Category: Engineering Entrance Exam

PHYSICS

SECTION A

Q. 1: An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle ‘α’ with the plates. It leaves the plates at angle ‘β’ with kinetic energy K₂. Then the ratio of kinetic energies K₁ : K₂ will be:

(A) sin²β/cos²α

(B) cos²β/cos²α

(C) cosβ/sinα

(D) cosβ/cosα

Answer: (B)

∵v₁cosα = v₂cosβ [electric field inside a parallel plate capacitor is perpendicular to the plates, hence, there will be no change in parallel component of velocity]
v₁/v₂ = cosβ/cosα
Then the ratio of kinetic energies
k₁/k₂ = ½ mv₁²/ ½ mv₂² = (v₁/v₂)² = (cosβ/cosα)²
k₁/k₂ = cos²β/cos²α

 

Q. 2: Two identical springs of spring constant ‘2K’ are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. Then, time period of oscillations of this system is:

(A) π√(m/k)

(B) π√(m/2k)

(C) 2π√(m/k)

(D) 2π√(m/2k)

Answer: (A)

Springs are in parallel combination.
Hence, K_eff = 2k + 2k = 4k
∵ T = 2π √(m/k_eff)
 = 2π√(m/4k)
T  = π √(m/k)

 

Q. 3: The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is:

(A) 194.8 nm

(B) 490.7 nm

(C) 913.3 nm

(D) 121.8 nm

Answer: (D)
∆E = 10.2 eV is the energy difference between the state n = 2 & n = 1
∆E = –3.4 – (–13.6) = 10.2 eV
hc/λ = 10.2 ev
λ = hc/(10.2)e (in meters) where ‘e’ = 1.6 × 10^–19 J/V
= 12400/10.2 Å (because hc = 12400 eV nm)
= 121.56 nm
≃ 121.8 nm

 

Q. 4: In a ferromagnetic material below the Curie temperature, a domain is defined as:

(A) a macroscopic region with consecutive magnetic dipoles oriented in opposite directions.

(B) a macroscopic region with zero magnetization.

(C) a macroscopic region with saturation magnetization.

(D) a macroscopic region with randomly oriented magnetic dipoles.

Answer: (C)
In a ferromagnetic material below the Curie temperature, the domain is defined as a macroscopic region with saturation magnetization.

 

Q. 5: The point A moves with a uniform speed along the circumference of a circle of radius 0.36m and covers 30^o in 0.1s. The perpendicular projection ‘P’ form ‘A’ on the diameter MN represents the simple harmonic motion of ‘P’. The restoring force per unit mass when P touches M will be:

(A) 100 N

(B) 50 N

(C) 9.87 N

(D) 0.49 N

Answer: (C)

The point covers 30^o in 0.1 sec.
Means
π/6 → 0.1sec.
1 → 0.1/(π/6)
2π = → 0.1×6×2π/π
T = 1.2 sec.
We know that ω = 2π/T
ω = 2π/1.2
Restoring force (F) = mω²A
Then, Restoring force per unit mass (F/m) = ω²A
F/m = (2π/1.2)²×0.36
≃ 9.87 N

 

Q. 6: Match List I with List II.

List I	List II
(a) Rectifier	(i) Used either for stepping up or stepping down the A.C.voltage
(b) Stabilizer	(ii) Used to convert A.C. voltage into D.C. voltage
(c) Transformer	(iii) Used to remove any ripple in the rectified output voltage
(d) Filter	(iv) Used for constant output voltage even when the input voltage or load current change

Choose the correct answer form the options given below:

(A) (a)-(ii), (b)- (i), (c)-(iv), (d)-(iii)

(B) (a)-(ii), (b)- (iv), (c)-(i), (d)-(iii)

(C) (a)-(ii), (b)- (i), (c)-(iii), (d)-(iv)

(D) (a)-(iii), (b)- (iv), (c)-(i), (d)-(ii)

Answer: (B)
(a) Rectifier:- used to convert A.C voltage into D.C. Voltage.
(b) Stabilizer:- used for constant output voltage even when the input voltage or load current change
(c) Transformer:- used either for stepping up or stepping down the A.C. voltage.
(d) Filter:- used to remove any ripple in the rectified output voltage.

 

Q. 7: Y = A sin(ωt + ϕ) is the time – displacement equation of an SHM, At t = 0, the displacement of the particle is Y = A/2 and it is moving along negative x-direction. Then, the initial phase angle ϕ will be.

(A) π/6

(B) π/3

(C) 2π/3

(D) 5π/6

Answer: (D)

y = A sin (ωt + ϕ)
t = 0, x = A/2
1/2 = sin ϕ
ϕ = π/6, 5π/6
v = dy/dt = A ωcos(ωt + ϕ)
t = 0, v = A ωcosϕ
ϕ = π/6, for v (positive)
ϕ = 5π/6, for v (negative)
∴ϕ = 5π/6

 

Q. 8: A sphere of radius ‘a’ and mass ‘m’ rolls along horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upwar(D) Assuming that it rolls without slipping how far up the sphere will travel (along the incline)?

(A) (⅖) v₀²/g sin θ

(B) 10v₀²/7gsin θ

(C) v₀²/5gsin θ

(D) 7v₀²/10gsin θ

Answer: (D)

From energy conservation
mgh = ½ mv² + ½ Iω²
mgh = ½ mv² + ½×(⅖)ma²× v₀²/a²
gh = ½ v₀²+ ⅕ v₀²
gh = (7/10)v₀²
h = (7/10)v₀²/g
From triangle, sinθ = h/l
Then, h = lsinθ
l sinθ = (7/10) (v₀²/g)
l = (7/10) (v₀²/g sin θ)

 

Q. 9: Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter 0.1 μm. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that:

(A) its size decreases, but intensity increases

(B) its size increases, but intensity decreases

(C) its size increases, and intensity increases

(D) its size decreases, and intensity decreases

Answer: (A)
For diffraction through a single slit, for first minimum.
sin θ = 1.22λ/D
If D is increased, then sinθ will decrease i.e θ will decrease
∴ size of circular fringe will decrease but intensity increases.

 

Q. 10: An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength λ_electron/λ_Proton will be:

(A) 918

(B) 1836

(C) 1/1836

(D) 1

Answer: (B)
Given mass of electron = m_e
Mass of proton = m_p
∴ given m_p = 1836 m_e
From de-Broglie wavelength
λ = h/p = h/mv
λ_e/λ_p = m_p/m_e [v is same]
= 1836m_e/m_e
λ_e/λ_p= 1836

 

Q. 11: The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is:

(A) 400 nm

(B) 382 nm

(C) 309 nm

(D) 329 nm

Answer: (B)
From the photoelectric effect equation
hc/λ = ϕ + ev_s
where Vs is the stopping potential and ϕ is work function of the metal.
So, evs1 = hc/λ₁ – ϕ ……(i)
evs2 = hc/λ₂ – ϕ …..(ii)
Subtract equation (i) from equation (ii)
ev_s1 – ev_s2 = hc/λ₁– hc/λ₂
v_s1 – v_s2 = (hc/e) (1/λ₁ – 1/λ₂)
(0.710 – 1.43) = 1240 (1/491 – 1/λ2)
(because hc = 1240 eV nm)
-0.72/1240 = 1/491 – 1/λ2
1/λ2 = 1/491 + 0.72/1240
1/λ2 = 0.00203 + 0.00058
1/λ2 = 0.00261
λ2 = 383.14
λ2 ≃ 382 nm

 

Q. 12: The truth table for the following logic circuit is:

Answer: (D)

If A = B = 0 then output y = 1
If A = B = 1 then output y = 1

 

Q. 13: If e is the electronic charged, c is the speed of light in free space and h is planck’s constant, the quantity (1/4πε₀) |e|²/hc has dimensions of:

(A) [ LC^-1]

(B) [M⁰ L⁰ T⁰]

(C) [ M L T⁰]

(D) [M L T^-1]

Answer: (B)
Given
e = electronic charge
c = speed of light in free space
h = planck’s constant
(1/4πε₀) e²/hc = (ke²/hc)× λ²/λ² [multiply and divide by lambda = wavelength]
= F×λ/E [ke²/λ² has dimensions of force and hc/λ = E]
= E/E [F× λ has dimension of E]
= dimensionless
= [ M⁰ L⁰ T⁰]

 

Q. 14: A charge ‘q’ is placed at one corner of a cube as shown in figure. The flux of electrostatic field E^→ through the shaded area is:

(A) q/48ε₀

(B) q/8ε₀

(C) q/24ε₀

(D) q/4ε₀

Answer: (C)
Total flux through the cube = (q/ε₀)× ⅛ = q/8ε₀
Total flux through one “outer” face of the cube = (q/8ε₀)×1/3 = q/24ε₀
[Because there is flux only through 3 faces]
Hence, total flux through shaded area
ϕ_T = (q/24ε₀ + q/24ε₀)×1/2 [half of each face is shaded]
ϕ_T = q/24ε₀

 

Q. 15: Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2 = 2V1 then the ratio of temperature T2/T1 is:

(A) 1/√2

(B) 1/2

(C) 2

(D) √2

Answer: (D)

From P-V diagram,
Given PV^1/2 = constant …..(i)
We know that
PV = nRT
P ∝ (T/V) [for 1 mole]
Put in equation (i)
(T/V) (V)^1/2 = constant
T ∝V^1/2
T₂/T₁ = √(V₂/V₁)
T₂/T₁ = √(2V₁/V₁)
T₂/T₁ = √2

 

Q. 16: Given below are two statements:

Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell’s distribution.

Statement II: In a diatomic molecule, the rotational energy at a given temperature equals the transnational kinetic energy for each molecule.

In the light of the above statements, choose the correct answer from the options given below:

(A) Both statement I and statement II are false.

(B) Both statement I and statement II are true.

(C) Statement I is false but statement II is true.

(D) Statement I is true but statement II is false.

Answer: (D)
The translational kinetic energy & rotational kinetic energy both obey Maxwell’s distribution independent of each other.
T.K.E. of diatomic molecules = (3/2) kT [3 translational D.O.F.]
R.K.E. of diatomic molecules = (2/2) kT = [2 rotational D.O.F.]
So statement I is true but statement II is false.

 

Q. 17: An LCR circuit contains resistance of 110 Ω and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 450. If on the other hand, only inductor is removed the current leads by 450 with the applied voltage. The rms current flowing in the circuit will be:

(A) 2.5 A

(B) 2 A

(C) 1 A

(D) 1.5 A

Answer: (B)
When L and C are connected with R in series the circuit will come in resonance So, current in the circuit will be:
I_rms = V_rms/R
= 220/110
= 2 A

 

Q. 18: For extrinsic semiconductors: when doping level is increased;

(A) Fermi–level of p and n-type semiconductors will not be affected.

(B) Fermi–level of p-type semiconductors will go downward and Fermi–level of n-type semiconductor will go upward.

(C) Fermi–level of both p–type and n–type semiconductors will go upward for T >TFK and downward for T<TFK, where TF is Fermi temperature.

(D) Fermi–level of p-type semiconductor will go upward and Fermi–level of n–type semiconductors will go downward.

Answer: (B)
In n-type semiconductor, pentavalent impurity is added Each pentavalent impurity donates a free electron. So the Fermi-level of n-type semiconductor will go upward and in p-type semiconductor, trivalent impurity is added. Each trivalent impurity creates a hole in the valence band So the Fermi-level of p-type semiconductor will go downward.

 

Q. 19: A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top, both stones reach the bottom of the building simultaneously. The height of the building is: [Take g = 10 m/s²]

(A) 45 m

(B) 35 m

(C) 25 m

(D) 50 m

Answer: (A)

Velocity of particle (1) at 5m below top u₁ = √2gh = √(2×10×5) =10 m/s
For particle (1), using 2^nd equation of motion
20+h = 10t + ½ gt² …..(i)
For particle (2), using 2^nd equation of motion
h = ½ gt² …..(ii)
Put equation (ii) in equation (i)
20 + ½ gt² = 10t + ½ gt²
t = 2 sec.
Put in equation (ii)
h = ½ gt²
= ½ ×10 × 22
h = 20m
The height of the building = 25 + 20 = 45m

 

Q. 20: If a message signal of frequency ‘fm’ is amplitude modulated with a carrier signal of frequency ‘fc’ and radiated through an antenna, the wavelength of the corresponding signal in air is:

[Given, C is the speed of electromagnetic waves in vacuum/air]

(A) c/(fc+fm)

(B) c/(fc-fm)

(C) c/fm

(D) c/fc

Answer: (A)
Equation of amplitude modulated wave y = (Ac + Am sin ω_mt) sin ω_ct
Here angular frequency of modulated signal = ω_c
Thus frequency of modulated signal = fc
Thus wavelength = c/fc

Section B

Q. 1: The initial velocity v1 required to project a body vertically upward from the surface of the earth to just reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that vi = √(x/y) v_e. The value of x will be_______.

Answer: 20

Here R = radius of the earth
From energy conservation
-Gm_em/R + ½ mv_i² = -Gm_em/11R + 0
½ mv_i² = (10/11)Gm_em/R
v_i = √(20Gm_e/11R)
v_i = √(20/11) v_e {∵escape  velocity  v_e = √(Gm_e/R)}
Then the value of x = 20

 

Q. 2: The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4% will be ___%.

Answer: 2
Speed of transverse wave is
V = √(T/μ)
Taking log on both sides
ln v = ½ lnT – ½ ln μ
Δv/v = ½ ΔT/T ⇒ (Δv/v)×100 = 1/2 ( ΔT/T)×100
= ½ ×4 [ μ is constant for a string]
(Δv/v)×100 = 2%

 

Q. 3: The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is (x/10) √(µ₀c/π) V/m. The efficiency of the bulb is 10% and it is a point source. The value of x is ___.

Answer: 2
Intensity I = ½ c∈₀E₀²
Intensity = Power / Area = 8 /(4π × 10²)
(8/4π×10²) ×½ =¼ × c ×(1/µ₀c²) ×E₀² [Multiply by ½ and ∈₀ =1/μ₀c²]
E₀ = (2/10)√(µ₀c/π)
⇒ x = 2

 

Q. 4: Two identical conducting spheres with negligible volume have 2.1nC and -0.1nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is _______× 10^-9 N. [Given : 4πε₀ = 1/9×10⁹ SI unit]

Answer: 36

When they are brought into contact & then separated by a distance = 0.5 m. Then charge distribution will be

The electrostatic force acting b/w the sphere is
Fe = kq₁q₂/r²
= 9×10⁹×1×10^-9×1×10^-9/(0.5)²
= (900/25) × 10^-9
Fe = 36 ×10^-9 N

 

Q. 5: A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 21Ω each and leaves by the corner R. The currents i1 in ampere is______.

Answer: 2

The current i₁ = (R₂/(R₁ + R₂))i
= (2/(4+2))×6
i₁ = 2A

 

Q. 6: The wavelength of an X-ray beam is 10 Å. The mass of a fictitious particle having the same energy as that of the X-ray photons is (x/3) h kg. The value of x is _______.

Answer: 10
Given wavelength of an X-ray beam = 10 Å
∵E = hc/λ = mc²
m = hc/λ will be the mass of a particle having the same energy
The mass of a fictitious particle having the same energy as that of the X-ray photons = (x/3)h kg
(x/3) h = h/cλ
x = 3/cλ
= 3/3×10⁸×10×10^-10
x = 10

 

Q. 7: A reversible heat engine converts one- fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled The temperature in Kelvin of the source will be_______.

Answer: 208
∵ n = W/Qin = 1/4
¼ = 1 – T₂/T₁ [for reversible heat engine]
T₂/T₁ = 3/4
When the temperature of the sink is reduced by 52k then its efficiency is doubled.
1/2 = 1 -(T₂ – 52)/T₁
(T₂– 52)/T₁ = 1/2
T₂/T₁ – 52/T₁ = 1/2
¾ – 52/T₁ = 1/2
52/T₁ = 1/4
T₁ = 208 K is the source temperature.

 

Q. 8: If P^￫ × Q^￫ = Q^￫ × P^￫ , the angle between P^￫  and Q^￫  is θ(0^o<θ< 360^o). The value of ‘θ’ will be ____.

Answer: 180

 

Q. 9: Two particles having masses 4g and 16g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n:2. The value of n will be_____.

Answer: 1
Relation b/w kinetic energy & momentum is
p = √(2mKE) (∵KE = same)
p₁/p₂ = √(m₁/m₂)
n/2 = √(4/16)
n = 1

 

Q. 10 : Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. Then charge on each of the sphere is (a/21)×10^-8C. The value of ‘a’ will be________. [Take g = 10 m/s²]

Answer: 20

From FBD, T sinθ = kq²/d² and T cosθ = mg
Dividing them,
mg tanθ = kq²/d²
q = √(mg tanθ)/25k
= √(10×10^-6×10×1/√24(25)9×10⁹)
= √(10^-4×4/√24×25×4×9×10⁹)
= (⅔)√(10^-4/√24×10¹¹)
Thus (a/21)×10^-8 = ⅔ √(10^-15/√24)
= ⅔ √(10^-16/0.49)
a = 2×21/3×0.7
= 20

CHEMISTRY

SECTION A

Q. 1. What is ‘X’ in the given reaction?

Answer: (A)

Q. 2. The major product of the following reaction:

Answer: (D)

Q. 3. Given below are two statements:

Statement I: The identification of Ni²⁺ is carried out by dimethylglyoxime in the presence of NH₄OH

Statement II: The dimethylglyoxime is a bidentate neutral ligand.

In the light of the above statements, choose the correct answer from the options given below:

(A) Both statement I and statement II are true

(B) Both statement I and statement II are false

(C) Statement I is false but statement II is true

(D) Statement I is true but statement II is false

Answer: (D)

Dimethylglyoxime is a negative bidentate legend.

 

Q. 4. Carbylamine test is used to detect the presence of a primary amino group in an organic compound Which of the following compounds is formed when this test is performed with aniline?

Answer: (B)

 

Q. 5. The correct order of bond dissociation enthalpy of halogen is:

(A) F₂>Cl₂>Br₂>I₂

(B) Cl₂>F₂>Br₂>I₂

(C) Cl₂>Br₂>F₂>I₂

(D) I₂>Br₂>Cl₂>F₂

Answer: (C)
Fact based
F₂ has F — F, F₂ involves repulsion of non-bonding electrons & more over its size is small and hence due to high repulsion its bond dissociation energy is very low.

Q. 6. Which one of the following statements is FALSE for hydrophilic sols?

(A) These sols are reversible in nature

(B) The sols cannot be easily coagulated

(C) They do not require electrolytes for stability

(D) Their viscosity is of the order of that of H2O

Answer: (D)
Fact based

Q. 7. Water does not produce CO on reacting with

(A) C₃H₈

(B) C

(C) CH₄

(D) CO₂

Answer: (D)
H₂O + CO₂ H₂CO₃

 

Q. 8: The correct sequence of reagents used in the preparation of 4-bromo-2-nitroethyl benzene from benzene is:

(A) CH₃COCl/AlCl₃, Br₂/AlBr₃, HNO₃/H₂SO₄, Zn/HCl

(B) CH₃COCl/AlCl₃, Zn-Hg/HCl, Br₂/AlBr₃, HNO₃/H₂SO₄

(C) Br₂/AlBr₃, CH₃COCl/AlCl₃, HNO₃/H₂SO₄, Zn/HCl

(D) HNO₃/H₂SO₄, Br₂/AlCl₃, CH₃COCl/AlCl₃, Zn-Hg/HCl

Answer: (B)

 

Q. 9. Given below are two statements:

Statement I: α and β forms of sulphur can change reversibly between themselves with slow heating or slow cooling.

Statement II: At room temperature the stable crystalline form of sulphur is monoclinic sulphur.

In the light of the above statements, choose the correct answer from the options given below.

(A) Both statement I and statement II are false

(B) Statement I is true but statement II is false

(C) Both statement I and statement II are true

(D) Statement I is false but statement II is true

Answer: (B)

 

Q. 10. Correct statement about the given chemical reaction is:

(A) Reaction is possible and compound (A) will be a major product.

(B) The reaction will form a sulphonated product instead of nitration.

(C) NH₂ group is ortho and para directive, so product (B) is not possible.

(D) Reaction is possible and compound (B) will be the major product.

Answer: (A)

Q. 11. Which of the following compounds is added to the sodium extract before addition of silver nitrate for testing of halogens?

(A) Nitric acid

(B) Sodium hydroxide

(C) Hydrochloric acid

(D) Ammonia

Answer: (A)
NaCN + HNO₃ → NaNO₃ + HCN ↑
Na₂S + HNO₃ → NaNO₃ + H₂S ↑
Nitric acid decomposes NaCN & Na₂S, else they precipitate in test & misguide the resolve.

 

Q. 12: Given below are two statements:

Statement I: The pH of rain water is normally ~5.6.

Statement II: If the pH of rain water drops below 5.6, it is called acid rain.

In the light of the above statements, choose the correct answer from the option given below.

