2𝑧 = α, where α>0, is 5, then

the foot of the perpendicular from 𝑃 to the

plane is

a) (

8

3,4

3, −

7

3) b) (

4

3,−

4

3,1

3)

c) (

1

3,2

3,10

3) d) (

2

3,−

1

3,5

2)

90. 𝑧 = 30𝑥 + 20𝑦, 𝑥 + 𝑦 ≤ 8, 𝑥 + 2𝑦 ≥ 4, 6𝑥 +

4𝑦 ≥ 12, 𝑥 ≥ 0, 𝑦 ≥ 0 has

a) Unique solution

b) Infinity many solution

c) Minimum at (4, 0)

d) Minimum 60 at point (0, 3)

P a g e | 8

THOMAStutorials

JEE (FINAL) Date : TEST NO: 18 Time : 03 HRS PCM MARKS: 360

: ANSWER KEY :

1) b 2) a 3) a 4) a 5) b 6) a 7) c

8) b 9) c 10) b 11) a 12) d 13) c 14) a

15) d 16) c 17) c 18) b 19) c 20) b 21) c

22) d 23) a 24) b 25) d 26) c 27) c 28) b

29) b 30) c 31) c 32) d 33) a 34) d 35) d

36) d 37) d 38) c 39) a 40) c 41) a 42) b

43) b 44) a 45) b 46) c 47) a 48) c 49) b

50) a 51) c 52) b 53) a 54) a 55) b 56) d

57) c 58) b 59) b 60) b 61) d 62) c 63) c

64) a 65) a 66) b 67) b 68) d 69) a 70) c

71) d 72) c 73) d 74) b 75) c 76) a 77) d

78) b 79) b 80) a 81) b 82) c 83) d 84) b

85) b 86) d 87) a 88) c 89) a 90) d

: HINTS AND SOLUTIONS :

1 (b)

Surface tension =Force

Length= 𝑛𝑒𝑤𝑡𝑜𝑛/𝑚𝑒𝑡𝑟𝑒

2 (a)

Average velocity =2𝑣1𝑣2

𝑣1+ 𝑣2

Given, 𝑣𝑎𝑣 = 40 km/h, 𝑣1 = 60 km/h and 𝑣2 =?

∴ 40 =2 × 60 × 𝑣260 + 𝑣2

80𝑣2 = 2400

𝑣2 = 30 km/h

3 (a)

Time of flight, 𝑇 =2𝑢 sinθ

g

∴ 𝑑𝑇 =2𝑑𝑢 sin θ

g

Now, ∴ 𝑑𝑇

𝑇=

𝑑𝑢

𝑢=

1

20

∴ % increase in 𝑇 =𝑑𝑇

𝑇× 100

=1

20× 100 = 5%

4 (a)

𝑣 = √𝜇𝑟𝑔 = √0.5 × 500 × 10 = 50 𝑚/𝑠

5 (b)

Mass per unit length=M

L

=4

2= 2kgm−1

The mass of 0.6 m of chain

= 0.6 × 2 = 1.2kg

∴Center of mass of hanging part

ℎ =0.6 + 0

2= 0.3m

Hence, work done in pulling the chain on the table

=work done against gravity force

𝑊 = 𝑚𝑔ℎ = 1.2 × 10 × 0.3 = 3.6 J

6 (a)

For collision between blocks 𝐴 and 𝐵,

𝑒 =𝑣𝐵 − 𝑣𝐴𝑢𝐴 − 𝑢𝐵

=𝑣𝐵 − 𝑣𝐴10 − 0

=𝑣𝐵 − 𝑣𝐴10

∴ 𝑣𝐵 − 𝑣𝐴 = 10𝑒 = 10 × 0.5 = 5 … . (i)

from principle of momentum conservation,

𝑚𝐴𝑢𝐴 +𝑚𝐵𝑢𝐵 = 𝑚𝐴𝑣𝐴 +𝑚𝐵𝑣𝐵

P a g e | 9

Or 𝑚× 10 + 0 = 𝑚𝑣𝐴 +𝑚𝑣𝐵

∴ 𝑣𝐴 + 𝑣𝐵 = 10 … . (ii)

Adding Eqs. (i) and (ii), we get

𝑣𝐵 = 7.5 ms−1 … (iii)

Similarly for collision between 𝐵 and 𝐶

𝑣𝐶 − 𝑣𝐵 = 7.5𝑒 = 7.5 × 0.5 = 3.75

∴ 𝑣𝐶 − 𝑣𝐵 = 3.75ms−1 …(iv)

Adding Eqs. (iii) and (iv) we get

2𝑣𝐶 = 11.25

∴ 𝑣𝐶 =11.25

2= 5.6ms−1

7 (c)

Mass of planet, 𝑀𝑝 = 10𝑀𝑒 , where 𝑀𝑒 is mass of

earth. Radius of planet,

𝑅𝑝 =𝑅𝑒10

,where 𝑅𝑒 is radius of earth.

