Final JEE-Main Exam August, 2021/27-08-2021/Evening Session
TEST PAPER WITH SOLUTION

PHYSICS

1. Curved surfaces of a plano-convex lens of refractive index μ1 and a plano-concave lens of refractive index μ2 have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.
(1) 1/(μ2 - μ1)
(2) μ1 - μ2
(3) 1/(μ1 - μ2)
(4) μ2 - μ1
Official Ans. by NTA (2)
Sol.
1/f1 = (μ1 - 1)(1/R)
1/f2 = (μ2 - 1)(-1/R)
1/f1 + 1/f2 = 1/feq = ((μ1 - 1) - (μ2 - 1))/R
1/feq = (μ1 - μ2)/R
R/feq = (μ1 - μ2)

2. The boxes of masses 2 kg and 8 kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8 kg to strike the ground starting from rest. (use g = 10 m/s²)
(1) 0.34 s
(2) 0.2 s
(3) 0.25 s
(4) 0.4 s
Official Ans. by NTA (4)
Sol.
(m1g - 2T) = m1a - (1)
T - m2g = m2(2a)
2T - 2m2g = 4m2a - (2)
m1g - 2m2g = (m1 + 4m2)a
a = (8 - 4)g/(8 + 8) = 4/16 g = g/4
a = 10/4 m/s²
S = 1/2 at²
0.2 × 2 × 4 / 10 = t²
t = 0.4 sec

3. For a transistor α and β are given as α = IC/IE and β = IC/IB. Then the correct relation between α and β will be :
(1) α = (1 - β)/β
(2) β = α/(1 - α)
(3) αβ = 1
(4) α = β/(1 - β)
Official Ans. by NTA (2)
Sol.
α = IC/IE, β = IC/IB, IE = IC + IB
α = IC/(IC + IB) = (IC/IB)/(IC/IB + 1) = β/(β + 1)
1 + 1/β = 1/α
1/β = (1 - α)/α
β = α/(1 - α)

4. Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.
(1) 4.18 m
(2) 2.94 m
(3) 2.45 m
(4) 7.35 m
Official Ans. by NTA (4)
Sol.
H = 1/2 gt²
9.8 × 2 / 9.8 = t²
t = √2 sec
Δt : time interval
h = 1/2 g(√2 - Δt)²
0 = 1/2 g(√2 - 2Δt)
Δt = 1/√2
h = 1/2 g(√2 - 1/√2)²
H - h = 9.8 - 2.45 = 7.35 m

5. Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω1 and ω2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :
(1) I1I2/(I1 + I2) (ω1 - ω2)²
(2) (I1 - I2)² ω1 ω2 / (2(I1 + I2))
(3) I1I2/(2(I1 + I2)) (ω1 - ω2)²
(4) (ω1 - ω2)² / (2(I1 + I2))
Official Ans. by NTA (3)
Sol.
From conservation of angular momentum
I1ω1 + I2ω2 = (I1 + I2)ω
ω = (I1ω1 + I2ω2)/(I1 + I2)
ki = 1/2 I1ω1² + 1/2 I2ω2²
kf = 1/2 (I1 + I2)ω²
ki - kf = 1/2 [I1ω1² + I2ω2² - (I1ω1 + I2ω2)²/(I1 + I2)]
Solving above we get
ki - kf = 1/2 (I1I2/(I1 + I2)) (ω1 - ω2)²

6. Three capacitors C1 = 2 μF, C2 = 6 μF and C3 = 12 μF are connected as shown in figure. Find the ratio of the charges on capacitors C1, C2 and C3 respectively:
(1) 2:1:1
(2) 2:3:3
(3) 1:2:2
(4) 3:4:4
Official Ans. by NTA (3)
Sol.
(VD - V)C2 + (VD - 0)C3 = 0
(VD - V)6 + (VD - 0)12 = 0
VD - V + 2VD = 0
VD = V/3
q2 = (V - VD)C2 = (V - V/3)(6 μF)
q2 = (4V) μF
q3 = (VD - 0)C3 = V/3 × 12 μF = 4V μF
q1 = (V - 0)C1 = V(2 μF)
q1:q2:q3 = 2:4:4
q1:q2:q3 = 1:2:2

7. The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is:
(1) (5700 ± 285) Ω
(2) (7500 ± 750) Ω
(3) (5700 ± 375) Ω
(4) (7500 ± 375) Ω
Official Ans. by NTA (4)
Sol.
R = 75 × 10² ± 5% of 7500
R = (7500 ± 375) Ω

8. An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km ?
(1) 37.8 m
(2) 605 m
(3) 75.6 m
(4) 302 m
Official Ans. by NTA (2)
Sol.
h : height of antenna
λ : wavelength of signal
h < λ
λ > h
λ > 400 m

9. If the rms speed of oxygen molecules at 0°C is 160 m/s, find the rms speed of hydrogen molecules at 0°C.
(1) 640 m/s
(2) 40 m/s
(3) 80 m/s
(4) 332 m/s
Official Ans. by NTA (1)
Sol.
Vrms = √(3KT/M)
(Vrms)O2/(Vrms)H2 = √(MH2/MO2) = √(2/32)
(Vrms)H2 = 4 × (Vrms)O2
= 4 × 160
= 640 m/s

10. A constant magnetic field of 1 T is applied in the x > 0 region. A metallic circular ring of radius 1 m is moving with a constant velocity of 1 m/s along the x-axis. At t = 0 s, the centre of O of the ring is at x = -1 m. What will be the value of the induced emf in the ring at t = 1 s ? (Assume the velocity of the ring does not change.)
(1) 1V
(2) 2π V
(3) 2V
(4) 0V
Official Ans. by NTA (3)
Sol.
emf = BLV
= 1.(2R).1
= 2V

11. A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V kg/m. The value of V is :
(1) -60 G
(2) +2 G
(3) -20 G
(4) -4 G
Official Ans. by NTA (4)
Sol.
VA = [-GM1/r - GM2/R]
= [-50/25 G - 100/50 G]
= -4G

12. For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :
(1) 1 Ω
(2) 5 Ω
(3) 4 Ω
(4) 2 Ω
Official Ans. by NTA (4)
Sol.
Imax = 50/2 = 25 mA
R = V/I = 50 mV/25 mA = 2 Ω

13. A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm ?
(1) 0.96 V
(2) 1.25 V
(3) 0.24 V
(4) 1.5 V
Official Ans. by NTA (2)
Sol.
kEmax = hc/λi + φ
eVo = hc/λi + φ
when λi = 670.5 nm; Vo = 0.48
when λi = 474.6 nm; Vo = ?
e(0.48) = 1240/670.5 + φ ... (1)
e(Vo) = 1240/474.6 + φ ... (2)
(2)-(1)
e(Vo - 0.48) = 1240(1/474.6 - 1/670.5) eV
Vo = 0.48 + 1240((670.5 - 474.6)/(474.6 × 670.5)) Volts
Vo = 0.48 + 0.76
Vo = 1.24 V ≈ 1.25 V