(A) Statement I is false but Statement II is true

(B) Both statement I and statement II are true

(C) Both statement I and statement II are false

(D) Statement I is true but statement II is false

Answer: (B)
Both statements are correct

 

Q. 13. The solubility of Ca(OH)₂ in water is:

[Given: The solubility product of Ca(OH)₂ in water = 5.5×10^-6]

(A) 1.11 ×10^-6

(B) 1.77 ×10^-6

(C) 1.77×10^-2

(D) 1.11×10^-2

Answer: (D)
Ca(OH)₂ ⇌ Ca⁺² + 2OH^–
s (2s + 10^-7)
s(2s+10^-7)² = 55×10^-7
4s³ = 55×10^-7
s³ = 5500 / 4 × 10^-9

s = (2250/2) ^1/3× 10^-3

s = (1125)⅓ × 10^-3
s = 1.11 × 10^-2

 

Q. 14. The major components of German Silver are:

(A) Cu, Zn and Mg

(B) Ge, Cu and Ag

(C) Zn, Ni and Ag

(D) Cu, Zn and Ni

Answer: (D)
Fact
German silver is alloy which does not have silver.
Cu-50%; Ni-30%; Zn-20%

 

Q. 15. The method used for the purification of Indium is:

(A) Van Arkel method

(B) Vapour phase refining

(C) Zone refining

(D) Liquation

Answer: (C)
Fact
Ga, In, Si, Ge are refined by zone refining or vacuum refining.

 

Q. 16. Which of the following is correct structure of α-anomer of maltose

Answer: (D)

Q. 17. The major product of the following reaction is:

(A) CH₃CH₂CH₂CHO

(B) CH₃CH₂CH=CH–CHO

(C) CH₃CH₂CH₂CH₂CHO

(D)

Answer: (C)

 

Q. 18. The correct order of acid character of the following compounds is:

(A) II>III>IV>I

(B) III>II>I>IV

(C) IV>III>II>I

(D) I>II>III>IV

Answer: (A)
Acidity of carboxylic acid α-R>-H>-I
1 / α + R > + H > + I

Q. 19. Which among the following species has unequal bond lengths?

(A) XeF₄

(B) SiF₄

(C) BF_4-

(D) SF₄

Answer: (D)

 

Q. 20. If which of the following orders the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?

(i) [FeF₆]^3-

(ii) [Co(NH₃)₆]³⁺

(iii) [NiCl₄]^2-

(iv) [Cu(NH₃)₄]²⁺

(A) (ii)>(i)>(iii)>(iv)

(B) (iii)>(iv)>(ii)>(i)

(C) (ii)>(iii)>(i)>(iv)

(D) (i)>(iii)>(iv)>(ii)

Answer: (D)
[FeF₆]^3-: Fe³⁺ 3d⁵ → 5-unpaired electrons as F- is weak field legend
[Co(NH₃)₆]³⁺: Co³⁺ 3d⁶→ No-unpaired electron as NH₃ is strong field light and causes pairing
[NiCl₄]^2-: Ni²⁺ 3d⁸ → 2-unpaired electrons
[Cu(NH₃)₄]²⁺: Cu²⁺ 3d⁹ → 1-unpaired electrons

 

SECTION B

Q. 1. Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.

(i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL

If the volume of oxalic acid taken was 10.0 ml. then the molarity of the NaOH solution is ____M. (Rounded-off to the nearest integer)

Answer: 6
Eq. of NaOH = Eq. of oxalic acid
[NaOH] × 1 × 4.4 = 5/4 × 2 ×10
[NaOH] = 100 / 4 × 4.4 = 25 / 4.4 = 5.68
Nearest integer = 6 M

 

Q. 2. Electromagnetic radiation of wavelength 663 nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol—1 is____. (Rounded off to the nearest integer)

[h=6.63×10^-34Js, c = 3.00×10⁸ms^-1, NA=6.02×10²³ mol^-1]

Answer: 180
Energy required to ionize an atom of metal ‘A’ = hc / λ = hc / 663 nm for 1 mole atoms of ‘A’
Total energy required = N_A × hc / λ
= 

[6.023 × 10²³  × 6.63 × 10^-34  × 3  × 10⁸]/[663 × 10^-9]

= 6.023 × 3 × 10^23-34+8+7
= 18.04 × 10⁴ J/mol
= 180.4 kJ/mol
Nearest Integer = 180 kJ/Mol.

 

Q. 3. Copper reduces into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed [Cu²⁺] and PNO=PNO₂), the HNO3 concentration at which the thermodynamic tendency for reduction of into NO and NO₂ by copper is same is 10x M.

The value of 2x is ______. (Rounded-off to the nearest integer)

[Given: 

E⁰(Cu²⁺ /Cu) = 0.34V, E⁰ (No₃^– / NO) = 0.96V, E⁰(No₃^– / NO₂ = 0.79V and at 298 K, 

RT/F(2.303)= 0.059

Answer: 4
Anode
Cu(s) → Cu⁺² + 2e^–
Cathode (1)



log (HNO₃) = 2.16
[HNO₃] = 102.16 = 10x
x = 2.16 ⇒ 2x = 4.32 ≈ 4

 

Q. 4. Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against a constant external 4.3 MP(A) The heat transferred in this process is ____kJ mol^-1. (Rounded-off of the nearest integer)

[Use R = 8.314 J mol^-1 K^-1]

Answer: 15
Moles (n) = 5
T = 293 K
Process = Iso T. → Irreversible
Pⁱⁿⁱ = 2.1 M Pa
P^t = 1.3 M Pa
P^ext = 4.3 mPa
Work = – P^ext Δv

=−4.3×(5×293R / 1.3 − 5×293/2.1)

=−5×293×8.314×43(1/13 −1/21)

=(5×293×8.314×43×8)/(21×13)

= -15347.7049 J
= – 15.34 kJ
Isothermal process, so ΔU = 0
w = – Q
Q = 15.34 kJ / mol
So, answer is 15.

 

Q. 5. Among the following, the number of metal/s which can be used as electrodes in the photoelectric cell is _____(Integer answer).

Answer: (A)
Cs is used in photoelectric cells due to its very low ionization potential.

 

Q. 6. The rate constant of a reaction increases by five times on increasing temperature from 27⁰ C to 52⁰C. The value of activation energy in kJ mol^-1 is ______.

(Rounded off to the nearest integer)

[R=8.314 J K^-1 mol^-1]

Answer: 52

Ea = 51524.96 J/mol
Ea = 51.524 kJ/mol
52 Ans.

 

Q. 7. If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ______molal.

Answer: 3
AB → A⁺ + B^–
1 – α α α
α = 3 / 4
N = 2
i = [1+(2-1)α]
2.5 = [1+(2-1)3/4] × 0.52 × m

m = 2.5/0.52 × 7/4 = 10/3.64 = 2.747
m= 2.747≈ 3 mol/kg

 

Q. 8. The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is ____BM.

Answer: 2
₂₉Cu⁺² → [Ar]¹⁸3d⁹

No. of unpaired e^– = 1
Magnetic moment = μ = √n(n + 2)
μ = √(1)(1 + 2) = √3B.M.
= 1.73 B.M

 

Q. 9. The number of compound/s given below which contain/s —COOH group is ______

(A) Sulphanilic acid

(B) Picric acid

(C) Aspirin

(D) Ascorbic acid

Answer: (A)

 

Q. 10. The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ______.

Answer: 9077
a = 3.596 Å

d = Z × GMM / N_A × a³

d = (Z4 × 63.54 × 10^-3)/ 6.022 × 10²³ ×  (3.956 × 10^-10)³

d = 0.9076 × 104 = 9076.2 kg/m3

MATHS 

SECTION A

Q. 1: If the curve x² + 2y² = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is:

(A) 𝜋/2 + tan^–1(1/4)

(B) 𝜋/2 – 𝑡𝑎𝑛^–1(1/4)

(C) 𝜋/2 + 𝑡𝑎𝑛^–1(1/3)

(D) 𝜋/2 – 𝑡𝑎𝑛^–1(1/3)

Answer: (A)

Using homogenization
x² + 2y² = 2(1) 2
⇒x² + 2y² = 2(x + y) ²
⇒x² + 2y² = 2x² + 2y² + 4xy
⇒x² + 4xy = 0
for ax² + 2hxy + by² = 0, obtuse angle between lines θ is
tan θ = ±(2√(h²–ab))/(a+b)
⇒tan θ = ±4
⇒tan θ = –4
cot θ = -1/4
θ = cot^–1(–1/4)
θ = π – cot^–1 (1/4)
θ = π – (π/2– tan^–1(1/4) )
θ = π/2 + tan-1(1/4)

 

Q. 2:

(A) loge|x² + 5x – 7| + c

(B) (1/4) loge |x² + 5x – 7| + c

(C) 4 loge|x² + 5x – 7| + c

(D) loge √(x² + 5x – 7) + c

Answer: (C)

 

Q. 3: A hyperbola passes through the foci of the ellipse x²/25 + y²/16 = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

(A) x²/9 – y²/4 = 1

(B) x²/9 – y²/16 = 1

(C) x² – y² = 9

(D) x²/9 – y²/25 = 1

Answer: (B)
For ellipse, e1 = √(1–16/25)=3/5
Foci = (±3,0)
Let equation of hyperbola be x²/A² – y²/B² = 1, passes through (±3, 0)
A² =9, A=3, e²=5/3
e₂² = 1 + B²/A²
25/9 = 1 + B²/9 ⇒B² = 16
x²/9 – y²/16 = 1

 

Q. 4: lim_n→∞[1/n + n/(n+1)² + n/(n+2)² + …….. + n/(2n-1)²] is equal to:

(A) 1

(B) 1/3

(C) 1/2

(D) 1/4

Answer: (C)

 

Q. 5: In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarians and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from chest disorder. The probability that the selected person is a smoker and non-vegetarian is:

(A) 7/45

(B) 8/45

(C) 14/45

(D) 28/45

Answer: (D)
Based on Bayes’ theorem
Probability = ((160/400×35/100))/((160/400×35/100)+(100/400×20/100)+(140/400×10/100) )
= 5600/9000
= 28/45

 

Q. 6: The following system of linear equations

2x + 3y + 2z = 9

3x + 2y + 2z = 9

x – y + 4z = 8

(A) does not have any solution

(B) has a unique solution

(C) has a solution (α, β, γ) satisfying α + β² + γ³ = 12

(D) has infinitely many solutions

Answer: (B)
Determinant of given system = -20 ≠ 0
It has a unique solution.

 

Q. 7: The minimum value of f(x) = a^a^x + a^(1-a^x) where a, x ∈ R and a > 0, is equal to:

(A) a + 1/a

(B) a + 1

(C) 2a

(D) 2√a

Answer: (D)
Using AM ≥ GM inequality, we get

 

Q. 8: A function f(x) is given by f(x) = 5^x/(5^x+5), then the sum of the series f(1/20) + f(2/20) + f(3/20) + …… + f(39/20) is equal to:

(A) 19/2

(B) 49/2

(C) 39/2

(D) 29/2

Answer: (C)
f(x) = 5^x/(5^x + 5)….(i)
⇒ f(2–x) = 5^2–x/(5^2–x +5)
⇒f(2–x) = 5/(5^x+5)……(ii)
Adding equation(i)and(ii)
f(x)+f(2–x)=1
f(1/20)+f(39/20)=1
f(2/20)+f(38/20)=1
……
…..
f(19/20) + f(21/20)=1
and f(20/20) = f(1) = 1/2
Therefore, f(1/20) + f(2/20) + f(3/20) + …… + f(39/20) = 19 + 1/2 = 39/2

 

Q. 9: A plane passes through the points A(1,2,3), B(2,3,1) and C(2,4,2). If O is the origin and P is (2,–1,1), then the projection of vector(OP) on this plane is of length:

(A) √(2/5)

(B) √(2/3)

(C) √(2/11)

(D) √(2/7)

Answer: (C)
A(1,2,3),B(2,3,1),C(2,4,2),O(0,0,0)
Equation of plane passing through A, B, C will be

(x – 1)(–1 + 4) – (y – 2)(–1 + 2) + (z – 3)(2 – 1) = 0
(x – 1)(3) – (y – 2)(1) + (z – 3)(1) = 0
3x – 3 – y + 2 + z – 3 = 0
3x – y + z – 4 = 0, is the required plane equation
Now, given O(0,0,0) & P(2,–1,1)
Plane is 3x – y + z – 4 = 0
Let O’ & P’ be the foot of perpendiculars.

 

Q. 10: The contra positive of the statement “If you will work, you will earn money” is:

(A) If you will not earn money, you will not work

(B) You will earn money, if you will not work

(C) If you will earn money, you will work

(D) To earn money, you need to work

Answer: (A)
Contrapositive of p → q is ~q →~p
p: you will work
q: you will earn money
~q: you will not earn money
~p: you will not work
~q →~p: if you will not earn money, you will not work

 

Q. 11: The shortest distance between the line x – y = 1 and the curve x² = 2y is:

(A) 1/2

(B) 0

(C) 1/(2√2)

(D) 1/√2

Answer: (C)
Shortest distance must be along common normal

m1 (slope of line x–y = 1) = 1 ⇒slope of perpendicular line =–1
m² = 2x/2 = x ⇒ m² = h ⇒slope of normal = –1/h
–1/h=–1 ⇒h=1
So, the point is (1,1/2)
Therefore, Shortest distance (D) =|(1–1/2–1)/√(1+1)|=1/(2√2)

 

Q. 12: Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is:

(A) 1/5

(B) 2/9

(C) 97/297

(D) 122/297

Answer: (C)
Total cases
= ⁴C₁ × 9 × 9 × 9 – ³C₁ × 9 × 9
(as 4 digit number having 0 at thousands place have to be excluded)
For a number to have a remainder 2 when divided by 5 it’s unit digit should be 2 or 7
Case 1: when unit digit is 2
Number of four-digit number = ³C₁×9×9 – ²C₁×9
Case 2: when unit digit is 7
Number of four digits number = 8 × 9 × 9
So total number favorable cases = 3×92 – 2×9 + 8×92
Required Probability = ((3×9×9)–(2×9)+(8×9×9))/((4×93) – (3×92))
= 97/297

 

Q. 13: cosec[2 cot^–1( 5) + cos^–1(4/5) ] is equal to:

(A) 75/56

(B) 65/56

(C) 56/33

(D) 65/33

Answer: (B)
cosec[2 cot^–1( 5) + cos^–1(4/5) ]

 

Q. 14: If 0 < x, y < π and cos x + cos y – cos (x + y) = 3/2, then sin x + cos y is equal to:

(A) (1+√3)/2

(B) (1–√3)/2

(C) √3/2

(D) 1/2

Answer: (A)

 

Q. 15: If α, β∈ R are such that 1 – 2i (here i² = –1) is a root of z² + αz + β = 0, then (α – β) is equal to:

(A) 7

(B) -3

(C) 3

(D) -7

Answer: (D)
1-2i is a root of z² + αz + β = 0.
(1-2i) ² + α(1-2i)+β=0
⇒1-4-4i+α-2iα+β=0
⇒(α+β-3)-i(4+2α)=0
⇒α+β-3=0 & 4+2α=0
So, α=-2, β=5
Therefore, α-β=-7

 

Q. 16: If , then

(A) 1/(I₂ + I₄ ) , 1/(I₃ + I₅ ), 1/(I₄ + I₆) are in G.P.

(B) 1/(I₂ + I₄ ) , 1/(I₃ + I₅ ), 1/(I₄ + I₆) are in A.P.

(C) I₂ + I₄ , I₃ + I₅ , I₄ + I₆) are in A.P.

(D) I₂ + I₄ , I₃ + I₅ , I₄ + I₆) are in G.P.

Answer: (B)

​

I_n+2 + Iⁿ = 1/(n+1)
I₂ + I₄ =1/3, I₃ + I₅ =1/4 and I₄ + I₆ =1/5
So, 1/(I₂ + I₄ ) , 1/(I₃ + I₅ ) and 1/(I₄ + I₆) are in A.P.

Q. 17: If for the matrix, A= AAT = I₂, then the value of α⁴ + β⁴ is:

(A) 1

(B) 3

(C) 2

(D) 4

Answer: (A)

 

Q. 18: Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions from the set A to the set A × B. Then:

(A) y = 273x

(B) 2y = 91x

(C) y = 91x

(D) 2y = 273x

Answer: (B)
Number of elements in A = 3
Number of elements in B = 5
Number of elements in A × B = 15
Number of one-one function is x = 5 × 4 × 3
⇒x = 60
Number of one-one function is y = 15 × 14 × 13
⇒y = 15 × 4 × 14/4 × 13
⇒y = 60 × 7/2 × 13
⇒2y = (13)(7x)
⇒2y = 91x

 

Q. 19: Let α and β be the roots of x² – 6x – 2 = 0. If an = αⁿ – βⁿ for n ≥ 1, then the value of (a¹⁰–2a⁸)/(3a⁹ ) is:

(A) 4

(B) 1

(C) 2

(D) 3

Answer: (C)
α and β be the roots of x²–6x–2=0.
So,
α²–6α–2=0⇒α²–2=6α
β²–6β–2=0⇒β²–2=6β
Now,

 

Q. 20: Let A be a 3 × 3 matrix with det(A) = 4. Let Ri denote the ith row of A. If a matrix B is obtained by performing the operation R²→ 2R² + 5R³ on 2A, then det(B) is equal to:

(A) 64

(B) 16

(C) 80

(D) 128

Answer: (A)

Section B

Q. 1: Let . If the area of the parallelogram whose adjacent sides are represented by the vectors a^￫ and a^￫ is 8√3 square units, then 

a^￫.b^￫is equal to ______.

Answer: 2
Area of parallelogram = 

|a^￫ × b^￫|

(64)(3) = 16α² + 64 + 16α²
⇒ α² = 4
Now, 

a^￫.b^￫ = 3 – α² + 3
= 6 – α²
= 6 – 4
= 2

 

Q. 2: If the curve y = y(x) represented by the solution of the differential equation (2xy² – y)dx + xdy = 0, passes through the intersection of the lines, 2x – 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to ______.

Answer:1
Given,
(2xy² – y)dx + xdy = 0
⇒ dy/dx + 2y² – y/x = 0
⇒ (–1/y²) . dy/dx + (1/y) (1/x) = 2
Let, 1/y = z
(–1/y²) dy/dx = dz/dx
⇒dz/dx + z(1/x) = 2

As it passes through P(2, 1)
[Point of intersection of 2x – 3y = 1 and 3x + 2y = 8]
Therefore, 2/1 = 4 + c
⇒c = –2
⇒x/y = x2 – 2
Put x = 1
1/y = 1 – 2 = –1
⇒y(1) = –1
⇒|y(1)| = 1

 

Q. 3: If lim_x→0 ax−(e^4x−1) / ax(e^4x−1) exists and is equal to b, then the value of a – 2b is ______.

Answer: 5

lim_x→0 ax−(e^4x−1) / ax(e^4x−1)

Applying L’Hospital Rule

lim_x→0 a−4e^4x / a(e^4x−1)+ax(4e^4x) So for limit to exist,a=4
Applying L’Hospital Rule

lim_x→0 −16e^4x / a(4e^4x) + a(4e^4x) + ax(16e^4x)

(-16)/(4a+4a)=(–16)/32=–1/2=b
a-2b = 4–2((–1)/2) = 4+1 = 5

 

Q. 4: A line L passing through origin is perpendicular to the lines

If the co-ordinates of the point in the first octant on L2 at the distance of √17 from the point of intersection of L and L1 are (a, b, c), then 18(a+b+c) is equal to ______.

Answer: 44

D.R. of L is parallel to (L₁ × L₂) ⇒ (-2, 3, -2)
Equation of l : x/2 = y/(–3) = z/2
Solving L & L₁
(2λ, –3λ, 2λ) = (µ + 3, 2µ – 1, 2µ + 4)
µ = – 1 , λ = 1
So, intersection point P(2, –3, 2)
Let, Q(2ν + 3, 2ν + 3, ν + 2) be required point on L₂

 

Q. 5: A function f is defined on [–3,3] as

where [x] denotes the greatest integer ≤ x. The number of points, where f is not differentiable in (–3,3) is ______.

Answer: (5)

Points of non-differentiable in (–3, 3) are at x = –2, –1, 0, 1, 2.
i.e. 5 points.

 

Q. 6: A line is a common tangent to the circle (x – 3)² + y² = 9 and the parabola y² = 4x. If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a+c) is equal to ______.

Answer: 9
Sol. Circle: (x – 3)²+ y²= 9
Parabola: y² = 4x
Let common tangent equation be y = mx + a/m
⇒y = mx + 1/m
⇒m²x – my + 1 = 0
The above line is also tangent to circle
(x – 3)² + y²= 9
Therefore, the perpendicular from (3, 0) to line = 3
⇒|(3m² – 0 + 1)/√(m²+m⁴ )| = 3
⇒(3m² + 1)² = 9(m² + m⁴)
⇒m = ±1/√3
Tangent is
y = 1/√3x + √3 (it will be used)
=> m = 1/√3
or y = –1/√3x – √3 (rejected)

For Parabola, point of contact is (a/m² , 2a/m)= (3, 2√3) = (c, d)
Solving Circle (x – 3)² + y² = 9 & line equation y = (1/√3) x + √3
(x – 3)² + ((1/√3) x + √3)² = 9
⇒x² + 9 – 6x + (1/3)x² + 3 + 2x = 9
⇒(4/3)x² – 4x + 3 = 0
⇒x = 3/2=a
∴ 2(a + c) = 2(3/2+3)= 9

 

Q. 7: The value of is ______.

Answer: 19

x²-x-2=(x–2)(x+1)

 

Q. 8: If the remainder when x is divided by 4 is 3, then the remainder when (2020+x)²⁰²² is divided by 8 is ______.

Answer: 1
Let x = 4k + 3
(2020 + x)²⁰²²
= (2020 + 4k + 3)²⁰²²
= (2024 + 4k – 1)²⁰²²
= (4A – 1)²⁰²²
=²⁰²²C₀(4A)²⁰²²(–1)₀ + ²⁰²²C₁(4A)²⁰²¹(–1)₁ + ……+²⁰²²C₂₀₂₁(4A)¹(–1)₂₀₂₁+ ²⁰²²C₂₀₂₂(4A)⁰(–1)²⁰²²
Which will be of the form 8λ+1
So, Remainder is 1.