Escape speed is given by,

𝑣 = √2𝐺𝑀

𝑅

So, for planet 𝑣𝑝 = √2 𝐺 ×𝑀𝑝

𝑅𝑝= √

100 × 2 𝐺𝑀𝑒

𝑅𝑒

= 10 × 𝑣𝑒

= 10 × 11kms−1 = 110kms−1

8 (b)

Young’s modulus 𝑌 =Stress

Strain=

𝐹

𝐴

Strain

or 𝑌𝑚g

𝐴×strain

or 𝑚 =𝑌×𝐴×strain

g

=2 × 1011 × 10−3 × 10−6

10= 60kg

9 (c)

From the equation of continuity, the amount of

mass that flows past any cross-section of a pipe

has to be the same as the amount of mass that

flows past any other cross-section.

𝑖𝑒, 𝑚1 = 𝑚2

⟹ ρ1𝐴1𝑣1 = ρ2𝐴2𝑣2

Given ρ1 = ρ2 𝐴2 =𝐴12

∴ 𝐴1𝑣1 =𝐴12𝑣2

⟹ 𝑣2 = 2𝑣1

10 (b)

From Newton’s law of cooling when a hot

body is cooled in air, the rate of loss of heat by

the body is proportional to the temperature

difference between the body and its

surroundings.

Given, θ1 = 60℃, θ2 = 50℃, θ = 25℃

∴ Rate of loss of heat=K

(Mean temp.-Atmosphere temp.)

Where K is coefficient of thermal conductivity

θ1−θ2

𝑡= 𝐾 (

θ1+θ2

2− θ)

60−50

10= 𝐾 (

60+50

2− 25)

⇒ 𝐾 =1

30

Also putting the value of 𝐾, we have

50−θ3

10=

1

30(50+θ3

2− 25)

⇒ θ3 = 42.85℃

11 (a)

In an adiabatic process,

𝑝𝑉γ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

⇒ 𝑝1𝑝2

= (𝑉2𝑉1)γ

⇒ 𝑝1𝑝2

= (1

8)5/3

⇒ 𝑝1𝑝2

= (1

23)5/3

=1

32

∴ 𝑝2𝑝1

= 32

12 (d)

Let 𝑇0 be the initial temperature of the black body

∴ 𝜆0𝑇0 = 𝑏 (Wien’s law)

Power radiated, 𝑃0 = 𝐶𝑇04, where, 𝐶 is constant.

If 𝑇 is new temperature of black body, then 3𝜆04

𝑇 = 𝑏 = 𝜆0𝑇0 or 𝑇 =4

3 𝑇0

Power radiated, 𝑃 = 𝐶𝑇4 = 𝐶𝑇04 (

4

3)4

𝑃 = 𝑃0 ×256

81 or

𝑃

𝑃0=256

81

13 (c)

As springs and supports (𝑀1 and 𝑀2) are having

negligible mass. Whenever springs pull the

massless supports, springs will be in natural

P a g e | 10

length. At maximum compression, velocity of 𝐵

will be zero

And by energy conservation 1

2(4𝐾)𝑦2 =

1

2𝐾𝑥2 ⇒

𝑦

𝑥=1

2

14 (a)

Velocity of sound 𝑣 ∝ √𝑇

Time

𝑡 ∝1

√𝑣

∴ 𝑡 ∝1

√𝑇

𝑡1𝑡2= √

𝑇1𝑇2

2

𝑡2= √

273 + 30

273 + 10

2

𝑡2= √

303

283= 1.03

𝑡2 =2

1.03= 1.9𝑠

15 (d)

Energy density 𝑢𝑒 =1

2𝜀0𝐸

2 =1

2× 8.86 × 10−12 ×

(𝑉

𝑟)2

= 2.83 𝐽/𝑚3

16 (c)

