PHYSICS 

1. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: A pendulum clock when taken to Mount Everest becomes fast.
Reason: The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.
In the light of the above statements, choose the most appropriate answer from the options given below
(1) Both A and R are correct but R is NOT the correct explanation of A
(2) A is correct but R is not correct
(3) Both A and R are correct and R is the correct explanation of A
(4) A is not correct but R is correct
Sol. 4
T = 2π√(l/g)
T ∝ 1/√g
on Everest g decreases, so T increases, so moves slow.

2. The frequency (ν) of an oscillating liquid drop may depend upon radius (r) of the drop, density (ρ) of liquid and the surface tension (s) of the liquid as : ν = r^a ρ^b s^c. The values of a, b and c respectively are
(1) (-3/2, 1/2, 1/2)
(2) (3/2, -1/2, 1/2)
(3) (-3/2, -1/2, 1/2)
(4) (3/2, 1/2, -1/2)
Sol. 3
f α r^a ρ^b s^c
T?¹ = [L]^a [ML?³]^b [MT?²]^c
T?¹ = L^(a-3b) M^(b+c) T^(-2c)
-2c = -1 ...(1)
c = 1/2
b + c = 0 ...(2)
b = -1/2
a - 3b = 0 ...(3)
a = 3b = -3/2

3. Given below are two statements:
Statement I : Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.
Statement II : Acceleration due to earth's gravity is same at a height 'h' and depth 'd' from earth's surface, if h = d.
In the light of above statements, choose the most appropriate answer form the options given below
(1) Both Statement I and Statement II are incorrect
(2) Statement I is incorrect but statement II is correct
(3) Both Statement I and II are correct
(4) Statement I is correct but statement II is incorrect
Sol. 4
g(1 - 2h/R) = g(1 - d/R)
h = d/2

4. A long solenoid is formed by winding 70 turns cm?¹. If 2.0 A current flows, then the magnetic field produced inside the solenoid is (μ? = 4π × 10?? T m A?¹)
(1) 88 × 10?? T
(2) 352 × 10?? T
(3) 176 × 10?? T
(4) 1232 × 10?? T
Sol. 3
B = μ?ni
= 4 × 22/7 × 10?? × 70 × 100 × 2
= 176 × 10??

5. The electric potential at the centre of two concentric half rings of radii R? and R?, having same linear charge density λ is :
(1) λ/(2ε?)
(2) λ/(4ε?)
(3) 2λ/ε?
(4) λ/ε?
Sol. 1
V_c = K/R? q? + kq?/R?
= 1/(4πε?) × λπR?/R? + 1/(4πε?) × λπR?/R?
= λ/(2ε?)

6. If the distance of the earth from Sun is 1.5 × 10? km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :
(1) 6 × 10? km
(2) 3 × 10? km
(3) 3 × 10? km
(4) 6 × 10? km
Sol. 2
T² ∝ R³
(T_E/T_P)^(2/3) = R_E/R_P
(1/2.83)^(2/3) = 1.5 × 10?/R
R = 1.5 × 10? × (2.83)^(2/3)
1.5 × 10? × (1.41 × 2)^(2/3)
1.5 × 10? × (2√2)^(2/3)
1.5 × 10? × (√8)^(2/3)
3 × 10? KM

7. A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 × 10?¹? eVs):
(1) 99.3 nm
(2) 941 nm
(3) 974 nm
(4) 94.1 nm
Sol. 4
ΔE = E? - E?
hc/λ = -0.85 - (-13.6)
(4 × 10?¹? × 3 × 10¹? nm)/λ_(nm) = 12.75
λ = 1200/nm
= 94.1 nm

8. A cell of emf 90 V is connected across series combination of two resistors each of 100Ω resistance. A voltmeter of resistance 400Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be:
(1) 90 V
(2) 45 V
(3) 80 V
(4) 40 V
Sol. 2
as Resistance are same so equal division of potential.
∴ 90/2 = 45 V

9. If two vectors P = i + 2m j + m k and Q = 4i - 2j + m k are perpendicular to each other. Then, the value of m will be:
(1) -1
(2) 3
(3) 2
(4) 1
Sol. 3
P·Q = 0
4 × 1 + 2m × -2 + m² = 0
m² - 4m + 4 = 0
(m - 2)² = 0
m = 2

10. The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by
E_x = E? sin(kz - ωt)
B_y = B? sin(kz - ωt)
Then the correct relation between E? and B? is given by
(1) E?B? = ωk
(2) E? = kB?
(3) kE? = ωB?
(4) ωE? = kB?
Sol. 3
by theory of EM wave
E?/B? = v = ω/K

11. The logic gate equivalent to the given circuit diagram is :
(1) NAND
(2) OR
(3) AND
(4) NOR
Sol.
by truth table
A1 B1 Y
0 0 1
0 1 1
1 0 1
1 1 0
NAND gate

12. Let γ? be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and γ? be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, γ?/γ? is :
(1) 25/21
(2) 35/27
(3) 21/25
(4) 27/35
Sol.
5/3 / 7/5 = 25/21

13. When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called:
(1) Spherical aberration
(2) Polarisation
(3) Chromatic aberration
(4) Scattering
Sol. Theory : Colors are due to chromatic aberration.