14. Match List-I with List-II.
List-I
(a) RH (Rydberg constant)
(b) h (Planck's constant)
(c) μB (Magnetic field energy density)
(d) η (coefficient of viscosity)
List-II
(i) kg m?¹ s?¹
(ii) kg m² s?¹
(iii) m?¹
(iv) kg m?¹ s?²
Choose the most appropriate answer from the options given below :
(1) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(2) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
(4) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
Official Ans. by NTA (2)
Sol.
SI unit of Rydberg const. = m?¹
SI unit of Planck's const. = kg m² s?¹
SI unit of Magnetic field energy density = kg m?¹ s?²
SI unit of coeff. of viscosity = kg m?¹ s?¹

15. If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density :
(1) [FL??T²]
(2) [FL?³T²]
(3) [FL?³T²]
(4) [FL?³T³]
Official Ans. by NTA (1)
Sol.
Density = [F¹L??T²]
[ML?³] = [M?L?T?]
[M¹L?³] = [M?L???T?²???]
a = 1; a + b = -3; -2a + c = 0
1 + b = -3, c = 2a
b = -4, c = 2
So, density = [F¹L??T²]

16. A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i0, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?
(1) x²/a²
(2) a²/x²
(3) x²/(b² - a²)
(4) (b² - a²)/x²
Official Ans. by NTA (1)
Sol.
when x < a
B1(2πx) = μ0(i0/(πa²))πx²
B1 = μ0i0x/(2πa²) ... (1)
when a < x < b
B2(2πx) = μ0i0
B2 = μ0i0/(2πx) ... (2)
B1/B2 = (μ0i0x/(2πa²))/(μ0i0/(2πx)) = x²/a²

17. The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall ? [Given 1 cal = 4.2 J and specific heat of water = 1 cal g?¹ °C?¹]
(1) 0.147°C
(2) 14.76°C
(3) 1.476°
(4) 0.014°C
Official Ans. by NTA (1)
Sol.
Change in P.E. = Heat energy
mgh = mSΔT
ΔT = gh/S
= 10 × 63 / 4200 J/kgC
= 0.147°C

18. A player kicks a football with an initial speed of 25 ms?¹ at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g = 10 ms?²)
(1) hmax = 10 m, T = 2.5 s
(2) hmax = 15.625 m, T = 3.54 s
(3) hmax = 15.625 m, T = 1.77 s
(4) hmax = 3.54 m, T = 0.125 s
Official Ans. by NTA (3)
Sol.
H = U² sin²θ / 2g
= (25)².(sin 45)² / (2 × 10)
= 15.625 m
T = U sinθ / g
= 25 × sin 45° / 10
= 2.5 × 0.7
= 1.77 s

19. The light waves from two coherent sources have same intensity I1 = I2 = I0. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima ?
(1) I0
(2) 2I0
(3) 5I0
(4) 4I0
Official Ans. by NTA (4)
Sol.
Imax = (√I1 + √I2)²
= 4I0

20. Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field E at the centre of curvature O ?
(1) 3√3 Q/(8π ε0 R²) (i)
(2) 3√3 Q/(8π² ε0 R²) (i)
(3) 3√3 Q/(16π² ε0 R²) (i)
(4) 3√3 Q/(8π² ε0 R²) (i)
Official Ans. by NTA (2)
Sol.
ε = 2kλ/R sin(θ/2)(-i)
λ = (-Q/(Rθ)) = (-Q/(R(2π/3)))
λ = -3Q/(2πR)
ε = 2k/R · (-3Q)/(2πR) · sin(60°)(-i)
ε = 3√3 Q/(8π² ε0 R²) (+i)

SECTION-B

1. A heat engine operates between a cold reservoir at temperature T2 = 400 K and a hot reservoir at temperature T1. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be K.
Official Ans. by NTA (500)
Sol.
Qin = 300 J; Qout = 240 J
Work done = Qin - Qout = 300 - 240 = 60 J
Efficiency = W/Qin = 60/300 = 1/5
efficiency = 1 - T2/T1
1/5 = 1 - 400/T1 ⇒ 400/T1 = 4/5
T1 = 500 k

2. Two simple harmonic motions are represented by the equations
y1 = 10 sin(3πt + π/3)
y2 = 5(sin 3πt + √3 cos 3πt)
Ratio of amplitude of y1 to y2 = x : 1. The value of x is
Official Ans. by NTA (1)
Sol.
y1 = 10 sin(3πt + π/3) ⇒ Amplitude = 10
y2 = 5(sin 3πt + √3 cos 3πt)
y2 = 10(1/2 sin 3πt + √3/2 cos 3πt)
y2 = 10(cos π/3 sin 3πt + sin π/3 cos 3πt)
y2 = 10 sin(3πt + π/3) ⇒ Amplitude = 10
So ratio of amplitudes = 10/10 = 1

3. X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n = 6 ? The value of X is
Official Ans. by NTA (15)
Sol.
No. of different wavelengths = n(n - 1)/2
= 6 × (6 - 1)/2 = 6 × 5/2 = 15

4. A zener diode of power rating 2 W is to be used as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate voltage fluctuated between 6 V and 14 V, the value of Rs for safe operation should be Ω.
Official Ans. by NTA (20)
Sol.
When unregulated voltage is 14 V voltage across zener diode must be 10 V
So potential difference across resistor ΔVRs = 4 V
and Pzener = 2 W
VI = 2
I = 2/10 = 0.2 A
ΔVRs = IRs
4 = 0.2 Rs ⇒ Rs = 40/2 = 20 Ω

5. Wires W1 and W2 are made of same material having the breaking stress of 1.25 × 10? N/m². W1 and W2 have cross-sectional area of 8 × 10?? m² and 4 × 10?? m², respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is kg. (Use g = 10 m/s²)
Official Ans. by NTA (40)
Sol.
B.S1 = T1max/(8×10??) ⇒ T1max = 8 × 1.25 × 100
= 1000 N
B.S2 = T2max/(4×10??) ⇒ T2max = 4 × 1.25 × 100
= 500 N
m = (500 - 100)/10 = 40 kg

6. A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = m/s so that the pendulum describes a circle. (Assume the string to be inextensible and g = 10 m/s²)
Official Ans. by NTA (400)
Sol.
V' = √(5gR) = √(5 × 10 × 0.5)
V' = 5 m/s
m1V = m2 × 5 - m1 × 100
10/1000 × V = 5 - 10/1000 × 100
V = 400 m/s

7. An ac circuit has an inductor and a resistor of resistance R in series, such that XL = 3R. Now, a capacitor is added in series such that XC = 2R. The ratio of new power factor with the old power factor of the circuit is √5 : x. The value of x is ______.
Official Ans. by NTA (1)
Sol.
cosφ = R/√(R² + 3R²) = 1/√10
cosφ' = R/√(R² + R²) = 1/√2
cosφ'/cosφ = √10/√2 = √5/1
∴ x = 1