 

Q. 9: If the curves x = y⁴ and xy = k cut at right angles, then (4k)⁶ is equal to ______.

Answer: 4

 

Q. 10: The total number of two digit numbers ‘n’, such that 3n+7n is a multiple of 10, is ______.

Answer: 45

Let n = 2t; t ∈ N
3n = 32t = (10 – 1)t
=10p + (–1)t
= 10p ± 1
If n = even then 7ⁿ + 3ⁿ will not be multiple of 10
So, if n is odd then only 7ⁿ + 3ⁿ will be a multiple of 10
Therefore, n = 11,13,15,………..,99
Therefore, the answer is 45

PHYSICS

SECTION A

Q. 1: Match List-I with List-II:

List-I	List-II
(A)h (Planck’s constant)	(i) [MLT–1]
(B)E (Kinetic energy)	(ii) [ML2T–1]
(C)V (Electric potential)	(iii) [ML2T–2]
(D)P (Linear momentum)	(iv) [ML2I–1T–3]

Choose the correct answer from the options given below:

(A) (A) → (ii), (B) →(iii), (C) → (iv), (D) → (i)

(B) (A) →(i), (B) → (ii), (C) →(iv), (D) → (iii)

(C) (A) → (iii), (B) → (ii), (C) →(iv), (D) →(i)

(D) (A) → (iii), (B) → (iv), (C) →(ii), (D) →(i)

Answer: (A)
K.E. = [ML²T^–2]
P (Linear momentum) = [MLT^–1]
h (Planck’s constant) = [ML²T^–1]
V (Electric potential) = [ML²T^–3I^–1]

 

Q. 2: The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lines 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while the 72^nd division on circular scale coincides with the reference line. The radius of the wire is:

(A) 1.64 mm

(B) 1.80 mm

(C) 0.82 mm

(D) 0.90 mm

Answer: (C)
Least count. = pitch/no. of div.
= 1 mm/100
= 0.01 mm
+ve zero error = 8 × L.C. = +0.08 mm
Measured reading = 1mm + 72 × L.C.
= 1mm + 0.72 mm
= 1.72 mm
True reading = 1.72 – 0.08
= 1.64 mm
Radius = 1.64/2 = 0.82 mm

 

Q. 3: If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

(A) 2π² ms^–2

(B) 16 m/s²

(C) 9.8 ms^–2

(D) π² ms^–2

Answer: (A)
T = 2π√(l/g)
T² = 4π²l/g
g = 4π²l/T²
= 4π²×2/(2)² = 2π² ms^–2

 

Q. 4: An α particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λp/λα is:

(A) 8

(B) 2.8

(C) 3.8

(D) 7.8

Answer: (B)
λ = h/p
= h/√(2mqV)
λp/λα = √(mαqα/mpqp)
= √(4×2/1×1)
= 2√2 = 2.8

 

Q. 5: Given below are two statements: one is labelled as Assertion A and the other is labelled as reason R.

Assertion A: The escape velocities of planet A and B are same. But A and B are of unequal masses.

Reason R: The product of their masses and radii must be same. M1R1 = M2R2

In the light of the above statements, choose the most appropriate answer from the options given below:

(A) Both A and R are correct but R is NOT the correct explanation of A

(B) A is correct but R is not correct

(C) Both A and R are correct and R is the correct explanation of A

(D) A is not correct but R is correct

Answer: (B)
Ve = escape velocity
v_e = √(2GM/R)
So, for same v_e, M₁/R₁ = M₂/R₂
A is true but R is false

 

Q. 6: A diatomic gas, having Cp = (7/2)R and Cv = (5/2)R, is heated at constant pressure. The ratio dU : dQ : dW

(A) 3 : 7 : 2

(B) 5 : 7 : 2

(C) 5 : 7 : 3

(D) 3 : 5 : 2

Answer: (B)
Since the gas is diatomic in nature and the process is isobaric, we have
Cp = (7/2)R
Cv = (5/2)R
dU = nCvdT
dQ = nCpdT
dW = nRdT
dU : dQ : dW
Cv : Cp : R
(5/2) R : (7/2) R : R
5 : 7 : 2

 

Q. 7: Statement I: A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.

Statement II: The side band frequencies are 1002 kHz and 998 kHz.

In the light of the above statements, choose the correct answer from the options given below:

(A) Both statement I and statement II are false

(B) Statement I is false but statement II is true

(C) Statement I is true but statement II is false

(D) Both statement I and statement II are true

Answer: (D)
Side band = (fc – fm) to (fc + fm)
= (1000 – 2) kHz to (1000 + 2) kHz
= 998 kHz to 1002 kHz
Band width = 2fm
= 2 ×2kHz
= 4 kHz
Both statements are true.

 

Q. 8: The current (i) at time t = 0 and t = ∞ respectively for the given circuit is:

(A) 18E/55,5E/18

(B) 5E/18,18E/55

(C) 5E/18,10E/33

(D) 10E/33,5E/18

Answer: (C)

At t = 0, inductor is open
Initial-state equivalent of the circuit shown in figure -1 is

Req = 6×9/(6+9) = 54/15
I (t = 0) = E×15/54 = 5E/18
At t = ∞, For steady state inductor is replaced by plane wire
Steady state equivalent of the circuit shown in figure-1 is

Equivalent circuit diagram is given by

Req =1×4/(1+4) + 5×5/(5+5)
= 4/5 + 5/2
= (8 + 25)/10
= 33/10
I = E/Req = 10E/33

 

Q. 9: Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.

If TA and TB are the time periods of A and B respectively then the value of TB – TA:

[Given: Radius of earth = 6400 km, mass of earth = 6×10²⁴ kg]

(A) 4.24× 10² s

(B) 3.33 × 10² s

(C) 1.33 × 10³ s

(D) 4.24 × 10³ s

Answer: (C)

V = √(GMe/r)

T = 2​𝜋r/ (√(GMe/r)) = 2​𝜋r/ (√(r/GMe))

T =√(4𝜋²r³/GMe) = √(4𝜋²r³/GMe)
T₂ – T₁ = √(4𝜋²(8000×10³)³/G×6×10²⁴ ) – √(4𝜋²(7000×10³)³/G×6×10²⁴ )
≅ 1.33 ×10³ s

 

Q. 10: An engine of a train, moving with uniform acceleration, passes the signal post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is:

(A) √(v² – u²)/2

(B) (v – u)/2

(C) √(v² + u²)/2

(D) (u + v)/2

Answer: (C)

a = uniform acceleration
u = velocity of first compartment
v = velocity of last compartment
l = length of train
v² = u² + 2as (3rd equation of motion)
v² = u² + 2al …..(1)
v² _middle = u² + 2al/2
∴ v ²_middle = u² + al ….(2)
From equation (1) and (2)
v²_middle = u² + (v² – u²)/2
= (v² + u²)/2
∴ v_middle = √(v² + u²)/2

 

Q. 11: A 5V battery is connected across the points X and Y. Assume D₁ and D₂ to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.

(A) ~0.86 A

(B) ~0.5 A

(C) ~0.43 A

(D) ~1.5 A

Answer: (C)
Since silicon diode is used so 0.7 Volt is drop across it, only D₁ will conduct so current through cell.
I = (5 – 0.7)/10 = 0.43 A

 

Q. 12: A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F₁. Now a spherical cavity of radius R/2 is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁: F₂ is:

(A) 41 : 50

(B) 36 : 25

(C) 50 : 41

(D) 25 : 36

Answer: (A)
Gravitational field intensity g₁ = GM/(3R)² = GM/9R² …(1)
Gravitational field intensity g₂ = GM/9R² – G(M/8)/(5R/2)²
= GM/9R² – GM/R²50
= (41/9×50) GM/R²….(2)
Implies , g₁/g₂ = 41/50
⇒ F₁/F₂ = mg₁/mg₂ = 41/50

 

Q. 13: A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:

(A) 13 cm

(B) 14.8 cm

(C) 16.6 cm

(D) 18.4 cm

Answer: (B)
λ = v/f
= 336/504 = 66.66cm
λ/4 = l + e = l + 0.3d
= l + 1.8
16.66 = l + 1.8 cm
l = 14.86 cm

 

Q. 14: A proton, a deuteron and an α particle are moving with the same momentum in a uniform magnetic fiel(D) The ratio of magnetic forces acting on them is _______ and their speeds are in the ratio______.

(A) 2 : 1 : 1 and 4 : 2 : 1

(B) 1 : 2 : 4 and 2 : 1 :1

(C) 1 : 2 : 4 and 1 : 1 : 2

(D) 4 : 2 : 1 and 2 : 1 : 1

Answer: (A)
As v = p/m & F = qvB
∴ F = qpB/m
F₁ = qpB/m, v₁ = p/m
F₂ = qpB/2m, v₂ = p/2m
F₃ = 2qpB/4m, v₃ = p/4m
F₁ : F₂ : F₃ & V₁ : V₂ : V₃
1 : ½ : ½  & 1 : ½ :¼
2 : 1 : 1 & 4 : 2 : 1

 

Q. 15: Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: When a rod lying freely is heated, no thermal stress is developed in it.

Reason R: On heating, the length of the rod increases

In the light of the above statements, choose the correct answer from the options given below:

(A) A is true but R is false

(B) Both A and R are true and R is the correct explanation of A

(C) Both A and R are true but R is NOT the correct explanation of A

(D) A is false but R is true

Answer: (C)
If a rod is free and it is heated then there is no thermal stress produced in it.
The rod will expand due to increase in temperature.
So, both A & R are true.

 

Q. 16: In an octagon ABCDEFGH of equal side, what is the sum of

Answer: (A)

 

Q. 17: Two radioactive substances X and Y originally have N₁ and N₂ nuclei respectively. Half-life of X is half of the half-life of Y. After three half-lives of Y, numbers of nuclei of both are equal. The ratio N₁/N₂ will be equal to:

(A) 8/1

(B) 1/8

(C) 3/1

(D) 1/3

Answer: (A)
After n half-life no of nuclei undecayed = No/2ⁿ
Given, t_(1/2)x= t_(1/2y)/2
So 3 half-life of y = 6 half-life of x
Given, Nx = Ny after 3t_(½)y
N₁/2⁶ = N₂/2³
N₁/N₂ = 2⁶/2³ = 2³= 8/1

 

Q. 18: The angular frequency of alternating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

(A) 0.8 H and 250 μF

(B) 0.8 H and 150 μF

(C) 1.33 H and 250 μF

(D) 1.33 H and 150 μF

Answer: (A)

Since key is open, circuit is series L-C-R circuit
15 = i_rms (60)
∴i_rms = ¼ A
Now, 20 = ¼ XL = ¼ (ωL)
∴ L = 4/5 = 0.8 H
& 10 = ¼ 1/(100C)
C = (1/4000) F
= 250 μF

 

Q. 19: Two coherent light sources having intensities in the ratio 2x produce an interference pattern. The ratio (I_max – I_min)/(I_max + I_min) will be:

(A) 2√(2x)/(x + 1)

(B) √(2x)/(2x + 1)

(C) 2√(2x)/(2x + 1)

(D) √(2x)/(x + 1)

Answer: (C)
Let I₁ = 2x
I₂ = 1
I_max = (√I₁ + √I₂)²
I_min = (√I₁ – √I₂)²

(I_max-I_min)/ (I_max+ I_min) = [(√2x + 1)² -(√2x – 1)²] / [(√2x + 1)² +​(√2x – 1)²]

= 4√(2x)/(2 + 4x)
= 2√(2x)/(2x + 1)

 

Q. 20: Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ______

(A) 0.15 m

(B) 0.2 m

(C) 0.1 m

(D) 1.0 m

Answer: (C)

B = μ₀NiR²/2(R² + x²)^3/2
at x₁ = 0.05m, B₁ = μ₀NiR²/2(R² + (0.05)²)^3/2
at x₂ = 0.2m, B₂ = μ₀NiR²/2(R² + (0.2)²)^3/2
B₁/B₂ = (R² + 0.04)^3/2/(R² + 0.0025)^3/2
(8/1)^2/3 = (R² + 0.04)/(R² + 0.0025)
4 (R² + 0.0025) = R² + 0.04
3R² = 0.04 – 0.01
R² = 0.03/3 = 0.01
R = √0.01 = 0.1 m

Section – B

Q. 1: The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as U= α/r¹⁰– β/r⁵ – 3. Where, a and b are positive constants. The equilibrium distance between two atoms will (2α/β)^a/(B) Where a =_______

Answer: 1
F = -dU/dr
F = -[-10α/r¹¹ + 5β/r⁶]
for equilibrium, F = 0
10α/r¹¹ = 5β/r⁶
2α/β = r⁵
r = (2α/β)^1/5
a = 1

 

Q. 2: A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ______ m/s. (take g = 10 m/s²)

Answer: 5

By conservation of energy,
v²_min = V² – 4gl ….(1)
T_max = mg + mv2/l ….(2)
T_min = mv²_min/l – mg ….(3)
from equation (1) and (3)
T_min = (m/l) (v² – 4gl) – mg
T_max/T_min = (v²/l + g)/(v²/l – 5g)
5/1 = (v²/1 + 10)/ (v²/1 – 50)
5v² – 250 = v² + 10
v² = 65 ….(4)
from equation (4) and (1)
v²_min = 65 – 40 = 25
v_min = 5 m/s

 

Q. 3: In a certain thermodynamic process, the pressure of a gas depends on its volume as kV³. The work done when the temperature changes from 1000 C to 3000 C will be ____ nR, where n denotes the number of moles of a gas.

Answer: 50
P = kv³
pv^–3 = k
x = –3
w = nR(T₁ – T₂)/(x – 1)
= nR(100-300)/(-3 -1)
= nR(-200)/-4
= 50nR

 

Q. 4: In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E₁ and E₂ as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E₁) is replaced by the second battery (E₂) for working by making K₁ open and E₂ close(D) The galvanometer gives then null deflection at J₂. The value of E₁/E₂ is the smallest fraction of a/b, Then the value of a is ____.

Answer: 1
E₁/E₂ = l₁/l₂
= 3×100 cm + (100 – 20)cm)/(7×100 cm + 60 cm)
= 380/760
= ½
= a/b
a = 1

 

Q. 5: The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of a convex lens is ______ cm.

Answer: 15
1/v – 1/u = 1/f …(1)
m = v/u ….(2)
from (1) and (2) we get
m = f/(f + u)
Given conditions
m₁ = -m₂
f/(f – 10) = -f/(f – 20)
f – 20 = –f + 10
2f = 30
f = 15 cm

 

Q. 6: A transmitting station releases waves of wavelength 960 m. A capacitor of 256 μF is used in the resonant circuit. The self inductance of coil necessary for resonance is _____ × 10^–8H.

Answer: 10
At resonance
ωr = 1/√(Lc)
∴ 2πf = 1/√(Lc)
∴ 4π²c²/λ² = 1/Lc
∴4π²×3×10⁸×3×10⁸/960×960 = 1/L×2.56×10^-6
L = 375×960/10^-6×4×π²×9×10¹⁶ = 10³/10¹⁰
= 10^–7 H
= 10 × 10^–8 H

 

Q. 7: The electric field in a region is given by . The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y-z plane) to that of the surface of area 0.3 m2(parallel to x-z plane) is a : 2, where a = ________

[Here i, j and k are unit vectors along x, y and z-axes respectively]

Answer: 1

Therefore, a = 1

 

Q. 8: 512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ____ V.

Answer: 128
Let charge on each drop = q
Radius = r
V = kq/r
2 = kq/r
Radius of bigger
4πR³/3 = 512 × (4/3) πr³
R = 8r
V = k(512)q/R = (512/8) (kq/r)
= (512/8) × 2
= 128 V

 

Q. 9: A monatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with a velocity 30 m/s. If the container is suddenly stopped then the change in temperature of the gas (R=gas constant) is x/3R. Value of x is ______.

Answer: 3600
ΔKE = ΔU
ΔU = nCVΔT
½ mv² = (3/2) nRΔT
mv²/3nR = ΔT
4×(30)²/3×1×R = ΔT
ΔT = 4×(30)²/3×1×R
x/3R = 1200/R
x = 3600

 

Q. 10: A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______ J.

Answer: 144
L di/dt = ε
= 3t
L ∫di = 3 ∫t dt
Li = 3t²/2
i = 3t²/2L
Energy stored in the coil, E = ½ Li²
= ½ L (3t²/2L)²
= ½ × 9t⁴/4L
= 9/8 × (4)⁴/2
= 144 J

Chemistry

SECTION A

Q. 1. The hybridization and magnetic nature of [Mn(CN)₆]^4– and [Fe(CN)₆]^3–, respectively are:

(A) d²sp³ and paramagnetic

(B) sp³d² and paramagnetic

(C) d²sp³ and diamagnetic

(D) sp³d² and diamagnetic

Answer: (A)

 

Q. 2. Identify A and B in the chemical reaction.

Answer: (D)

 

Q. 3. Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are:

(A) 2 and 3 only

(B) 1 only

(C) 2 only

(D) 3 only

Answer: (A)
Compounds that are more acidic than H₂CO₃, gives CO₂ gas in reaction with NaHCO₃. Compound B i.e. Benzoic acid and compound C i.e. picric acid both are more acidic than H₂CO₃.

 

Q. 4. Ellingham diagram is a graphical representation of:

(A) ΔG vs T

(B) (ΔG – TΔS) vs T

(C) ΔH vs T

(D) ΔG vs P

Answer: (A)
Ellingham diagram tells us about the spontaneity of a reaction with temperature.

 

Q. 5. Which of the following equations depicts the oxidizing nature of H2O2?

(A) Cl₂ + H₂O₂ → 2HCl + O₂

(B) KlO₄ + H₂O₂ → KlO₃ + H₂O + O₂

(C) 2l^– + H₂O₂ + 2H⁺ → I₂ + 2H₂O

(D) I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂

Answer: (C)
2l^– + H₂O₂ + 2H⁺ → I₂ + 2H₂O
Oxygen reduces from –1 to –2
So, its reduction will take place. Hence it will behave as oxidising agent or it shows
oxidising nature.

 

Q. 6. In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption (x/m) is directly proportional to Px. The value of x is:

(A) ∞

(B) 1

(C) zero

(D) 1/n

Answer: (D)
x / m = p^x
The formula is x / m = p^x
Hence, x= 1 / n. The value of ‘n’ is any natural number.

 

Q. 7. According to molecular orbital theory, the species among the following that does not exist is:

(A) He₂^–

(B) He₂⁺

(C) O₂^2-

(D) Be₂

Answer: (D)
B.O. of Be₂ is zero, so it does not exist.

 

Q. 8. In which of the following pairs, the outermost electronic configuration will be the same?

(A) Fe²⁺ and Co⁺

(B) Cr⁺ and Mn²⁺

(C) Ni²⁺ and Cu⁺

(D) V²⁺ and Cr⁺

Answer: (B)
Cr⁺ → [Ar]3d⁵
Mn²⁺ → [Ar]3d⁵

 

Q. 9. Given below are two statements:

Statement-I: An allotrope of oxygen is an important intermediate in the formation of reducing smog.

Statement-II: Gases such as oxides of nitrogen and Sulphur present in the troposphere contribute to the formation of photochemical smog.

(A) Statement I and Statement II are true

(B) Statement I is true about Statement II is false

(C) Both Statement I and Statement II are false

(D) Statement I is false but Statement II is true

Answer: (C)
Reducing smog acts as a reducing agent and the reducing character is due to presence of sulphur dioxide and carbon particles.

 

Q. 10. The plots of radial distribution functions for various orbitals of hydrogen atom against ‘r’ are given below:

(A) (4)

(B) (2)

(C) (1)

(D) (3)

Answer: (A)
3s orbital
Number of radial nodes = n – λ – 1
For 3s orbital n = 3, λ = 0
Number of radial nodes = 3 – 0 – 1 = 2
It is correctly represented in graph of option 4

 

Q. 11. Identify A in the given chemical reaction.

Answer: (D)

Aromatization reaction or hydroforming reaction.

 

Q. 12. Given below are two statements:

Statement-I: CeO₂ can be used for oxidation of aldehydes and ketones.

Statement-II: Aqueous solution of EuSO₄ is a strong reducing agent.

(A) Statement I is true, statement II is false

(B) Statement I is false, statement II is true

(C) Both Statement I and Statement II are false

(D) Both Statement I and Statement II are true

Answer: (D)
CeO₂ can be used as an oxidizing agent like seO₂. Similarly, EuSO₄ is used as a reducing agent.

