JEE MAINS Previous year solved question paper 2019 - phase 2 -
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Physics
Two particles move at right angle to each other. Their de Broglie wavelengths are λ1 and λ2 respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength λ of the final particle, is given by:
(1) λ = √(λ1λ2)
(2) λ = (λ1 + λ2)/2
(3) 2/λ = 1/λ1 + 1/λ2
(4) 1/λ^2 = 1/λ1^2 + 1/λ2^2
Answer (4)
Sol.
p1 = h/λ1
p2 = h/λ2
∴ p = √(p1^2 + p2^2)
⇒ h/λ = √(h^2/λ1^2 + h^2/λ2^2)
⇒ 1/λ^2 = 1/λ1^2 + 1/λ2^2
In SI units, the dimensions of √(ε0/μ0) is:
(1) AT^2M^-1L^-1
(2) AT^-3ML^{3/2}
(3) A^-1TML^3
(4) A^2T^3M^-1L^-2
Answer (4)
Sol. [√(ε0/μ0)] = [LT^-1] × [ε0]
∴ F = q^2/(4πε0r^2)
⇒ [ε0] = [AT]^2/([MLT^-2] × [L^2])
∴ [√(ε0/μ0)] = [LT^-1] × [A^2M^-1L^-3T^4]
= [M^-1L^-2T^3A^2]
A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled 3 m is
(1) 4 J
(2) 2.5 J
(3) 5 J
(4) 6.5 J
Answer (4)
Sol. = Area under F-x graph
ΔK.E = W = 1/2 × (3 + 2) × (3 - 2) + 2 × 2
= 2.5 + 4
= 6.5 J
Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is
(1) n = 2 → n = 4
(2) n = 2 → n = 5
(3) n = 2 → n = 3
(4) n = 1 → n = 4
Answer (1)
Sol. Energy released by hydrogen atom
ΔE1 = 13.6 × (1/1 - 1/4) = 3/4 × 13.6 eV = 10.2 eV
Also, energy absorbed by He+ ion in transition n = 2 → n = 4
ΔE2 = 13.6 × 4 × (1/4 - 1/16) = 10.2 eV
So, possible transition is n = 2 → n = 4
Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of : (density of water = 1000 kg/m^3, coefficient of viscosity of water = 1 mPa s)
(1) 10^2
(2) 10^4
(3) 10^3
(4) 10^6
Answer (2)
Sol. Flow rate of water (Q) = 100 lit/min
= (100 × 10^-3)/60 = 5/3 × 10^-3 m^3
Velocity of flow (v) = Q/A = (5 × 10^-3)/(3 × π × (5 × 10^-2)^2)
= 10/(15π) = 2/(3π) m/s = 0.2 m/s
Reynold number (Re) = Dvρ/η
= (10 × 10^-2) × (2/(3π)) × 1000 / 1 = 2 × 10^4
Order of Re = 10^4
In figure, the optical fiber is l = 2 m long and has a diameter of d = 20 μm. If a ray of light is incident on one end of the fiber at angle θ1 = 40°, the number of reflections it makes before emerging from the other end is close to: (refractive index of fiber is 1.31 and sin 40° = 0.64)
(1) 66000
(2) 55000
(3) 45000
(4) 57000
Answer (4)
Sol. 1 × sin 40° = 1.31 sin θ
⇒ sin θ = 0.64/1.31 ⇒ θ = 30°
l = 20 μm × cot θ
∴ N = 2/(20 × 10^-6 × cot θ)
= (2 × 10^6)/(20 × √3) = 57735
N = 57000
A 200 Ω resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be:
(1) 400 Ω
(2) 500 Ω
(3) 300 Ω
(4) 100 Ω
Answer (2)
Sol. 200 Ω = Red + Black + Brown
Green = 5
So, Green + Black + Brown = 500 Ω
A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is:
(1) 2/ln 2
(2) 2 ln 2
(3) ln 2
(4) 1/2 ln 2
Answer (2)
Sol. PR = i^2 × R
PB = V × i
∴ PL = Vi - i^2R
⇒ Vi - i^2R = i^2R
⇒ i = V/2R and i = V/R · (1 - e^{-t/τ})
∴ V/2R = V/R (1 - e^{-t/τ})
⇒ t = τ ln(2) = (20/10) ln(2) = 2 ln(2)
Four particles A, B, C and D with masses mA = m, mB = 2m, mC = 3m and mD = 4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is :
(1) Zero
(2) a(i + j)
(3) a/5 (i + j)
(4) a/5 (i - j)
Answer (4)
Sol. aCM = ((2m)a j + 3m × a i + m a(-i) + 4m × a(-j))/(2m + 3m + 4m + m)
= (2a i - 2a j)/10 = a/5 (i - j)
A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E = 6 Vm^-1 along y-direction. Its corresponding magnetic field component, B would be :
(1) 2 × 10^-8 T along y-direction
(2) 6 × 10^-8 T along z-direction
(3) 2 × 10^-8 T along z-direction
(4) 6 × 10^-8 T along x-direction
Answer (3)
Sol. B0 = E0/c = 6/(3 × 10^8) = 2 × 10^-8 T
Propagation direction = E × B
i = j × B
⇒ B = k
Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 10^6 V/m. The plate area is 10^-4 m^2. What is the dielectric constant if the capacitance is 15 pF?
(given ε0 = 8.86 × 10^-12 C^2/Nm^2)
(1) 3.8
(2) 4.5
(3) 8.5
(4) 6.2
Answer (3)
Sol. C = kε0A/d
E = V/d
15 × 10^-12 = (k × 8.86 × 10^-12 × 10^-4 × 10^6)/500
k = 8.5
The wavelength of the carrier waves in a modern optical fiber communication network is close to :
(1) 600 nm
(2) 900 nm
(3) 1500 nm
(4) 2400 nm
Answer (3)
Sol. Fact Based
Wavelength of carrier waves in modern optical fiber communication is most widely used near about 1500 nm.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1π ms^-2, what will be the tensile stress that would be developed in the wire?