8. The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x : 8. The value of x is ______.
Official Ans. by NTA (9)

9. A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. The magnetic field at this point will be x × 10?? T. The value of x is
Official Ans. by NTA (2)
Sol.
|B| = |E|/C = 6/(3×10?)
= 2 × 10?? T
∴ x = 2

10. A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be cm. (Take speed of sound in air as 340 ms?¹)
Official Ans. by NTA (34)
Sol.
λ/4 = ? ⇒ λ = 4?
f = V/λ = V/(4?)
⇒ 250 = 340/(4?)
⇒ ? = 34/(4 × 25) = 0.34 m
? = 34 cm

CHEMISTRY

1. Choose the correct statement from the following :
(1) The standard enthalpy of formation for alkali metal bromides becomes less negative on descending the group.
(2) The low solubility of CsI in water is due to its high lattice enthalpy.
(3) Among the alkali metal halides, LiF is least soluble in water.
(4) LiF has least negative standard enthalpy of formation among alkali metal fluorides.
Official Ans. by NTA (3)
Sol.
1. Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.
2. In case of CsI, lattice energy is less, but Cs? is having less hydration enthalpy due to which it is less soluble in water.
3. For alkali metal fluorides, the solubility in water increases from lithium to caesium. LiF is least soluble in water.
4. Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

2. The addition of dilute NaOH to Cr³? salt solution will give :
(1) a solution of [Cr(OH)4]?
(2) precipitate of Cr2O3(H2O)n
(3) precipitate of [Cr(OH)4]³?
(4) precipitate of Cr(OH)3
Official Ans. by NTA (2)
Sol.
Cr³? + NaOH → Cr2O3(H2O)n

5. The compound/s which will show significant intermolecular H-bonding is/are :
(a)
(b)
(c)
(1) (b) only
(2) (c) only
(3) (a) and (b) only
(4) (a), (b) and (c)
Official Ans. by NTA (1)
Sol.
(a) Shows intra molecular H-bonding
(b) Shows significant intermolecular H-bonding
(c) It do not show intermolecular H-bonding due to steric hindrance.

6. Which one of the following chemicals is responsible for the production of HCl in the stomach leading to irritation and pain?
(1)
(2)
(3)
(4)
Official Ans. by NTA (2)
Sol.
Histamine stimulate the secretion of HCl
Histamine structure

7. The oxide that gives H2O2 most readily on treatment with H2O is :
(1) PbO2
(2) Na2O2
(3) SnO2
(4) BaO2·8H2O
Official Ans. by NTA (2)
Sol.
1. PbO2 + 2H2O → Pb(OH)4
2. Na2O2 + 2H2O → 2NaOH + H2O2
this reaction is possible at room temperature
3. SnO2 + 2H2O → Sn(OH)4
4. Acidified BaO2·8H2O gives H2O2 after evaporation.

8. Which one of the following reactions will not yield propionic acid?
(1) CH3CH2COCH3 + OI?/H2O
(2) CH3CH2CH3 + KMnO4(Heat), OH?/H3O?
(3) CH3CH2CCl3 + OH?/H2O
(4) CH3CH2CH2Br + Mg, CO2 dry ether/H3O?
Official Ans. by NTA (4)
Sol.
All gives propanoic acid as product but option 4 gives butanoic acid as product
CH3CH2CH2Br → Mg dry ether → CH3CH2CH2MgBr
→ CO2 → CH3CH2CH2COOMgBr
→ H3O? → CH3CH2CH2COOH
Butanoic acid

9. The correct order of ionic radii for the ions, P³?, S²?, Ca²?, K?, Cl? is :
(1) P³? > S²? > Cl? > K? > Ca²?
(2) Cl? > S²? > P³? > Ca²? > K?
(3) P³? > S²? > Cl? > Ca²? > K?
(4) K? > Ca²? > P³? > S²? > Cl?
Official Ans. by NTA (1)
Sol.
P³? > S²? > Cl? > K? > Ca²?
(Correct order of ionic radii)
all the given species are isoelectronic species.
In isoelectronic species size increases with increase of negative charge and size decreases with increase in positive charge.

10. Which one of the following is the major product of the given reaction?
Official Ans. by NTA (1)
Sol.

11. The major product (A) formed in the reaction given below is :
Official Ans. by NTA (2)
Sol.

12. Which one of the following is used to remove most of plutonium from spent nuclear fuel?
(1) ClF3
(2) O2F2
(3) I2O5
(4) BrO3
Official Ans. by NTA (2)
Sol.
O2F2 oxidises plutonium to PuF6 and the reaction is used in removing plutonium as PuF6 from spent nuclear fuel.

13. Lyophilic sols are more stable than lyophobic sols because :
(1) there is a strong electrostatic repulsion between the negatively charged colloidal particles.
(2) the colloidal particles have positive charge.
(3) the colloidal particles have no charge.
(4) the colloidal particles are solvated.
Official Ans. by NTA (4)
Sol.
In the lyophilic colloids, the colloidal particles are extensively solvated.

14. The major product of the following reaction, if it occurs by SN2 mechanism is :
Official Ans. by NTA (4)
Sol.

15. Potassium permanganate on heating at 513 K gives a product which is :
(1) paramagnetic and colourless
(2) diamagnetic and green
(3) diamagnetic and colourless
(4) paramagnetic and green
Official Ans. by NTA (4)
Sol.
2KMnO4 → Δ → 2K2MnO4 + MnO2 + O2
Green Black
In K2MnO4, manganese oxidation state is +6 and hence it has one unpaired e.

16. Which one of the following tests used for the identification of functional groups in organic compounds does not use copper reagent ?
(1) Barfoed's test
(2) Selivanoff's test
(3) Benedict's test
(4) Biuret test for peptide bond
Official Ans. by NTA (2)
Sol.
In Selivanoff's reagent, Cu is not present. In Barfoed, Biuret and in Benedict reagent Cu is present.

17. Hydrolysis of sucrose gives :
(1) α-D-)-Glucose and β-D-)-Fructose
(2) α-D-+)-Glucose and α-D-)-Fructose
(3) α-D-)-Glucose and α-D-+)-Fructose
(4) α-D-+)-Glucose and β-D-)-Fructose
Official Ans. by NTA (4)
Sol.
Sucrose is formed by α-D+).Glucose + β-D-Fructose. we obtain these monomers on hydrolysis.

