JEE MAINS 2014 QUESTION PAPER (PHYSICS CHEMISTRY MATHEMATICS )
1. The pressure that has to be applied to the ends of a steel wire of length 10cm to keep its length constant when its temperature is raised by 100°C is : (For steel Young's modulus is 2×10¹¹ Nm⁻² and coefficient of thermal expansion is 1.1×10⁻⁵ K⁻¹)
(1) 2.2×10⁷ Pa
(2) 2.2×10⁶ Pa
(3) 2.2×10⁸ Pa
(4) 2.2×10⁹ Pa
Ans. (3)
Sol. Thermal strain = α ΔT (by ℓ = ℓ₀(1 + α ΔT))
⇒ Thermal stress in Rod (Pressure due to Thermal strain) = YαΔT
= 2×10¹¹ × 1.1×10⁻⁵ × 100
= 2.2×10⁸ Pa
2. A conductor lies along the z-axis at −1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A in −a_z direction (see figure). For a field B̄ = 3.0×10⁻⁴ e^{−0.2x} a_y T find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5×10⁻³ s. Assume parallel motion along the x-axis.
(1) 14.85 W
(2) 29.7 W
(3) 1.57 W
(4) 2.97 W
Ans. (4)
Sol. Force on Conductor F = iℓB
F = (10)(3)(3×10⁻⁴ e^{−0.2x})
F̄ = 90×10⁻⁴ (e^{−0.2x}) â_x
Work done W = ∫{x=0}^{x=2} F dx
W = 90×10⁻⁴ ∫₀² e^{−0.2x} dx
W = 90×10⁻⁴ [e^{−0.2x}/(−0.2)]{x=0}^{x=2}
W = 90×10⁻⁴ [(e^{−0.4} − 1)/(−0.2)]
Now Average power
P_avg. = work/time
P_avg. = 90×10⁻⁴ × (1 − e^{−0.4})/(5×10⁻³ × 0.2)
P_avg. = 2.97 Watt
3. A bob of mass m attached to an inextensible string of length ℓ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension :
(1) Angular momentum changes in direction but not in magnitude
(2) Angular momentum changes both in direction and magnitude
(3) Angular momentum is conserved
(4) Angular momentum changes in magnitude but not in direction.
Ans. (1)
Sol. τ̄ of mg is perpendicular to L̄. Hence magnitude of L̄ is constant but direction will change.
4. The current voltage relation of diode is given by I = (e^{1000V/T} − 1) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ±0.01 V while measuring the current of 5 mA at 300 K, what will be error in the value of current in mA ?
(1) 0.5 mA
(2) 0.05 mA
(3) 0.2 mA
(4) 0.02 mA
Ans. (3)
Sol. Given current I = (e^{1000V/T} − 1) mA ⇒ I + 1 = e^{1000V/T}
dI = (1000/T)[e^{1000V/T}] dV
dI = (1000/T)[1 + 1] dV
= (1000/300)[6] × (0.01)
dI = 0.2 mA
5. An open glass tube is immersed in mercury in such a way that a length of 8cm extends above the mercury level. The open of the tube is then closed and sealed and the tube is raised vertically up by addition 46cm. What will be length of the air column above mercury in the tube now ? (Atmospheric pressure = 76cm of Hg)
(1) 38cm
(2) 6cm
(3) 16cm
(4) 22cm
Ans. (3)
Sol. P₀A(8) = P′A(54 − x)
P₀ 8 = P′(54 − x) ....(1)
P′ = P₀ − ρgx ....(2)
Comparing (P₀)8 = (P₀ − ρgx)(54 − x)
(76)8 = (76 − x)(54 − x)
Solving we get x = 38cm
Air column = 54 − 38 = 16cm hence 3 option is correct.
6. Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists :
List-I List-II
(a) Infrared waves (i) To treat muscular strain
(b) Radio waves (ii) For broadcasting
(c) X-rays (iii) To detect fracture of bones
(d) Ultraviolet rays (iv) Absorbed by the ozone layer of the atmosphere
(a) (b) (c) (d)
(1) (iii) (ii) (i) (iv)
(2) (i) (ii) (iii) (iv)
(3) (iv) (iii) (ii) (i)
(4) (i) (ii) (iv) (iii)
Ans. (2)
Sol. Factual question
7. A parallel plate capacitor is made of two circular plates separated by a distance of 5mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric field in the dielectric is 3×10⁴ V/m the charge density of the positive plate will be close to :
(1) 3×10⁴ C/m²
(2) 6×10⁴ C/m²
(3) 6×10⁻⁷ C/m²
(4) 3×10⁻⁷ C/m²
Ans. (3)
Sol. σ/(Kε₀) = 3×10⁴
σ/(2.25×8.86×10⁻¹²) = 3×10⁴
σ = 6×10⁻⁷ C/m²
8. A student measured the length of a rod and wrote it as 3.50cm. Which instrument did he use to measure it ?
(1) A screw gauge having 100 divisions in the circular scale and pitch as 1mm
(2) A screw gauge having 50 divisions in the circular scale and pitch as 1mm
(3) A meter scale
(4) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1cm
Ans. (4)
Sol. Least count of varnier calliper is 0.01cm Hence it matches with the reading.
9. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :
(1) √(GM/R (1 + 2√2))
(2) (1/2)√(GM/R (1 + 2√2))
(3) √(GM/R)
(4) √(2√2 GM/R)
Ans. (2)
Sol. Net force on one particle F_net = F₁ + 2F₂ cos45° = Centripetal force
⇒ GM²/(2R)² + [2GM²/(√2R)² cos45°] = MV²/R
V = (1/2)√(GM/R (1 + 2√2))
10. In a large building, there are 15 bulbs of 40W, 5 bulbs of 100W, 5 fans of 80W and 1 heater of 1kW. The voltage of the electric mains is 220V. The minimum capacity of the main fuse of the building will be :
(1) 12 A
(2) 14 A
(3) 8 A
(4) 10 A
Ans. (1)
Sol. All devices are in parallel so total current drawn is gives as
i_net = Total Power/220
i_net = (15×40 + 5×100 + 5×80 + 1000)/220
i_net = 2500/220 ≈ 11.36 A
minimum capacity of main fuse should be more than 11.36 A
Ans is ≈ 12 A Hence (1)
11. A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in same direction, then :
(1) Amplitude of motion is 4a
(2) Time period of oscillation is 6τ
(3) Amplitude of motion is 3a
(4) Time period of oscillation is 8τ
Ans. (2)
Sol.
x = A sin(ωt + π/2)
x = A cos ωt
A − a = A cos ωτ
cos ωτ = (A − a)/A
cos 2ωτ = (A − 3a)/A
2cos²ωτ − 1 = (A − 3a)/A = 2((A − a)/A)² − 1
2(A − a)² − A² = (A)(A − 3a)
2A² + 2a² − 4Aa − A² = A² − 3aA
2a² − 4Aa = −3aA
2a² = Aa
A = 2a
cos ωτ = a/2a = 1/2
ωτ = π/3
2π/ω = T
T = 2π·3τ/π
T = 6τ
12. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3×10³ Am⁻¹. The current required to be passed in a solenoid of length 10cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is :
(1) 3A
(2) 6 A
(3) 30 mA
(4) 60 mA
Ans. (1)
Sol. Coercivity = B/μ₀ = 3×10³ = nI
3×10³ = 1000I
I = 3A
13. The forward biased diode connection is :
(1)
2V
−2V
+2V
−3V
−3V
−3V
−3V
−3V
−3V
(2)
(3)
(4)
Ans. (3)
Sol. By convention
14. During the propagation of electromagnetic waves in a medium :
(1) Electric energy density is equal to the magnetic energy density
(2) Both electric magnetic energy densities are zero
(3) Electric energy density is double of the magnetic energy density
(4) Electric energy density is half of the magnetic energy density.
Ans. (1)
Sol. Factual question
15. In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to :
(1) −1
(2) (1 − e)/e
(3) e/(1 − e)
(4) 1
Ans. (1)
Sol. V_R + V_L = 0
V_R/V_L = −1
16. A mass 'm' is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
(1) 5g/6
(2) g
(3) 2g/3
(4) g/2
Ans. (4)
Sol. mg − T = ma
T×R = (mR²/2)×(a/R)
⇒ a = g/2
17. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400K, 800K and 600K respectively. Choose the correct statement :
(1) The change in internal energy in the process AB is −350R
(2) The change in internal energy in the process BC is −500R
(3) The change in internal energy in whole cyclic process is 250R
(4) The change in internal energy in the process CA is 700R
Ans. (2)
Sol. Change in internal energy = f n R ΔT / 2
= 5/2 × 1 × R × (−200)
= −500R
18. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is :
(1) 2gH = nu²(n − 2)
(2) gH = (n − 2)u²
(3) 2gH = n²u²
(4) gH = (n − 2)²u²
Ans. (1)
Sol. Time to reach highest point t = u/g
time to reach ground = nt
S = ut + 1/2 at²
−H = u(nt) − 1/2 g(nt)²
⇒ 2gH = nu²(n − 2)
19. A thin convex lens made from crown glass (μ = 3/2) has focal length f. When it is measured in two different liquids having refractive indices 4/3 and 5/3, it has the focal length f₁ and f₂ respectively. The correct relation between the focal lengths is :
(1) f₂ > f and f₁ becomes negative
(2) f₁ and f₂ both become negative
(3) f₁ = f₂ < f
(4) f₁ > f and f₂ become negative
Ans. (4)
20. Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = 4cm². End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :
(1) 4.8 cal/s
(2) 6.0 cal/s
(3) 1.2 cal/s
(4) 2.4 cal/s
Ans. (1)
Sol.