14. A metallic rod of length 'L' is rotated with an angular speed of 'ω' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be:
(1) 1/4 BL²ω
(2) 1/2 B²L²ω
(3) 1/4 B²Lω
(4) 1/2 BL²ω
Sol.
e = ∫?? Bω dx
= 1/2 BωL²

15. An α-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:
(1) λ_α < λ_p < λ_e
(2) λ_α = λ_p = λ_e
(3) λ_α > λ_p > λ_e
(4) λ_α > λ_p < λ_e
Sol. 1
λ = h/√(2mkE) ∝ 1/√m
m_α > m_p > m_e
∴ λ_α < λ_p < λ_e

16. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason
Assertion A: Steel is used in the construction of buildings and bridges.
Reason R: Steel is more elastic and its elastic limit is high.
In the light of above statements, choose the most appropriate answer from the options given below
(1) Both A and R are correct and R is the correct explanation of A
(2) Both A and R are correct but R is NOT the correct explanation of A
(3) A is correct but R is not correct
(4) A is not correct but R is correct
Sol. 1
Steel is more elastic.

17. In an Isothermal change, the change in pressure and volume of a gas can be represented for three different temperature; T? > T? > T? as:
Sol. 3
PV = nRT const.

18. Match List I with List II
LIST I LIST II
A. AM Broadcast I. 88 – 108 MHz
B. FM Broadcast II. 540 – 1600 kHz
C. Television III. 3.7 – 4.2 GHz
D. Satellite Communication IV. 54 MHz – 890 MHz
Choose the correct answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-I, B-III, C-II, D-IV
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-III, C-I, D-IV
Sol.
by concept of AM & FM freq. range

19. A body of mass 200 g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be :
(1) 2:5
(2) 1:1
(3) 2:3
(4) 1:2
Sol.
kx = mω²(?? + x)

20. The velocity time graph of a body moving in a straight line is shown in figure.
The ratio of displacement to distance travelled by the body in time 0 to 10 s is :
(1) 1:1
(2) 1:2
(3) 1:3
(4) 1:4
Sol.
disp. = area = 8 × 2 + (4 × 4) - 2 × 4 - 2 × 4
= 32 - 16 = 16
distance = 32 + 16 = 48

21. A body of mass 1 kg begins to move under the action of a time dependent force F = (t i + 3t² j) N, where i and j are the unit vectors along x and y axis. The power developed by above force, at the time t = 2s will be W.
Sol. 100

22. A convex lens of refractive index 1.5 and focal length 18 cm in air is immersed in water. The change in focal length of the lens will be cm (Given refractive index of water = 4/3)
Sol. 54
Div eq. by eq. 2
f/f = 0.5 × 8 / 1
f = 72 cm
change = 72 - 18 = 54

23. The energy released per fission of nucleus of ²??X is 200 MeV. The energy released if all the atoms in 120 g of pure ²??X undergo fission is × 10²? MeV (Given N_A = 6 × 10²³)
Sol. 6
no. of atoms = 120/240 × 6 × 10²³
= 3 × 10²³
Energy released = 200 × 3 × 10²³
= 6 × 10²?

24. A uniform solid cylinder with radius R and length L has moment of inertia I?, about the axis of the cylinder. A concentric solid cylinder of radius R' = R/2 and length L' = L/2 is carved out of the original cylinder. If I? is the moment of inertia of the carved out portion of the cylinder then I?/I? = (Both I? and I? are about the axis of the cylinder)
Sol. 32
I? = MR²/2
mass = ρπ R²/4 · L/2
m? = M/8
I? = m?R?²/2 = MR²/(8 × 4 × 2)
I?/I? = 32

25. A parallel plate capacitor with air between the plate has a capacitance of 15 pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes x/4 pF. The value of x is
Sol. 105
C = Aε?/d
C = KAε?/2d
= KC/2
= 3.5 × 15/2
= 105/4
= 105

26. A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be x/130 N. The value of x is
Sol. 9
Force on 5 cm length = i ∫ dl × B
= i × (5/100) × 0.75 × sin(90 + θ)
= 2 × 5/100 × 0.75 × cos θ
= 10/100 × 0.75 × 12/13 = x/130
⇒ x = 9

27. A mass m attached to free end of a spring executes SHM with a period of 1 s. If the mass is increased by 3 kg the period of oscillation increases by one second, the value of mass m is kg.
Sol. 1
2π√(m/k) = 1 ...(1)
2π√((m + 3)/k) = 2 ...(2)
(2) ÷ (1)
√((m + 3)/m) = 2/1
(m + 3)/m = 4
4m = m + 3
m = 1 kg.

28. If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is %.
Sol. 44
Length becomes = 1.2 times
?' = 1.2?
R' = n²R
= (1.2)²R
= 1.44R
ΔR = 0.44R
ΔR/R × 100% = 44%

29. Three identical resistors with resistance R = 12Ω and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be A.
Sol. 3
Short all inductor
Req. = R/3 = 12/3 = 4Ω
I = 12/4 = 3 A

30. A Spherical ball of radius 1 mm and density 10.5 g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g/cc. Viscous force on the ball when it attains constant velocity is 3696 × 10?? N. The value of x is (Given, g = 9.8 m/s² and π = 22/7)
Sol. 7
V_T = 2r²g(σ_s - ρ_f)/a_n
= (2 × 10?? × 9.8 × (10.5 - 1.5) × 10³)/(9.8 × 0.1 × 9)
= 2 × 10?² m/s
F = 6πxrV_T
= 6 × 22/7 × 9.8 × 0.1 × 10?³ × 18 × 10?²
= 3696 × 10??
= 7

31. Identify the correct statements about alkali metals.
A. The order of standard reduction potential (M?|M) for alkali metal ions is Na > Rb > Li
B. CsI is highly soluble in water.
C. Lithium carbonate is highly stable to heat.
D. Potassium dissolved in concentrated liquid ammonia is blue in colour and paramagnetic.
E. All the alkali metal hydrides are ionic solids.
Choose the correct answer from the options given below:
(1) C and E only
(2) A, B and E only
(3) A, B, D only
(4) A and E only
Sol. 4
(i) These standard potentials of
Element Li Na Rb
SRP -3.237 -2.898 -3.079
(ii) All the alkali metal hydrides are ionic solids with high M.P.

32. Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Beryllium has less negative value of reduction potential compared to the other alkaline earth metals.
Reason : Beryllium has large hydration energy due to small size of Be²? but relatively large value of atomization enthalpy
In the light of the above statements, choose the most appropriate answer from the options given below
(1) A is not correct but R is correct
(2) A is correct but R is not correct
(3) Both A and R are correct and R is the correct explanation of A
(4) Both A and R are correct but R is NOT the correct explanation of A
Sol. 3
Be has least negative SRP value. Value in alkaline earth metal group as it has high hydration enthalpy and high enthalpy of atomization.

33. A student has studied the decomposition of a gas AB? at 25°C. He obtained the following data.
p(mmHg) 50 100 200 400
relative t?/?(s) 4 2 1 0.5
The order of the reaction is
(1) 0 (zero)
(2) 0.5
(3) 1
(4) 2
Sol. 4
t½ ∝ (Co)^(1-n)
= (t½)?/(t½)? = (P?/P?)^(1-n)
= 4/2 = (50/100)^(1-n) ⇒ 2 = (1/2)^(1-n)
2 = (2)^(n-1)
n = 2

34. K?Cr?O? paper acidified with dilute H?SO? turns green when exposed to
(1) Carbon dioxide
(2) Sulphur trioxide
(3) Sulphur dioxide
(4) Hydrogen sulphide
Sol. 3
K?Cr?O? + 2H? + SO? → 2Cr?³ + 3SO??² + H?O

35. Which will undergo deprotonation most readily in basic medium?
(1) c only
(2) only
(3) Both a and c
(4) b only
Sol. 2
order of -m effect = -C-CH? > -C-OCH?
strong -m effect of both ketone

36. The hybridization and magnetic behaviour of cobalt ion in [Co(NH?)?]³? complex, respectively is
(1) d²sp³ and paramagnetic
(2) sp³d² and diamagnetic
(3) d²sp³ and diamagnetic
(4) sp³d² and paramagnetic
Sol. 3
[Co(NH?)?]³?
Co³? → [Ar]3d? 4s?
NH? → SFL, Pairing of e?
hybridisation d²sp³
μ = 0
diamagnetic

37. Given below are two statements:
Statement I : H?N-C(=O)-CH?-CH?-C(=O)-CH? under Clemmensen reduction conditions will give HOOC-CH?-CH?-CH?-CH?
Statement II : (CH?)?CH-CH?-CH(Cl)-C(=O)-CH? under Wolff-Kishner reduction condition will give (CH?)?CH-CH?-CH=CH-CH?
In the light of the above statements, choose the correct answer from the options given below:
(1) Statement I is false but Statement II is true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false
Sol. 2

38. Which of the following cannot be explained by crystal field theory?
(1) The order of spectrochemical series
(2) Stability of metal complexes
(3) Magnetic properties of transition metal complexes
(4) Colour of metal complexes
Sol. 1
Crystal field theory introduce spectrochemical series based upon the experimental value of Δ but can't explain it's order. While other three points are explained by CFT. Specially when the CFSE increases thermodynamic stability of the complex increases.

39. The number of s-electrons present in an ion with 55 protons in its unipositive state is
Sol. 10
Cs?_(55) = 1s², 2s², 2p?, 3s², 3p?, 4s², 3d¹?, 4p?, 5s², 4d¹?, 5p?
no. of s-electron = 10

40. Which one amongst the following are good oxidizing agents?
(A) Sm²? (B) Ce²? (C) Ce?? (D) Tb??
Choose the most appropriate answer from the options given below:
(1) D only
(2) C only
(3) C and D only
(4) A and B only
Sol. 3
Ce?? & Tb?? are good oxidizing agent.

41. Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.
Sol. 1
C?H?COOH + NaOH → C?H?COONa + H?O
when weak acid C?H?COOH titrated against strong base NaOH in the beginning the conductance Inc. slowly and after equivalent point it increase rapidly.