 

Q. 13. The major product of the following chemical reaction is:

(A) (CH₃CH₂CO)₂O

(B) CH₃CH₂CHO

(C) CH₃CH₂CH₃

(D) CH₃CH₂CH₂OH

Answer: (B)

Q. 14. Complete combustion of 1.80 g of an oxygen-containing compound (CxHyOz) gave 2.64 g of CO₂ and 1.08 g of H₂O. The percentage of oxygen in the organic compound is:

(A) 63.53

(B) 51.63

(C) 53.33

(D) 50.33

Answer: (C)
n(CO₂) = 2.64 / 44 = 0.06
n_c = 0.06
weight of carbon = 0.06 × 12 = 0.72 gm
n(H₂O) = 1.08 / 1.8 = 0.06
n_H = 0.06 × 2 = 0.12
weight of H = 0.12 gm
∴ Weight of oxygen in CxHyOz
= 1.8 – (0.72 + 0.12)
= 0.96 gram
% weight of oxygen = 0.96/18 × 100 = 53.3%

 

Q. 15.The correct statement about B₂H₆ is:

(A) All B–H–B angles are 120°.

(B) Its fragment, BH₃, behaves as a Lewis base.

(C) Terminal B–H bonds have less p-character when compared to bridging bonds.

(D) The two B–H–B bonds are not of the same length.

Answer: (C)
The terminal bond angle is greater than that of bridge bond angle
Bond angle ∝ S-character

α  1/ p−character

 

Q. 16. Which of the glycosidic linkage galactose and glucose is present in lactose?

(A) C-1 of glucose and C-6 of galactose

(B) C-1 of galactose and C-4 of glucose

(C) C-1 of glucose and C-4 of galactose

(D) C-1 of galactose and C-6 of glucose

Answer: (B)

 

Q. 17. Which one of the following reactions will not form acetaldehyde?

Answer: (A)

 

Q. 18. Which of the following reactions will not give p-amino azobenzene?

(A) 2 only

(B) 1 and 2

(C) 3 only

(D) 1 only

Answer: (A)

 

Q. 19. The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is: [Assume: No cyano complex is formed; Ksp(AgCN) = 2.2 × 10^–16 and Ka (HCN) = 6.2 × 10^–10]

(A) 0.625 × 10^-6

(B) 1.6 × 10^-6

(C) 2.2 × 10^-16

(D) 1.9 × 10^-5

Answer: (D)
Let solubility is x
AgCN ⇌ Ag⁺ + CN^–Ksp = 2.2 × 10^–16
x x
H⁺ + CN^– ⇌ HCN

K = 1/ K_a

K= 1/ 6.2 × 10^–10​

Ksp × 1/ka = [Ag⁺] [CN^–] × [HCN] / [H⁺][CN^–]
2.2 × 10^-16 × 1 / 6.2 × 10^-10 = [S][S] / 10^-3
S² = 2.2 / 6.2 × 10^-9
S² = 3.55 × 10^–10
S = √3.55 × 10^–10
S = 1.88 × 10^–5 ⇒ 1.9 × 10^–5

 

Q. 20. Which statement is correct?

(A) Buna-S is a synthetic and linear thermosetting polymer

(B) Neoprene is an addition copolymer used in plastic bucket manufacturing

(C) Synthesis of Buna-S needs nascent oxygen

(D) Buna-N is a natural polymer

Answer: (C)
Synthesis of Buna-S needs nascent oxygen.

SECTION B

Q. 1. A car tire is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi. The temperature in °C at which the car tyre will burst is ______. (Rounded-off to the nearest integer)

Answer: 69.85°C ≃ 70°C
P₁ / T₁ = P₂ / T₂
35 / 300 = 40 / T₂
T₂ = 40 × 300 / 35
= 342.86 K
= 69.85°C ≃ 70°C

 

Q. 2. The reaction of cyanamide, NH₂CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be –742.24 kJ mol–1. The magnitude of ΔH298 for the reaction NH₂CN (s) + 3/2 O₂ (g) → N₂ (g) + O₂ (g) + H₂O(l) is______kJ. (Rounded off to the nearest integer). [Assume ideal gases and R = 8.314 J mol–1 K–1]

Answer: 741 kJ/mol
NH₂CN(s) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(λ)
Δng = (1 + 1) – 3/2 = ½
ΔH = ΔU + Δn_g RT
= -742.24 + ½ × 8.314 × 298 / 1000
= -742.24 + 1.24
= 741 kJ/mol

 

Q. 3. The ionization enthalpy of Na⁺ formation from Na(g) is 495.8 kJ mol^–1, while the electron gain enthalpy of Br is –325.0 kJ mol^–1. Given the lattice enthalpy of NaBr is –728.4 kJ mol^–1. The energy for the formation of NaBr ionic solid is (–)_____ × 10^–1 kJ mol^–1.

Answer: 5576 kJ
Na(s) → Na+(g) ΔH = 495.8
½ Br2(λ) + e^– → Br^–(g) ΔH = 325
Na⁺(g) + Br^–(g) → NaBr(s) ΔH = –728.4
Na(s) + ½ Βr2(λ) → NaBr(s). ΔH =?
ΔH = 495.8 – 325 – 728.4
–557.6 kJ = –5576 × 10^–1 kJ

 

Q. 4. Consider the following chemical reaction.

Answer: 7

All carbon atoms in benzaldehyde are sp² hybridized.

 

Q. 5. Among the following, the number of halide(s) which is/are inert to hydrolysis is ______.

(A) BF₃

(B) SiCl₄

(C) PCl₅

(D) SF₆

Answer: 1
Due to crowding, SF₆ is not hydrolyzed.

 

Q. 6. In basic medium CrO₄^2– oxidizes S₂O₃^2– to form SO²₄ and itself changes into Cr(OH)^4–. The volume of 0.154 M CrO₄^2– required to react with 40 mL of 0.25 M S₂O₃^2– is ______ mL. (Rounded-off to the nearest integer)

Answer: 173 mL
17H₂O + 8CrO₄ + 3S₂O₃ → 6SO₄ + 8Cr(OH)^4– + 2OH^–
Applying mole-mole analysis
0.154 × v / 8 = 40 × 0.25 / 3
V = 173 mL

 

Q. 7. 1 molal aqueous solution of an electrolyte A₂B₃ is 60% ionise(D) The boiling point of the solution at 1 atm is _____ K. (Rounded-off to the nearest integer). [Given Kb for (H₂O) = 0.52 K kg mol^–1]

Answer: 375 K
A2B3 → 2A⁺³ + 3B^–2
No. of ions = 2 + 3 = 5
wi = 1 + (n – 1) ∝
= 1 + (5 – 1) × 0.6
= 1 + 4 × 0.6 = 1 + 2.4 = 3.4
ΔTb = Kb × m × i
= 0.52 × 1 × 3.4 = 1.768°C
ΔTb = (Tb)solution – [(TbH₂O)]solution
1.768 = (Tb)solution – 100
(Tb)solution = 101.768 °C
= 375 K

 

Q. 8. Using the provided information in the following paper chromatogram:

The calculated Rf value of A ______ × 10^–1.

Answer: 4

Rr = Distance Travelled By Solvent / Distance Travelled By Compound

On chromatogram distance travelled by compound is → 2 cm
Distance travelled by solvent = 5 cm
So Rf = 2 / 5 = 4 × 10–1 = 0.4

 

Q. 9. For the reaction, aA + bB → cC + dD, the plot of log k vs 1/T is given below:

The temperature at which the rate constant of the reaction is 10^–4s^–1 is ________ K. [Rounded off to the nearest integer)

[Given: The rate constant of the reaction is 10^–5 s^–1 at 500 K]

Answer: 526 K

log₁₀ K = log₁₀ A – E₀/2.303RT

Slope = E₀/2.203 = -10000

log₁₀ K₁/K₂ = ​E₀/2.203 × (1/T₁ – 1/T₂)

log₁₀ 10^-4/10^-5 = 1000 × (1/500 -1/T)

1 = 1000 × (1/500 -1/T)

1/10000 = 1/500 = 1/T

1/T = 1/500 = 1/10000

1/T = 20-1 / 10000 = 19/10000

T = 10000 / 19
T= 526 K

 

Q. 10. A 0.4g mixture of NaOH, Na₂CO₃ and some inert impurities was first titrated with N / 10HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrate(D) 1.5 mL of the same HCl was required for the next end point. The weight percentage of Na₂CO₃ in the mixture is ______. (Rounded-off to the nearest integer)

Answer: 3%
1st end point reaction
NaOH + HCl → NaCl + H₂O
nf = 1
NaCO₃ + HCl → NaHCO₃
nf = 1
Eq of HCl used = n(NaOH) × 1 + n(Na₂CO₃) × 1
17.5 × 1/10 × 10^-3 = n(NaOH) + n(Na₂CO₃)

2nd end point
NaHCO₃ + HCl → H₂CO₃
1.5 × 1/10 × 10^-3 = n(NaHCO₃) × 1 = n(NaHCO₃)
0.15 mmol = n(Na₂CO₃)
0.15 = n(Na₂CO₃)

W(Na₂CO₃) = 0.15×106×10^-3 × 100 × 10
= 3 × 106 × 10^–2
= 3 × 1.06 = 3.18%

Maths

SECTION A

Q. 1: A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform spee(D) At that point, the angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

(A) 10(√3-1)

(B) 10√3

(C) 10

(D) 10(√3+1)

Answer: (D)

x + y = √3h ….…….(1)
Also,
h/y = tan 45^o
h = y …….(2)
put in (1)
x + y = √3y
x = (√3 – 1)y
x/20=v (speed)
Therefore, time taken to reach
Foot from B
= y/V
= x/(√3-1)x . 20
= 10 (√3 + 1)

 

Q. 2:

(A) xyz = 4

(B) xy – z = (x + y)z

(C) xy + yz + zx = z

(D) xy + z = (x+y)z

Answer: (D)

 

Q. 3: The equation of the line through the point (0,1,2) and perpendicular to the line (x-1)/2 = (y+1)/3 = (z-1)/-2 is :

(A) ​x/-3 = (y-1)/ 4 = (z-2)/3

(B) x/3 = (y-1)/ 4 = (z-2)/3

(C) x/3 = (y-1)/ -4 = (z-2)/3

(D) x/3 = (y-1)/ 4 = (z-2)/-3

Answer: (A)

(x-1)/2 = (y+1)/3 = (z-1)/-2 =λ
Any point on this line (2λ + 1, 3λ – 1, -2λ + 1)
Direction ratio of given line d₁ ≡ (2, 3, -2)
Direction ratio of the line to be found  d₂ ≡(2λ+1,3λ−2,−2λ−1)

 ∴ d₁ d₂ = 0

λ = 2 / 17
Direction ratio of line (21, -28, -21) ≡ (3, -4, -3) ≡ (-3, 4, 3)

 

Q. 4: The coefficients a,b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :

(A) 1/54

(B) 1/72

(C) 1/36

(D) 5/216

Answer: (D)
ax² + bx + c = 0
a,b,c ∈ {1,2,3,4,5,6}
n(s) = 6 × 6 × 6 = 216
D=0 ⇒ b² = 4ac
ac = b²/4, If b = 2, ac = 1 ⇒ a = 1,c = 1
If b = 4, ac = 4 ⇒ a = 1, c = 4
a = 4, c = 1
a = 2, c = 2
If b = 6, ac = 9 ⇒ a = 3, c = 3
Therefore, probability = 5/216

 

Q. 5: Let α be the angle between the lines whose direction cosines satisfy the equations l +m – n = 0 and l² + m² – n² = 0. Then the value of sin⁴ α + cos⁴ α is :

(A) 3/4

(B) 1/2

(C) 5/8

(D) 3/8

Answer: (C)
Given that l+m=n ….(1)
l² + m² – n² = 0 ….(2)
Squaring equation (1)
l² + m² + 2lm = n² ….(3)
From equations (2) and (3)
lm=0 ⇒ l = 0 or m = 0
Case (1) : l = 0
⇒ m=n
⇒ l² + m² + n² = 0
⇒ m=n = ±1/(√2 )
∴(l,m,n) = (0, 1/√2, 1/√2) or (0, -1/√2, -1/√2)
Case (2): m = 0
⇒ l = n
⇒ l² + m² + n² = 0

 

Q. 6: The value of the integral

(where c is a constant of integration)

(A) 1/18[9 – 2 sin⁶θ – 3 sin⁴θ – 6 sin²θ ]^(3/2) + c

(B) 1/18[11 – 18 sin²θ + 9 sin⁴θ – 2 sin⁶θ ]^(3/2) + c

(C) 1/18[11 – 18 cos²θ + 9 cos⁴θ – 2 cos⁶θ ]^(3/2) + c

(D) 1/18[9 – 2 cos⁶θ – 3 cos⁴θ – 6 cos²θ ]^(3/2) + c

Answer: (C)
Using trig identities, sin2A = 2sinAcosA and 1-cos2A = 2sin²A

 

Q. 7: The statement A→ (B → A) is equivalent to:

(A) A → (A ᐱ B)

(B) A → (A ᐯ B)

(C) A → (A → B) 

(D) A → (A ↔ B)

Answer: (B)
A → (B → A)
⇒ A → (∼ B ᐯ A)
⇒ ∼ A ᐯ (∼ B ᐯ A)
⇒ ∼ B ᐯ (∼ A ᐯ A)
⇒ ∼ B ᐯ t
= t (tautology)
From options:
(B) A → (A ᐯ B)
⇒ ∼ A ᐯ (A ᐯ B)
⇒ (∼ A ᐯ A) ᐯ B
⇒ t ᐯ B
⇒ t

 

Q. 8: The integer k, for which the inequality x² – 2(3k – 1)x + 8k² – 7 >0 is valid for every x in R is :

(A) 3

(B) 2

(C) 4

(D) 0

Answer: (A)
D < 0
(2(3k-1))² – 4(8k² – 7) < 0
4(9k² – 6k + 1) – 4(8k² – 7) < 0
k² – 6 k + 8 < 0
(k–4)(k–2)<0
2< k < 4
then k = 3

 

Q. 9: All possible values of θ ∈ [0, 2π] for which sin2θ + tan2θ > 0 lie in

a.(0,π/2)∪(π,3π/2)

(B) (0,π/4)∪(π/2,3π/4)∪(π,5π/4)∪(3π/2,7π/4)

(C) (0,π/2)∪(π/2,3π/4)∪(π,7π/4)

(D) (0,π/4)∪(π/2,3π/4)∪(3π/2,11π/6)

Answer: (B)

 

Q. 10: The image of the point (3,5) in the line x – y + 1 = 0, lies on :

(A) (x – 2)² + (y – 4) ² =4

(B) (x – 4) ² + (y + 2)² =16

(C) (x – 4)² + (y – 4)² = 8

(D) (x – 2)² + (y – 2)² =12

Answer: (A)
Image of P(3, 5) on the line x – y + 1 = 0 is

(x-3)/1 = (y-5)/-1 = (-2(3-5+1))/ 2 = 1

x = 4, y = 4
Image is (4, 4)
Which lies on (x – 2)² + (y – 4)² = 4

 

Q. 11: If Rolle’s theorem holds for the function f(x) = x³ – ax² + bx – 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :

(A) (–5, 8)

(B) (5, 8)

(C) (5, –8)

(D) (–5, –8)

Answer: (B)
f(1) = f(2)
⇒ 1 – a + b –4 = 8 – 4a + 2b –4
3a – b = 7 …(1)
f’(x) = 3x² – 2ax + b
⇒f’(4/3) = 0 ⇒ 3 x 16/9 – 8a/3 + b =0
⇒–8a + 3b = –16 …(2)
a = 5, b = 8

 

Q. 12: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is (x²-4x+y+8)/(x-2), then this curve also passes through the point :

(A) (4, 5)

(B) (5, 4)

(C) (4, 4)

(D) (5, 5)

Answer: (D)

dy/dx = [(x-2)² +y+4]/(x-2) = (x-2) + (y+4)/(x-2)

Let x – 2 = t ⇒ dx = dt
and y + 4 = u ⇒dy = du
dy/dx = du/dt
du/dt = t + u/t ⇒ du/dt – u/t = t
Here, IF = 1/t
u. (1/t) = ∫ t.(1\t) dt
⇒ u/t = t + c
⇒ (y+4)/(y-2) = (x – 2) + c
Passing through (0, 0)
c = 0
⇒ (y + 4) = (x – 2)²

 

Q. 13: The value of  ∫¹_-1 x² e^{x^3} dx where [t] denotes the greatest integer ≤ t,is :

(A) (e+1)/3

(B) (e-1)/3e

(C) (e+1)/3e

(D) 1/3e

Answer: (C)

 

Q. 14: When a missile is fired from a ship, the probability that it is intercepted is 1/3 and the probability that the missile hits the target, given that it is not intercepted, is 3/4. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:

(A) 1/8

(B) 1/27

(C) 3/4

(D) 3/8

Answer:(A)
Probability of not getting intercepted = 2/3
Probability of missile hitting target = 3/4
Probability that all 3 hit the target = (2/3 x3/4)³ = ⅛

 

Q. 15: If the curves, x²/a + y²/b  and x²/c + y²/d intersect each other at an angle of 90°, then which of the following relations is true ?

(A) a + b = c + d

(B) a- b = c – d

(C) ab = (c+d)/(a+b)

(D) a-c = b+d

Answer: (B)

x²/a + y²/b ………(1)

diff : 

2x/a +2y/b dx/dy =0

y/b . dx/dy = -x/a

dx/dy = -bx/ay……(2)

x²/c + y²/d………(3)

Diff : 

dy/dx = -dx/cy ………(4)

m₁m₂ = –1 

⇒ -bx/ay × -dx/cy​ =−1

⇒ bdx² = – acy²…….(5)
(1)–(3) ⇒ (1/a – 1/c)x² + (1/b – 1/d) y² = 0
Solve above equation using 5
⇒ (c – a) – (d – b) = 0
⇒ c – a = d – b
⇒ c – d = a – b

 

Q. 16: A tangent is drawn to the parabola y² = 6x which is perpendicular to the line 2x + y =1. Which of the following points does NOT lie on it?

(A) (0,3)

(B) (-6,0)

(C) (4,5)

(D) (5,4)

Answer: (D)
Equation of tangent : y = mx + 3/(2m)
mT = (1/2) (Because perpendicular to line 2x + y = 1)
Therefore, tangent is: y = x/2 + 3 ⇒ x – 2y + 6 = 0

 

Q. 17: Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true ?

(A) f is one-one

(B) If fog is one-one, then g is one-one

(C) If g is onto, then fog is one-one

(D) If f is onto, then f(n) = n for all n ∈ N

Answer: (C)
f(n + 1) = f(n) +f(1)
⇒ f(n + 1)- f(n)= f(1) → A.P. with common difference = f(1)
General term = Tn = f(1) + (n-1)f(1) = nf(1)
⇒f(n) = nf(1)
Clearly f(n) is one-one.
For fog to be one-one, g must be one-one.
For f to be onto, f(n) should take all the values of natural numbers.
As f(x) is increasing, f(1)=1
⇒f(n)=n
If g is many one, then fog is many one. So “if g is onto, then fog is one-one” is incorrect.

 

Q. 18: Let the lines (2 – i)z = (2 + i)zˉ and (2 + i)z + (i – 2)zˉ– 4i = 0, (here i² = –1) be normal to a circle C. If the line iz + zˉ+ 1 + i = 0 is tangent to this circle C, then its radius is :

(A) 3/√2

(B) 3√2

(C) 3/(2√2)

(D) 1/(2√2)

Answer: (C)
(2–i)z = (2+i)zˉ

⇒(2–i)(x+ iy)=(2+i)(x–iy)
⇒2x–ix + 2iy + y=2x+ ix–2iy+y
⇒2ix –4iy=0
L1 ∶ x–2y=0
⇒(2+i)z+(i –2)zˉ–4i=0.
⇒(2+i) (x + iy)+(i –2)(x–iy)– 4i = 0.
⇒2x +ix + 2iy – y + ix – 2x + y +2iy – 4i =0
⇒2ix + 4iy – 4i =0
L2 ∶ x+2y–2 = 0
Solve L₁ and L₂
4y=2 or y=1/2
∴ x = 1
Centre(1, 1/2)
L3∶ iz + zˉ+ 1 + i = 0
⇒i(x+iy)+x–iy+1+i=0
⇒ix–y+x–iy+1+i=0
⇒(x–y+1)+i(x–y+1)=0
Radius = distance from (1, 1/2) to x – y + 1 = 0
r = (1-1/2+1)/ √2
r = 3/2√2

 

Q. 19:

(A) 1/2

(B) 1/e

(C) 1

(D) 0

Answer: (C)

 

Q. 20: The total number of positive integral solutions (x, y, z) such that xyz = 24 is

(A) 36

(B) 45

(C) 24

(D) 30

Answer: (D)
x.y.z = 24
x.y.z= 2³ × 3¹
x=2(a1) ⋅3(b1)
y = 2(a2) ⋅3(b2)
z = 2(a3)⋅3(b3)
a1, a2, a3 ∈ {0,1,2,3}
b1, b2, b3 ∈ {0,1}
Case 1: a1 + a2 + a3 =3
Non negative solution = ^(3+3-1) C_(3-1) = ⁵C₂ = 10
Case 2: b1 + b2 + b3 = 1
Non negative solution = ^(1+3-1) C_(3-1) = ³C₂ = 3
∴ Total solutions =10×3=30

Section B

Q. 1: Let A is a matrix where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A² = l₃, then the value of x³ + y³ + z³ is ______.