(100 − θ)/R_c = (θ − 0)/R_B + (θ − 0)/R_S
where R = ℓ/KA
on solving we get θ = 40
Heat flow per unit time through copper rod
= (100 − 40)/ℓ_c (K_c A_c)
= 60/46 × 0.92 × 4
= 4.8 cal/s
21. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
(1) 6
(2) 4
(3) 12
(4) 8
Ans. (1)
Sol. Fundamental frequency of closed organ pipe
f₀ = v/4ℓ = 340/(4×0.85) = 100 Hz
So possible frequencies below 1250 Hz are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz
⇒ No. of frequencies = 6
22. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d₁ and d₂ are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α with vertical. Ratio d₁/d₂ is :
(1) (1 + tan α)/(1 − tan α)
(2) (1 + sin α)/(1 − cos α)
(3) (1 + sin α)/(1 − sin α)
(4) (1 + cos α)/(1 − cos α)
Ans. (1)
Sol.
1/f = (3/2 / 1 − 1)(1/R₁ − 1/R₂) …(1)
1/f₁ = (3/2 / 4/3 − 1)(1/R₁ − 1/R₂) …(2)
1/f₂ = (3/2 / 5/3 − 1)(1/R₁ − 1/R₂) …(3)
23. A green light is incident from the water to the air-water interface at the critical angle (θ) Select the correct statement.
(1) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.
(2) The entire spectrum of visible light will come out of the water at various angles to the normal
(3) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal.
(4) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.
Ans. (4)
Sol. Frequency of light (ν) > frequency of green light (ν_G)
μ is also greater than μ_G and critical angle of light is less than green light therefore light will got total internal reflection and not come out to the air.
For frequency of light (ν) < ν_G ; light will not suffer T.I.R. Therefore light come out to the air
(1) λ₁ = λ₂ = 4λ₃ = 9λ₄
(2) λ₁ = 2λ₂ = 3λ₃ = 4λ₄
(3) 4λ₁ = 2λ₂ = 2λ₃ = λ₄
(4) λ₁ = 2λ₂ = 2λ₃ = λ₄
Ans. (1)
Sol. In Bohr model
1/λ = Rz²[1/n₁² − 1/n₂²]
⇒ λ ∝ 1/z²
λ₁ : λ₂ : λ₃ : λ₄ : 1/1 : 1/1 : 1/4 : 1/9
⇒ λ₁ = λ₂ = 4λ₃ = 9λ₄
24. Hydrogen (₁H¹), Deuterium (₁H²), singly ionised Helium (₂He⁴)⁺ and doubly ionised lithium (₃Li⁶)⁺ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are λ₁, λ₂, λ₃ and λ₄ respectively then approximately which one of the following is correct ?
Sol.
Let Radius of circular tube is R
Also P_A = P_C = P₀
Pressure at point B
P_B = P₀ + d₁(R − R sin α)g
= P₀ + d₂(R sin α + R cos α)g
⇒ d₁(cos α − sin α) = d₂(sin α + cos α)
d₁/d₂ = (cos α + sin α)/(cos α − sin α) = (1 + tan α)/(1 − tan α)
25. The radiation corresponding to 3→2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3×10⁻⁴T. If the radius of the largest circular path followed by these electrons is 10.0mm the work function of the metal is close to :-
(1) 0.8eV
(2) 1.6eV
(3) 1.8eV
(4) 1.1eV
Ans. (4)
26. A block of mass m is placed on a surface with a vertical cross section given by y = x³/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is :-
(1) 1/3 m
(2) 1/2 m
(3) 1/6 m
(4) 2/3 m
Ans. (3)
For equilibrium under limiting friction mg sin θ = μ mg cos θ ⇒ tan θ = μ
From the equation of surface y = x³/6
slope = dy/dx = 3x²/6 = tan θ
⇒ x²/2 = μ = 0.5 ⇒ x = 1
So y = 1/6
27. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx² where a and b are constants. The work done in stretching the unstretched rubber-band by L is:-
K.E. = q²B²R²/2m
K.E. = 0.80eV
Energy of photon for transition from 3→2 in hydrogen atom
E = 13.6[1/2² − 1/3²] = 1.88eV
From Einstein photoelectric equation
E = K.E_max + φ
⇒ 1.88 = 0.8 + φ ⇒ φ = 1.08eV
φ ≈ 1.1eV
(1) aL²/2 + bL³/3
(2) 1/2(aL²/2 + bL³/3)
(3) aL² + bL³
(4) 1/2(aL² + bL³)
Ans. (1)
Sol. Work done = ∫₀¹ Fdx
= ∫₀¹ (ax + bx²)dx
= aL²/2 + bL³/3
28. On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is:- (density of water is ρ_w)
(1) R²√(ρ_w g/T)
(2) R²√(3ρ_w g/T)
(3) R²√(ρ_w g/3T)
(4) R²√(ρ_w g/6T)
Ans. (Bonus)
Sol.