(1) 4.8 × 10^6 Nm^-2
(2) 3.1 × 10^6 Nm^-2
(3) 5.2 × 10^6 Nm^-2
(4) 6.2 × 10^6 Nm^-2
Answer (2)
Sol. Stress = F/A = (4 × 3.1π)/(π × (2 × 10^-3)^2) = 3.1 × 10^6 N/m^2
A thin circular plate of mass M and radius R has its density varying as ρ(r) = ρ0 r with ρ0 as constant and r is the distance from its center. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = aMR^2. The value of the coefficient a is
(1) 3/2
(2) 3/5
(3) 3/5
(4) 8/5
Answer (4)
Sol. M = ∫_0^R ρ0 r × 2πr dr = 2πρ0R^3/3
IC = ∫_0^R ρ0 r × 2πr dr × r^2 = 2πρ0R^5/5
∴ I = IC + MR^2 = 2πρ0R^5(1/3 + 1/5) = 16πρ0R^5/15
= 8/5 [2/3 π ρ0R^3]R^2 = 8/5 MR^2
An alternating voltage v(t) = 220 sin 100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is
(1) 2.2 ms
(2) 7.2 ms
(3) 5 ms
(4) 3.3 ms
Answer (4)
Sol. I = Im sin(100πt)
⇒ Im/2 = Im sin(100πt1)
⇒ π/6 = 100πt1
⇒ t1 = 1/600 s
T = 2π/100π = 1/50 s
t2 = 1/4 × 1/50 = 1/200 s
Δt = 1/200 - 1/600 = 2/600 = 1/300 s = 3.3 ms
Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of others' gravitational field in a circular orbit circumscribing the square?
(1) 1.41√(GM/a)
(2) 1.16√(GM/a)
(3) 1.21√(GM/a)
(4) 1.35√(GM/a)
Answer (2)
Sol. r = a/√2, F = GM^2/a^2 (√2) + GM^2/(2a^2)
∴ Mv^2/(a/√2) = GM^2/a^2 (√2 + 1/2)
⇒ v^2 = GM/a (1 + 1/(2√2))
⇒ v = √(GM/a (1 + 1/(2√2))) = 1.16√(GM/a)
The bob of a simple pendulum has mass 2 g and a charge of 5.0 μC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is
(take g = 10 m/s^2)
(1) tan^-1(0.2)
(2) tan^-1(0.5)
(3) tan^-1(2.0)
(4) tan^-1(0.8)
Answer (4)
Sol. T cos θ = mg
T sin θ = qE
tan θ = qE/mg
tan θ = (5 × 10^-6 × 2000)/(2 × 10^-3 × 10) = 1/2
⇒ tan^-1(1/2)
A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is
(1) 4 : 9
(2) 1 : 2
(3) 3 : 5
(4) 1 : 4
Answer (2)
Sol. f = p/(2L) √(T/(πr^2ρ)) = q/(2L) √(T/(π(2r)^2ρ))
p/q = r/(2r) = 1/2
p : q = 1 : 2
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.
The current Iz through the Zener is
(1) 15 mA
(2) 7 mA
(3) 10 mA
(4) 17 mA
Answer (3)
Sol. I800Ω = 5.6/800 A = 7 mA
I200Ω = (9 - 5.6)/200 = 17 mA
∴ Iz = 17 - 7 = 10 mA
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 Nm^-1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 Ω and air drag negligible, N will be close to
(1) 1000
(2) 5000
(3) 50000
(4) 10000
Answer (2)
Sol. F = -kx - ilB = -kx - (Blv/R) × lB
F = -kx - (B^2 l^2/R) × v
So, it is case of damped oscillation
⇒ A = A0 e^{-bt/(2m)}
⇒ A0/e = A0 e^{-bt/(2m)}
⇒ t = 2m/(B^2 l^2/R) = (2 × 50 × 10^-3 × 10)/(0.01 × 0.01) = 10000 s
Time period, T = 2π√(m/k) = 2 s
∴ Number, N = 10000/2 = 5000
Ship A is sailing towards north-east with velocity v = 30i + 50j km/hr where i points east and j, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in:
(1) 2.2 hrs.
(2) 4.2 hrs.
(3) 3.2 hrs.
(4) 2.6 hrs.
Answer (4)
Sol. vA = 30i + 50j km/hr
rBA = (80i + 150j) km
vB = (-10i) km/hr
vBA = vB - vA = -10i - 30i - 50j = -40i - 50j
Projection of (rBA) on vBA = (rBA · vBA)/|vBA|
= ((80i + 150j) · (-40i - 50j))/(10√41)
= (10 × 107)/(√41 × 10√41) = 107/41 = 2.6 Hrs.
For the circuit shown, with R1 = 1.0 Ω, R2 = 2.0 Ω, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V)
(1) 2.7
(2) 3.7
(3) 2.3
(4) 3.3
Answer (4)
Sol. Vab = (E1r2r3 + E2r3r1 + E3r1r2)/(r1r2 + r2r3 + r3r1)
= (2(2 × 2) + 4(2 × 2) + 4(2 × 2))/(4 + 4 + 4)
= 40/12 = 10/3 = 3.3 Volt
Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of 8 × 10^2 kg/m^3 and specific heat of 2000 kg^-1 K^-1 while liquid in B has density of 10^3 kg/m^-3 and specific heat of 4000 kg^-1 K^-1. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)
(1) A above B
(2) B above A
(3) A and B same
(4) A above B (different)
Answer (2)
Sol. ms(-dT/dt) = eσAT^4
dT/dt = (eσ × A × T^4)/(ρ × Vol × S)
(-dT/dt)A = (ρB/ρA) × (2) > 1
So, A cools down at faster rate.
A solid conducting sphere, having a charge Q is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of -4Q, the new potential difference between the same two surfaces is:
(1) -2V
(2) V
(3) 2V
(4) 4V
Answer (2)
Sol. In case-1
Electric field between spherical surface
E = KQ/r^2
In case-2
Electric field between surfaces remain unchanged.
Potential difference between them remain unchanged too.