18. Match List-I with List-II :
List-I (Name of ore/mineral)
(a) Calamine
(b) Malachite
(c) Siderite
(d) Sphalerite
List-II (Chemical formula)
(i) ZnS
(ii) FeCO3
(iii) ZnCO3
(iv) CuCO3·Cu(OH)2
Choose the most appropriate answer from the options given below :
(1) (a)-iii), (b)-iv), (c)-ii), (d)-i)
(2) (a)-iii), (b)-iv), (c)-i), (d)-ii)
(3) (a)-iv), (b)-iii), (c)-i), (d)-ii)
(4) (a)-iii), (b)-ii), (c)-iv), (d)-i)
Official Ans. by NTA (1)
Sol.
(a) Calamine ZnCO3
(b) Malachite CuCO3·Cu(OH)2
(c) Siderite FeCO3
(d) Sphalerite ZnS

19. Which one of the following is formed (mainly) when red phosphorus is heated in a sealed tube at 803 K ?
(1) White phosphorus
(2) Yellow phosphorus
(3) β-Black phosphorus
(4) α-Black phosphorus
Official Ans. by NTA (4)
Sol.
When red phosphorus is heated in a sealed tube at 803 K, α-black phosphorus is formed.

20. The correct structures of A and B formed in the following reactions are :
Official Ans. by NTA (4)
Sol.

SECTION-B

1. The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36 × 10?³ s?¹ and activation energy is 209 kJ mol?¹. Its rate constant (in s?¹) at 600 K is x × 10??. The value of x is (Nearest integer)
[Given R = 8.31 J K?¹ mol?¹; log 6.36 × 10?³ = -2.19; 10??·?? = 1.62 × 10??]
Official Ans. by NTA (16)
Sol.
K700 = 6.36 × 10?³ s?¹
K600 = x × 10?? s?¹
Ea = 209 kJ/mol
Applying ;
log(KT2/KT1) = -Ea/(2.303R)(1/T2 - 1/T1)
log(K700/K600) = -Ea/(2.303R)(1/700 - 1/600)
log(6.36×10?³/K600) = +209×1000/(2.303×8.31)(100/(700×600))
log(6.36×10?³) - log K600 = 2.6
⇒ log K600 = -2.19 - 2.6 = -4.79

2. The number of optical isomers possible for [Cr(C2O4)3]³? is
Official Ans. by NTA (2)
Sol.
The number of optical isomers for [Cr(C2O4)3]³? is two.

3. Two flasks I and II shown below are connected by a valve of negligible volume.
When the valve is opened, the final pressure of the system in bar is x × 10?². The value of x is (Integer answer)
[Assume-Ideal gas; 1 bar = 10? Pa; Molar mass of N2 = 28.0 g mol?¹; R = 8.31 J mol?¹ K?¹]
Official Ans. by NTA (84)
Sol.
Applying ; (ni + nII)initial - (ni + nII)final
⇒ Assuming the system attains a final temperature of T (such that 300 < T < 60
⇒ [Heat lost by N2 of container I] = [Heat gained by N2 of container I]
⇒ n1Cm(300 - T) = n1Cm(T - 60)
⇒ [2.8/28](300 - T) = 0.2/28 (T - 60)
⇒ 14(300 - T) = T - 60
⇒ (14×300 + 60)/15 = T
⇒ T = 284 K (final temperature)
⇒ If the final pressure = P
⇒ (nI + nII)final = (3.0/28)
⇒ P/RT (VI + VII) = 3.0 gm/(28 mg/mol)
P = (3/28 mol) × 8.31 J/(mol - K) × 284 K/(3×10?³ m³) × 10?? bar/Pa
⇒ 0.84287 bar
⇒ 84.28 × 10?² bar
⇒ 84

4. 100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x × 10?². The value of x is . (Nearest integer) [Atomic weight : H = 1.008, C = 12.00, O = 16.00]
Official Ans. by NTA (19)
Sol.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
t = 0: 2.27 mol, 31.25 mol
t = ∞: 0, 19.9 mol, 6.81 mol, 9.08 mol
mole fraction of CO2 in the final reaction mixture (heterogeneous)
X_CO2 = 6.81/(19.9 + 6.81 + 9.08)
= 0.1902 = 19.02 × 10?²
= 19

5. 40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is K. (Nearest integer) [Given : Kf = 1.86 K kg mol?¹; Density of water = 1.00 g cm?³; Freezing point of water = 273.15 K]
Official Ans. by NTA (271)
Sol.
molality = (40/180) mol / 0.2 Kg = (10/9) molal
⇒ ΔTf = Tf - Tf' = 1.86 × 10/9
⇒ Tf' = 273.15 - 1.86 × 10/9
= 271.08 K
= 271 K (nearest integer)

6. The resistance of a conductivity cell with cell constant 1.14 cm?¹, containing 0.001 M KCl at 298 K is 1500 Ω. The molar conductivity of 0.001 M KCl solution at 298 K in S cm² mol?¹ is (Integer answer)
Official Ans. by NTA (760)
Sol.
K = 1/R × ?/A = ([1/1500] × 1.14) S cm?¹
⇒ Λm = 1000 × [1.14/1500]/0.001 S cm² mol?¹
= 760 S cm² mol?¹
⇒ 760

7. The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm in 0.1 second is x × 10¹³. The value of x is (Nearest integer) (h = 6.63 × 10?³? Js, c = 3.00 × 10? ms?¹)
Official Ans. by NTA (50)
Sol.
Energy emitted in 0.1 sec.
= 0.1 sec × 10?³ J/s
= 10?? J
If 'n' photons of λ = 1000 nm are emitted,
then: 10?? = n × hc/λ
⇒ 10?? = n × 6.63 × 10?³? × 3 × 10? / (1000 × 10??)
⇒ n = 5.02 × 10¹? = 50.2 × 10¹³
⇒ 50 (nearest integer)

8. When 5.1 g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27°C is x × 10?². The value of x is (Integer answer) [Given R = 0.082 L atm K?¹ mol?¹]
Official Ans. by NTA (6)
Sol.
moles of NH4HS initially taken = 5.1 g / 51 g/mol
= 0.1 mol
volume of vessel = 2 ?
NH4HS(s) ? NH3(g) + H2S(g)
t = 0: 0.1 mol
t = ∞: 0.1(1 - 0.2), 0.1 × 0.2, 0.1 × 0.2
⇒ partial pressure of each component
P = nRT/V = 0.1 × 0.2 × 0.082 × 300 / 2
= 0.246 atm
⇒ kp = P_NH3 × P_H2S = (0.246)² = 0.060516
= 6.05 × 10?²
⇒ 6

9. The number of species having non-pyramidal shape among the following is
(A) SO3
(B) NO3?
(C) PCl3
(D) CO3²?
Official Ans. by NTA (3)
Sol.
Trigonal planar, Trigonal planar, Pyramidal, Trigonal planar
Hence non-pyramidal species are SO3, NO3? and CO3²?