42. Match List I with List II
LIST I Type LIST II Name
A. Antifertility drug I. Norethindrone
B. Tranquilizer II. Meprobamate
C. Antihistamine III. Seldane
D. Antibiotic IV. Ampicillin
Choose the correct answer from the options given below:
(1) A-I, B-III, C-II, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-I, B-II, C-III, D-IV
(4) A-II, B-I, C-III, D-IV
Sol. 3
LIST I Type LIST II Name
A. Antifertility drug I. Norethindrone
B. Tranquilizer II. Meprobamate
C. Antihistamine III. Seldane
D. Antibiotic IV. Ampicillin

43. Find out the major products from the following reaction
Sol. 3

44. Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion : Benzene is more stable than hypothetical cyclohexatriene
Reason : The delocalized π electron cloud is attracted more strongly by nuclei of carbon atoms.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both A and R are correct and R is the correct explanation of A
(2) Both A and R are correct but R is NOT the correct explanation of A
(3) A is false but R is true
(4) A is true but R is false
Sol. 1
Both A and R are correct and R is the correct explanation of A

45. In which of the following reactions the hydrogen peroxide acts as a reducing agent?
(1) PhS + 4H?O? → PhSO? + 4H?O
(2) Mn²? + H?O? → Mn?? + 2OH?
(3) HOCl + H?O? → H?O? + Cl? + O?
(4) 2Fe²? + H?O? → 2Fe³? + 2OH?
Sol. 3
HOCl + H?O? → H?O? + Cl? + O? hydrogen peroxide acts as a reducing agent

46. Given below are two statements:
Statement I : Pure Aniline and other arylamines are usually colourless.
Statement II : Arylamines get coloured on storage due to atmospheric reduction
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Both Statement I and Statement II are incorrect
(2) Statement I is incorrect but Statement II is correct
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are correct
Sol. 3

47. Correct statement is:
(1) An average human being consumes nearly 15 times more air than food
(2) An average human being consumes 100 times more air than food
(3) An average human being consumes equal amount of food and air
(4) An average human being consumes more food than air
Sol. 1
An average human being requires. hourly 12 –15 times more air than the food.

48. What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : N? ; N?? ; O? ; O???
(1) 2,1,0,1
(2) 0, 1, 0, 1
(3) 0,1,0,1
(4) 2,1,2,1
Sol. 2
N? = σ1s², σ*1s², σ2s², σ*2s², π2px² = π2py², σ2pz²
no. of e? present in HOMO = 0
N?? = σ1s², σ*1s², σ2s², σ*2s², π2px² = π2py², σ2pz¹
no. of unpaired e? present in HOMO = 1
O? = σ1s², σ*1s², σ2s², σ*2s², σ2pz², π2px² = π2py², π*2px¹ = π*2py¹
no. of unpaired e? present in HOMO = 2
O?? = σ1s², σ*1s², σ2s², σ*2s², σ2pz², π2px² = π2py², π*2px¹ = π*2py?
no. of unpaired e? present in HOMO = 1

49. The metal which is extracted by oxidation and subsequent reduction from its ore is:
(1) Ag
(2) Fe
(3) Cu
(4) Al
Sol. 1
4Ag(s) + 8CN?(aq) + 2H?O(aq) + O?(g) → 4[Ag(CN)?]?(aq) + 4OH?(aq)
2[Ag(CN)?]?(aq) + Zn(s) → 2Ag(s) + [Zn(CN)?]²?(aq)

50. Choose the correct colour of the product for the following reaction.
(1) White
(2) Red
(3) Blue
(4) Yellow
Sol. 2
Red

Section:B

51. Following figure shows spectrum of an ideal black body at four different temperatures. The number of correct statement/s from the following is
A. T? > T? > T? > T?
B. The black body consists of particles performing simple harmonic motion.
C. The peak of the spectrum shifts to shorter wavelength as temperature increases.
D. T?/v? = T?/v? = T?/v? ≠ constant
E. The given spectrum could be explained using quantisation of energy.
Sol. 2
(A) T? > T? > T? > T?
(C) The peak of the spectrum shift to shorter wavelength of temp. Inc.

52. The number of units, which are used to express concentration of solutions from the following is
Mass percent, Mole, Mole fraction, Molarity, ppm, Molality
Sol. 5
Conc. Express in → mass percentage
→ mole fraction
→ molarity
→ PPM
→ molality

53. The number of statement/s which are the characteristics of physisorption is
A. It is highly specific in nature
B. Enthalpy of adsorption is high
C. It decreases with increase in temperature
D. It results into unimolecular layer
E. No activation energy is needed
Sol. 2
(C) It decreases with increase in temperature
(E) No activation energy is needed

54. Sum of π-bonds present in peroxodisulphuric acid and pyrosulphuric acid is:
Sol. 8
Peroxodisulphuric acid (H?S?O?)
Pyrosulphuric acid (H?S?O?)
π bond = 4
total π bond = 4 + 4 = 8

55. If the pKa of lactic acid is 5, then the pH of 0.005M calcium lactate solution at 25°C is × 10?¹ (Nearest integer)
Sol. 85
Ca(Lac)? → Ca²? + 2Lac?
5 × 10?³ 5 × 10?³ 10?² M
Salt of strong base weak acid salt
pH = 7 + 1/2 pka + 1/2 log c
= 7 + 1/2 × 5 + 1/2 log 10?²
= 7 + 2.5 - 1 = 8.5
= 85 × 10?¹

56. The total pressure observed by mixing two liquids A and B is 350 mmHg when their mole fractions are 0.7 and 0.3 respectively. The total pressure become 410 mmHg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is mmHg. (Nearest integer)
Consider the liquids and solutions behave ideally.
Sol. 314
XaP_A? + X_BP_B? = P_S
0.7P_A? + 0.3P_B? = 350
0.2P_A? + 0.8P_B? = 410
∴ P_A? = 314 torr