A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field Bi. The torque on the coil due to the magnetic field is:
(1) Bπr^2I/N
(2) Br^2I/πN
(3) Bπr^2IN
(4) Zero
Answer (3)
Sol. |τ| = |μ × B|
= NIA × B
⇒ τ = Nπr^2B
An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be:
(1) 20 cm from the convergent mirror, twice the size of the object
(2) 20 cm from the convergent mirror, same size as the object
(3) 40 cm from the convergent lens, twice the size of the object
(4) 40 cm from the convergent mirror, same size as the object
Answer Bonus)
Sol. v1 = (40 × 20)/(40 - 20) = 40 cm
u2 = 60 - 40 = 20 cm
∴ v2 = (20 × 10)/(20 - 10) = 20 cm
∴ Image traces back to object itself as image formed by lens is a centre of curvature of mirror
If 10^22 gas molecules each of mass 10^-26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m^2 with a speed 10^4 m/s, the pressure exerted by the gas molecules will be of the order of:
(1) 10^8 N/m^2
(2) 10^3 N/m^2
(3) 10^16 N/m^2
(4) 10^4 N/m^2
Answer Bonus)
Sol. P = N × (2mv)/(Δt × A)
= (10^22 × 2 × 10^-26 × 10^4)/(1 × 1) = 2 N/m^2
A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to:
Latent heat of vaporization of water = 2.10 × 10^6 J kg^-1 and Latent heat of Fusion of water = 3.36 × 10^5 J kg^-1
(1) 130 g
(2) 150 g
(3) 20 g
(4) 35 g
Answer (3)
Sol. Let amount of water evaporated be m gram.
m × Lv = (150 - m) × Ls
m × 540 = (150 - m) × 80
⇒ m = 20 g
In an interference experiment the ratio of amplitudes of coherent waves is a1/a2 = 1/3. The ratio of maximum and minimum intensities of fringes will be:
(1) 4
(2) 18
(3) 9
(4) 2
Answer (1)
Sol. a1/a2 = 3/1
∴ Imax/Imin = ((a2 + a1)/(a2 - a1))^2 = ((3 + 1)/(3 - 1))^2 = 4
A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to:
(1) 10^4 Nm^-2
(2) 10^3 Nm^-2
(3) 10^8 Nm^-2
(4) 10^6 Nm^-2
Answer (4)
Sol. 1/2 (YA/L)(Δl)^2 = 1/2 mv^2
⇒ Y = mv^2L/(A(Δl)^2)
= (0.02 × 400 × 0.42 × 4)/(π × 36 × 10^-6 × 0.04)
= 2.3 × 10^6 N/m^2
So, order is 10^6.
PART-B : CHEMISTRY
With respect to an ore, Ellingham diagram helps to predict the feasibility of its
(1) Zone refining
(2) Vapour phase refining
(3) Thermal reduction
(4) Electrolysis
Answer (3)
Sol. Ellingham diagram is used to select reducing agent so it help to predict feasibility of its thermal reduction.
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
(1) Isothermal process : q = -w
(2) Cyclic process : q = -w
(3) Isochoric process : ΔU = q
(4) Adiabatic process : ΔU = -w
Answer (4)
Sol. ΔU = q + W
Adiabatic process q = 0
ΔU = W
For isothermal, ΔU = 0
For cyclic, ΔU = 0
For isochoric, W = 0
The quantum number of four electrons are given below:
(I) n = 4, l = 2, m1 = -2, ms = -1/2
(II) n = 3, l = 2, m1 = 1, ms = +1/2
(III) n = 4, l = 1, m1 = 0, ms = +1/2
(IV) n = 3, l = 1, m1 = 1, ms = -1/2
The correct order of their increasing energies will be:
(1) IV < II < III < I
(2) I < III < II < IV
(3) IV < III < II < I
(4) I < II < III < IV
Answer (1)
Sol. n + l
(I) n = 4 l = 2 4d 6
(II) n = 3 l = 2 3d 5
(III) n = 4 l = 1 4p 5
(IV) n = 3 l = 1 3p 4
more is n + l value, more is energy
3p < 3d < 4p < 4d
The major product of the following reaction is :
(1) COOH with Cl
(2) COOH with Cl
(3) COOH with Cl
(4) COOH with Cl
Answer (3)
Sol.
Chlorobenzene + phthalic anhydride → (i) AlCl3, Heat (ii) H2O → 2-(4-chlorobenzoyl)benzoic acid
The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, [V(CN)6]4-, [Fe(CN)6]4-, [Ru(NH3)6]3+ and [Cr(NH3)6]2+, is:
(1) V2+ > Cr2+ > Ru3+ > Fe2+
(2) Cr2+ > V2+ > Ru3+ > Fe2+
(3) V2+ > Ru3+ > Cr2+ > Fe2+
(4) Cr2+ > Ru3+ > Fe2+ > V2+
Answer (1)
Sol.
No. of unpaired electrons
[V(CN)6]4- 1
[Ru(NH3)6]3+ 1
[Fe(CN)6]4- 0
[Cr(NH3)6]2+ 2
Order of spin magnetic moment
V2+ > Cr2+ > Ru3+ > Fe2+
Which is wrong with respect to our responsibility as a human being to protect our environment?
(1) Using plastic bags
(2) Restricting the use of vehicles
(3) Avoiding the use of floodlighted facilities
(4) Setting up compost tin in gardens
Answer (1)
Sol. Use of plastic bags is hazardous to our environment
Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of log x/m versus log p is shown in the given graph. x/m is proportional to
(1) p^{3/2}
(2) p^3
(3) p^{2/3}
(4) p^2
Answer (3)
Sol. x/m ∝ p^{1/n}, x/m = kp^{1/n}
Slope = 2/3
log x/m = log k + 1/n log p
Slope = 1/n = 2/3
x/m ∝ p^{2/3}
Given that E°(H2O) = +1.23 V; E°(H2O) = 2.05 V; E°(H2O) = +1.09 V; E°(H2O) = +1.4 V. The strongest oxidizing agent is
(1) Br2
(2) Au3+
(3) S2O6^2-
(4) O2
Answer (3)
Sol. More positive is the reduction potential stronger is the oxidizing agent.
Reduction potential is maximum for S2O6^2-
Coupling of benzene diazonium chloride with 1-naphthol in alkaline medium will give
Answer (3)
Sol.
Benzene diazonium chloride + 1-naphthol → OH- → 1-phenylazo-2-naphthol
Maltose on treatment with dilute HCl gives
(1) D-Galactose
(2) D-Glucose and D-Fructose
(3) D-Glucose
(4) D-Fructose
Answer (3)
Sol. Hydrolysis of maltose give glucose as maltose is composed of two α-D glucose units.