Answer: 7

 

Q. 2: Let A₁, A₂, A₃,…. be squares such that for each n ≥ 1, the length of the side of An equals the length of diagonal of A_n+1. If the length of A₁ is 12 cm, then the smallest value of n for which area of An is less than one is ____________.

Answer: (9)

 

Q. 3: The locus of the point of intersection of the lines (√3)kx + ky – 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is _________.

Answer: (2)
(√3)kx + ky – 4√3 = 0 …(1)
√3kx -ky= 4√3 k2 …(2)
Adding equation (1) & (2)
2√3 kx = 4√3(k^2 + 1)
x = 2 (k + 1/k) ….(3)
Subtracting equation (1) & (2)
y = 2√3(1/k – k) ………(4)
x²/4 – y2/12 = 4
x²/16 – y²/48 = 1 – Hyperbola
e² = 1 + 48/16
or e = 2

 

Q. 4:

Answer: 13

|I₂ + A|=|I₂ – A|
From equation (1)
∴a² + b² = 1
⇒13(a² + b²) = 13

 

Q. 5: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A⁴ is equal to ___________

Answer: (64)

 

Q. 6: The number of points, at which the function f(x) = |2x + 1| – 3|x+2|+|x² + x–2|, x ∈ R is not differentiable, is ________.

Answer: 2

Check at 1, –2 and -1/2
Non-Differentiable at x = 1 and -1/2

 

Q. 7: If the system of equations

kx + y + 2z = 1

3x – y – 2z = 2

–2x – 2y – 4z = 3

has infinitely many solutions, then k is equal to __________.

Answer: 21
D = D₁ = D₂ = D₃ = 0
Choose D₂ = 0
k(–8+6)–1(–12–4)+2(9+4)=0
–2k + 16 + 26 = 0
2k = 42
k = 21

 

Q. 8: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is ________.

Answer: 32

 

Q. 9: Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x⁶ is unity and it has extrema at x = –1 and x = 1. If lim_x→0 f(x)/x³ =1, then 5.f(2) is equal to _________.

Answer: (144)
f(x) = x⁶ + ax⁵ + bx⁴ + x³
f’(x) = 6x⁵ + 5ax⁴ + 4bx³ + 3x²
Roots 1 & – 1
Therefore, 6 + 5a + 4b + 3 = 0 & – 6 + 5a – 4b + 3 = 0 solving
a = -3/5 and b = -3/2
f(x) = x⁶ – (3/5)x⁵ – (3/2)x⁴ + x³
5.f(2) = 5 [64 – 96/5 – 24 + 8] = 144

 

Q. 10: Let  and be three given vectors. If ^→r is a vector such that and ^→r.^→b =0, then 

^→r.^→a =0 is equal to _______

Answer: 12

[tabs title=”JEE MAINS Previous Year Paper” type=”centered”]

[tab title=”PHYSICS”]

PHYSICS

SECTION A

Q 1: A body weighs 49 N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[Use g = GM/R2 = 9.8 ms-2 and radius of earth, R = 6400 km.]

(A) 49 N

(B) 49.83 N

(C) 49.17 N

(D) 48.83 N

Answer: (D)
At North Pole, weight
Mg = 49
Now, at equator
g’ = g – ω²R
⇒ Mg’ = M(g – ω²R)
⇒ weight will be less than Mg at equator.
Alter:
g is maximum at the poles.
Hence from options only (D) has lesser value than 49N.

Q 2: The logic circuit shown below is equivalent to :

Answer: (2)

Q 3: If one mole of an ideal gas at (P₁, V₁) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B→C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to:

(A) 0

(B) -RT/2( γ – 1)

(C) RT [ln2 -1/2( γ -1)]

(D) RT ln2

Answer: (C)
AB → Isothermal process
W_AB = nRT ln2 = RT ln2
BC → Isochoric process
W_BC = 0
CA → Adiabatic process
W_CA = P₁V₁ – (P₁/4)X²V₁)/(1 – γ)
= P₁V₁/2(1 – γ)
= RT/2(1 – γ )
W_ABCA = RT ln2 + RT/2(1 – γ)
= RT [ln2 – 1/2(γ – 1)]

Q 4: Match List – I with List – II.

List – I	List – II
(a) Source of microwave frequency	(i) Radioactive decay of nucleus
(b) Source of infrared frequency	(ii) Magnetron
(c) Source of Gamma Rays	(iii) Inner shell electrons
(d) Source of X-rays	(iv) Vibration of atoms and molecules
–	(v) LASER
–	(vi) RC circuit

Choose the correct answer from the options given below:

(A) (a)-(ii),(b)-(iv),(c)-(i),(d)-(iii)

(B) (a)-(vi),(b)-(iv),(c)-(i),(d)-(v)

(C) (a)-(ii),(b)-(iv),(c)-(vi),(d)-(iii)

(D) (a)-(vi),(b)-(v),(c)-(i),(d)-(iv)

Answer: (A)
(a) Source of microwave frequency – (ii) Magnetron
(b) Source of infrared frequency – (iv) Vibration of atom and molecules
(c) Source of gamma ray – (i) Radioactive decay of nucleus
(d) Source of X-ray – (iii) inner shell electron

Q 5: The period of oscillation of a simple pendulum is T = 2π√(L/g). Measured value of ‘L’ is 1.0 m from meter-scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of ‘g’ will be:

(A) 1.33 %

(B) 1.30 %

(C) 1.13 %

(D) 1.03 %

Answer: (C)
T = 2π√(L/g)
T² = 4𝜋² [L/g]
g = 4𝜋² [L/T2]
Δg/g = ΔL/L + 2ΔT/T
= [1mm/1m + 2(10×10-3)/1.95]×100
= 1.13 %

Q 6: When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:

(A) elliptical

(B) parabolic

(C) straight line

(D) circular

Answer: (A)
We know that in SHM;
V = ω√(A² – x²)

Elliptical
Alternate:
x = A sinωt ⇒ sin ωt = x/A
v = Aω cos ωt ⇒ cos ωt = v/Aω
Hence (x/A)² + (v/Aω)² = 1
Which is the equation of an ellipse.

Q 7: In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:

(A) (1/2π) √(2K/Mg sin α)

(B) (1/2π) √(K/Mg sin α)

(C) (1/2π) √(2K/M)

(D) (1/2π) √(K/2M)

Answer: (A)
Equivalent K = K + K = 2K
Now, T = 2π √(M/K_eq)
⇒ T = 2π √(M/2K)
∴f = (1/2π) √(2K/M)

Q 8: Given below are two statements:

Statement I: PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal.

Statement II: In the study of transistor, the amplification factor β indicates ratio of the collector current to the base current.

In the light of the above statements, choose the correct answer from the options given below.

(A) Statement I is false but Statement II is true.

(B) Both Statement I and Statement II are true.

(C) Statement I is true but Statement II is false.

(D) Both Statement I and Statement II are false.

Answer: (A)
S-1
Statement 1 is false because in case of two discrete back to back connected diodes, there are four doped regions instead of three and there is nothing that resembles a thin base region between an emitter and a collector.
S-2
Statement-2 is true, as
β = I_C/I_B

Q 9: Figure shows a circuit that contains four identical resistors with resistance R = 2.0 Ω. Two identical inductors with inductance L = 2.0 mH and an ideal battery with emf E = 9.V. The current ‘i’ just after the switch ‘s’ is closed will be:

(A) 9A

(B) 3.0 A

(C) 2.25 A

(D) 3.37 A

Answer: (C)
Just when switch S is closed, inductor will behave like an infinite resistance. Hence, the circuit will be like

Given: V = 9 V
From V = IR
I = V/R
R_eq. = 2 + 2 = 4 Ω
i = 9/4 = 2.25 A

Q 10: A circular hole of radius (a/2) is cut out of a circular disc of radius ‘a’ shown in figure. The centroid of the remaining circular portion with respect to point ‘O’ will be:

(A) (10/11)a

(B) (⅔)a

(C) (⅙)a

(D) (⅚)a

Answer: (D)
Let σ be the surface mass density of disc.

Xcom= (m₁x₁ – m₂x₂)/(m₁ – m₂)
Where m = σπr²
Xcom= (σ×πa²×a) – (σπa²/4) × 3a/2)/(σπa² – σπa²/4)
Xcom= (a – 3a/8)/(1 – ¼)
Xcom= (5a/8)/(¾)
Xcom= 5a/6

Q 11: The de Broglie wavelength of a proton and α-particle are equal. The ratio of their velocities is:

(A) 4:2

(B) 4:1

(C) 1:4

(D) 4:3

Answer: (B)
From De-Broglie’s wavelength:-
λ = h/mv
Given λP = λα
v ∝ 1/m
v_p/v_α = m_α/m_p = 4m_p/m_p
= 4/1

Q 12: On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

(A) suffer change in momentum when impinge on the walls of container.

(B) continuously stick to the walls of container.

(C) continuously lose their energy till it reaches wall.

(D) are attracted by the walls of container.

Answer: (A)
Based on kinetic theory of gases, molecules suffer change in momentum when impinge on the walls of container. Due to this they exert a force resulting in exerting pressure on the walls of the container.

Q 13: Two electrons each are fixed at a distance ‘2d’. A third charge proton placed at the midpoint is displaced slightly by a distance x (x<<d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:

(m = mass of charged particle)

(A) (q²/2πε₀md³)^1/2

(B) (πε₀md³/2q²)^1/2

(C) (2πε₀md³/q²)^1/2

(D) (2q²/πε₀md³)^1/2

Answer: (A)

Restoring force on proton:-
Fr = 2F₁ sinθ where F₁ = kq²/(d² + x²)
Fr = 2Kq²x/[d² + x²]^3/2
x <<< d
Fr = 2kq²x/d³
= q²x/2πε₀d³
= kx
K = q²/2πε₀d³
Angular Frequency:-
ω = √(k/m)
ω = √(q²/2πε₀md³)

Q 14: An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:

(A) 10^-2 nm

(B) 10^-3 nm

(C) 10^-4 nm

(D) 10^-1 nm

Answer: (B)

The minimum wavelength of photon will correspond to the maximum energy due to accelerating by V volts in the tube.
λmin = hc/eV
λmin = 1240nm-eV/1.24×10⁶
λmin = 10^-3 nm

Q 15: A soft ferromagnetic material is placed in an external magnetic fiel(D) The magnetic domains:

(A) decrease in size and changes orientation.

(B) may increase or decrease in size and change its orientation.

(C) increase in size but no change in orientation.

(D) have no relation with external magnetic field.

Answer: (B)
Atoms of ferromagnetic material in unmagnetized state form domains inside the ferromagnetic material. These domains have large magnetic moment of atoms. In the absence of magnetic field, these domains have magnetic moment in different directions. But when the magnetic field is applied, domains aligned in the direction of the field grow in size and those aligned in the direction opposite to the field reduce in size and also its orientation changes.

Q 16: According to Bohr atom model, in which of the following transitions will the frequency be maximum?

(A) n = 2 to n = 1

(B) n = 4 to n = 3

(C) n = 5 to n = 4

(D) n = 3 to n = 2

Answer: (A)

Since, ΔE is maximum for the transition from n = 2 to n = 1
f is more for transition from n = 2 to n = 1.

Q 17: Which of the following equations represents a travelling wave?

(A) y = Ae^{x^2} (vt + θ )

(B) y = A sin(15x – 2t)

(C) y = Aex cos(ωt – θ)

(D) y = A sin x cos ωt

Answer: (B)
Y = F(x, t)
For travelling wave y should be linear function of x and t and they must exist as (x ± vt)
Y = A sin (15x – 2t) which is a linear function in x and t.

Q 18: Zener breakdown occurs in a p-n junction having p and n both:

(A) lightly doped and have wide depletion layer.

(B) heavily doped and have narrow depletion layer.

(C) heavily doped and have wide depletion layer.

(D) lightly doped and have narrow depletion layer.

Answer: (B)
The zener breakdown occurs in the heavily doped p-n junction diode. Heavily doped p-n junction diodes have narrow depletion region. The narrow depletion layer width leads to a high electric field which causes the p-n junction breakdown.

Q 19: A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = -αx². The distance at which the particle stops:

(A) (2v₀/3α)^1/3

(B) (3v₀²/2α)^1/2

(C) (3v₀²/2α)^1/3

(D) (2v₀²/3α)^1/2

Answer: Bonus
a = vdv/dx

Given:- v_i = v₀
V_f = 0
X_i = 0
X_f = x
From Damping Force: a = -αx²/m

-v₀²/2 = (-α/m) [x³/3]
x = [3mv₀²/2α]^1/3
Most suitable answer could be (3) as mass ‘m’ is not given in any options.

Q 20: If the source of light used in a Young’s double slit experiment is changed from red to violet:

(A) the fringes will become brighter.

(B) consecutive fringe lines will come closer.

(C) the central bright fringe will become a dark fringe.

(D) the intensity of minima will increase.

Answer: (B)
β = λD/d
As λv<λR
βv< βR
⇒ Consecutive fringe lines will come closer.
⇒ (2)

Section – B

Q 1: A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is ______×10^-1 kg m².

Answer: 8

MOI of AB about P : IABP = (M/6)(l/6)²/12
MOI of AB about O,

= (6/100) [(24×24/12×36) + (24×24/36)×3/4]
= 0.8 kg m²
= 8×10^-1 kg m²

Q 2: Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be A/1. So the value of A will be ________.

Answer: 2
Given that, M₁/M₂ = 1/2
we know that
K = p²/2M
K₁/K₂ = (p²/2M₁ )×2M₂/p²
K₁/K₂ = M₂/M₁ = 2/1
A/1 = 2/1
⇒∴ A = 2

Q 3: An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _______ ×10^-2 cm.

Answer: 667
ε_r = 2.25
Assuming non-magnetic material ⇒ μ_r = 1
Hence refractive index of the medium
n = √(μ_rε_r )= √2.25 = 1.5
∴λ_v/λ_m=n
λ_m = C/f.n
= 3×10⁸/3×10⁹×1.5
= (⅔) 10^-1 m
= 667×10^-2 cm

Q 4: The root mean square speed of molecules of a given mass of a gas at 270C and 1 atmosphere pressure is 200 ms-1. The root mean square speed of molecules of the gas at 1270C and 2 atmosphere pressure is x/√3 ms-1. The value of x will be __________.

Answer: 400 m/s
Vrms = √(3RT₁/M₀)
200 = √(3R×300/M₀) ….(1)
Also, x/√3 = √(3R×400/M₀) …(2)
(1)÷(2)
200/(x/√3) = √(300/400) = √(3/4)
⇒x = 400 m/s

Q 5: A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5 dB per km and cable length is 20 km. The power received at the receiver is 10^-xW. The value of x is ______.

[Gain in dB = 10 log₁₀(P₀/P_i)]

Answer: 8
Power of signal transmitted: P_i = 0.1 Kw = 100w
Rate of attenuation = -5 dB/Km
Total length of path = 20 km
Total loss suffered = -5×20 = -100dB
Gain in dB = 10 log₁₀(P₀/P_i)
-100 = 10log₁₀(P₀/P_i)
log₁₀(P_i/P₀) = 10
log₁₀(P_i/P₀) = log101010
100/P₀ = 1010
⇒ P₀ = 1/10⁸ = 10^-8
⇒ x = 8

Q 6: A series LCR circuit is designed to resonate at an angular frequency ω0 = 10⁵rad/s. The circuit draws 16W power from 120 V source at resonance. The value of resistance ‘R’ in the circuit is _______ Ω.

Answer: 900
P = V²/R
16 = 120²/R
⇒ R = 14400/16
⇒ R = 900 Ω

Q 7: A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force, will be ________ cm.

Answer: 2

y = Fl/A∆l
F/A = y∆l/l
F/A = y×0.04/l …(1)
When length & diameter is doubled.
⇒ F/4A = y × ∆l/2l …(2)
(1)÷(2)
(F/A)/(F/4A) = (y×0.04/l)y×∆l/2l
4 = 0.04×2/∆l
∆l = 0.02
∆l = 2×10^-2
∴ x = 2

Q 8: A cylindrical wire of radius 0.5 mm and conductivity 5×107 S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x3π mA. The value of x is ____.

Answer: 5
We know that
J = σE
⇒ J = 5×10⁷×10×10^-3
⇒ J = 50×104 A/m²
Current flowing;
I = J × πR²
I = 50×10⁴ ×π(0.5×10^-3)²
I = 50×10⁴×π×0.25×10^-6
I = 125×10^-3π
X = 5

Q 9: A point charge of +12 μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be _____ ×10³ Nm²/C.

Answer: 226
Using Gauss law, it is a part of cube of side 12 cm and charge at centre,
ϕ = Q/6ε₀= 12μc/6ε₀

12×10⁻⁶ /6×8.85×10⁻¹²​

= 226×103 Nm²/C

Q 10: Two cars are approaching each other at an equal speed of 7.2 km/hr. When they see each other, both blow horns having frequency of 676 Hz. The beat frequency heard by each driver will be ________ Hz. [Velocity of sound in air is 340 m/s.]

Answer: 8

Speed = 7.2 km/h = 2 m/s
Frequency as heard by A
f’A= fB(v + v₀)/(v – v_s)
f’A = 676(340 + 2)/(340 – 2)
f’A = 684Hz
∴ fBeat = f’A- fB
= 684 – 676
= 8 Hz

[/tab]

[tab title=”CHEMISTRY”]

Chemistry

SECTION A

Q 1. Given below are two statements:

(A) Both Statement I and Statement II are false

(B) Statement I is false but Statement II is true

(C) Statement I is true but Statement II is false

(D) Both Statement I and Statement II are true

Answer: (C)
For survival of aquatic life dissolved oxygen is responsible for its optimum limit 6.5 ppm and optimum limit of BOD ranges from 10-20 ppm & BOD stands for biochemical oxygen demand.

Q 2. Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of HgSO₄ and H₂SO₄?

Answer: (A)
Reaction of Alkyne with HgSO₄ & H₂SO₄ follow as

Hence, by this process preparation of CH₃CH₂CHO can’t be possible.

Q 3. Which one of the following compounds is non-aromatic?

Answer: (B)

(Not planar)
Hence, it is non-aromatic.

Q 4. The incorrect statement among the following is:

(A) VOSO₄ is a reducing agent

(B) Red color of ruby is due to the presence of CO³⁺

(C) Cr₂O₃ is an amphoteric oxide

(D) RuO₄ is an oxidizing agent

Answer: (B)
Red color of ruby is due to presence of CrO₃ or Cr⁺⁶ not CO³⁺

Q 5. According to Bohr’s atomic theory:

(a) Kinetic energy of electron is ∝ Z² / n²

(b) The product of velocity (v) of electron and principal quantum number (n). ‘vn’ ∝ z²

(c) Frequency of revolution of electron in an orbit is ∝ Z³ / n³

(d) Coulombic force of attraction on the electron is ∝ Z³ / n⁴

Choose the most appropriate Answer from the options given below

(A) (C) only

(B) (A) and (D) only

(C) (A) only

(D) (A), (C) and (D) only

Answer: (B)
(a) KE = –TE = 13.6 × Z² / n² eV
KE α Z² / n²
(b) V = 2.188 × 10⁶ × z / n m/s
So, V_n ∝ Z
Frequency = V / 2πr
F α Z² / n² [∴ r α Z² / n² and v α Z / n ]
(d) Force ∝ Z² / r²
So, F ∝ Z³ / n⁴
So, only statement (A) is correct.

Q 6. Match List-I with List-II

List- I	List-II
(a) Valium	(iv) Tranquilizer
(b) Morphine	(iii) Analgesic
(c) Norethindrone	(i) Antifertility drug
(d) Vitamin B12	(ii) Pernicious anemia

(A) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)

(B) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)

(C) (A)-(ii), (B)-(iv), (C)-(iii), (D)-(i)

(D) (A)-(iv), (B)-(iii), (C)-(i), (D)-(ii)

Answer: (D)
(a) Valium (iv) Tranquilizer
(b) Morphine (iii) Analgesic
(c) Norethindrone (i) Antifertility drug
(d) Vitamin B₁₂ (ii) Pernicious anemia

Q 7. Match List-I with List-II

List- I (Salt)	List-II(Flame colour wavelength)
(a) LiCl	(i) 455.5 nm
(b) NaCl	(ii) 970.8 nm
(c) RbCl	(iii) 780.0 nm
(d) CsCl	(iv) 589.2 nm

Choose the correct Answer from the options given below:

(A) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

(B) (A)-(ii), (B)-(iv), (C)-(iii), (D)-(i)

(C) (A)-(iv), (B)-(ii), (C)-(iii), (D)-(i)

(D) (A)-(i), (B)-(iv), (C)-(ii), (D)-(iii)

Answer: (B)
Range of visible region: 390 nm – 760 nm
VIBGYOR
Violet – Red
LiCl Crimson Red
NaCl Golden yellow
RbCl Violet
CsCl Blue
So, LiCl which is crimson have wavelengths close to red in the spectrum of visible region which is as per given data.