10. Data given for the following reaction is as follows:
FeO(s) + C(graphite) → Fe(s) + CO(g)
Substance | ΔH° (kJ mol-1) | ΔS° (J mol-1K-1)
FeO(s) | -266.3 | 57.49
C(graphite) | 0 | 5.74
Fe(s) | 0 | 27.28
CO(g) | -110.5 | 197.6
The minimum temperature in K at which the reaction becomes spontaneous is (Integer answer)
Official Ans. by NTA (964)
Sol.
Tmin = [Δ°H/Δ°S]
Δ°Hrxn = [Δf°H(Fe) + Δf°H(CO)] - [Δf°H(FeO) + Δf°H(Cgraphite)]
= [0 - 110.5] - [-266.3 + 0]
= 155.8 kJ/mol
Δ°Srxn = [Δ°S(Fe) + Δ°S(CO)] - [Δ°S(FeO) + Δ°S(Cgraphite)]
= [27.28 + 197.6] - [57.49 + 5.74]
= 161.65 J/mol-K
Tmin = 155.8 × 10³ J/mol / 161.65 J/mol-K = 963.8 K
= 964 K (nearest integer)

MATHEMATICS

1. The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m - n = 0 and mn + nl + lm = 0, is :
(1) π/2
(2) π - cos?¹(4/9)
(3) cos?¹(8/9)
(4) π/3
Official Ans. by NTA (1)
Sol.
n = 2(? + m)
?m + n(? + m) = 0
?m + 2(? + m)² = 0
2?² + 2m² + 5m? = 0
2(?/m)² + 2 + 5(?/m) = 0
2t² + 5t + 2 = 0
(t + 2)(2t + 1) = 0
⇒ t = -2; -1/2
?/m = -2, ?/m = -1/2
n = -2?
(?, -2?, -2?)
cos θ = (-2 - 2 + 4)/√9 × 1/√9 = 0 ⇒ θ = π/2

2. Let A = [ [x + 1] [x + 2] [x + 3] ; [x] [x + 3] [x + 3] ; [x] [x] [x + 4] ] where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval:
(1) [68, 69)
(2) [62, 63)
(3) [65, 66)
(4) [60, 61)
Official Ans. by NTA (2)
Sol.
[ [x + 1] [x + 2] [x + 3] ; [x] [x + 3] [x + 3] ; [x] [x] [x + 4] ] = 192
Ri → Ri - Ri & Ri → Ri - Ri
[ 1 0 -1 ; 0 1 -1 ; [x] [x] + 2 [x] + 4 ] = 192
2[x] + 6 + [x] = 192 ⇒ [x] = 62

3. Let M and m respectively be the maximum and minimum values of the function f(x) = tan?¹(sin x + cos x) in [0, π/2]. Then the value of tan(M - m) is equal to:
(1) 2 + √3
(2) 2 - √3
(3) 3 + 2√2
(4) 3 - 2√2
Official Ans. by NTA (4)
Sol.
Let g(x) = sin x + cos x = √2 sin(x + π/4)
g(x) ∈ [1, √2] for x ∈ [0, π/2]
f(x) = tan?¹(sin x + cos x) ∈ [π/4, tan?¹√2]
tan(tan?¹√2 - π/4) = (√2 - 1)/(1 + √2) × (√2 - 1)/(√2 - 1) = 3 - 2√2

4. Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :
(1) 1/8
(2) 5/8
(3) 5/16
(4)
Official Ans. by NTA (3)
Sol.
C-I '0' Head: TTT (1/2)³(1/2)³ = 1/64
C-II '1' head: HTT (3/8)(3/8) = 9/64
C-III '2' Head: HHT (3/8)(3/8) = 9/64
C-IV '3' Heads: HHH (1/8)(1/8) = 1/64
Total probability = 5/16

5. A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, -3) from the line 3x + 4y = 5 is given by :
(1) 10 d²y/dx² = 11
(2) 11 d²x/dy² = 10
(3) 10 d²x/dy² = 11
(4) 11 d²y/dx² = 10
Official Ans. by NTA (4)
Sol.
α.R = |3(2) + 4(-3) - 5|/5 = 11/5
(x - h)² = 11/5 (y - k)
differentiate w.r.t 'x': -
2(x - h) = 11/5 dy/dx
again differentiate
2 = 11/5 d²y/dx²
11 d²y/dx² = 10

6. If two tangents drawn from a point P to the parabola y² = 16(x - 3) are at right angles, then the locus of point P is :
(1) x + 3 = 0
(2) x + 1 = 0
(3) x + 2 = 0
(4) x + 4 = 0
Official Ans. by NTA (2)
Sol.
Locus is directrix of parabola
x - 3 + 4 = 0 ⇒ x + 1 = 0

7. The equation of the plane passing through the line of intersection of the planes r.(i + j + k) = 1 and r.(2i + 3j - k) + 4 = 0 and parallel to the x-axis is:
(1) r.(j - 3k) + 6 = 0
(2) r.(i + 3k) + 6 = 0
(3) r.(i - 3k) + 6 = 0
(4) r.(j - 3k) - 6 = 0
Official Ans. by NTA (1)
Sol.
Equation of planes are
r.(i + j + k) - 1 = 0 ⇒ x + y + z - 1 = 0
and r.(2i + 3j - k) + 4 = 0 ⇒ 2x + 3y - z + 4 = 0
equation of planes through line of intersection of these planes is :-
(x + y + z - 1) + λ(2x + 3y - z + 4) = 0
⇒ (1 + 2λ)x + (1 + 3λ)y + (1 - λ)z - 1 + 4λ = 0
But this plane is parallel to x-axis whose direction are (1, 0, 0)
∴ (1 + 2λ)1 + (1 + 3λ)0 + (1 - λ)0 = 0
λ = -1/2
Required plane is
0x + (1 - 3/2)y + (1 + 1/2)z - 1 + 4(-1/2) = 0
⇒ -y/2 + 3/2 z - 3 = 0
⇒ y - 3z + 6 = 0
⇒ r.(j - 3k) + 6 = 0

8. If the solution curve of the differential equation (2x - 10y³)dy + ydx = 0, passes through the points (0, 1) and (2, β), then β is a root of the equation:
(1) y? - 2y - 2 = 0
(2) 2y? - 2y - 1 = 0
(3) 2y? - y² - 2 = 0
(4) y? - y² - 1 = 0
Official Ans. by NTA (4)
Sol.
(2x - 10y³)dy + ydx = 0
⇒ dx/dy + (2/y)x = 10y²
I. F. = e^{∫(2/y)dy} = e^{2ln(y)} = y²
Solution of D.E. is
∴ x.y = ∫(10y²)y² dy
xy² = 10y?/5 + C ⇒ xy² = 2y? + C
It passes through (0,1) → 0 = 2 + C ⇒ C = -2
∴ Curve is xy² = 2y? - 2
Now, it passes through (2, β)
2β² = 2β? - 2 ⇒ β? - β² - 1 = 0
∴ β is root of an equation y? - y² - 1 = 0