The vapour pressures of pure liquids A and B are 400 and 600 mmHg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :
(1) 500 mmHg, 0.4, 0.6
(2) 500 mmHg, 0.5, 0.5
(3) 450 mmHg, 0.4, 0.6
(4) 450 mmHg, 0.5, 0.5
Answer (1)
Sol. P = xB PB° + xA PA°
= 0.5 × 600 + 0.5 × 400 = 300 + 200 = 500
PB = yB PTotal
yB = PB/PTotal = 300/500 = 3/5 = 0.6
yA = PA/PTotal = 200/500 = 2/5 = 0.4
For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is
[A] (mol L-1) [B] (mol L-1) Initial Rate (mol L-1 s-1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72
(1) Rate = k[A]^2[B]^2
(2) Rate = k[A][B]
(3) Rate = k[A]^2[B]
(4) Rate = k[A][B]^2
Answer (4)
Sol. 2A + B → P Rate = k[A]^x[B]^y
Exp-1, 0.045 = k[0.05]^x[0.05]^y ...(i)
Exp-2, 0.090 = k[0.1]^x[0.05]^y ...(ii)
Exp-3, 0.72 = k[0.2]^x[0.1]^y ...(iii)
Divide equation (i) by equation (ii)
0.045/0.090 = (1/2)^x ⇒ x = 1
Divide equation (i) by equation (iii)
0.045/0.72 = (0.05/0.1)^x (0.05/0.2)^y
(1/2)^2 = (1/2)^y ⇒ y = 2
Rate law = k[A]^1[B]^2
For silver, Cp (JK^-1 mol^-1) = 23 + 0.01T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH will be close to
(1) 21 kJ
(2) 13 kJ
(3) 62 kJ
(4) 16 kJ
Answer (3)
Sol. n = 3
T1 = 300
T2 = 1000
Cp = 23 + 0.01T
ΔH = ∫{T1}^{T2} nCp dT
= n ∫{300}^{1000} (23 + 0.01T)dT
= 3[23T + 0.01T^2/2]_{300}^{1000}
= 3[16100 + 4550]
= 3 × 20650 = 61950 J
= 61.95 kJ
The lanthanide ion that would show colour is
(1) Gd3+
(2) Lu3+
(3) La3+
(4) Sm3+
Answer (4)
Sol. Sm3+ = Partially filled f orbital = 4f^5
Sm = 4f^6 6s^2
Sm3+ = Yellow.
Lu3+ = 4f^14 colourless.
The correct order of hydration enthalpies of alkali metal ions is
(1) Na+ > Li+ > K+ > Rb+ > Cs+
(2) Li+ > Na+ > K+ > Cs+ > Rb+
(3) Na+ > Li+ > K+ > Cs+ > Rb+
(4) Li+ > Na+ > K+ > Rb+ > Cs+
Answer (4)
Sol. Smaller is size more is hydration energy.
Li+ < Na+ < K+ < Rb+ < Cs+ Size
Li+ > Na+ > K+ > Rb+ > Cs+ Hydration energy
The major product of the following reaction is
(1) OH with Br
(2) OH with OMe
(3) epoxide
(4) alkene with OMe
Answer (3)
Sol.
Phenacyl bromide + NaBH4/MeOH, 25°C → styrene oxide
Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is
(1) A2B2O
(2) AB2O4
(3) A4B2O
(4) A2BO4
Answer (2)
Sol. Lattice formed by B(ccp) = 4
A = 50% of octahedral voids = 2
O = tetrahedral voids = 8
Formula = AB2O4
If solubility product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then which of the following relation between S and Ksp is correct?
(1) S = (Ksp/929)^{1/9}
(2) S = (Ksp/216)^{1/7}
(3) S = (Ksp/144)^{1/6}
(4) S = (Ksp/6912)^{1/7}
Answer (4)
Sol. Zr3(PO4)4 ? 3Zr4+ + 4PO4^3-
Ksp = [Zr4+]^3[PO4^3-]^4 = (3S)^3(4S)^4
Ksp = 6912 S^7
S = (Ksp/6912)^{1/7}
In order to oxidise a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 required is
(1) 1.5
(2) 2
(3) 3
(4) 1
Answer (2)
Sol. 5e + MnO4^- → Mn^2+
FeC2O4 → Fe3+ + 2CO2 + 3e
1 mole of FeC2O4 react with 3/5 moles of acidified KMnO4
Fe2(C2O4)3 → Fe3+ + CO2 + 6e
1 mole of Fe2(C2O4)3 react with 6/5 moles of KMnO4
FeSO4 → Fe3+ + e
1 mole of FeSO4 react with 1/5 moles of KMnO4
∴ Total moles required = 3/5 + 6/5 + 1/5 = 2
100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is (molar mass of calcium bicarbonate is 162 g mol^-1 and magnesium bicarbonate is 146 g mol^-1)
(1) 5,000 ppm
(2) 100 ppm
(3) 10,000 ppm
(4) 1,000 ppm
Answer (3)
Sol. Moles of Ca(HCO3)2 = 0.005
Moles of Mg(HCO3)2 = 0.005
Hardness in terms of CaCO3 ppm
= ((0.005 + 0.005) × 100)/100 × 10^6
= 10^4 ppm
The following ligand is
(1) Tetradentate
(2) Tridentate
(3) Bidentate
(4) Hexadentate
Answer (1)
Sol. It has four lone pairs but maximum it will be able to donate three lone pairs.
Maximum denticity is 3.
Diborane (B2H6) reacts independently with O2 and H2O to produce, respectively:
(1) H3BO3 and B2O3
(2) HO2 and H3BO3
(3) B2O3 and H3BO3
(4) B2O3 and [BH4]^-
Answer (3)
Sol. B2H6 + 3O2 → B2O3 + 3H2O
B2H6 + 6H2O → 2H3BO3 + 6H2
The size of the iso-electronic species Cl^-, Ar and Ca2+ is affected by
(1) Nuclear charge
(2) Principal quantum number of valence shell
(3) Azimuthal quantum number of valence shell
(4) Electron-electron interaction in the outer orbitals
Answer (1)
Sol. Iso-electronic species differ in size due to different effective nuclear charge.
The IUPAC name of the following compound is:
CH3OH
H3C - CH - CH - CH2 - COOH
(1) 3-Hydroxy-4-methylpentanoic acid
(2) 4-Methyl-3-hydroxypentanoic acid
(3) 2-Methyl-3-hydroxypentan-5-oic acid
(4) 4,4-Dimethyl-3-hydroxubutanoic acid
Answer (1)
Sol. IUPAC name
3-Hydroxy-4-methylpentanoic acid
Which of the following amines can be prepared by Gabriel phthalimide reaction?