57. The number of statement/s, which are correct with respect to the compression of carbon dioxide from point (a) in the Andrews isotherm from the following is
A. Carbon dioxide remains as a gas upto point (b)
B. Liquid carbon dioxide appears at point (c)
C. Liquid and gaseous carbon dioxide coexist between points (b) and (c)
D. As the volume decreases from (b) to (c), the amount of liquid decreases
Sol. 4
At
(a) → CO? exist as gas
(b) → liquefaction of CO? starts
(c) → liquefaction ends
(d) → CO? exist as liquid
Between (b) & (c) → liquid and gaseous CO? co-exist. As volume changes from (b) to (c) gas decreases and liquid increases.
(A), (C) → Correct

58. Maximum number of isomeric monochloro derivatives which can be obtained from 2, 2, 5, 5 tetramethylhexane by chlorination is
Sol. 3

59. Total number of tripeptides possible by mixing of valine and proline is
Sol. 8
(1) P-P-P
(2) V-V-V
(3) P-V-V
(4) V-P-V
(5) V-V-P
(6) V-P-P
(7) P-V-P
(8) P-P-V

60. One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is J (nearest integer)
Sol. 6
I→II → Isobaric
II→III → Isochoric
III→I → Isothermal
W_{I-II} = -1[40 - 20] = -20 Lit atm
W_{II-III} = 0
W_{III-I} = 2.303 nRT log(V?/V?)
= 2.303 PV log(V?/V?)
= 2.303(1 × 20) log 2
= 2.303 × 20 × 0.3010 = 13.818
W total = -20 + 13.818 = (-6.182 lit atm) = 6.182 lit atm

61. If, f(x) = x³ - x² f'(1) + x f''(2) - f'''(3), x ∈ R then
(1) f(1) + f(2) + f(3) = f(0)
(2) 2f(0) - f(1) + f(3) = f(2)
(3) 3f(1) + f(2) = f(3)
(4) f(3) - f(2) = f(1)
Sol. 2
f(x) = x³ - x² f'(1) + x f''(2) - f'''(3)
f(x) = x³ - ax² + bx - c
f'(x) = 3x² - 2ax + b
f''(x) = 6x - 2a
f'''(x) = 6
f'''(3) = 6
f'(1) = 3 - 2a + b = a ⇒ 3a = b + 3
f''(2) = 12 - 2a = b ⇒ 2a = 12 - b
a = 3, b = 6
f'''(3) = 6 = c
f(x) = x³ - 3x² + 6x - 6
f(0) = -6, f(2) = 2
f(1) = -2, f(3) = 12

62. If the system of equations
has infinitely many solutions, then the ordered pair (λ, μ) is equal to :
(1) (-72/5, 21/5)
(2) (-72/5, -21/5)
(3) (72/5, -21/5)
(4) (72/5, 21/5)
Sol. 3
Planes are not parallel
∴ (x + 2y + 3z - 3) + a(4x + 3y - 4z - 4) = 8x + 4y - λz - 9 - μ = 0
(1 + 4a)/8 = (2 + 3a)/4 = (3 - 4a)/(-λ) = (-3 - 4a)/(-9 - μ)
i) 1 + 4a = 4 + 6a ⇒ a = -3/2
(ii) (2 - 9/2)/4 = (3 + 6)/(-λ)
36/2 = 5/λ ⇒ 72/5
(iii) 5/3 - 4a = 8/(-9 - μ)
5/3 - 6 = 8/(-9 - μ)
-9 - μ = -24/5
μ = -45 + 24/5 = -21/5

63. If, then f(x) = 2^{2x}/(2^{2x} + 2), x ∈ R, then f(1/2023) + f(2/2023) + ... + f(2022/2023) is equal to
(1) 1011
(2) 2010
(3) 1010
(4) 2011
Sol. 1
f(x) = 4^x/(4^x + 2)
f(1 - x) = 4/(4^x)/(4/4^x + 2) = 4/(4 + 2 · 4^x) = 2/(2 + 4^x)
f(x) + f(1 - x) = 1

64. Let α = 4i + 3j + 5k and b = i + 2j - 4k. Let b? be parallel to α and b? be perpendicular to α. If b = b? + b?, then the value of 5b? · (i + j + k) is
(1) 7
(2) 9
(3) 6
(4) 11
Sol. 1
b? = (α · b)/|α| α
= ((4 + 6 - 20)/√50)(4, 3, 5)/√50
= -10/50 (4, 3, 5)
b? = (-4, -3, -5)/5
b? + b? = (1, 2, -4)
b? = (1 + 4/5, 2 + 3/5, -4 + 1)
b? = (9/5, 13/5, -3)
∴ 5b? = (9, 13, -15)
∴ 5b? · (1, 1, 1) = 9 + 13 - 15 = 7

65. Let y = y(x) be the solution of the differential equation (x² - 3y²)dx + 3xy dy = 0, y(1) = 1. Then 6y²(e) is equal to
(1) 2e²
(2) 3e²
(3) e²
(4) 3/2 e²
Sol. 1
(x² - 3y²)dx + 3xy dy = 0
dy/dx = (3y² - x²)/3xy
y = vx
dy/dx = v + x dv/dx
v + x dv/dx = (3v²x² - x²)/3vx²
v + x dv/dx = (3v² - 1)/3v
x dv/dx = (3v² - 1)/3v - v ⇒ -1/3v
3v dv = -dx/x
3v²/2 = -ln x + C [y(1) = 1]
3y²/2x² = -ln x + C
C = 3/2
3y²/2x² = 3/2 ln e = ln x
∴ 3y² = 3x² ln e - 2x² ln x
x = e, 3y² = 3e² ln e - 2e² ln e
= e² ln e = e²
6y² = 2e²