Q 8. Match List-I and List-II.

List-I	List-II
(A)	(i) Br2 / NaOH
(B)	(ii) H2 / Pd-BaSO4
(C)	(iii) Zn (Hg) / Conc. HCl
(D)	(iv) Cl2 / Red P, H2O

(A) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

(B) (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)

(C) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(D) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

Answer: (D)

Q 9. In polymer Buna-S: ‘S’ stands for

(A) Styrene

(B) Sulphur

(C) Strength

(D) Sulphonation

Answer: (A)
Buna-S is the co-polymer of buta-1,3-diene & styrene

Q 10. Most suitable salt which can be used for efficient clotting of blood will be:

(A) Mg(HCO₃)₂

(B) FeSO₄

(C) NaHCO₃

(D) FeCl₃

Answer: (D)
Blood is a negative sol, according to Hardy-Schulz’s rule, the cation with high charge has high coagulation power. Hence, FeCl₃ can be used for clotting blood.

Q 11. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere.

Reason R: Hydrogen is the lightest element.

In the light of the above statements, choose the correct Answer from the given below

(1) A is false but R is true

(2) Both A and R are true and R is the correct explanation of A

(3) A is true but R is false

(4) Both A and R are true but R is NOT the correct explanation of A

(A) A is false but R is true

(B) Both A and R are true and R is the correct explanation of A

(C) A is true but R is false

(D) Both A and R are true but R is NOT the correct explanation of A

Answer: (B)
Hydrogen is the most abundant element in the universe because all luminous bodies of the universe i.e. stars and nebulae are made up of hydrogen which acts as nuclear fuel and fusion reaction is responsible for their light.

Q 12. What is the correct sequence of reagents used for converting nitrobenzene into m- dibromobenzene?

Answer: (D)

Q 13. The correct shape and I-I-I bond angles respectively in I^–₃ ion are:

(A) Trigonal planar; 120^o

(B) Distorted trigonal planar; 135^o and 90^o

(C) Linear; 180º

(D) T-shaped; 180º and 90º

Answer: (C)
I^–₃ has sp³d hybridization (2 BP + 3 LP) and linear geometry.

Q 14. What is the correct order of the following elements with respect to their density?

(A) Cr < Fe < Co < Cu < Zn

(B) Cr < Zn < Co < Cu < Fe

(C) Zn < Cu < Co < Fe < Cr

(D) Zn < Cr < Fe < Co < Cu

Answer: (D)
Fact Based
Density depends on many factors like atomic mass. atomic radius and packing efficiency.

Q 15. The Correct set from the following in which both pairs are in correct order of melting point is

(A) LiF > LiCl ; NaCl > MgO

(B) LiF > LiCl ; MgO > NaCl

(C) LiCl > LiF ; NaCl > MgO

(D) LiCl > LiF ; MgO > NaCl

Answer: (B)
Generally,
M.P. ∝ Lattice energy = KQ₁Q₂ / r⁺ + r^–​

∝ (packing efficiency)

Q 16. The calculated magnetic moments (spin only value) for species

[FeCl₄]^2–, [Co(C₂O₄)³]^3– and MnO_2–⁴ respectively are:

Answer: (C)

[FeCl₄]^2–
Fe²+ 3d⁶ → 4 unpaired electrons. as Cl^– in a weak field liquid.

Q 17. Which of the following reagent is suitable for the preparation of the product in the following reaction?

(A) Red P + Cl₂

(B) NH₂-NH₂/ C₂H₅O^–Na⁺

(C) Ni/H₂

(D) NaBH₄

Answer: (B)

It is a wolf-kishner reduction of carbonyl compounds.

Q 18. The diazonium salt of which of the following compounds will form a coloured dye on reaction with β-Naphthol in NaOH?

Answer: (C)

Orange bright dye.

Q 19. The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is:

(A) (iv) < (i) < (iii) < (ii)

(B) (iv) < (i) < (ii) < (iii)

(C) (i) < (ii) < (iii) < (iv)

(D) (iv) < (iii) < (ii) < (i)

Answer: (C)

Reactivity ∝ – M group present at o/p position.

Q 20. Match List-I with List-II

List- I(Metal)	List-II(Ores)
(a) Aluminum	(i) Siderite
(b) Iron	(ii) Calamine
(c) Copper	(iii) Kaolinite
(d) Zinc	(iv) Malachite

(A) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)

(B) (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)

(C) (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

(D) (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

Answer: (C)
Siderite FeCO₃
Calamine ZnCO₃
Kaolinite Si₂Al₂O₅(OH)₄ or Al₂O₃.2SiO₂.2H₂O
Malachite CuCO₃.Cu(OH)₂

Section B

Q 1. The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O₂ for complete oxidation and produces 4 times its own volume of CO₂ is C_xH_y. The value of y is

Answer: 8
C_xH_y + 6O₂ 4CO₂ + y/2 H₂O
Applying POAC on ‘O’ atoms
6 × 2 = 4 × 2 + y/2 × 1
y/2 = 4 ⇒ y = 8

Q 2. The volume occupied by 4.75 g of acetylene gas at 50°C and 740 mmHg pressure is _______L. (Rounded off to the nearest integer)

(Given R = 0.0826 L atm K^–1 mol^–1)

Answer: 5
T = 50C = 323.15 K
P = 740 mm of Hg = 740 / 760 atm
V = ?
moles (n) = 4.75 / 26 atm
V = 4.75 / 26 × 0.0821 × 323.15 / 740 × 760
V = 4.97 5 Lit

Q 3. Sucrose hydrolysis in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25ºC. After 9h, the fraction of sucrose remaining is f. The value of is _________× 10^–2 (Rounded off to the nearest integer)

[Assume: ln10 = 2.303, ln2 = 0.693]

Answer: 81

Sucrose  ^__(Hydrolysis)→  Glucose + Fructose
t_1/2 = 3.33h = 10 / 3h

C_t = C₀ / {2 × T/t_1/2}

Fraction of sucrose remaining = f = Ct/Co = 1/ {2 × T/t_1/2}

1/f = 2^(t/t_1/2)

Log(1/f) = log(2^(t/t_1/2)) = t/t_1/2 log(2)

9/ (10/3) × 0.3 – 8.1/10 – 0.81

= x × 10^–2

x = 81

Q 4. The total number of amines among the following which can be synthesized by Gabriel synthesis is _______

Answer: 3
Only 1^o amines can be prepared by Gabriel synthesis.

Q 5. 1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is ____× 10^–2.

Answer: 243

93 g Aniline produce 135 g acetanilide
1.86 g produce 135 × 1.86 / 93 = 2.70 g
At 10% loss, 90% product will be formed after purification.

Q 6. Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is ______.

(a) α-sulphur (B) β-sulphur (C) S₂-form

Answer: (A)
S₂ is like O₂ i.e. paramagnetic as per molecular orbital theory.

Q 7. C₆H₆ freezes at 5.5ºC. The temperature at which a solution of 10 g of C₄H₁₀ in 200 g of C₆H₆ freeze is ________ C. (The molal freezing point depression constant of C₆H₆ is 5.12C/m).

Answer: 1
ΔT_f = i × K_f × m
= 1 × 5.12 × 10 / 58200 × 1000
∆T_f = 5.12 × 50 / 58 = 4.414
T_f(solution) = T_k(solvent) – ΔT = 5.5 – 4.414 = 1.086oC
≈ 1.09°C = 1 (nearest integer)

Q 8. Assuming ideal behavior, the magnitude of log K for the following reaction at 25ºC is x × 10^–1. The value of x is __________. (Integer Answer)

3HC ≡ CH(g) ⇌ C₆H₆ (l)

[Given: Δ_fG° (HC = CH) = – 2.04 × 10⁵] mol^–1; 

Δ_fG°(C₆H₆) = – 1.24 × 105 J mol^–1;

R = 8.314 J K^–1 mol^–1] 

Answer: 855
ΔG°_r = ΔG°_f [C₆H₆ (l)] – 3 × ΔG°_f [HC = CH]

Q 9. The magnitude of the change in oxidising power of the MnO^4-/ Mn²⁺ couple is x × 10^–4 V, if the H+ concentration is decreased from 1M to 10^–4 M at 25°C. (Assume concentration of MnO^4- and Mn²⁺ to be same on change in H⁺ concentration). The value of x is _____. (Rounded off to the nearest integer).

[Given: 230RT / F = 0.059]

Answer: 3776
5e- + MnO^–₄ + 8H → Mn⁺² + 4H₂O

= 3776 × 10^–4
So, x = 3776

Q 10. The solubility product of PbI₂ is 8.0 × 10^–9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^–6 mol/L. The value of x is ________ (Rounded off to the nearest integer)

Given: √2 = 1.41

Answer: 141
PbI₂(s) ⇌ Pb²⁺ (aq) + 2I^– (aq)
S+0.1 2s
K_SP (PbI₂) = 8 × 109
K_SP = [Pb⁺²][I^–]²
8 × 10^–9 = (S + 0.1) (2S)² ⇒ (8 × 10^–9 + 0.1) × 4S²
⇒ S² = 2 × 10^–8
S = 1.414 × 10^–4 mol/Lit
= x × 10^–6 mol/Lit
∴ x = 141.4 141

[/tab]

[tab title=”MATHS”]

Maths

SECTION A

Q 1: Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (10/3, 7/3). If α, β are the roots of the equation ax² + bx + 1 = 0, then the value of α² + β² – αβ is:

(A) 71/256

(B) -69/256

(C) 69/256

(D) -71/256

Answer: (D)
2b = a + c
(2a + 2)/3 = 10/3 and (2b + c)/3 = 7/3
⇒ a = 4
2b + c = 7, 2b – c = 4 },solving
b = 11/4 and c = 3/2
∴ Quadratic equation is 4x²+ (11/4)x + 1 = 0
∴ The value of (α + β)² – 3αβ = (121/256) – (¾)
= -71/256

Q 2: The value of the integral, where [x] denotes the greatest integer less than or equal to x , is :

(A) –4

(B) –5

(C) -√2 – √3 – 1

(D) -√2 – √3 + 1

Answer: (C)

I =

Put x – 1 = t ; dx = dt

I = -6 + (√2 – 1) + 2√3 – 2√2 + 6 – 3√3
I = -1 – √2 – √3

Q 3: Let f: R → R be defined as

Let A = {x ∈ R ∶ f is increasing}. Then A is equal to :

(A) (-5, -4) ∪ (4,∞)

(B) (-5, ∞)

(C) (-∞, -5) ∪ (4, ∞)

(D) (-∞, -5) ∪ (-4, ∞)

Answer: (A)
f’(x) = {-55 ; x< -5 6(x² – x – 20) ; -5 < x < 4 6(x²– x – 6) ; x > 4
⇒ f’(x) = {-55 ; x< -5 6(x – 5)(x + 4) ; –5 < x < 4 6(x – 3)(x + 2) ; x > 4
Hence, f(x) is monotonically increasing in (-5, -4) ∪ (4, ∞)

Q 4: If the curve y = ax² + bx + c,x ∈ R passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are:

(A) a = 1, b = 1, c = 0

(B) a = -1, b = 1, c = 1

(C) a = 1, b = 0, c = 1

(D) a = 1/2, b = 1/2 ,c = 1

Answer: (A)
2 = a + b + c
dy/dx = 2ax + b, (dy/dx)(0,0) = 1
⇒ b = 1 and a + c = 1
Since (0, 0) lies on curve,
∴ c = 0, a = 1
TRICK: (0, 0) lies on the curve. Only option (1) has c = 0

Q 5: The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is:

(A) 65/2⁷

(B) 135/2⁹

(C) 65/2⁸

(D) 35/2⁷

Answer: (B)
Let A and B be two subsets.
For each x∈{1,2,3,4,5} , there are four possibilities:
x ∈ A∩B, x∈ A’∩B, x ∈ A∩B’, x ∈ A’∩B’
So, the number of elements in sample space =45
Required probability
= (⁵C₂ × 3³)/4⁵
= (10×27)/2¹⁰
= 135/2⁹

Q 6: The vector equation of the plane passing through the intersection of the planes 

and

and the point (1, 0, 2) is:

Answer: (B)
Family of planes passing through the intersection of planes is

The above curve passes through

(3 − 1) + λ(1 + 2) = 0
⇒ λ = -2/3
Hence, equation of plane is

⇒

TRICK: Only option (2) satisfies the point (1, 0, 2)

Q 7: If P is a point on the parabola y = x² + 4 which is closest to the straight line y = 4x − 1, then the coordinates of P are :

(A) (–2, 8)

(B) (1, 5)

(C) (3, 13)

(D) (2, 8)

Answer: (D)
Tangent at P is parallel to the given line.
dy/dx|P = 4
⇒2x₁ = 4

⇒x₁ = 2
Required point is (2, 8)

Q 8: Let A and B be 3×3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A2B2 − B2A2)X = O, where X is a 3×1 column matrix of unknown variables and O is a 3×1 null matrix, has:

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

Answer: (C)
AT = A, BT = –B
Let A²B² − B²A²= P
PT = (A²B² – B²A²)T
= (A²B²)T – (B²A²)T
= (B²)T (A²)T– (A²)T (B²)T
= B²A² – A²B
⇒ P is a skew-symmetric matrix.

∴ ay + bz = 0 …(1)
–ax + cz = 0 …(2)
–bx – cy =0 …(3)
From equation (1), (2), (3)
Δ = 0 and Δ₁ = Δ₂= Δ₃ = 0
∴ System of equations has infinite number of solutions.

Q 9: If n ≥ 2 is a positive integer, then the sum of the series n+¹C₂ + 2(²C₂ + ³C₂ + ⁴C₂ + …. + ⁿC₂) is

(A) n(n + 1)² (n + 2)/12

(B) n(n – 1)(2n + 1)/6

(C) n(n + 1)(2n + 1)/6

(D) n(2n + 1)(3n + 1)/6

Answer: (C)
²C₂ = ³C₃
Let S = ³C₃ + ³C₂ + ……. + ⁿC₂ = n+¹C₃ (∵ ⁿC_r + ⁿC_r–1 = ⁿ⁺¹C_r)
∴ ⁿ⁺¹C₂ + ⁿ⁺¹C₃ + ⁿ⁺¹C₃
= ⁿ⁺²C₂ + ⁿ⁺¹C₃
=(n + 2)!/3!(n – 1)! + (n + 1)!/3!(n – 2)!
=(n + 2)(n + 1)n/6 + (n + 1)(n)(n – 1)/6 = (n(n + 1)(2 + 1))/6
TRICK : Put n = 2 and verify the options.

Q 10: If a curve y = f(x) passes through the point (1, 2) and satisfies xdy/dx + y = bx4, then for what value of b,  

(A) 5

(B) 62/5

(C) 31/5

(D) 10

Answer: (D)
dy/dx + y/x = bx³
I.F. =

= x
∴ yx = ∫bx⁴ dx = bx⁵/5 + c
The above curve passes through (1, 2).
2 = b/5 + c
Also,

⇒ (b/25) ×32 + c ln 2 – b/25 = 62/5
⇒ c = 0 and b = 10

Q 11: The area of the region: R{(x, y): 5x² ≤ y ≤ 2x² + 9} is:

(A) 9√3 square units

(B) 12√3 square units

(C) 11√3 square units

(D) 6√3 square units

Answer: (B)

Required area
​

= 12√3

Q 12: Let f(x) be a differentiable function defined on [0,2] such that f’(x) =f ‘(2 – x) for all x∈(0, 2), f(0) = 1 and f(2) = e². Then the value of is:

(A) 1 + e²

(B) 1 – e²

(C) 2(1 – e²)

(D) 2(1+e²)

Answer: (A)
f'(x) = f'(2 – x)
On integrating both sides, we get
f(x) = -f(2 – x)+c
Put x = 0
f(0) + f(2) = c
⇒ c = 1+ e²
⇒ f(x) + f(2 – x) = 1 + e²
I =

= 1+e²

Q 13: The negation of the statement ~p∧(p∨q) is∶

(A) ~p∧q

(B) p∧~q

(C) ~p∨q

(D) p∨∼q

Answer: (D)
Negation of ~p∧(p∨q) is
∼[~p∧(p∨q)]
≡ p∨ ∼(p∨q)
≡ p∨(∼p∧∼q)
≡ (p∨∼p)∧(p∨∼q)
≡ T∧(p∨∼q), where T is tautology.
≡ p∨∼q

Q 14: For the system of linear equations: x − 2y = 1, x − y + kz = −2, ky + 4z = 6,k ∈ R

Consider the following statements:

(A) The system has unique solution if k ≠ 2,k ≠ −2.

(B) The system has unique solution if k = -2.

(C) The system has unique solution if k = 2.

(D) The system has no-solution if k = 2.

(E) The system has infinite number of solutions if k ≠ -2.

Which of the following statements are correct?

(A) (B) and (E) only

(B) (C) and (D) only

(C) (A) and (D) only

(D) (A) and (E) only

Answer: (C)
x -2y + 0.z = 1
x – y + kz = -2
0.x + ky + 4z = 6
 

= 4 – k2
For unique solution, 4 – k² ≠ 0
k ≠ ±2
For k = 2,
x – 2y + 0.z =1
x – y + 2z = -2
0.x + 2y + 4z = 6

= -8 + 2 (-20)
⇒Δx= -48 ≠ 0
For k = 2,Δx ≠ 0
So, for k = 2, the system has no solution.

Q 15: For which of the following curves, the line x + √3y = 2√3 is the tangent at the point (3√3/2, 1/2)?

(A) x² + 9y² = 9

(B) 2x² – 18y² = 9

(C) y² = x/(6√3)

(D) x² + y² = 7

Answer: (A)
Tangent to x² + 9y² = 9 at point (3√3/2, 1/2) is x(3√3)/2 + 9y(1/2) = 9
⇒3√3 x + 9y = 18
⇒x +√3 y = 2√3
⇒ Option (A) is true.

Q 16: The angle of elevation of a jet plane from a point A on the ground is 600. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 300. If the jet plane is flying at a constant height, then its height is:

(A) 1200√3 m

(B) 1800√3 m

(C) 3600√3 m

(D) 2400√3 m

Answer: (A)

v = 432 × 1000/(60×60) m/sec
= 120 m/sec
Distance PQ = v × 20 = 2400 m
In ΔPAC
tan 60⁰ = h/AC
⇒ AC = h/(√3)
In ΔAQD
tan 30⁰ = h/AD
⇒AD = √3h
AD = AC + CD
⇒√3 h = h/√3 + 2400
⇒2h/√3 = 2400
⇒ h = 1200√3 m

Q 17: For the statements p and q, consider the following compound statements:

(a) (~q∧(p→q)) → ~p

(b) ((p∨q))∧~p) → q

Then which of the following statements is correct?

(A) (A) is a tautology but not (B)

(B) (A) and (B) both are not tautologies

(C) (A) and (B) both are tautologies

(D) (B) is a tautology but not (A)

Answer: (C)

(A) is tautology.

(B) is tautology.
∴ (A) and (B) both are tautologies.

Q 18: Let a, b∈R. If the mirror image of the point P(a, 6, 9) with respect to the line (x – 3)/7 = (y – 2)/5 = (z – 1)/(-9) is (20, b, -a, -9), then |a + b| is equal to:

(A) 86

(B) 88

(C) 84

(D) 90

Answer: (B)
P(a, 6, 9), Q (20, b, -a-9)
Midpoint of PQ=((a+20)/2, (b+6)/2, -a/2) lie on the line.

=> (a + 20 – 6)/14 = (b + 6 – 4)/10 = (-a – 2)/(-18)
=> (a + 14)/14 = (a + 2)/18
=> 18a + 252 = 14a + 28
=> 4a = -224
a = -56
(b + 2)/10 = (a + 2)/18
=> (b + 2)/10 = (-54)/18
=>(b + 2)/10 = -3
=> b = -32
|a + b| = |-56 – 32|
= 88

Q 19: Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x∈R. If |f(x) f'(x) f'(x) f”(x)| = 0, for all x∈R, then the value of f(1) lies in the interval:

(A) (9, 12)

(B) (6, 9)

(C) (3, 6)

(D) (0, 3)

Answer: (B)
Given f(x) f”(x) – f'(x)² =0
Let h(x) = f(x)/f'(x)
Then h'(x) = 0
⇒ h(x) = k
⇒ f(x)/f'(x) = k
⇒ f(x) = k f’(x)
⇒ f(0) = k f'(0)
⇒ k = 1/2
Now, f(x) = ½ f'(x)
⇒∫ 2dx = ∫ f'(x)/f(x) dx
⇒ 2x = ln |f(x)| + C
As f(0) = 1 ⇒ C = 0
⇒ 2x = ln |f(x)|
⇒ f(x) = ±e^2x
As f(0) = 1 ⇒ f(x) = e^2x
∴ f(1) = e² ≈ 7.38

Q 20: A possible value of tan (¼ sin^-1 √63/8) is:

(A) 1/(2√2)

(B) 1/√7

(C) √7 – 1

(D) 2√2 – 1

Answer: (B)
tan (¼ sin^-1 √63/8)
Let sin^-1(√63/8) = θ
sin θ = √63/8

cos θ = 1/8
2 cos²(θ/2) – 1 = 1/8
⇒ cos² θ/2 = 9/16
cos θ/2 = 3/4
⇒(1- tan² θ/4 )/(1 + tan² θ/4) = 3/4
tan θ/4 = 1/√7

Section B

Q 1: If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x – 2)²+ (y – 3)² = 25 at the point (5, 7) is A, then 24A is equal to ___.