9. Let A(a, 0), B(b, 2b + 1) and C(0, b), b ≠ 0, |b| ≠ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :
(1) -2b/(b + 1)
(2) 2b/(b + 1)
(3) 2b²/(b + 1)
(4) -2b²/(b + 1)
Official Ans. by NTA (4)
Sol.
|1/2 a b 2b + 1 1| = 1
⇒ |a/b 2b + 1 1| = ±2
⇒ a(2b + 1 - b) - 0 + 1(b² - 0) = ±2
⇒ a = (±2 - b²)/(b + 1)
∴ a = (2 - b²)/(b + 1) and a = (-2 - b²)/(b + 1)
sum of possible values of 'a' is
= -2b²/(b + 1)

10. Let [λ] be the greatest integer less than or equal to λ. The set of all values of λ for which the system of linear equations x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + (28 + [λ])z = [λ] has a solution is:
(1) R
(2) (-∞, -9) ∪ (-9, ∞)
(3) [-9, -8)
(4) (-∞, -9) ∪ [-8, ∞)
Official Ans. by NTA (1)
Sol.
D = |1 1 1; 3 2 5; 9 4 28 + [λ]| = -24 - [λ] + 15 = -[λ] - 9
if [λ] + 9 ≠ 0 then unique solution
if [λ] + 9 = 0 then D1 = D2 = D3 = 0 so infinite solutions
Hence λ can be any real number.

11. The set of all values of k > -1, for which the equation (3x² + 4x + 3)² - (k + 1)(3x² + 4x + 3)(3x² + 4x + 2) + k(3x² + 4x + 2)² = 0 has real roots, is :
(1) [1/5]
(2) [2, 3)
(3) [-1/2, 1)
(4) (1/2, 3/2) [-1]
Official Ans. by NTA (1)
Sol.
Let 3x² + 4x + 3 = a
and 3x² + 4x + 2 = b ⇒ b = a - 1
Given equation becomes
⇒ a² - (k + 1)ab + kb² = 0

12. A box open from top is made from a rectangular sheet of dimension a × b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :
(1) (a + b - √(a² + b² - ab))/12
(2) (a + b - √(a² + b² + ab))/6
(3) (a + b - √(a² + b² - ab))/6
(4) (a + b + √(a² + b² - ab))/6
Official Ans. by NTA (3)
Sol.
V = ?.b.h = (a - 2x)(b - 2x)x
⇒ V(x) = (2x - a)(2x - b)x
⇒ V(x) = 4x³ - 2(a + b)x² + abx
⇒ d/dx V(x) = 12x² - 4(a + b)x + ab
d/dx (V(x)) = 0 ⇒ 12x² - 4(a + b)x + ab = 0
⇒ x = (4(a + b) ± √(16(a + b)² - 48ab))/(2(12))
= ((a + b) ± √(a² + b² - ab))/6
Let x = α = ((a + b) + √(a² + b² - ab))/6
β = ((a + b) - √(a² + b² - ab))/6
Now, 12(x - α)(x - β) = 0
∴ x = β = (a + b - √(a² + b² - ab))/6

13. The Boolean expression (p ∧ q) ⇒ ((r ∧ q) ∧ p) is equivalent to :
(1) (p ∧ q) ⇒ (r ∧ q)
(2) (q ∧ r) ⇒ (p ∧ q)
(3) (p ∧ q) ⇒ (r ∨ q)
(4) (p ∧ r) ⇒ (q ∧ p)
Official Ans. by NTA (1)
Sol.
(p ∧ q) ⇒ ((r ∧ q) ∧ p)
~ (p ∧ q) ∨ ((r ∧ q) ∧ p)
~ (p ∧ q) ∨ ((r ∧ p) ∧ (p ∧ q))
⇒ [~ (p ∧ q) ∨ (p ∧ q)] ∧ [~ (p ∧ q) ∨ (r ∧ p)]
⇒ t ∧ [~ (p ∧ q) ∨ (r ∧ p)]
⇒ ~ (p ∧ q) ∨ (r ∧ p)
⇒ (p ∧ q) ⇒ (r ∧ p)
Aliter :
given statement says
" if p and q both happen then p and q and r will happen"
it Simply implies
" If p and q both happen then 'r' too will happen "
i.e.
" if p and q both happen then r and p too will happen
i.e.
(p ∧ q) ⇒ (r ∧ p)

14. Let Z be the set of all integers,
A = {(x, y) ∈ Z × Z : (x - 2)² + y² ≤ 4},
B = {(x, y) ∈ Z × Z : x² + y² ≤ 4} and
C = {(x, y) ∈ Z × Z : (x - 2)² + (y - 2)² ≤ 4}
If the total number of relation from A ∩ B to A ∩ C is 2^p, then the value of p is :
(1) 16
(2) 25
(3) 49
(4) 9
Official Ans. by NTA (2)
Sol.
(x - 2)² + y² ≤ 4
x² + y² ≤ 4
No. of points common in C1 & C2 is 5.
(0, 0), (1, 0), (2, 0), (1, 1), (1, -1)
Similarly in C2 & C3 is 5.
No. of relations = 2^{5×5} = 2^{25}

15. The area of the region bounded by the parabola (y - 2)² = (x - 1), the tangent to it at the point whose ordinate is 3 and the x-axis is :
(1) 9
(2) 10
(3) 4
(4) 6
Official Ans. by NTA (1)
Sol.
y = 3 ⇒ x = 2
Point is (2, 3)
Diff. w.r.t x
2(y - 2) y' = 1
⇒ y' = 1/(2(y - 2))
⇒ y'(2,3) = 1/2
⇒ (y - 3)/(x - 2) = 1/2 ⇒ x - 2y + 4 = 0
Area = ∫?³ ((y - 2)² + 1 - (2y - 4)) dy = 9 sq. units

16. If y(x) = cot?¹((√(1 + sin x) + √(1 - sin x))/(√(1 + sin x) - √(1 - sin x))), x ∈ (π/2, π), then dy/dx at x = 5π/6 is:
(1) -1/2
(2) -1
(3) 1/2
(4) 0
Official Ans. by NTA (1)
Sol.
y(x) = cot?¹[(cos(x/2) + sin(x/2) + sin(x/2) - cos(x/2))/(cos(x/2) + sin(x/2) - sin(x/2) + cos(x/2))]
y(x) = cot?¹(tan(x/2)) = π/2 - x/2
y'(x) = -1/2

17. Two poles, AB of length a metres and CD of length a + b (b ≠ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan|∠ACB = 1/2, then:
(1) x² + 2(a + 2b)x - b(a + b) = 0
(2) x² + 2(a + 2b)x + a(a + b) = 0
(3) x² - 2ax + b(a + b) = 0
(4) x² - 2ax + a(a + b) = 0
Official Ans. by NTA (3)
Sol.
tan θ = 1/2
tan(θ + α) = x/b, tan α = x/(a + b)
⇒ (1/2 + x/(a + b))/(1 - (1/2) × x/(a + b)) = x/b
⇒ x² - 2ax + ab + b² = 0