66. The locus of the mid points of the chords of the circle C? : (x - 4)² + (y - 5)² = 4 which subtend an angle θ? at the centre of the circle C?, is a circle of radius r?. If θ? = π/3, θ? = 2π/3 and r?² = r?² + r?², then θ? is equal to
(1) π/4
(2) π/2
(3) π/6
(4) 3π/4
Sol. 2

67. The number of real solutions of the equation 3(x² + 1/x²) - 2(x + 1/x) + 5 = 0, is
(1) 0
(2) 3
(3) 4
(4) 2
Sol. 1

68. Let A be a 3 × 3 matrix such that |adj(adj(adj A))| = 12?. Then |A?¹ adj A| is equal to
(1) √6
(2) 2√3
(3) 12
(4) 1
Sol. 2
adj(adj(adj A)) = |A|^((n-1)³) = 12?
|A|? = (12)?
|A| = (12)^(1/2)
∴ |A?¹ · adj(A)| = |A?¹| × |adj A|
= 1/|A| × |A|^(n-1)
= 1/|A| × |A|² = |A| = √12 = 2√3

69. ∫_{3√2/4}^{3√3/4} 48/√(9 - 4x²) dx is equal to
(1) 2π
(2) π/6
(3) π/3
(4) π/2
Sol. 1
48 ∫_{3√2/4}^{3√3/4} dx/√(9 - 4x²) = 48/2 sin?¹(2x/3)|_{3√2/4}^{3√3/4}
= 24[sin?¹(√3/2) - sin?¹(1/√2)]
= 24[π/3 - π/4] = 2π

70. The number of square matrices of order 5 with entries form the set {0,1}, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is
(1) 125
(2) 225
(3) 150
(4) 120
Sol. 4
?C? × ?C? × ³C? × ²C? × ¹C? = 120

71. If (³?C?)² + 2(³?C?)² + 3(³?C?)² + ... + 30(³?C??)² = α 60!/(30!)² then α is equal to :
(1) 30
(2) 10
(3) 60
(4) 15
Sol. 4
C?² + 2C?² + 3C?² + ... + 30C??²
S = 0 C?² + 1 C?² + ... + 30 C??²
S = 30 C??² + 29 C??² + ... + 0 C?²
S = 30 C?² + C?² + ... + C??²
S = 15 × C??² = α · 60!/(30!)² ⇒ α = 15

72. Let the plane containing the line of intersection of the planes P1 : x + (λ + 4)y + z = 1 and P2 : 2x + y + z = 2 pass through the points (0,1,0) and (1,0,1). Then the distance of the point (2λ, λ, -λ) from the plane P2 is
(1) 4√6
(2) 3√6
(3) 5√6
(4) 2√6
Sol. 2
[x + (λ + 4)y + z - 1] + μ[2x + y + z - 2] = 0
(0,1,0)
(i) (λ + 4 - 1) + μ[-1] = 0
λ - μ = -3
(1,0,1) (ii) 1 + μ[1] = 0 ⇒ μ = -1, λ = -4
∴ point (-8, -4, 4); 2x + y + z - 2 = 0
d = |-16 - 4 + 4 - 2|/√6 = 18/√6 = 3√6

73. Let f(x) be a function such that f(x + y) = f(x) · f(y) for all x, y ∈ N. If f(1) = 3 and Σ_{k=1}^n f(k) = 3279 then the value of n is
(1) 9
(2) 6
(3) 8
(4) 7
Sol. 4
f(x + y) = f(x) · f(y), x, y ∈ N
f(2) = 3²
f(3) = 3³ ∴ 3[3? - 1]/2 = 3279
3? - 1 = 1093 × 2
3? - 1 = 2186
3? = 2187
n = 7

74. Let the six numbers a?, a?, a?, a?, a?, a?, be in A.P. and a? + a? = 10. If the mean of these six numbers is 19/2 and their variance is σ², then 8σ² is equal to :
(1) 210
(2) 220
(3) 200
(4) 105
Sol. 1
a + (a + 2d) = 10 ⇒ a + d = 5 ...(1)
Mean ⇒ (6/2[2a + 5d])/6 = 19/2
2a + 5d = 19 ...(2)
from (1) and (2)
3d = 9 ⇒ d = 3; a = 2
∴ σ² = Σx?²/n - (x?)²
= (2² + 5² + 8² + 11² + 14² + 17²)/6 - (19/2)²
= 699/6 - 361/4
= 233/2 - 361/4
8σ² = 932 - 722 = 210

75. The equations of the sides AB and AC of a triangle ABC are (λ + 1)x + λy = 4 and λx + (1 - λ)y + λ = 0 respectively. Its vertex A is on the y-axis and its orthocentre is (1,2). The length of the tangent from the point C to the part of the parabola y² = 6x in the first quadrant is :
(1) 4
(2) 2
(3) √6
(4) 2√2
Sol. 4
(λ + 1)(1 - λ)x + λ(1 - λ)y = 4(1 - λ)
λ²x + λ(1 - λ)y = -λ²
- - +
(1 - λ² - λ²)x = 4 - 4λ + λ²
(1 - 2λ²)x = 4 - 4λ + λ²
x = 0 ⇒ λ = 2
AB : 3x + 2y = 4
AC : 2x - y + 2 = 0
A(0, 2)
CH ⊥ AB
((b - 2)/(a - 1)) × (-3/2) = -1
3b - 6 = 2a - 2
Also 2a - b + 2 = 0
3b - 2a = 4
b = 2a + 2
6a + 6 = 2a + 4
c(-1/2, 1)
4a = -2
a = -1/2, b = 1
∴ y² = 6x
ty = x + 3/2 t²
t = -1/2 + 3/2 t²
3t² - 1 = 2t
3t² - 2t - 1 = 0
3t² - 3t + t - 1 = 0
(3t + 1)(t - 1) = 0
t = 1
P(3/2, 3)
∴ d = √((3/2 + 1/2)² + (3 - 1)²)
= √(4 + 4) = 2√2