𝑈 =1

4𝜋𝜀0

𝑞1𝑞2𝑟

∴ 𝑈 =9 × 109 × (1.6 × 10−19)(−1.6 × 10−19)

10−10

= −9 × 109 × 1.6 × 10−19 × 1010eV

= −14.4eV

17 (c)

For semiconductors, resistance decreases on

increasing the temperature

18 (b)

Here 𝑃1 = 𝑖2 ×𝑟

3, 𝑃2 = 𝑖2 × 3𝑟,

𝑃3 = 𝑖2 (𝑟

2+ 𝑟) =

3

2𝑖2𝑟

and 𝑃4 = 𝑖2 [2𝑟×2𝑟

2𝑟+2𝑟] = 𝑖2𝑟

So it is obvious that 𝑃2 > 𝑃3 > 𝑃4 > 𝑃1

19 (c)

𝑟 =𝑚𝑣

𝑞𝐵⇒ 𝑟 ∝ 𝑣

20 (b)

Pole strength = 𝑚 =𝑀

𝑙. When the wire is bent at

its middle point 𝑂 at 60°, then as is clear from

figure.

60° + θ + θ = 180°

2θ = 180° − 60° = 120°,

∴ 𝑂𝐴𝐵 is an equilateral triangle.

∴ 𝐴𝐵 = 2𝑙′′ = 𝑙/2

New magnetic moment

𝑀′′ = 𝑚(2𝑙′′) =𝑚𝑙

2=𝑀

2

21 (c)

The DC generator must be mixed wound to

withstand the load variation.

22 (d)

The instantaneous values of emf and current in

inductive circuit are given by 𝐸 = 𝐸0 sin𝜔𝑡 and

𝑖 = 𝑖0 sin (𝜔𝑡 −𝜋

2) respectively

So, 𝑃inst = 𝐸𝑖 = 𝐸0 sin𝜔𝑡 × 𝑖0 sin (𝜔𝑡 −𝜋

2)

= 𝐸0𝑖0 sin𝜔𝑡 (sin𝜔𝑡 cos𝜋

2− cos𝜔𝑡 sin

𝜋

2)

= 𝐸0𝑖0 sin𝜔𝑡 cos𝜔𝑡

=1

2𝐸0𝑖0 sin 2𝜔𝑡 (sin 2𝜔𝑡 = 2 sin𝜔𝑡 cos𝜔𝑡)

Hence, angular frequency of instantaneous power

is 2𝜔

23 (a)

Solar radiations are transverse Electromagnetic

waves. The central core of the sun emits a

continuous Electromagnetic Spectrum.

24 (b)

Wavelength of a certain colour in air λair =

600 nm.

Wavelength of a certain colour in glass of

refractive index μ = 1.5

∴ λglass =λairμglass

=600

1.5

λglass = 400 nm

Also, 𝑣glass =𝑣airμglass

=3 × 108

1.5

P a g e | 11

𝑣glass = 2.0 × 108 ms−1

26 (c)

𝑣 =𝐸

𝐵;where 𝐸 =

𝑉

𝑑=

1000

1 × 10−2= 105𝑉/𝑚

⇒ 𝑣 =105

1= 105𝑚/𝑠

27 (c)

According to the Bohr’s theory the wavelength of

radiations emitted from hydrogen atom given by

1

𝜆= 𝑅 [

1

𝑛12 −

1

𝑛22] ⇒ 𝜆 =

𝑛12𝑛2

2

(𝑛22𝑛1

2)𝑅

For maximum wavelength if𝑛1 = 𝑛,then

𝑛2 = 𝑛 + 1

∴ 𝜆 𝑖𝑠 maximumfor𝑛2 = 3 and 𝑛1 = 2.

28 (b)

Let nucleus be 𝑍𝑋𝐴. Nuclear radius, 𝑅 = 𝑅0𝐴

1/3

where 𝑅0 is a constant whose value is found to be

1.2 × 10−15𝑚 and 𝐴 is the mass number

∴𝑅𝑋𝑅𝐶𝑠

= (𝐴

189)1/3

, ∴1

3= (

𝐴

189)1/3

𝐴 =189

33=189

27= 7

The given nucleus is 𝐿𝑖7

29 (b)

When p-n junction diode is forward biased, more

number of charge carriers (electrons in 𝑛-side

and holes in 𝑝-side) moves through the junction.