Answer: Q is wrong

Equation of normal at P is
(y – 7) = ((7 – 3)/(5 – 2))(x – 5)
⇒ 3y – 21 = 4x – 20
⇒ 4x – 3y + 1 = 0
⇒ M is (-1/4, 0)
Equation of tangent at P is
(y – 7) = (-¾) (x – 5)
⇒ 4y – 28 = -3x + 15
⇒ 3x + 4y = 43
⇒ N is (43/3, 0)
The question is wrong. The normal cuts at a point on the negative axis.

Q 2: If a + α = 1, b + β = 2 and af(x) + αf(1/x) = bx + β/2, x ≠ 0 then the value of the expression [f(x) + f(1/x) ]/(x + 1/x)

Answer: 2
af(x) + αf(1/x) = bx + β/x …(i)
Replace x by 1/x
af(1/x) + αf(x) = b/x + βx …(ii)
(i) + (ii)
(a + α)[f(x) + f(1/x) ]
= (x + 1/x)(b + β)
⇒ ( f(x) + f(1/x))/(x + 1/x)
= 2/1
= 2

Q 3: If the variance of 10 natural numbers 1,1,1,…,1,k is less than 10, then the maximum possible value of k is ________.

Answer: 11

⇒ 10(9 + k²) – (81 + k² + 18k) < 1000
⇒ 90 + 10k² – k² – 18k – 81 < 1000
⇒ 9k²– 18k + 9 < 1000
⇒ (k – 1)² < 1000/9
⇒ k – 1< (10√10)/3
⇒ k < (10√10)/3 + 1
Maximum possible integral value of k is 11.

Q 4: Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point (-5, 0). If the locus of the point P is a circle of radius r, then 4r² is equal to ___.

Answer: 56.25
Let P be (h, k), A(5, 0) and B(-5, 0)
Given PA = 3PB
⇒ PA² = 9PB²
⇒ (h-5)² + k² = 9[(h + 5)² + k²]
⇒ 8h² + 8k² + 100h + 200 = 0
∴ Locus of P is x² + y² + (25/2)x + 25 = 0
Centre = (-25/4, 0)
∴ r² = (-25/4)2 – 25
= 625/16 – 25
= 225/16
∴4r² = 4×225/16
= 225/4
= 56.25

Q 5: The number of the real roots of the equation (x + 1)² + |x – 5| = 27/4 is ___.

Answer: 2
For x ≥ 5,
(x + 1)²+ (x – 5) = 27/4
⇒ x² + 3x – 4 = 27/4
⇒ x² + 3x – 43/4 = 0
⇒ 4x² + 12x – 43 = 0
x = (-12 ± √(144 + 688))/8
x = (-12 ± √832)/8
= (-12 ± 28.8)/8
= (-3 ± 7.2)/2
= (-3 + 7.2)/2, (-3 – 7.2)/2 (therefore, no solution)
For x < 5,
(x + 1)²– (x – 5) = 27/4
⇒ x² + x + 6 – 27/4=0
⇒ 4x² + 4x – 3=0
x = (-4 ± √(16 + 48))/8
x = (-4 ± 8)/8
⇒ x = -12/8, 4/8
∴ 2 real roots.

Q 6: The students S₁, S₂,… S₁₀ are to be divided into 3 groups A, B, and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ___.

Answer: 31650

Number of ways
= ¹⁰C₁ [2⁹ – 2] + ¹⁰C₂ [2⁸– 2] + ¹⁰C₃ [2⁷ – 2]
= 27 [¹⁰C₁×4 + ¹⁰C₂ ×2 + ¹⁰C₃] – 20 – 90 – 240
= 128 [40 + 90 + 120 ] – 350
= (128 × 250) – 350
= 10 [3165]
= 31650

Q 7: For integers n and r, let

The maximum value of k for which the sum

exists, is equal to ____.

Answer: Bonus
(1+x)¹⁰ = ¹⁰C₀ + ¹⁰C₁x + ¹⁰C₂x² + …… + ¹⁰C₁₀x¹⁰
(1+x)¹⁵= ¹⁵C₀ + ¹⁵C₁x +¹⁵C₂x² + …… + ¹⁵C_k-1x^k-1 + ¹⁵C_k+1x^k+1 + …¹⁵C₁₅x¹⁵

Coefficient of x^k+1 in (1+x)²⁵
= ²⁵C_k+1
²⁵C_k + ²⁵C_k+1 = ²⁶C_k+1
For maximum value
as per given, k can be as large as possible.

Q 8: Let λ be an integer. If the shortest distance between the lines x – λ = 2y – 1 = -2z and x = y + 2λ = z – λ is √7/2√2, then the value of |λ| is __

Answer: 1
(x – λ)/1 = (y – 1/2)/(1/2) = z/(-1/2)
(x – λ)/2 = (y-1/2)/1 = z/(-1) …(1) Point on line = (λ, 1/2, 0)
x/1 = (y + 2λ)/1 = (z – λ)/1 …(2) Point on line = (0, -2λ, λ)

= |-5λ – 3/2|/√14
= √7/(2√2) (Given)
⇒ |10λ + 3| = 7
⇒ λ = -1 as λ is an integer.
⇒ |λ| = 1

Q 9: If i = √-1. If [(-1 + i√3)²¹/(1 – i)²⁴ + (1 + i√3)²¹/(1 + i)²⁴ ] = k, and n = [|k| ] be the greatest integral part of |k|. Then

is equal to ___

Answer: 310

= 2⁹ e^i(20π) +2⁹ e^iπ
= 2⁹ + 2⁹ (-1)
= 0 = k
∴ n = 0
∑(j=0)⁵ (j + 5)²– ∑(j=0)⁵ (j + 5)
= [5² + 6² + 7² + 8² + 9² + 10² ] – [5 + 6 + 7 + 8 + 9 + 10]
= [(1²+ 2² +… + 10²) – (1² + 2²+ 3²+ 4²) ] – [(1 + 2 + 3 + … + 10) – (1 + 2 + 3 + 4) ]
= (385 – 30) -[55 – 10]
= 355 – 45
= 310

Q 10: The sum of first four terms of a geometric progression (G.P.) is 65/12 and the sum of their respective reciprocals is 65/18. If the product of first three terms of the G.P. is 1, and the third term is α then 2α is____

Answer: 3
Let terms of GP are a, ar, ar², ar³
a + ar + ar² + ar³ = 65/12 …….(1)
1/a + 1/ar + 1/(ar²) + 1/ar³) = 65/18
⇒1/a (r³ + r² + r + 1)/r³) = 65/18 …….(2)
(1)/(2), we get
a²r³ = 18/12
= 3/2
Also, a³ r³ = 1
⇒ a(3/2) = 1
⇒a = 2/3
(4/9) r³ = 3/2
⇒ r³ = 33/23
⇒ r = 3/2
α = ar²
= 2/3.(3/2)² = 3/2
∴2α = 3

[/tab]

[/tabs]

[tabs title=”JEE MAINS EXAM Previous Year Paper” type=”centered”]

[tab title=”Physics”]

PHYSICS

SECTION A

Q 1. A current through a wire depends on time as i = α₀t + βt² where α₀ = 20 A/s and β = 8 As^-2. Find the charge crossed through a section of the wire in 15 s.

(A) 2100 C

(B) 260 C

(C) 2250 C

(D) 11250 C

Answer: (D) 

 

Q 2. Each side of a box made of metal sheet in cubic shape is ‘a’ at room temperature ‘T’, the coefficient of linear expansion of the metal sheet is ‘α’. The metal sheet is heated uniformly, by a small temperature ΔT, so that its new temperature is T+ΔT. Calculate the increase in the volume of the metal box:

(A) 4/3 πa³ α ΔT

(B) 4πa³ α ΔT

(C) 3a³ α ΔT

(D) 4a³ α ΔT

Answer: (C)

 

Q 3. Given below are two statements:

Statement-I: Two photons having equal linear momenta have equal wavelengths.

Statement-II: If the wavelength of a photon is decreased, then the momentum and energy of a photon will also decrease.

In the light of the above statements, choose the correct answer from the options given below.

(A) Statement-I is false but Statement-II is true

(B) Both Statement-I and Statement-II are true

(C) Both Statement-I and Statement-II are false

(D) Statement-I is true but Statement-II is false

Answer: (D) 

 

Q 4. A cube of side ‘a’ has point charges +Q located at each of its vertices except at the origin where the charge is –Q. The electric field at the centre of cube is:

Answer: (C) 

 

Q 5. If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

Answer: (A)

 

Q 6. In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:

Answer: (A) 

 

Q 7. Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be

​

Answer: (A)

 

Q 8. In Young’s double-slit experiment, the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

(A) 4: 1

(B) 2: 1

(C) 3: 1

(D) 1: 4

Answer: (A) 

 

Q 9. Consider two satellites S₁ and S₂ with periods of revolution 1 hr and 8 hr respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S₁ to the angular velocity of satellite S₂ is

(A) 8: 1

(B) 1: 8

(C) 2: 1

(D) 1: 4

Answer: (A)

 

Q 10. If an emitter current is changed by 4mA, the collector current changes by 3.5 mA. The value of β will be:

(A) 7

(B) 0.875

(C) 0.5

(D) 3.5

Answer: (A)

 

Q 11. n moles of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes:

A→ B: Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ and pressure changes from P₁ to P₂.

B → C: Isobaric compression at pressure P₂ to initial volume V₁.

C → A: Isochoric change leading to change of pressure from P₂ to P₁.

Total work done in the complete cycle ABCA is –

(A) 0

(B) nRT(ln₂ + 1/2)

(C) nRTln₂

(D) nRT (ln₂ – 1/2)

Answer: (D)

 

Q 12. The work done by a gas molecule in an isolated system is given by, , where x is the displacement, k is the Boltzmann constant and T is the temperature α and β are constants. Then the dimensions of β will be:

 

(A) [M⁰LT⁰]

(B) [M²LT²]

(C) [MLT^–2]

(D) [ML²T^–2]

Answer: (C)

 

Q 13. Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be

(A) 2: 1

(B) 1: 4

(C) 4: 1

(D) 1: 2

Answer: (B)

 

Q 14. Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as:

I₁ = M.I. of thin circular ring about its diameter,

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Then:

(A) I₁ = I₂ = I₃< I₄

(B) I₁ + I₂ = I₃ + 5/2 I₄

(C) I₁ + I₃< I₂ + I₄

(D) I₁ = I₂ = I₃> I₄

Answer: (D)

 

Q 15. A cell E₁ of emf 6V and internal resistance 2Ω is connected with another cell E₂ of emf 4V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is

(A) 3.6V

(B) 10.0V

(C) 5.6V

(D) 2.0V

Answer: (C) 

 

Q 16. The focal length f is related to the radius of curvature r of the spherical convex mirror by:

(A) f = r

(B) f = – ½ r

(C) f = +½ r

(D) f = – r

Answer: (C) 

 

Q 17. If Y, K and η are the values of Young’s modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.

Answer: (A) 

 

Q 18. In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.

The transition A, B and C respectively represents:

(A) The series limit of Lyman series, third member of Balmer series and second member of Paschen series

(B) The first member of the Lyman series, third member of Balmer series and second member of Paschen series

(C) The ionization potential of hydrogen, second member of Balmer series and third member of Paschen series

(D) The series limit of Lyman series, second member of Balmer series and second member of Paschen series

Answer: (A)

 

Q 19. Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is

Answer: (A) 

 

Q 20. Match List-I with List-II

List – I	List – II
(A)  Isothermal	(i) Pressure constant
(B)  Isochoric	(ii) Temperature constant
(C)  Adiabatic	(iii) Volume constant
(D)  Isobaric	(iv) Heat content is constant

Choose the correct answer from the options given below:

(A) (A)  – (ii), (B)  – (iv), (C)  – (iii), (D)  – (i)

(B) (A)  – (ii), (B)  – (iii), (C)  – (iv), (D)  – (i)

(C) (A)  – (i), (B)  – (iii), (C)  – (ii), (D)  – (iv)

(D) (A)  – (iii), (B)  – (ii), (C)  – (i), (D)  – (iv)

Answer: (B) 

SECTION-B

Q 1. In connection with the circuit drawn below, the value of current flowing through the 2kΩ resistor is _______ × 10–4 A.

Answer: 25
In zener diode there will be o change in current after 5V
Zener diode breakdown
⇒ i = 5 / 2 × 103
⇒ i = 2.5 × 10^–3 A
⇒ i = 25 × 10^–4 A

 

Q 2. An inclined plane is bent in such a way that the vertical cross-section is given by y = x² / 4 where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction μ = 0.5, the maximum height in cm at which a stationary block will not slip downward is _____ cm.

Answer: 25

Given,
y = x² / 4
μ = 0.5
Condition for block will not slip downward
mg sin θ = μmg cos θ

⇒ tan θ = μ
And we know that
⇒ tanθ = dv / dx
⇒ dv / dx = μ ⇒ x/2 = 0.5 [y = x² / 4 dy / dx = x / 2]
⇒ x = 1,
put x = 1 in equation y = x²/4
⇒ y = (1)2 / 4 ⇒ y = ¼ ⇒ y = 0.25
y = 25 cm

 

Q 3. An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of the light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30° in clockwise direction, the intensity of emerging light will be________ Lumens.

 

Answer: 75
Given: I₀ = 100 lumens, θ = 30^o
Inet = I₀ cos2θ
Inet = 100 × (√3/2)2 = 100 × 3 / 4
Inet = 75 lumens

 

Q 4. The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be ______ N.

[g = 10 ms-2]

Answer: 25

Given: μs = 0.2
m = 0.5 kg
g = 10 m/s2
We know that
fs = μsN and …. (1)
To keep the block adhere to the wall
Here, N = F … (2)
fs = mg …. (3)
From equation (1), (2), and (3), we get
⇒mg = μs F
⇒ F = mg / μs ⇒ F = 0.5 × 10 / 0.2
F = 25 N (for all values of F greater than or equal to 25 N this case is possible)

 

Q 5. A hydraulic press can lift 100 kg when a mass ‘m’ is placed on the smaller piston. It can lift _______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass ‘m’ on the smaller piston.

Answer: 25600

Atmospheric pressure P0 will be acting on both the limbs of the hydraulic lift.
Applying pascal’s law for the same liquid level
⇒ P₀ + mg / A₁ = P_o + (100)g / A₂
⇒ mg / A₁ = (100)g / A₂ ⇒ m / 100 = A₁/ A₂ …(1)
Diameter of piston on side of 100 kg is increased by 4 times so new area = 16A₂
Diameter of piston on side of (m) kg is decreasing
A₁ = A₁ / 16
(In order to increasing weight lifting capacity, diameter of smaller piston must be reduced)
Again, mg / (A₁/16) = M’g / 16A₂ ⇒ 256m / M’ = A₁/ A₂
From equation (1) = 256m / M’ = m / 100 ⇒ M’ = 25600 kg

 

Q 6. An audio signal υm = 20sin2π(1500t) amplitude modulates a carrier υc =80 sin 2π (100,000t). The value of percent modulation is ________.

Answer: 25
We know that, modulation index = A_m / A_c
From given equations, A_m = 20 and A_c = 80
Percentage modulation index = A_m / A_c × 100
⇒ 20 / 80 × 100 = 25%
The value of percentage modulation index is
= 25

 

Q 7. A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of the secondary coil are 24, then the number of turns of the primary are ______.

Answer: 440
Given,
Primary voltage, Vp = 220 V
Secondary voltage, vs = 12 V
No. of turns in secondary coil is Ns = 24
No. of turns in primary coil, Np = ?
We know that for a transformer
⇒ Np / Ns = Vp / Vs
⇒ Np = Vp × Ns / Vs = 220 × 24 / 12
⇒ Np = 440

 

Q 8. A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30° with the original direction. The ratio of velocities of the balls after collision is x : y, where x is _________.

Answer: 1

Momentum is conserved just before and just after the collision in both x-y direction.
In y-direction,
pi = 0
pf = mv₁ sin30^o – mv₂ sin30^o
Pf = m × ½ v₁ – m × ½ v₂
pi = pf, so
= mv₁ / 2 – mv₂ / 2 = 0
⇒ mv₁ / 2 = mv₂ / 2 ⇒ v₁ = v₂
v₁ / v₂ = 1

 

Q 9. An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is _______ × 107 m/s.

Answer: 15
Given: f = 5 GHz
εr =2
μr = 2
Velocity of wave ⇒ v = c / n ….(1)
Where, n = √μrεr and c = speed of light = 3 × 108 m/s
n = √2 × 2 = 2
put the value of n in we get
⇒ v = 3 × 108 / 2 = 15 × 107 m/s
⇒ X × 107 = 15 × 107
X = 15

 

Q 10. A resonance circuit having inductance and resistance 2 × 10–4 H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of the quality factor of this resonator is________. [π = 3.14]

Answer: 2000
Given: R = 6.28 Ω
f = 10 MHz
L = 2 × 10-4 Henry
We know that quality factor Q is given by
⇒ Q = XL / R = ωL / R
also, ω = 2πf, so
⇒ Q = 2πfL / R
⇒ Q = 2π × 10 × 106 × 2 × 10-4 / 6.28 = 2000
Q = 2000

[/tab]

[tab title=”Chemistry”]

CHEMISTRY

SECTION-A 

Q 1. Which reagent (A) is used for the following given conversion?

(A) Cu / ∆ / high pressure

(B) Molybdenum Oxide

(C) Manganese Acetate

(D) Potassium Permanganate

Answer: (B)

 

Q 2. S-1: Colourless cupric metaborate is converted into cuprous metaborate in a luminous flame.

S-2: Cuprous metaborate is formed by reacting copper sulphate with boric anhydride heated in non luminous flame.

(A) S-1 is true and S-2 is false

(B) S-1 is false and S-2 is true

(C) Both are true

(D) Both are false

Answer: (D)

 

Q 3. Find A and B.

Answer: (B)

 

Q 4. EoM2+/ M has a positive value for which of the following elements of 3d transition series?

(A) Cu

(B) Zn

(C) Cr

(D) Co

Answer: (A)

 

Q 5. What is the reason for the formation of a meta product in the following reaction?

(A) Aniline is ortho/para directing

(B) Aniline is meta directing

(C) In acidic medium, aniline is converted into anilinium ion, which is ortho/para directing

(D) In acidic medium, aniline is converted into anilinium ion which is meta directing

Answer: (D) 

 

Q 6. Identify X, Y, Z in the given reaction sequence.

(A) X = Na[Al(OH)₄] ; Y = CO₂ ; Z = Al₂O₃.xH₂O

(B) X = Na[Al(OH)₄] ; Y = SO₂ ; Z = Al₂O₃.xH₂O

(C) X = Al(OH)₃ ; Y = CO₂ ; Z = Al₂O₃

(D) X = Al(OH)₃ ; Y = SO₂ ; Z = Al₂O₃

Answer: (A) 

 

Q 7. The missing reagent P is:

Answer: (A) 

Q 8. Arrange Mg, Al, Si, P and S in the correct order of their ionisation potentials.

 

Answer: P > S > Si > Mg > Al

 

Q 9. Which force is responsible for the stacking of the α-helix structure of protein? Hydrogen (¹H, ²H, ³H) is ________.

(A) H-bond

(B) Ionic bond

(C) Covalent bond

(D) Van der Waals forces

Answer: (A) 

 

Q 10. The composition of gun metal is:

(A) Cu, Zn, Sn

(B) Al, Mg, Mn, Cu

(C) Cu, Ni, Fe

(D) Cu, Sn, Fe

Answer: (A)

 

Q 11. The gas evolved due to anaerobic degradation of vegetation causes:

(A) Global warming and cancer

(A) Global warming and cancer

(B) Acid rain

(C) Ozone hole

(D) Metal corrosion

Answer: (A) 

 

Q 12. The slope of the straight line given in the following diagram for adsorption is:

(A) 1/n (0 to 1)

(B) 1/n (0.1 to 0.5)

(C) log n

(D) log (1/n)

Answer: (A)

 

Q 13. Match the following:

(i) Caprolactam	(a)  Neoprene
(ii) Acrylonitrile	(b)  Buna N
(iii) 2-chlorobuta-1,3-diene	(c)  Nylon – 6
(iv) 2-Methylbuta-1,3-diene	(d)  Natural rubber

(A) (i) →(b) , (ii) → (c) , (iii) → (a) , (iv) → (d) 

(B) (i) → (a) , (ii) → (c) , (iii) → (b) , (iv) → (d) 

(C) (i) → (c) , (ii) → (b) , (iii) → (a) , (iv) → (d) 

(D) (i) → (c) , (ii) → (a) , (iii) → (b) , (iv) → (d) 

Answer: (C)

 

Q 14. In the given reactions,

1) I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂

2) H₂O₂ + HOCl → Cl^– + H₃O+ + O₂

(A) H₂O₂ acts as an oxidising agent in both the reactions

(B) H₂O₂ acts as a reducing agent in both the reactions

(C) H₂O₂ acts as an oxidising agent in reaction (1) and as a reducing agent in reaction (2)

(D) H₂O₂ acts as a reducing agent in reaction (1) and as an oxidizing agent in reaction (2)

Answer: (B) 

 

Q 15. What is the major product of the following reaction

Answer: (A) 

 

Q 16. Which of the following ores are concentrated by cyanide of group 1st element?