(1) Neo-pentylamine
(2) n-butylamine
(3) t-butylamine
(4) Triethylamine
Answer (2)
Sol. Primary amines are prepared by Gabriel phthalimide synthesis
Pot. phthalimide + C4H9Cl → N-C4H9 phthalimide → NaOH(aq) → n-C4H9NH2
An organic compound 'X' showing the following solubility profile is
X + water → insoluble
X + 5% HCl → insoluble
X + 10% NaOH → soluble
X + 10% NaHCO3 → insoluble
(1) Benzamide
(2) Oleic acid
(3) o-Toluidine
(4) m-Cresol
Answer (4)
Sol. X is Meta cresol
Benzamide is amphoteric
Oleic acid will dissolve in NaOH as well as NaHCO3 due to acidic nature.
The major product of the following reaction is
(1) OH with Br-CHCH3
(2) Br with Br-CHCH3
(3) OH with CH2CH2Br
(4) Br with CH2CH2Br
Answer (1)
Sol.
m-Methoxystyrene + Conc HBr (excess)/heat → m-(1-bromoethyl)phenol
Assertion : Ozone is destroyed by CFCs in the upper stratosphere.
Reason : Ozone holes increase the amount of UV radiation reaching the earth.
(1) Assertion and reason are both correct, and the reason is the correct explanation for the assertion.
(2) Assertion is false, but the reason is correct.
(3) Assertion and reason are correct, but the reason is not the explanation for the assertion.
(4) Assertion and reason are incorrect.
Answer (3)
Sol. CFC's are responsible for depletion of ozone layer
CF2Cl2 → Cl(g) + F2ClCl
Cl + O3 → ClO + O2
ClO + O → Cl + O2
Both statements are correct.
An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is
(1) ketone with C2H5
(2) aldehyde with OCH3
(3) alcohol with CH3 and C2H5
(4) ketone with OH
Answer (3)
Sol.
Reacts with Grignard's reagent due to acidic hydrogen. Fehling solution test is negative as there is no -CHO group. Neutral FeCl3 test is negative as there is no phenolic group.
In the following compounds, the decreasing order of basic strength will be
(1) NH3 > C2H5NH2 > (C2H5)2NH
(2) C2H5NH2 > NH3 > (C2H5)2NH
(3) (C2H5)2NH > NH3 > C2H5NH2
(4) (C2H5)2NH > C2H5NH2 > NH3
Answer (4)
Sol. Correct order of Kb value
(C2H5)2NH > (C2H5)3N > NH3
In aqueous medium sec. amines are most basic.
3° amines are more basic than NH3 as +I factor dominate over steric factor.
Mathematics
The shortest distance between the line y = x and the curve y^2 = x - 2 is :
(1) 11/(4√2)
(2) 7/8
(3) 2
(4) 7/(4√2)
Answer (4)
Sol. The shortest distance between line y = x and parabola = the distance between line y = x and tangent of parabola having slope 1.
Let equation of tangent of parabola having slope 1 is,
y = m(x - 2) + a/m
where m = 1 and a = 1/4
Equation of tangent y = x - 7/4
Distance between the line y = x and the tangent
The sum of the co-efficients of all even degree terms in x in the expansion of (x + √(x^3 - 1))^6 + (x - √(x^3 - 1))^6, (x > 1) is equal to :
(1) 24
(2) 32
(3) 26
(4) 29
Answer (1)
Sol. (x + √(x^3 - 1))^6 + (x - √(x^3 - 1))^6
= 2[^6C0 x^6 + ^6C2 x^4(x^3 - 1) + ^6C4 x^2(x^3 - 1)^2 + ^6C6(x^3 - 1)^3]
= 2[x^6 + 15x^7 - 15x^4 + 15x^8 - 30x^5 + 15x^2 + x^9 - 3x^6 + 3x^3 - 1]
Sum of coefficients of even powers of x = 2[1 - 15 + 15 + 15 - 3 - 1] = 24
If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, f(x) = 9x^4 + 12x^3 - 36x^2 + 25, x ∈ R then:
(1) S1 = {-2} ; S2 = {0, 1}
(2) S1 = {-2, 1} ; S2 = {0}
(3) S1 = {-1} ; S2 = {0, 2}
(4) S1 = {-2, 0} ; S2 = {1}
Answer (2)
Sol. f(x) = 9x^4 + 12x^3 - 36x^2 + 25
f'(x) = 36[x^3 + x^2 - 2x] = 36(x - 1)(x + 2)
-2 0 1
Whenever derivative changes sign from negative to positive, we get local minima, and whenever derivative changes sign from positive to negative, we get local maxima (while moving left to right on x-axis)
S1 = {-2, 1}
S2 = {0}
If 2y = (cot^-1((√3 cos x + sin x)/(cos x - √3 sin x)))^2, x ∈ (0, π/2) then dy/dx is equal to :
(1) 2x - π/3
(2) x - π/6
(3) π/3 - x
(4) π/6 - x
Answer (2)
Sol. 2y = ((√3/2 cos x + 1/2 sin x)/(1/2 cos x - √3/2 sin x))^2
⇒ 2y = (cos(π/6 - x)/sin(π/6 - x))^2
⇒ 2y = [cot^-1(cot(π/6 - x))]^2
π/6 - x ∈ (-π/3, π/6)
= [7π/6 - x]^2 if π/6 - x ∈ (-π/3, 0)
= [π/6 - x]^2 if π/6 - x ∈ (0, π/6)
dy/dx = -7π/6 if x ∈ (π/6, π/2)
= -π/6 if x ∈ (0, π/6)
If the sum of the first 15 terms of the series (3/4)^3 + (1 1/2)^3 + (2 1/4)^3 + 3^3 + (3 3/4)^3 + ... is equal to 225k, then k is equal to :
(1) 27
(2) 9
(3) 108
(4) 54
Answer (1)
Sol. Series = (3/4)^3 + (6/4)^3 + (9/4)^3 + (12/4)^3 + ...