Therefore, in forward biasing barrier potential is

reduced. Thus, increasing the current in the

circuit.

30 (c)

For commercial UHF, TV broadcasts, the range

allotted is 470-960 MHz

31 (c)

∵ Mass of 22400 cm3 of CH4 at STP = 16 g

∴ Mass of 1 cm3of CH4 at STP = 16

22400g

∴ Mass of 112 cm3 of CH4 at STP =16

22400 × 112

= 0.08 g

32 (d)

The threshold frequency (𝑣0) is the lowest

frequency that photons may possess to produce

the photoelectric effect. The energy

corresponding to this frequency is the minimum

energy (𝐸)

𝐸 = ℎ𝑣0

= (6.625 × 10−27erg s) (1.3 × 1015 × s−1)

= 8.6 × 10−12 erg

33 (a)

basic nature of oxides Al2O3 < 𝑀𝑔𝑂 < 𝑁a2O <

K2O

34 (d)

Phosphorus atom is 𝑠𝑝3 hybridised in 𝑃4 usually.

Therefore, p-character 75%

36 (d)

PbO2 + Pb → 2PbO

Oxidation state +4 0 +2

Since, ∆𝑟𝐺° < 0, hence +2 state of lead is

favourable.

SnO2 + Sn → 2SnO

Oxidation state +4 0 +2

Since, ∆𝑟𝐺° > 0, it means forward reaction is

not spontaneous.

2SnO → SnO2 + Sn

+2 +4 0

For this ∆𝑟𝐺° < 0, thus +4 state of tin is

favourable.

37 (d)

Aspirin is a weak acid. Due to common ion effect,

it is unionised in acid medium but completely

ionised in alkaline medium

38 (c)

Only this reaction involves oxidation and

reduction.

39 (a)

Calgon is represented by sodium hexa

metaphosphate, (NaPO3)6 or Na2[Na4(PO3)6].

40 (c)

Lithium and magnesium shows diagonal

relationship. Some points of similarity are

(i) Polarising power of Li+ and Mg+ are almost

same.

(ii) Like Li, Mg decomposes water very slowly.

P a g e | 12

(iii) LiCl and MgCl2 are deliquescent.

(iv) Like Li, Mg do not form solid bicarbonates.

41 (a)

Chain silicates Double chain silicates can be

formed when two simple chains are joined

together by shared oxygens. These minerals are

called amphiboles, and they are well known. The

most numerous and best known amphiboles are

the asbestos minerals. These are based on the

structural unit (Si4O11)𝑛6𝑛−. The structure of

amphiboles is

42 (b)

The IUPAC name of this molecule is 2-bromo-3-

ethyl-1, 4-pentadiene.

45 (b)

In fcc unit cell

√2𝑎 = 4𝑟 ⇒ 𝑟 =√2 𝑎

4

=√2 × 361

4= 127 pm

46 (c)

∆𝑇𝑏 = 𝑖𝑚 𝑘𝑏 = 0.52 × 1 × 2 = 1.04

∴ 𝑇𝑏 = 𝑇 + ∆𝑇𝑏 = 100 + 1.04 = 101.04℃

47 (a)

We know from Kohlrausch’s law

⋀CH3COOH° = ⋀CH3COONa

° + ⋀HCl° − ⋀NaCl

°

48 (c)

2𝐴 + 𝐵 → 𝐶

Rate of reaction,

= −1

2

𝑑[𝐴]

𝑑𝑡= −

𝑑[𝐵]

𝑑𝑡=𝑑[𝐶]

𝑑𝑡

∴ −𝑑[𝐴]

𝑑𝑡= 2

𝑑[𝐶]

𝑑𝑡

= 2 × 2.2 × 10−3

= 4.4 × 10−3𝑚𝑜𝑙 𝐿−1min−1

49 (b)

Micelles show lower colligative properties as that

of common colloidal solution

50 (a)

SiO2 + CaO → CaSiO3

acidic impurity basic flux slag

51 (c)

The true peroxide contains O22−(O − O)2− ion.

∵ Out of given choices only BaO2 has O22− in its

structure.

∴ BaO2 is true peroxide.