76. Let p and q be two statements. Then ~(p ∧ (p ⇒ ~q)) is equivalent to
(1) p ∨ (p ∧ q)
(2) p ∨ (p ∧ (~q))
(3) (~p) ∨ q
(4) p ∨ ((~p) ∧ q)
Sol. 3

77. The set of all values of a for which lim_{x→a} ([x - 5] - [2x + 2]) = 0, where [α] denotes the greatest integer less than or equal to α is equal to
(1) [-7.5, -6.5)
(2) [-7.5, -6.5]
(3) (-7.5, -6.5]
(4) (-7.5, -6.5)
Sol. 4
lim_{x→a} ([x] - 5 - [2x] - 2) = 0
lim_{x→a} ([x] - [2x]) = 7
[a] - [2a] = 7
If a = -7.5, [-7.5] = -8
[-15] = -15 ∴ -8 + 15 = 7
If a = -6.5, [-6.5] = -7, -7 + 13 = 6
[-13] = -13
∴ a ∈ (-7.5, -6.5)

78. If the foot of the perpendicular drawn from (1,9,7) to the line passing through the point (3,2,1) and parallel to the planes x + 2y + z = 0 and 3y - z = 3 is (α, β, γ), then α + β + γ is equal to
(1) 3
(2) 1
(3) -1
(4) 5
Sol. 4
n? = (1, 2, 1) n? = (0, 3, -1)
P = |i j k; 1 2 1; 0 3 -1| = i(-5) - j(-1) + k(3)
= (-5, 1, 3)
∴ line : (x - 3)/(-5) = (y - 2)/1 = (z - 1)/3 = λ
(2 - 5λ), -5 + (λ - 7) · 1 + (3λ - 6) · 3 = 0
25λ - 10 + λ - 7 + 9λ - 18 = 0
35λ = 35
λ = 1
∴ Point is (-2, 3, 4) α + β + γ = 5

79. The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition, is
(1) 168
(2) 220
(3) 120
(4) 48
Sol. 1
C-1 2 × 4 × 3 × 2 = 48
C-2 5! (5 digit nos) = 120
168

80. The value of ((1 + sin(2π/9) + i cos(2π/9))/(1 + sin(2π/9) - i cos(2π/9)))³ is
(1) -1/2(√3 - i)
(2) -1/2(1 - i√3)
(3) 1/2(1 - i√3)
(4) 1/2(√3 + i)
Sol. 1
π/2 - 2π/9
= (9π - 4π)/18 = 5π/18
⇒ (1 + cos(5π/18) + i sin(5π/18))/(1 + cos(5π/18) - i sin(5π/18))
= (2cos²(5π/36) + 2i sin(5π/36)cos(5π/36))/(2cos²(5π/36) - 2i sin(5π/36)cos(5π/36)) ⇒ (e^(i5π/36)/e^(-i5π/36))³
= e^(i5π/12) = e^(i5π/6)
(cos(5π/6) + i sin(5π/6)) = -√3/2 + i/2

SECTION B

81. If the shortest distance between the lines (x+√6)/2 = (y-√6)/3 = (z-√6)/4 and (x-λ)/3 = (y-2√6)/4 = (z+2√6)/5 is 6, then the square of sum of all possible values of λ is
Sol. 384
P(-√6, √6, √6) Q(λ, 2√6, -2√6)
n? = (2, 3, 4) n? = (3, 4, 5)
n? × n? = |i j k; 2 3 4; 3 4 5| = i(-1) - j(-2) + k(-1) = (-1, 2, -1)
∴ S_d = |PQ · (-1, 2, -1)|/√6 = |(λ + √6, √6, -3√6) · (-1, 2, -1)|/√6
= |-λ - √6 + 2√6 + 3√6|/√6 = 6
⇒ |-λ + 4√6| = 6√6
(+) -λ + 4√6 = 6√6 ⇒ λ = -2√6
(-) λ - 4√6 = 6√6 ⇒ λ = 10√6
∴ (8√6)² = 384

82. Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and λ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y² = λx with one vertex at the vertex of the parabola, is
Sol. 432
A: 4R, 6B
B: 5R, 5B
C: λR, 4B
P(Red from C) = (1/3 × λ/(λ+4))/(1/3 · λ/(λ+4) + 1/3 · 4/10 + 1/3 · 5/10)
= λ/(λ+4) / (λ/(λ+4) + 9/10)
10/4 = 10/10
100/40 = 36/144
24/144 = 6
m = 2/t = 1/√3
t = 2√3
P(12a, 4√3a)
(Side)² = 144a² + 48a² = 192 × 9/4 = 432