= (3/4)^3 [1^3 + 2^3 + 3^3 + ... + 15^3]
= 27/64 [15 × 16/2]^2
= 27/64 × 120^2
= 27/64 × 14400 = 27 × 225
225k = 27 × 225 ⇒ k = 27
Let S be the set of all real values of λ such that a plane passing through the points (-λ^2, 1, 1), (1, -λ^2, 1) and (1, 1, -λ^2) also passes through the point (-1, -1, 1). Then S is equal to :
(1) {√3}
(2) {√3, -√3}
(3) {1, -1}
(4) {3, -3}
Answer (2)
Sol. Plane through (-λ^2, 1, 1), (1, -λ^2, 1), (1, 1, -λ^2)
Equation of plane: x + y + z = 3 - λ^2
Passes through (-1, -1, 1): -1 - 1 + 1 = 3 - λ^2
-1 = 3 - λ^2 ⇒ λ^2 = 4? Wait check: -1 = 3 - λ^2 ⇒ λ^2 = 4? But answer says {√3, -√3}. Let's recalc: x+y+z = 3 - λ^2. For (-1,-1,1): -1-1+1 = -1 = 3 - λ^2 ⇒ λ^2 = 4, so λ = ±2. But options don't include 2. Maybe original problem has different point? The given options {√3, -√3}. Hmm. Let's see if plane equation maybe x+y+z = 1 - λ^2? If (-1,-1,1): -1 = 1 - λ^2 ⇒ λ^2 = 2, λ=±√2. Not options. If x+y+z = -1 - λ^2? Then -1 = -1 - λ^2 ⇒ λ=0. Not. Wait maybe the points are (-λ^2,1,1), (1,-λ^2,1), (1,1,-λ^2) and (-1,-1,1). The plane equation: determinant. Let's compute normal: vectors (1+λ^2, -λ^2-1, 0), (1+λ^2, 0, -λ^2-1). Cross product = (λ^2+1)^2 (1,1,1) so plane x+y+z = constant. Plug first point: -λ^2+1+1 = 2 - λ^2. So plane x+y+z = 2 - λ^2. Plug (-1,-1,1): -1 = 2 - λ^2 ⇒ λ^2 = 3 ⇒ λ = ±√3. Yes I earlier wrote 3 - λ^2 incorrectly. So answer (2).
If the system of linear equations
x - 2y + kz = 1
2x + y + z = 2
3x - y - kz = 3
has a solution (x, y, z), z ≠ 0 then (x, y) lies on the straight line whose equation is
(1) 3x - 4y - 4 = 0
(2) 3x - 4y - 1 = 0
(3) 4x - 3y - 1 = 0
(4) 4x - 3y - 4 = 0
Answer (4)
Sol. x - 2y + kz = 1, 2x + y + z = 2, 3x - y - kz = 3
Δ = |1 -2 k; 2 1 1; 3 -1 -k| = 1(-k + 1) + 2(-2k - 3) + k(-2 - 3)
= -k + 1 - 4k - 6 - 5k = -10k - 5 = -5(2k + 1)
Δ1 = |1 -2 k; 2 1 1; 3 -1 -k| = -5(2k + 1)
Δ2 = |1 1 k; 2 2 1; 3 3 -k| = 0, Δ3 = |1 -2 1; 2 1 2; 3 -1 3| = 0
∴ z ≠ 0
⇒ Δ = 0
⇒ k = -1/2
∴ System of equation has infinite many solutions.
Let z = λ ≠ 0 then x = (10 - 3λ)/10 and y = -2λ/5
∴ (x, y) must lie on line 4x - 3y - 4 = 0
The sum of the squares of the lengths of the chords intercepted on the circle, x^2 + y^2 = 16 by the lines, x + y = n, n ∈ N, where N is the set of all natural numbers, is
(1) 105
(2) 160
(3) 320
(4) 210
Answer (4)
Sol. Let the chord x + y = n cuts the circle x^2 + y^2 = 16 at A and B length of perpendicular from O on AB = |(0 + 0 - n)/√(1^2 + 1^2)| = n/√2
Length of chord AB = 2√(4^2 - (n/√2)^2)
= 2√(16 - n^2/2)
Here possible values of n are 1, 2, 3, 4, 5.
Sum of square of length of chords
= ∑_{n=1}^5 4(16 - n^2/2)
= 64 × 5 - 2 × (5×6×11)/6 = 210
If f(x) = log_e((1 - x)/(1 + x)), |x| < 1 then f(2x/(1 + x^2)) is equal to:
(1) 2f(x)
(2) 2f(x^2)
(3) -2f(x)
(4) (f(x))^2
Answer (1)
Sol. ∴ f(x) = ln((1 - x)/(1 + x))
f(2x/(1 + x^2)) = ln((1 - 2x/(1 + x^2))/(1 + 2x/(1 + x^2)))
= ln((1 + x^2 - 2x)/(1 + x^2 + 2x))
= ln((1 - x)/(1 + x))^2
= 2 ln((1 - x)/(1 + x))
= 2f(x)
Let y = y(x) be the solution of the differential equation, (x^2 + 1)^2 dy/dx + 2x(x^2 + 1)y = 1 such that y(0) = 0. If √a y(1) = π/32, then the value of 'a' is
(1) 1/2
(2) 1/4
(3) 1
(4) 1/16
Answer (4)
Sol. (1 + x^2)^2 dy/dx + 2x(1 + x^2)y = 1
⇒ dy/dx + (2x/(1 + x^2))y = 1/(1 + x^2)^2
It is a linear differential equation
IF = e^{∫ 2x/(1 + x^2) dx} = e^{ln(1 + x^2)} = 1 + x^2
⇒ y(1 + x^2) = ∫ dx/(1 + x^2) + c
⇒ y(1 + x^2) = tan^-1 x + c
If x = 0 then y = 0
So, 0 = 0 + c
⇒ c = 0
⇒ y(1 + x^2) = tan^-1 x
put x = 1
2y = π/4
⇒ 2(π/(32√a)) = π/4
⇒ √a = 1/4
⇒ a = 1/16
If α = cos^-1(3/5), β = tan^-1(1/3), where 0 < α, β < π/2, then α - β is equal to
(1) tan^-1(9/14)
(2) cos^-1(9/(5√10))
(3) sin^-1(9/(5√10))
(4) tan^-1(9/(5√10))
Answer (3)
Sol. ∴ cos α = 3/5
⇒ tan α = 4/3
⇒ and tan β = 1/3
tan(α - β) = (tan α - tan β)/(1 + tan α tan β)
= (4/3 - 1/3)/(1 + 4/9) = (1)/(13/9) = 9/13
α - β = tan^-1(9/13) = sin^-1(9/(5√10)) = cos^-1(13/(5√10))
Let f : [0, 2] → R be a twice differentiable function such that f''(x) > 0, for all x ∈ (0, 2). If ?(x) = f(x) + f(2 - x), then ? is
(1) Decreasing on (0, 2)
(2) Increasing on (0, 2)
(3) Decreasing on (0, 1) and increasing on (1, 2)
(4) Increasing on (0, 1) and decreasing on (1, 2)
Answer (3)
Sol. ?(x) = f(x) + f(2 - x)
differentiating w.r.t. x
?'(x) = f'(x) - f'(2 - x)
For ?(x) to be increasing ?'(x) > 0
⇒ f'(x) > f'(2 - x)
(? f''(x) > 0 then f'(x) is an increasing function)
⇒ x > 2 - x
⇒ x > 1
So ?(x) is increasing in (1, 2) and decreasing in (0, 1)
The sum of the series 2·20C0 + 5·20C1 + 8·20C2 + 11·20C3 + ... + 62·20C20 is equal to
(1) 2^23
(2) 2^25
(3) 2^24
(4) 2^26
Answer (2)
Sol. 2·20C0 + 5·20C1 + 8·20C2 + ... + 62·20C20
= ∑{r=0}^{20} (3r + 2)·20Cr = 3∑{r=0}^{20} r·20Cr + 2∑{r=0}^{20} 20Cr
= 60 ∑{r=1}^{20} 19Cr-1 + 2 × 2^20
= 60 × 2^19 + 2 × 2^20
= 2^21 [15 + 1] = 2^25
All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is :
(1) 180
(2) 175
(3) 162
(4) 160
Answer (1)
Sol. There are total 9 digits; out of which only 3 digits are odd.