52 (b)

Basic copper acetate (verdigris – (CH3COO)2Cu ∙

Cu(OH)2) is blue green powder used in green

pigment and in dyes. Also in manufacture of

insecticides and fungicides

54 (a)

C2H5Cl Aq.KOH → C2H5OH

AgOH ← C2H5 Cl

55 (b)

When one H2SO4 reacts with ethyl alcohol at

room temperature, ethyl hydrogen sulphate is

formed

CH3CH2OH+ H2SO4Room temp → CH3CH2HSO4

+ H2O

Ethyl hydrogen

sulphate

56 (d)

The given reaction is an example of Diels-Alder

reaction, which is a cycloaddition

58 (b)

Although D-alanine is a constituent of a bacterial

cell walls, it is not found in proteins

59 (b)

Out of these statements, statement (b) is wrong.

61 (d)

Given figure clearly represents

(𝐴 − 𝐵) ∪ (𝐵 − 𝐴)

62 (c)

Given, 𝑓 (𝑝

𝑞) = √𝑝2 − 𝑞2, for

𝑝

𝑞= 𝒬

If 𝑝 < 𝑞, then 𝑓 (𝑝

𝑞) is not real.

Hence, statement I is false while statement II is

true.

63 (c)

2 tan(𝐴 − 𝐵) = 2(tan𝐴 − tan𝐵

1 + tan𝐴 tan𝐵)

P a g e | 13

= 2(2 tan𝐵 + cot𝐵 − tan𝐵

1 + (2 tan𝐵 + cot𝐵) tan𝐵)

[∵ tan𝐴 = 2 tan𝐵 + cot𝐵]

=2(tan𝐵 + cot𝐵)

2(1 + tan2 𝐵)= cot 𝐵

64 (a)

Let 𝑃(𝑛 ) ≡ 𝑥2𝑛−1 + 𝑦2𝑛−1 = 𝜆(𝑥 + 𝑦)

𝑃(1) ≡ 𝑥 + 𝑦 = 𝜆1(𝑥 + 𝑦)

𝑃(2) ≡ 𝑥3 + 𝑦3 = 𝜆2(𝑥 + 𝑦)

Hence, for ∀ 𝑛 ∈ 𝑁, 𝑃(𝑛) is true.

65 (a)

Since, 𝑧−1

𝑧+1 is purely imaginary

∴ 𝑧 − 1

𝑧 + 1= −(

𝑧 − 1

𝑧 + 1)

⇒ 𝑧 − 1

𝑧 + 1=𝑧 − 1

𝑧 + 1

⇒ 2𝑧

−2=

2

−2𝑧 ⇒ 𝑧𝑧 = 1

⇒ |𝑧|2 = 1 ⇒ |𝑧| = 1

66 (b)

∵ 𝑓(−1) < 1 ⇒ 𝑎 − 𝑏 + 𝑐 < 1 …(i)

and 𝑓(1) > −1, 𝑓(3) < −4, then

𝑎 + 𝑏 + 𝑐 > −1 …(ii)

9𝑎 + 3𝑏 + 𝑐 < −4 …(iii)

From Eq. (ii),

−𝑎 − 𝑏 − 𝑐 < 1 ….(iv)

On solving Eqs. (i), (iii) and (iv), we get

𝑎 < −1

8⇒ 𝑎 is negative

67 (b)

∵ Each letter can be posted in 3 ways

∴ Total number of ways = 36

68 (d)

∵ 𝑎, 𝑏, 𝑐 are in AP

⇒ 2𝑏 = 𝑎 + 𝑐

⇒ 𝑎 − 2𝑏 + 𝑐 = 0

On putting 𝑥 = 1, we get

Required sum = (1 + (𝑎 − 2𝑏 +

𝑐)2)1973 = (1 + 0)1973 = 1

69 (a)

∵ (𝑥 + 1) + (𝑥 + 4) + (𝑥 + 7)+. . . +(𝑥 + 28)

= 155

Let 𝑛 be the number of terms in the AP on LHS.

∴ 𝑥 + 28 = (𝑥 + 1) + (𝑛 − 1)3

⇒ 𝑛 = 10

∴ 10

2[(𝑥 + 1) + (𝑥 + 28)] = 155

⇒ 𝑥 = 1

70 (c)

For the greatest distance, both points lie on a

straight line.

∴ Required equation of line is

𝑦 − 2 =1 − 2

3 − 1(𝑥 − 1)

⇒ 𝑥 + 2𝑦 = 5

71 (d)

The equation of any tangent to the

parabola 𝑦2 = 4𝑥 is

𝑦 = 𝑚𝑥 +1

𝑚 … (i)

This touches the parabola 𝑥2 = −32𝑦,

therefore the equation 𝑥2 =

−32 (𝑚𝑥 +1

𝑚) has equal roots.