18. If 0 < x < 1 and y = 1/2 x² + 2/3 x³ + 3/4 x? + ..., then the value of e^{1+y} at x = 1/2 is:
(1) 1/2 e²
(2) 2e
(3) 1/2 √e
(4) 2e²
Official Ans. by NTA (1)
Sol.
y = (1 - 1/2)x² + (1 - 1/3)x³ + ...
= (x² + x³ + x? + ...) - (x²/2 + x³/3 + x?/4 + ...)
= x²/(1 - x) + ln(1 - x)
x = 1/2 ⇒ y = 1 - ln 2
e^{1+y} = e^{1+1-ln2} = e^{2-ln2} = e²/2

19. The value of the integral ∫?¹ √x dx/((1 + x)(1 + 3x)(3 + x)) is:
(1) π/8 (1 - √3/2)
(2) π/4 (1 - √3/6)
(3) π/8 (1 - √3/6)
(4) π/4 (1 - √3/2)
Official Ans. by NTA (1)
Sol.
I = ∫?¹ √x/((1 + x)(1 + 3x)(3 + x)) dx
Let x = t² ⇒ dx = 2t dt
I = ∫?¹ t(2t)/((t² + 1)(1 + 3t²)(3 + t²)) dt
I = ∫?¹ ((3t² + 1) - (t² + 1))/((3t² + 1)(t² + 1)(3 + t²)) dt
I = ∫?¹ dt/((t² + 1)(3 + t²)) - ∫?¹ dt/((1 + 3t²)(3 + t²))
= 1/2 ∫?¹ ((3 + t²) - (t² + 1))/((t² + 1)(3 + t²)) dt + 1/8 ∫?¹ ((1 + 3t²) - 3(3 + t²))/((1 + 3t²)(3 + t²)) dt
= 1/2 ∫?¹ dt/(1 + t²) - 1/2 ∫?¹ dt/(t² + 3) + 1/8 ∫?¹ dt/(t² + 3) - 3/8 ∫?¹ dt/(1 + 3t²)
= 1/2 ∫?¹ dt/(t² + 1) - 3/8 ∫?¹ dt/(t² + 3) - 3/8 ∫?¹ dt/(1 + 3t²)
= 1/2 (tan?¹(t))?¹ - 3/(8√3)(tan?¹(t/√3))?¹ - 3/(8√3)(tan?¹(√3 t))?¹
= 1/2 (π/4) - √3/8 (π/6) - √3/8 (π/3)
= π/8 - √3/16 π
= π/8 (1 - √3/2)

20. If lim_{x→∞}(√(x² - x + 1) - ax) = b, then the ordered pair (a, b) is:
(1) (1, 1/2)
(2) (1, -1/2)
(3) (-1, 1/2)
(4) (-1, -1/2)
Official Ans. by NTA (2)
Sol.
lim_{x→∞}(√(x² - x + 1) - ax) = b
⇒ a > 0
Now, lim_{x→∞} (x² - x + 1 - a²x²)/(√(x² - x + 1) + ax) = b
⇒ lim_{x→∞} ((1 - a²)x² - x + 1)/(√(x² - x + 1) + ax) = b
⇒ lim_{x→∞} ((1 - a²)x² - x + 1)/(x(√(1 - 1/x + 1/x²) + a)) = b
⇒ 1 - a² = 0 ⇒ a = 1
Now, lim_{x→∞} (-x + 1)/(x(√(1 - 1/x + 1/x²) + a)) = b
⇒ -1/(1 + a) = b ⇒ b = -1/2
(a, b) = (1, -1/2)

SECTION-B

1. Let S be the sum of all solutions (in radians) of the equation sin?θ + cos?θ - sin θ cos θ = 0 in [0, 4π]. Then 8S/π is equal to
Official Ans. by NTA (56)
Sol.
Given equation sin?θ + cos?θ - sin θ cos θ = 0
⇒ 1 - sin²θ cos²θ - sin θ cos θ = 0
⇒ 2 - (sin 2θ)² - sin 2θ = 0
⇒ (sin 2θ)² + (sin 2θ) - 2 = 0
⇒ (sin 2θ + 2)(sin 2θ - 1) = 0
⇒ sin 2θ = 1 or sin 2θ = -2 (not possible)
⇒ 2θ = π/2, 5π/2, 9π/2, 13π/2
⇒ θ = π/4, 5π/4, 9π/4, 13π/4
⇒ S = π/4 + 5π/4 + 9π/4 + 13π/4 = 7π
⇒ 8S/π = 8 × 7π/π = 56.00

2. Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x - y + z + 3 = 0 and let R (3, 5, γ) be a point of this plane. Then the square of the length of the line segment SR is
Official Ans. by NTA (72)
Sol.
Since R (3, 5, γ) lies on the plane 2x - y + z + 3 = 0
Therefore, 6 - 5 + γ + 3 = 0
⇒ γ = -4
Now, dr's of line QS are 2, -1, 1
equation of line QS is
(x - 1)/2 = (y - 3)/(-1) = (z - 4)/1 = λ (say)
⇒ F(2λ + 1, -λ + 3, λ + 4)
F lies in the plane
⇒ 2(2λ + 1) - (-λ + 3) + (λ + 4) + 3 = 0
⇒ 4λ + 2 + λ - 3 + λ + 7 = 0
⇒ 6λ + 6 = 0 ⇒ λ = -1
⇒ F(-1, 4, 3)
Since, F is mid-point of QS.
Therefore, co-ordinated of S are (-3, 5, 2).
So, SR = √(36 + 0 + 36) = √72
SR² = 72

3. The probability distribution of random variable X is given by:
X | 1 | 2 | 3 | 4 | 5
P(X) | K | 2K | 2K | 3K | K
Let p = P(1 < X < 4 | X < 3). If 5p = λK, then λ equal to
Official Ans. by NTA (30)
Sol.
∑P(X) = 1 ⇒ k + 2k + 2k + 3k + k = 1
⇒ k = 1/9
Now, p = P(kX < 4 / X < 3) = P(X = 2)/P(X < 3) = (2k/9k)/(k/9k + 2k/9k) = 2/3
⇒ p = 2/3
Now, 5p = λk
⇒ (5)(2/3) = λ(1/9)
⇒ λ = 30