Number of ways to arrange odd digits first
= ^4C3 · 3!/2!
Total number of 9 digit numbers = (^4C3 · 3!/2!) · 6!/(2!4!)
= 180
The greatest value of c ∈ R for which the system of linear equations
x - cy - cz = 0
cx - y + cz = 0
cx + cy - z = 0
has a non-trivial solution, is :
(1) -1
(2) 0
(3) 2
(4) 1/2
Answer (4)
Sol. For non-trivial solution, determinant = 0
|1 -c -c; c -1 c; c c -1| = 0
1(1 - c^2) + c(-c - c^2) - c(c^2 + c) = 0
1 - c^2 - c^2 - c^3 - c^3 - c^2 = 0
1 - 3c^2 - 2c^3 = 0
2c^3 + 3c^2 - 1 = 0
(c + 1)(2c^2 + c - 1) = 0
(c + 1)(2c - 1)(c + 1) = 0
c = -1, 1/2
Greatest value = 1/2
Let A and B be two non-null events such that A ⊂ B. Then, which of the following statements is always correct?
(1) P(A|B) = P(B) - P(A)
(2) P(A|B) ≤ P(A)
(3) P(A|B) ≥ P(A)
(4) P(A|B) = 1
Answer (3)
Sol. ∴ A ⊂ B ; so A ∩ B = A
Now, P(A/B) = P(A ∩ B)/P(B)
⇒ P(A/B) = P(A)/P(B)
∴ P(B) ≤ 1
So, P(A/B) ≥ P(A)
If α and β be the roots of the equation x^2 - 2x + 2 = 0, then the least value of n for which (α/β)^n = 1 is :
(1) 4
(2) 5
(3) 3
(4) 2
Answer (1)
Sol. x^2 - 2x + 2 = 0
roots of this equation are (2 ± √-4)/2 = 1 ± i
Then α/β = (1 + i)/(1 - i) = (1 + i)^2/(1 - i^2) = i
or α/β = (1 - i)/(1 + i) = (1 - i)^2/(1 - i^2) = -i
So, α/β = ±i
Now, (α/β)^n = 1 ⇒ (±i)^n = 1
⇒ n must be a multiple of 4.
minimum value of n = 4
The area (in sq. units) of the region A = {(x, y) ∈ R × R | 0 ≤ x ≤ 3, 0 ≤ y ≤ 4, y ≤ x^2 + 3x} is:
(1) 26/3
(2) 59/6
(3) 8
(4) 53/6
Answer (2)
Sol. y ≤ x^2 + 3x represents region below the parabola.
Area of the required region
= ∫_0^1 (x^2 + 3x)dx + ∫_1^3 4 dx
= 1/3 + 3/2 + 8
= 59/6
If f(x) = (2 - x cos x)/(2 + x cos x) and g(x) = log_e x (x > 0) then the value of the integral ∫{-√4}^{√4} g(f(x))dx is:
(1) log_e 1
(2) log_e 3
(3) log_e 2
(4) log_e e
Answer (1)
Sol. g(f(x)) = ln((2 - x cos x)/(2 + x cos x))
Let I = ∫{-√4}^{√4} ln((2 - x cos x)/(2 + x cos x))dx ...(i)
Using property ∫a^b f(x)dx = ∫a^b f(a + b - x)dx
I = ∫{-√4}^{√4} ln((2 + x cos x)/(2 - x cos x))dx ...(ii)
Adding (i) and (ii)
2I = ∫{-√4}^{√4} ln(1)dx = 0
= I = 0 = ln 1
The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:
(1) 2x - z = 2
(2) x - 3y - 2z = -2
(3) x - y - z = 0
(4) x + 3y + z = 4
Answer (3)
Sol. Let the equation of required plane be;
(2x - y - 4) + λ(y + 2z - 4) = 0
∴ This plane passes through (1, 1, 0) then
(2 - 1 - 4) + λ(1 + 0 - 4) = 0 ⇒ λ = -1
Equation of required plane will be
(2x - y - 4) - (y + 2z - 4) = 0
⇒ 2x - 2y - 2z = 0
⇒ x - y - z = 0
If cos(α + β) = 3/5, sin(α - β) = 5/13 and 0 < α, β < π/4, then tan(2α) is equal to:
(1) 21/16
(2) 63/52
(3) 33/52
(4) 63/16
Answer (4)
Sol. ? α + β and α - β both are acute angles.
cos(α + β) = 3/5
tan(α + β) = 4/3
And sin(α - β) = 5/13
tan(α - β) = 5/12
Now, tan 2α = tan((α + β) + (α - β))
= (tan(α + β) + tan(α - β))/(1 - tan(α + β)·tan(α - β))
= (4/3 + 5/12)/(1 - 4/3 × 5/12) = 63/16
Let A = (cos α -sin α; sin α cos α), (α ∈ R) such that
A^32 = (0 -1; 1 0).