83. Let S = {θ ∈ [0,2π) : tan(π cos θ) + tan(π sin θ) = 0}. Then Σ_{θ∈S} sin²(θ + π/4) is equal to
Sol. 2
tan(π cos θ) = tan[-π sin θ]
π cos θ = nπ - π sin θ (n ∈ I)
cos θ + sin θ = n
n ∈ [-√2, √2] n ∈ {-1, 0, 1}
cos(θ - π/4) = 1/√2, cos(θ - π/4) = 0, cos(θ - π/4) = -1/√2
θ - π/4 = 2mπ ± π/4, θ - π/4 = 2mπ + π/2, θ - π/4 = 2mπ ± 3π/4
θ = 2mπ + π/2, θ = 2mπ + 3π/4, θ = 2mπ + π
θ = 2mπ, θ = 2mπ - π/2
θ = {π/2, 0, π/4, π, 3π/2, 3π/4, 7π/4}

84. If (1³ + 2³ + 3³ + ... + up to n terms)/(1·3 + 2·5 + 3·7 + ... + up to n terms) = 9/5, then the value of n is
Sol. 5
(n(n+1)/2)² / Σr(2r+1)
⇒ (n²(n+1)²/4) / (2n(n+1)(2n+1)/6 + n(n+1)/2)
⇒ (n(n+1)/4) / ((2n+1)/3 + 1/2) ⇒ (n(n+1)/4) / ((4n+5)/6) = 9/5
⇒ 3(n+1)n / (2(4n+5)) = 9/5
⇒ 5n² + 5n = 24n + 30
⇒ 5n² - 19n - 30 = 0
5n² - 25n + 6n - 30 = 0
(5n + 6)(n - 5) = 0
n = 5

85. Let the sum of the coefficients of the first three terms in the expansion of (x - 3/x²)?, x ≠ 0, n ∈ N, be 376. Then the coefficient of x? is
Sol. 405

86. The equations of the sides AB, BC and CA of a triangle ABC are : 2x + y = 0, x + py = 21a, (a ≠ 0) and x - y = 3 respectively. Let P(2, a) be the centroid of ΔABC. Then (BC)² is equal to
Sol. 122
A(1, -2)
G(2, a)
B(α, -2α) C(β+3, β)
(α + β + 4)/3 = 2, (-2α - 2 + β)/3 = a
α + β = 2, -2α + β = 3a + 2
-2α + 2 - α = 3a + 2
α = -a
put 'B' in BC
α - 2pα = 21a
α(1 - 2p) = 21a
2p - 1 = 21
p = 11
put 'C' in BC
β + 3 + 11β = 21a
21α + 12β + 3 = 0
also β = 2 - α
Solving α = -3, β = 5
∴ BC = √122
BC² = 122

87. Let a = i + 2j + λk, b = 3i - 5j - λk, a · c = 7, 2b · c + 43 = 0, a × c = b × c. Then |a · b| is equal to
Sol. 8
a = (1, 2, λ) b = (3, -5, -λ)
a · c = 7
b · c = -43/2
(a - b) × c = 0
c = x[-2, 7, 2λ] = (-2x, 7x, 2λx)
now, a · c = -2x + 14x + 2λ²x = 7
2λ²x + 12x = 7 ...(1)
b · c = -6x - 35x - 2λ²x = -43/2
-41x - 2λ²x = -43/2 ...(2)
by adding (1) + (2)
-29x = -29/2 ⇒ x = 1/2
∴ λ² + 6 = 7 ⇒ λ² = 1
|a · b| = |3 - 10 - λ²| = |-8| = 8

88. The minimum number of elements that must be added to the relation R = {(a, b), (b, c), (b, d)} on the set {a, b, c, d} so that it is an equivalence relation, is
Sol. 13
a b c d
a 1 ? 8 9
b 5 2 ? ?
c 10 6 3 11
d 12 7 13 4
1, 2, 3, 4 → for reflexive
5, 6, 7 → for symmetric
8, 9, 10, 11, 12, 13 → for transitive

89. If the area of the region bounded by the curves y² - 2y = -x, x + y = 0 is A, then 8A is equal to
Sol. 36
y² = y + 2
y² - y - 2 = 0
(y - 2)(y + 1) = 0
y = 2, y = -1
Area = ∫_{-1}^2 (-y²) - (-2 - y) dy
= (-y³/3 + 2y + y²/2)|_{-1}^2
= (-8/3 + 4 + 2) - (1/3 - 2 + 1/2)
= -8/3 + 6 - 1/3 + 2 - 1/2
A = -3 + 8 - 1/2 ⇒ 9/2
∴ 8A = 36

90. Let f be a differentiable function defined on [0, π/2] such that f(x) > 0 and
f(x) + ∫_0^x f(t)√(1 - (log_e f(t))²) dt = e, ∀ x ∈ [0, π/2]. Then (6 log_e f(π/6))² is equal to
Sol. 27
f'(x) + f(x) · √(1 - log² f(x)) = 0
dy/dx = -y√(1 - log² y)
f(0) = e
dy/(y√(1 - log² y)) = -dx
log y = t
(1/y) dy = dt
dt/√(1 - t²) = -dx
Sin?¹(t) = -x + C
Sin?¹[log y] = -x + C
x = 0 Sin?¹(1) = C ⇒ π/2
log(y) = Sin(π/2 - x)
x = π/6 log_e[f(π/6)] = Sin(π/3) = √3/2
∴ (6 × √3/2)² = 27

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