4. Let z1 and z2 be two complex numbers such that arg(z1 - z2) = π/4 and z1, z2 satisfy the equation |z - 3| = Re(z). Then the imaginary part of z1 + z2 is equal to ______.
Official Ans. by NTA (6)
Sol.
|z - 3| = Re(z)
Let Z = x + iy
⇒ (x - 3)² + y² = x²
⇒ x² + 9 - 6x + y² = x²
⇒ y² = 6x - 9
⇒ y² = 6(x - 3/2)
⇒ z1 and z2 lie on the parabola mentioned in eq.(1)
arg(z1 - z2) = π/4
⇒ Slope of PQ = 1.
Let P(3/2 + 3/2 t1², 3t1) and Q(3/2 + 3/2 t2², 3t2)
Slope of PQ = 3(t2 - t1)/((3/2)(t2² - t1²)) = 1
⇒ 2/(t1 + t2) = 1
⇒ t1 + t2 = 2
Im(z1 + z2) = 3t1 + 3t2 = 3(t1 + t2) = 3(2) = 6
Aliter :
Let z1 = x1 + iy1; z2 = x2 + iy2
z1 - z2 = (x1 - x2) + i(y1 - y2)
∴ arg (z1 - z2) = π/4 ⇒ tan?¹((y1 - y2)/(x1 - x2)) = π/4
y1 - y2 = x1 - x2 ______(1)
|z1 - 3| = Re(z1) ⇒ (x1 - 3)² + y1² = x1² __(2)
|z2 - 3| = Re(z2) ⇒ (x2 - 3)² + y2² = x2² __(2)
sub (2) & (3)
(x1 - 3)² - (x2 - 3)² + y1² - y2² = x1² - x2²
(x1 - x2)(x1 + x2 - 6) + (y1 - y2)(y1 + y2) = (x1 - x2)(x1 + x2)
x1 + x2 - 6 + y1 + y2 = x1 + x2 ⇒ y1 + y2 = 6.

5. Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in the set T = {A ⊆ S : A ≠ φ and the sum of all the elements of A is not a multiple of 3} is ______.
Official Ans. by NTA (80)
Sol.
3n type → 3, 6, 9 = P
3n - 1 type → 2, 5 = Q
3n - 2 type → 1, 4 = R
number of subset of S containing one element which are not divisible by 3 = ³C? + ³C? = 4
number of subset of S containing two numbers whose some is not divisible by 3
= ³C? × ²C? + ³C? × ²C? + ²C? + ²C? = 14
number of subsets containing 3 elements whose sum is not divisible by 3
= ³C? × ?C? + (²C? × ²C?)2 + ³C?(²C? + ²C?) = 22
number of subsets containing 4 elements whose sum is not divisible by 3
= ³C? × ?C? + ³C?(²C? + ²C?) + (³C?²C? × ²C?)2 = 4 + 6 + 12 = 22.
number of subsets of S containing 5 elements whose sum is not divisible by 3.
= ³C?(²C? + ²C?) + (³C?²C? × ²C?) × 2 = 2 + 12 = 14
number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4
⇒ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.

6. Let A (secθ, 2tanθ) and B (secφ, 2tanφ), where θ + φ = π/2, be two points on the hyperbola 2x² - y² = 2. If (α, β) is the point of the intersection of the normals to the hyperbola at A and B, then (2β)² is equal to ______.
Official Ans. by NTA (36)
ALLEN Ans. (Bonus)
Sol.
Since, point A (sec θ, 2 tan θ) lies on the hyperbola
2x² - y² = 2
⇒ 2 + 2tan²θ - 4tan²θ = 2
⇒ tan θ = 0 ⇒ θ = 0
Similarly, for point B, we will get φ = 0.
but according to question θ + φ = π/2
which is not possible.
Hence it must be a 'BONUS'.

7. Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, and C1(α, β) and C2(γ, δ), C1 ≠ C2 are their centres, then |(α + β)(γ + δ)| is equal to ______.
Official Ans. by NTA (40)
Sol.
Slope of line joining centres of circles = 4/3 = tan θ
⇒ cos θ = 3/5, sin θ = 4/5
Now using parametric form
(x - 1)/cos θ = (y - 2)/sin θ = ±5
⊕ (x,y) = (1 + 5cos θ, 2 + 5sin θ)
(α, β) = (4, 6)
Θ (x,y) = (γ, δ) = (1 - 5cos θ, 2 - 5sin θ)
(γ, δ) = (-2, -2)
⇒ |(α + β)(γ + δ)| = |10 × -4| = 40

8. 3 × 7²² + 2 × 10²² - 44 when divided by 18 leaves the remainder ______.
Official Ans. by NTA (15)
Sol.
3(1 + 6)²² + 2·(1 + 9)²² - 44 = (3 + 2 - 44) = 18.I
= -39 + 18.I
= (54 - 39) + 18(I - 3)
= 15 + 18I1
⇒ Remainder = 15.

9. An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If μ is the average marks of girls and σ² is the variance of marks of 50 candidates, then μ + σ² is equal to ______.
Official Ans. by NTA (25)
Sol.
σb² = 2 (variance of boys)
nb = 20
x?b = 12, ng = 30
σg² = 2
x?g = (50 × 15 - 12 × 20)/30 = (750 - 240)/30 = 17 = μ
variance of combined series
σ² = (n1σb² + n2σg²)/(n1 + n2) + n1·n2/(n1 + n2)² (x?b - x?g)²
σ² = (20 × 2 + 30 × 2)/(20 + 30) + (20 × 30)/(20 + 30)² (12 - 17)²
σ² = 8
⇒ μ + σ² = 17 + 8 = 25

10. If ∫ (2e^x + 3e^{-x})/(4e^x + 7e^{-x}) dx = 1/14 (ux + v log4(4e^x + 7e^{-x})) + C, where C is a constant of integration, then u + v is equal to
Official Ans. by NTA (7)
Sol.
∫ 2e^x/(4e^x + 7e^{-x}) dx + 3∫ e^{-x}/(4e^x + 7e^{-x}) dx
= ∫ 2e^{2x}/(4e^{2x} + 7) dx + 3∫ e^{-2x}/(4 + 7e^{-2x}) dx
Let 4e^{2x} + 7 = T, Let 4 + 7e^{-2x} = t
8e^{2x} dx = dT, -14e^{-2x} dx = dt
2e^{2x} dx = dT/4, e^{-2x} dx = dt/14
∫ dT/(4T) - 3/14 ∫ dt/t
= 1/4 log T - 3/14 log t + C
= 1/4 log(4e^{2x} + 7) - 3/14 log(4 + 7e^{-2x}) + C
= 1/14 [1/2 log(4e^x + 7e^{-x}) + 13/2 x] + C
u = 13/2, v = 1/2 ⇒ u + v = 7
Aliter :
2e^x + 3e^{-x} = A(4e^x + 7e^{-x}) + B(4e^x - 7e^{-x}) + λ
2 = 4A + 4B ; 3 = 7A - 7B ; λ = 0
A + B = 1/2
A - B = 3/7
A = 1/2(1/2 + 3/7) = (7 + 6)/28 = 13/28
B = A - 3/7 = 13/28 - 3/7 = (13 - 12)/28 = 1/28
∫ 13/28 dx + 1/28 ∫ (4e^x - 7e^{-x})/(4e^x + 7e^{-x}) dx
13/28 x + 1/28 ln|4e^x + 7e^{-x}| + C
u = 13/2; v = 1/2
⇒ u + v = 7

JEE MAINS 2021 FREE PDF DOWNLOAD LINK ALL SOLVED QUESTION PAPER

2021 JEE MAINS SOLVED QUESTION PAPER MORNING SESSION

JEE MAINS 2021 SOLVED QUESTION PAPER AUGUST 2021