(A) Sphalerite

(B) Malachite

(C) Calamine

(D) Siderite

Answer: (A)

 

Q 17. What is the major product of the following reaction?

Answer: (C) 

 

Q 18. Which of the following pairs are isostructural

a) TiCl₄, SiCl₄

b) SO^2-₃, CrO^2-₃

c) NH₃, NO^–₃

d) ClF₃, BCl₃

(A) B, C

(B) A, C

(C) A, B

(D) A, D

Answer: (C)

 

Q 19. Identify the major product:

Answer: (B) 

Q 20. The products A and B are:

 

Answer: (A)

Section B

Q 1. Cu²⁺ + NH₃ ⇌ [Cu(NH₃)]²⁺ K₁ = 10⁴

[Cu(NH₃)]²⁺ + NH₃ ⇌ [Cu(NH₃)]²⁺ K₂ = 1.58 × 10³
[Cu(NH₃)₂]²⁺ + NH₃ ⇌ [Cu(NH₃)₃]²⁺ K₃ = 5 × 10²
[Cu(NH₃)₃]²⁺ + NH₃ ⇌ [Cu(NH₃)₄]²⁺ K₄ = 10²
If the dissociation constant of [Cu(NH₃)₄]²⁺ is X × 10¹². 

 

Determine X.
Answer: 1.26
Overall reaction constant (β):
β = K₁× K₂ × K₃ × K₄
= 104 (1.58 ×103) × 5 × 102 × 102 = 7.9 × 10¹¹
Cu₂+ + 4NH₃ ⇌ [Cu(NH₃)₄]²⁺ ; β = 7.9 × 10¹¹
So, the dissociation constant (K_Disso.) will be:

1/β = 1/ 7.9 × 10¹¹

​K_Disso = 1 / β = 1.26 × 10^-12
Hence, the value of X = 1.26

 

Q 2. What is the coordination number in Body Centered Cubic (BCC) arrangement of identical particles?

Answer: 8
The easiest way is to look at the atom at the body center. It lies at the center of the body diagonal and touches the eight corner atoms. So, coordination number = 8.

 

Q 3. Cl₂(g)⇌ 2Cl(g)

For the given reaction at equilibrium, moles of Cl₂(g) is equal to the moles of Cl(g) and the equilibrium pressure is 1atm. If Kp of this reaction is x ×10^–1, find x.

Answer: 5
According to the question: Cl₂ ⇌ 2Cl [P_total = 1 atm].
Moles of Cl₂ = Moles of Cl (at equilibrium).
Given: Kp = x ×10^-1
nCl₂ = nCl.
Therefore, P Cl₂ = P Cl
Hence, P Cl₂ = P Cl = 0.5 atm
kp= (P Cl)₂/ P Cl₂ = 0.5 = 5 x 10^-1
So, x = 5

 

Q 4. Among the following compounds, how many are amphoteric in nature?

Be(OH)₂, BeO, Ba(OH)₂, Sr(OH)₂

Answer: 2
The oxide and hydroxide of Be are amphoteri(C) Hence BeO and Be(OH)2 is amphoteri(C) Ba(OH)₂ and Sr(OH)₂ are basic.

 

Q 5. S₈ + bOH^– → cS^2- + sS₂O^2-₃ + H₂O. Find the value of c.

Answer: 4
Let us look at the half reactions i.e. oxidation and reduction separately.
Oxidation:
S₈ → S₂O^2-₃
S₈ + 24OH^– → 4S₂O^2-₃ + 12H₂O +16e- ….(1)
Reduction:
S₈ → S^2-
S₈ + 16e- → 8S^2- ….(2)
Adding both the reactions (1) and (2),
2S₈ + 24OH^– → 4S₂O^2-₃ + 8S^2- + 12H₂O
Dividing the whole equation by 2,
S₈ + 12OH^– → 2S₂O^2-₃ + 4S^2- + 6H₂O
So, c = 4

 

Q 6. 4.5 g of a solute having molar mass of 90 g/mol is dissolved in water to make a 250 mL solution. Calculate the molarity of the solution.

Answer: 0.2
W_B (given weight of solute) = 4.5 g
M_B (Molar mass of solute) = 90 g/mol
V_S (Volume of solution) = 250 mL
= 250 / 1000 L = 1/4 L
Molarity (M) = nB / V_S(L) = W_B / M_B.V_S(L) = 4.5 / (90 ×1/4) = 0.2 molar

 

Q 7. Calculate the time taken in seconds for 40% completion of a first order reaction, if its rate constant is 3.3× 10-4 sec-1.

Answer: 1518

= 30 x 103 x 0.506 = 1518 sec

 

Q 8. 9.45g of CH₂ClCOOH is dissolved in 500 mL of H₂O solution and the depression in freezing point of the solution is 0.5°C. Find the percentage dissociation.

(Kf)H₂O = 1.86K kg mole^-1.

Answer: 34.4%
CH₂Cl COOH CH₂ClCOO^– + H⁺
Initial 100
Dissociated α α α
Left (1-α) α α
i = final moles / initial moles = 1 – α + α + α / 1 = 1 + α
ΔTf = i × K_f × m
0.5 = (1 + α) × 1.86 × m
Molality (m) = nB / WA (kg) = WBMB x WA (kg) = 9.45 × 1000 / 94.5 × 500 = 0.2
Here, A is for solute and B is for solvent
W_A=W_H2O = 500g (Density of H₂O = 1g/mL)
0.5 = (1+ α) ×1.86 × 0.2
α = 0.344
Percentage of dissociation = 34.4%

 

Q 9. For a chemical reaction, Keq is 100 at 300K, the value of ΔGo is –xR Joule at 1 atm pressure. Find the value of x. (Use ln 10 = 2.3)

Answer: 1382
Given: K_eq= 100 at 300K and ΔG^o = –xR Joule
ΔGo = -RTln(K_eq) = -2.303RTlog(K_eq) = -1381.8R
Therefore, x = 1381.8 or 1382

 

Q 10. The mass of Li³⁺ is 8.33 times the mass of a proton. If Li3+ and proton are accelerated through the same potential difference, then the ratio of de Broglie’s wavelength of Li³⁺ to proton is x ×10^–1. Find x

Answer: 2

m (Li³⁺) = 8.33 × m_p+ (given)
Debroglie’s wavelength (λ) = h / p
KE = ½ mv²
Multiplying by m on both sides
m. KE = ½ mv²
2m. KE = (mv)² = p²
P = √2m. KE
Also KE= q × V
So, P = 2mq.V

Comparing with x × 10 ^-1 = 2 × 10 ^-1
x = 2

[/tab]

[tab title=”Maths”]

MATHS 

SECTION A

Q 1: The statement among the following that is a tautology is:

(A) A∧(A∨B)

(B) B→[A∧(A→B)]

(C) A∨(A∧B)

(D) [A∧(A→B)]→B

Answer: (D)
A∧ (~ A∨B)→B
= [(A∧~A)∨(A∧B)]→ B
= (A∧ B)→ B
= ~ (A∧B)∨B
= t

 

Q 2: Let f :R→R be defined as f(x) = 2x-1 and g:R – {1} →R be defined as g(x) = (x-½)/(x-1).

Then the composition function f(g(x)) is:

(A) Both one-one and onto

(B) onto but not one-one

(C) Neither one-one nor onto

(D) one-one but not onto

Answer: (D)
f(g(x)) = 2g(x) – 1
= 2(x – 1/2)/(x – 1) – 1
= x/(x – 1)
f(g(x)) = 1 + 1/(x – 1)

∴ one-one, into

 

Q 3: If f:R→ R is a function defined by f(x) = [x – 1] cos ((2x – 1)/2)𝜋 , where [.] denotes the greatest integer function, then f is:

(A) discontinuous only at x = 1

(B) discontinuous at all integral values of x except at x = 1

(C) continuous only at x = 1

(D) continuous for every real x

 

Answer: (D)
Doubtful points are x = n, n∈I

f(n) = 0
Hence continuous.

 

Q 4: If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

(A) –2t³

(B) –t³

(C) 0

(D) 2t³

Answer: (A)
Equation of tangent at P(t, t³)
(y – t³) = 3t²(x – t) ⋯(1)
Now solve the above equation with
y = x³ ⋯(2)
By (1) & (2)
x³ – t³ = 3t² (x – t)
x² + xt + t² = 3t²
x² + xt – 2t² = 0
⇒(x – t)(x + 2t) = 0
⇒x = – 2t
⇒Q(-2t, -8t3)
Ordinate of required point = (2t³ + (-8t)³)/3
= -2t³

 

Q 5: The value of – ¹⁵C₁ + 2. ¹⁵C₂ – 3.¹⁵C₃+ ….-15.¹⁵C₁ + ¹⁴C₁+ ¹⁴C₃ + ¹⁴C₅ + …¹⁴C₁₁ is

(A) 2¹⁴

(B) 2¹³ – 13

(C) 2¹⁶ – 1

(D) 2¹³ – 14

Answer: (D) 

= (¹⁴C₁+ ¹⁴C₃ + ¹⁴C₅ + …¹⁴C₁₁ + ¹⁴C₁₃) – ¹⁴C₁₃
= 2¹³ – 14
∴ S₁ + S₂ = 2¹³ – 14

 

Q 6: An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:

(A) 3/16

(B) 1/2

(C) 5/16

(D) 1/32

Answer: (B)
P(odd no. twice) = P(even no. thrice)
ⁿC₂ (1/2)ⁿ = ⁿC₃ (1/2)ⁿ
⇒ n = 5
Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)
= ⁵C₂ (1/2)⁵ + ⁵C₂ (1/2)⁵ + ⁵C₂ (1/2)⁵
= 16/2⁵
= 1/2

 

Q 7: Let p and q be two positive number such that p + q = 2 and p4 + q4 = 272. Then p and q are roots of the equation:

(A) x² – 2x + 2 = 0

(B) x² – 2x + 8 = 0

(C) x² – 2x + 136 = 0

(D) x² – 2x + 16 = 0

Answer: (D)
(p² + q²)² – 2p²q² = 272
((p + q)² – 2pq)² – 2p²q² = 272
⇒16 – 16pq + 2p²q² = 272
(pq)² – 8pq –128 = 0
⇒pq = (8±24)/2 = 16, – 8
⇒pq = 16
Now
x² – (p + q)x + pq = 0
x² – 2x + 16 = 0

 

Q 8: The area (in sq.units) of the part of the circle x² + y² = 36, which is outside the parabola y² = 9x, is:

(A) 24π + 3√3

(B) 12π + 3√3

(C) 12π – 3√3

(D) 24π – 3√3

 

Answer: (D)
The curves intersect at point (3, ± 3√3)

Required area

 

Q 9: If ∫(cos x -sin x)/√(8-sin 2x) dx = a sin-1(sin x + cos x)/b + c where c is a constant of integration, then the ordered pair (a, b) is equal to:

(A) (1, –3)

(B) (1, 3)

(C) (–1, 3)

(D) (3, 1)

 

Answer: (B)
Put sin x + cos x = t ⇒1 + sin 2x = t²
(cos x -sin x ) dx = dt
⇒I = ∫dt/√(8-(t²-1))
= ∫dt/(9-t²)
= sin^-1 (t/3) + c = sin^-1(sin x + cos x)/3 + c
⇒ a = 1 and b = 3

 

Q 10: The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is:

(A) x = a

(B) x = 0

(C) x = -a/2

(D) x = a/2

Answer: (B)

h = (at²+a)/2, k = (2at+0)/2
⇒ t² =(2h-a)/a and t=k/a
⇒ k²/a² =(2h-a)/a
⇒ Locus of (h, k) is y² = a (2x – a)
⇒ y² = 2a(x- a/2)
Its directrix is x – a/2 = -a/2
⇒ x = 0

 

Q 11: A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is 1/4. Three stones A, B and C are placed at the points (1,1),(2,2), and (4,4) respectively. Then which of these stones is/are on the path of the man?

(A) B only

(B) A only

(C) the three

(D) C only

 

Answer: (A)
x/a + y/b = 1
h/a + k/b = 1 ⋯⋯(1)
and (1/a + 1/b)/2 = 1/4
∴1/a + 1/b = 1/2 ⋯⋯(2)
(From (1) and (2))
Line passes through fixed point B(2, 2)

 

Q 12: The function f(x) = (4×3- 3×2)/6 – 2sin x + (2x – 1)cos x:

(A) increases in [1/2, ∞)

(B) decreases (-∞, 1/2]

(C) increases in (-∞, 1/2]

(D) decreases [1/2, ∞)

Answer: (A)
f’(x) = (2x – 1) (x – sin x )
⇒ f’x ≥ 0 in x∈(-∞, 0] ⋃ [1/2, ∞)
and f’x ≤ 0 in x∈(0, ½)

 

Q 13: The distance of the point (1, 1, 9) from the point of intersection of the line (x – 3)/1 = (y – 4)/2 = (z – 5)/2 and the plane x + y + z = 17 is:

(A) 38

(B) 19√2

(C) 2√19

(D) √38

Answer: (D)
(x – 3)/1 = (y – 4)/2 = (z – 5)/2 = λ
x = λ + 3, y = 2λ+ 4, z = 2λ+5
Which lies on given plane hence
⇒ λ+3+2λ +4+2λ+5 = 17
⇒ λ = 5/5 = 1
Hence, point of intersection is Q (4, 6, 7)
∴ Required distance =PQ
= √(9+25+4)
= √38

 

Q 14:

(A) 2/3

(B) 0

(C) 1/15

(D) 3/2

 

Answer: (A)

 

Q 15: Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

(A) 25

(B) 20√3

(C) 30

(D) 25√3

 

Answer: (D)

tan θ = h/75 = 75/3h
h² = 75²/3
h = 25√3m

 

Q 16: A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

(A) 560

(B) 1050

(C) 1625

(D) 575

Answer: (C)
(2I,4F)+ (3I,6F) + (4I,8F)
= ⁶C₂ ⁸C₄ + ⁶C₃ ⁸C₆ + ⁶C₄ ⁸C₈​

= 15 × 70 + 20 × 28 + 15 × 1
= 1050 + 560 + 15 = 1625

 

Q 17: The equation of the plane passing through the point (1,2,–3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is:

(A) 3x – 10y – 2z + 11 = 0

(B) 6x – 5y – 2z – 2 = 0

(C) 11x + y + 17z + 38 = 0

(D) 6x – 5y + 2z + 10 = 0

Answer: (C)

 

Q 18: The population P = P(t) at time ‘t’ of a certain species follows the differential equation dP/dt = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is:

(A) ½ log_e 18

(B) 2log_e 18

(C) log_e 9

(D) log_e 18

Answer: (B)
dp/dt = (p-900)/2

ln |900| – ln |50| = t/2
t/2 = ln |18|
⇒ t = 2ln 18

 

Q 19: If e^{(cos²x + cos⁴x + cos⁶x + …..∞ )log_e2 satisfies the equation t² – 9t + 8 = 0, then the value of 2sin x/(sin x + √3cos x) 0< x<π/2 is:

(A) 3/2

(B) 2√3

(C) 1/2

(D) √3

 

Answer: (C)

0 < x < π/2 ⇒ cot⁡x = √3
⇒(2 sin⁡x)/(sin⁡x+√3 cos⁡x )
= 2/(1+√3 cot ⁡x)
= 1/2

 

Q 20: The system of linear equations

3x – 2y – kz = 10

2x – 4y – 2z = 6

x + 2y – z = 5m

is inconsistent if :

(A) k = 3, m = 4/5

(B) k ≠ 3,m∈R

(C) k ≠ 3, m ≠ 4/5

(D) k = 3, m ≠ 4/5

 

Answer: (D)

⇒ 3(4 + 4) + 2(–2 + 2) -k(4 + 4) = 0
⇒ k = 3

⇒ 10(4 + 4) +2(-6 + 10m) -3(12 + 20m) ≠ 0
⇒ m ≠ 4/5

⇒ 3(-6 + 10m) – 10(- 2 + 2) – 3(10m – 6) ≠ 0
⇒ 0

⇒3(-20m – 12) +2(10m – 6) +10(4 + 4) ≠ 0
⇒ m ≠ 4/5

Section B

Q 1:

Solution:

Answer: 1

= tan 𝜋/4
= 1

Q 2: If and [x] denotes the greatest integer ≤ x, then 

is equal to

Answer: 3

 

Q 3: If one of the diameters of the circle x²+y² – 2x – 6y + 6 = 0 is a chord of another circle ‘C’ whose center is at (2,1), then its radius is ________

Solution:

Answer: 3

distance between (1,3) and (2,1) is √5
∴ (√5)2+(2)2= r2
⇒r = 3

 

Q 4: Let three vectors a,  b and c be such that is coplanar with a and b, a.c=7 and is perpendicular to c, where and , then the value of is _______

 

Answer: 75

 

Q 5: The minimum value of α for which the equation 4/sin⁡x +1/(1-sin⁡x )=α has at least one solution in (0,π/2) is _______

Solution:

Answer: 9
f(x⁡) = 4/sin⁡x + 1/(1 – sin⁡x )
Let sin⁡x = t ∵x∈(0, π/2) ⇒ 0 < t < 1
f(t⁡)= 4/t + 1/(1-t)
f'(t⁡) = (-4)/t2 + 1/(1 – t⁡)2 = 0
⇒(t2-4(1-t⁡)2/t2(1-t⁡)2 = 0
⇒t = 2/3
f_min at t = 2/3
α_min = f(2/3) =4/(2/3) + 1/(1 – 2/3)
= 6 + 3
= 9

 

Q 6: Let A = {n∈N∶ n is a 3-digit number}, B = {9k + 2∶ k∈N} and C = {9k + l∶ k∈N} for some l(0 < l < 9). If the sum of all the elements of the set A∩(B∪C) is 274×400, then l is equal to ___________

Answer: 5
3 digit number of the form 9K+2 are {101,109,⋯,992}
⇒ Sum equal to (100/2)(1093) = S₁= 54650
Now 274 × 400 =S₁+S₂
⇒274 × 400 = (100/2) [101 + 992] + S₂
⇒274 × 400 = 50 × 1093 + S₂
⇒S₂ = 109600 – 54650
∴S₂ = 54950
S₂ = 54950 = (100/2) [(99+l) + (990+l)]
⇒ 2l + 1089 = 1099
⇒ l = 5

 

Q 7: Let

where α∈R. Suppose Q = [q_ij ] is a matrix satisfying PQ = kI₃ for some non-zero k∈R. If q₂₃= – k/8 and |Q⁡| = k²/2, then α² + k² is equal to ___

 

Answer: 17

 

Q 8: Let M be any 3×3 matrix with entries from the set {0,1,2}. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is __________

Answer: 540

⇒ a² + b² + c² + d² + e² + f² + g² + h² + i² = 7
Case I∶ Seven (1’s) and two (0’s)
⁹C₂ = 36
Case II∶ One (2) and three (1’s) and five (0’s)
9!/5!3! = 504
∴Total = 540

 

Q 9: If the least and the largest real values of α, for which the equation z + α|z⁡-1| + 2 i⁡ = 0 (z∈C and i =√(-1)) has a solution, are p and q respectively; then 4(p2 + q2) is equal to _________

 

Answer: 10
x + iy + α√((x–1)² +y²) + 2i=0
⇒y + 2 = 0 and x + α√((x-1)²+y²)=0
y = –2 & x² = α²(x² – 2x + 1 + 4)
α2 = x²/(x²– 2x + 5)
⇒ x² (α² – 1) – 2xα² + 5α² = 0
∵x∈R ⇒ D≥0
⇒ 4α⁴ – 4(α² – 1)5α² ≥ 0
⇒ α² [4α² – 20α² + 20] ≥ 0
⇒ α² [-16α² + 20] ≥ 0
⇒ α² [α² – 5/4] ≤ 0
⇒ α² ∈ [0, 5/4]
⇒ α ∈ [-√5/2, √5/2]
then 4[p² + q²] = 4[5/4 + 5/4]
= 10

 

Q 10: Let Bi (i = 1,2,3) be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α – 2β)p ⁡= αβ and (β – 3γ)p⁡= 2βγ (All the probabilities are assumed to lie in the interval (0, 1)). Then (P(B1⁡))/(P(B3⁡) )is equal to _____

 

Answer: 6
Let x,y,z be probability of B1, B2, B3 respectively.
⇒ x(1 – y) (1 – z) = α
⇒ y(1 – x)(1 – z) = β
⇒ z(1 – x)(1 – y ) = γ
⇒ (1 – x)(1 – y)(1 – z) = p
Now (α – 2β)p = αβ
⇒ (x(1–y)(1–z)-2y(1-x)(1–z)) (1–x)(1–y)(1–z) = xy(1–x)(1–y) (1–z)2
⇒ x+ xy – 2y = xy
∴x = 2y⋯(1)
Similarly, (β–3γ) p = 2βγ
⇒ y = 3z ⋯(2)
From (1) & (2)
⇒x = 6z
Hence x/z = P(B1)/P(B3) = 6

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