∴ (32 𝑚)2 = 4 (32

𝑚) [∴ 𝐷2 = 4𝑎𝑐]

⇒ 8𝑚3 = 1 ⇒ 𝑚 =1

2

On putting the value of 𝑚 is Eq. (i). we

get

𝑥 − 2𝑦 + 4 = 0

72 (c)

lim𝑥→0 [2𝑥−1

√1+𝑥−1] = lim𝑥→0

2𝑥 log𝑒 21

2√1+𝑥

[by L’

Hospital’s rule]

= 2 log𝑒 2 = log𝑒 4

73 (d)

𝑆(𝑝, 𝑞, 𝑟) = (~𝑝) ∨ [∼ (𝑞 ∧ 𝑟)]

= (∼ 𝑝) ∨ [∼ 𝑞 ∨∼ 𝑟]

⇒ 𝑆 (~𝑝,~𝑞, ~𝑟) = 𝑝 ∨ (𝑞 ∨ 𝑟)

74 (b)

�� =1 + 2 + 3 +⋯+ 𝑛

𝑛=(𝑛 + 1)

2

P a g e | 14

∴ 𝜎2 =Σ(𝑥𝑖)

2

𝑛− (��)2

=Σ𝑛2

𝑛− (

𝑛 + 1

2)2

=𝑛(𝑛 + 1)(2𝑛 + 1)

6𝑛− (

𝑛 + 1

2)2

=𝑛2 − 1

12

75 (c)

The last digit of the product will be 1, 2, 3, 4, 5, 6,

7, 8 or 9 if and only if each of the 𝑛 positive

integers ends in any of these digits. Now the

probability of an integer ending

in 1, 2, 3, 4, 5, 6, 7, 8 or 9 is8

10. Therefore the probability that the last digit of the

product of 𝑛 integer in

1, 2, 3, 4, 5, 6, 7, 8 or 9 is (4

5)𝑛.The probability for

an integer to

end in 1, 3, 7 or 9 is4

10=

2

5

Therefore the probability for the product of 𝑛

positive integers to end in 1, 3, 7 or 9 is (2

5)𝑛

Hence the required probability= (4

5)𝑛− (

2

5)𝑛=

4𝑛−2𝑛

5𝑛

76 (a)

Here, 𝑎 = √(16 − 5)2 + (12 − 12)2 = 11

𝑏 = √(16 − 0)2 + (12 − 0)2 = 20

And 𝑐 = √(5 − 0)2 + (12 − 0)2 = 13

∴

Incentre= (11×0+20×5+13×16

11+20+13,11×0+20×12+13×12

11+20+13) =

(7, 9)

78 (b)

Let 𝑓(𝑥) = cos 𝑥, 𝑓′(𝑥) = sin 𝑥

⇒ 𝑓′′(𝑥) = − cos𝑥

⇒ 𝑓′′′(𝑥) = sin𝑥

Since, sin 𝑥 is an odd function.

∴ 𝑓′′′ is an odd function.

79 (b)

Given, tan−1 (1+𝑥

1−𝑥) =

𝜋

4+ tan−1 𝑥

RHS=𝜋

4+ tan−1 𝑥 = tan−1 1 + tan−1 𝑥

= tan−1 (1+𝑥

1−𝑥), if 𝑥 < 1

∴ 𝑥 ∈ (−∞, 1)

80 (a)

Given, 𝑥 [−34] + 𝑦 [

43] = [

10−5

]

∴ −3𝑥 + 4𝑦 = 10 …(i)

and 4𝑥 + 3𝑦 = −5 …(ii)

On solving Eqs. (i) and (ii), we get

𝑥 = −2, 𝑦 = 1

81 (b)

Since, |𝐴| = −1, |𝐵| = 3

∴ |𝐴𝐵| = |𝐴||𝐵| = −3

Now, |3𝐴𝐵| = (3)3(−3) = −81

82 (c)

LHL= limℎ→0

𝑓(0 − ℎ) = limℎ→0

sin5(0−ℎ)

(0−ℎ)2+2(0−ℎ)

= −limℎ→0

sin5ℎ

5ℎ1

5(ℎ − 2)