(1) π/32
(2) π/64
(3) 0
(4) π/16
Answer (2)
Sol. A = [cos α -sin α; sin α cos α]
A^2 = [cos α -sin α; sin α cos α][cos α -sin α; sin α cos α]
A^2 = [cos 2α -sin 2α; sin 2α cos 2α]
Then A^4 = A^2 A^2 = [cos 4α -sin 4α; sin 4α cos 4α]
similarly A^8 = A^4 A^4 [cos 8α -sin 8α; sin 8α cos 8α]
and so on A^32 = [cos 32α -sin 32α; sin 32α cos 32α] = [0 -1; 1 0]
So sin 32α = 1 and cos 32α = 0
⇒ 32α = 2nπ + π/2 ⇒ α = nπ/16 + π/64 where n ∈ Z
put n = 0
α = π/64
The contrapositive of the statement "If you are born in India, then you are a citizen of India", is:
(1) If you are born in India, then you are not a citizen of India.
(2) If you are not born in India, then you are not a citizen of India.
(3) If you are a citizen of India, then you are born in India.
(4) If you are not a citizen of India, then you are not born in India.
Answer (4)
Sol. S: "If you are born in India, then you are a citizen of India."
Contrapositive of p → q is ¬q → ¬p
So contrapositive of statement S will be:
"If you are not a citizen of India, then you are not born in India."
The sum of the solutions of the equation |√x - 2| + √x(√x - 4) + 2 = 0 (x > 0) is equal to:
(1) 4
(2) 10
(3) 9
(4) 12
Answer (2)
Sol. Let √x = t
|t - 2| + t(t - 4) + 2 = 0
⇒ |t - 2| + t^2 - 4t + 4 - 2 = 0
⇒ |t - 2| + (t - 2)^2 - 2 = 0
Let |t - 2| = z (Clearly z ≥ 0)
⇒ z + z^2 - 2 = 0
⇒ z = 1 or -2 (rejected)
⇒ |t - 2| = 1 ⇒ t = 1, 3
if √x = 1 ⇒ x = 1
if √x = 3 ⇒ x = 9
Sum of solutions = 10
The magnitude of the projection of the vector 2i + 3j + k on the vector perpendicular to the plane containing the vectors i + j + k and i + 2j + 3k, is:
(1) √(3/2)
(2) 3√6
(3) √(3/2)
(4) √6
Answer (1)
Sol. Let a = i + j + k and b = i + 2j + 3k vector perpendicular to a and b is a × b
Projection of vector c = 2i + 3j + k on a × b is
= c·(a × b)/|a × b| = |2 - 6 + 1|/√6
= 3/√6 = √(3/2)
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is:
(1) 45
(2) 40
(3) 48
(4) 49
Answer (3)
Sol. Let the remaining numbers are x and y
Mean(x?) = ∑xi/N = (2 + 4 + 10 + 12 + 14 + x + y)/7 = 8
⇒ x + y = 14 ...(i)
Variance(σ^2) = ∑xi^2/N - (x?)^2 = 16
⇒ (2^2 + 4^2 + 10^2 + 12^2 + 14^2 + x^2 + y^2)/7 - (8)^2 = 16
⇒ x^2 + y^2 = 100 ...(ii)
From (i) and (ii) (x, y) = (6, 8) or (8, 6)
xy = 48
lim_{x→0} sin^2 x/(√2 - √(1 + cos x)) equals
(1) √2
(2) 2√2
(3) 4
(4) 4√2
Answer (4)
Sol. lim_{x→0} sin^2 x/(√2 - √(1 + cos x)) = lim_{x→0} sin^2 x/(√2 - √(2 cos^2(x/2)))
= lim_{x→0} sin^2 x/(√2[1 - cos(x/2)])
= lim_{x→0} sin^2 x/(2√2 sin^2(x/4))
= lim_{x→0} (sin x/x)^2 / (2√2 (sin(x/4)/(x/4))^2)
= 16/(2√2) = 4√2
The sum of all natural numbers 'n' such that 100 < n < 200 and H.C.F. (91, n) > 1 is:
(1) 3303
(2) 3121
(3) 3203
(4) 3221
Answer (2)
Sol. ? 91 = 13 × 7
So the required numbers are either divisible by 7 or 13
Sum of such numbers = Sum of no. divisible by 7 + sum of the no. divisible by 13 - Sum of the numbers divisible by 91
= (105 + 112 + ... + 196) + (104 + 117 + ... + 195) - 182
= 2107 + 1196 - 182
= 3121
The length of the perpendicular from the point (2, -1, 4) on the straight line,
(x + 3)/10 = (y - 2)/(-7) = z/1 is:
(1) Greater than 3 but less than 4
(2) Greater than 2 but less than 3
(3) Greater than 4
(4) Less than 2
Answer (1)
Sol. Let P be the foot of perpendicular from point A(2, -1, 4) on the given line. So P can be assumed as P(10λ - 3, -7λ + 2, λ)
DR's of AP ∝ to 10λ - 5, -7λ + 3, λ - 4
AP and given line are perpendicular, so
10(10λ - 5) - 7(-7λ + 3) + 1(λ - 4) = 0
⇒ λ = 1/2
AP = √((10λ - 5)^2 + (-7λ + 3)^2 + (λ - 4)^2)
= √(0 + 1/4 + 49/4)
= √12.5 ; √12.5 ∈ (3, 4)
If the fourth term in the binomial expansion of (√(1/x^{1 + log10 x}) + x^{1/12})^6 is equal to 200, and x > 1, then the value of x is:
(1) 10
(2) 10^3
(3) 100
(4) 10^4
Answer (1)
Sol. T4 = 6C3 (√(1/x^{1 + log10 x}))^3 (x^{1/12})^3 = 200
⇒ 3/(20 x^{2(1 + log10 x)/4} · x^{1/4}) = 200
x^{1/4 - 3/(2(1 + log10 x))} = 10
Taking log10 on both sides and put log10 x = t
(1/4 + 3/(2(1 + t)))t = 1
((1 + t) + 6)/(4(1 + t)) t = 1 ⇒ t^2 + 7t = 4 + 4t
t^2 + 3t - 4 = 0 ⇒ t^2 + 4t - t - 4 = 0
⇒ t(t + 4) - 1(t + 4) = 0
⇒ t = 1 or t = -4
log10 x = 1 ⇒ x = 10 or if log10 x = -4 ⇒ x = 10^-4
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