=5

2

Since, it is continuous at 𝑥 = 0, therefore

LHL= 𝑓(0)

⇒ 5

2= 𝑘 +

1

2 ⇒ 𝑘 = 2

83 (d)

The point of intersection of given curve is

(−2,−1)

On differentiating the given curve respectively, we

get

𝑥𝑑𝑦

𝑑𝑥+ 𝑦 = 0 ⇒

𝑑𝑦

𝑑𝑥= −

𝑦

𝑥

⇒ 𝑚1 = (𝑑𝑦

𝑑𝑥)(−2,−1)

= −1

2

And 2𝑥 + 4𝑑𝑦

𝑑𝑥= 0

⇒ 𝑑𝑦

𝑑𝑥= −

𝑥

2

⇒ 𝑚2 = (𝑑𝑦

𝑑𝑥)(−2,−1)

= 1

∴ tan 𝜃 = |𝑚1 −𝑚2

1 +𝑚1𝑚2| = |

−1

2− 1

1 −1

2

| = 3

84 (b)

∫𝑑𝑥

𝑥(𝑥 + 1)= ∫

1

𝑥𝑑𝑥 − ∫

1

𝑥 + 1𝑑𝑥

= log 𝑥 − log(𝑥 + 1) + 𝑐

= log |𝑥

𝑥 + 1| + 𝑐

85 (b)

∫ |𝑥3 + 𝑥2 + 3𝑥|𝑑𝑥3

0= ∫ (𝑥3 + 𝑥2 + 3𝑥)𝑑𝑥

3

0

= [𝑥4

4+𝑥3

3+3𝑥2

2]0

3

=81

4+27

3+27

2=171

4

86 (d)

Required area = ∫ (√𝑥 − 𝑥2)𝑑𝑥1

0

P a g e | 15

= [2𝑥3/2

3−𝑥3

3]0

1

= (2

3−1

3) =

1

3 sq unit

87 (a)

Let 𝑥2 + 𝑦2 − 2𝑘𝑦 = 0

⇒ 2𝑥 + 2𝑦 𝑑𝑦

𝑑𝑥− 2𝑘

𝑑𝑦

𝑑𝑥= 0

⇒ 𝑘 =𝑘

(𝑑𝑦

𝑑𝑥)+ 𝑦

From Eq. (i),

𝑥2 + 𝑦2 − 2(𝑥

(𝑑𝑦 𝑑𝑥⁄ )+ 𝑦) 𝑦 = 0

⇒ (𝑥2 − 𝑦2)𝑑𝑦

𝑑𝑥− 2𝑥𝑦 = 0

88 (c)

Let α = 𝜆 �� + μ𝐛 + 𝑡𝐜 …(i)

Now, �� ∙ �� = 𝐛 ∙ �� = 𝐜 ∙ 𝐫 = 1

⇒ α ∙ �� = λ ( �� ∙ �� ) + 0 + 0

⇒ 𝜆 = α ∙ ��

Similarly, 𝜇 = α ∙ ��

and 𝑡 = α ∙ 𝐫

From Eq. (i), we get

α = ( α ∙ �� )�� + ( α ∙ �� )𝐛 + (α ∙ 𝐫 )𝐜

89 (a)

Distance of point 𝑃 from plane=5

∴ 5 |1 − 4 − 2 − α

3|

α =10

Foot perpendicular

𝑥 − 1

1=𝑦 + 2

2=𝑧 − 1

−2−(1 − 4 − 2 − 10)

1 + 4 + 4=5

3

⟹ 𝑥 =8

3, 𝑦 =

4

3, 𝑧 −

7

3

Thus, the foot of the perpendicular is

𝐴 (8

3,4

3,−

7

3)

90 (d)

Feasible region is 𝐴𝐵𝐶𝐷𝐹𝐴 and 𝑧 = 30𝑥 + 20𝑦

Now, at 𝐴(4, 0), 𝑧 = 30 × 4 + 0 = 120

𝐵(8, 0), 𝑧 = 30 × 8 + 0 = 240

𝐶(0, 8), 𝑧 = 0 + 20 × 8 = 160

𝐷(0, 3), 𝑧 = 0 + 20 × 3 = 60

And 𝐹 (1,3

2) , 𝑧 = 30 × 1 + 20 ×

3

2= 60

It is clear that minimum value of 𝑧 is 60 at points

𝐷(0, 3) and 𝐹 (1,3

2)