Force due to Surface Tension = T(2πr) sin θ
= T(2πr) × r/R
This force will balance the force of Bouyancy
T(2πr) × r/R = ρ₀ × 4/3 πR³g
r = R³√(2ρ₀g/3T)
29. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are I_A and I_B respectively, then I_A/I_B equals :
(1) 1
(2) 1/3
(3) 3
(4) 3/2
Ans. (2)
Sol. When polaroid is at Angle 30° with beam A, it makes 60° with beam B
by malus law
I_A cos²30° = I_B cos²60°
⇒ I_A/I_B = 1/3
30. Assume that an electric field Ē = 30x² exists in space. Then the potential difference V_A − V_O where V_O is the potential at the origin and V_A the potential at x = 2 m is :-
(1) −80 J
(2) 80 J
(3) 120 J
(4) −120 J
Ans. (1)
Sol. Given unit in options is wrong.
By using dv = −Ē·dx̄
ΔV = −∫Ē·dx̄
V_A − V₀ = −∫{x=0}^{x=2} 30x²dx = −10x³|{x=0}^{x=2}
V_A − V₀ = −10[8 − 0] = −80 V
31. The image of the line (x − 1)/3 = (y − 3)/1 = (z − 4)/−5 in the plane 2x − y + z + 3 = 0 is the line :
(1) (x + 3)/3 = (y − 5)/1 = (z − 2)/−5
(2) (x + 3)/−3 = (y − 5)/−1 = (z + 2)/5
(3) (x − 3)/3 = (y + 5)/1 = (z − 2)/−5
(4) (x − 3)/−3 = (y + 5)/−1 = (z − 2)/5
Ans. (1)
Sol.
L: (x − 1)/3 = (y − 3)/1 = (z − 4)/−5
P: 2x − y + z + 3 = 0
It can be observed given line is parallel to given plane.
Image of (1, 3, 4) in given plane can be calculated as
(x − 1)/2 = (y − 3)/−1 = (z − 4)/1 = −2(2×1 − 3 + 4 + 3)/6 = −2
⇒ x = −3; y = 5; z = 2
Required line is
(x + 3)/3 = (y − 5)/1 = (z − 2)/−5
32. If the coefficients of x³ and x⁴ in the expansion of (1 + ax + bx²)(1 − 2x)¹⁸ in powers of x are both zero, then (a, b) is equal to :-
(1) (16, 251/3)
(2) (14, 251/3)
(3) (14, 272/3)
(4) (16, 272/3)
Ans. (4)
Sol. In the expansion of (1 + ax + bx²)(1 − 2x)¹⁸
General term = (1 + ax + bx²) · ¹⁸C_r(−2x)^r
Cofficient of x³ = ¹⁸C₃(−2)³ + a · ¹⁸C₂(−2)² + b · ¹⁸C₁(−2) = 0
Coefficient of x⁴ = ¹⁸C₄(−2)⁴ + a · ¹⁸C₃(−2)³ + b · ¹⁸C₂(−2)² = 0
on solving the equations we get
153a − 9b = 1632 ...(i)
3b − 32a = −240 ...(ii)
on solving we get a = 16 & b = 272/3
33. If a ∈ ℝ and the equation −3(x − [x])² + 2(x − [x]) + a² = 0 (where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval :
(1) (−1,0) ∪ (0,1)
(2) (1,2)
(3) (−2, −1)
(4) (−∞, −2) ∪ (2,∞)
Ans. (1)
Sol. Given equation is
−3(x − [x])² + 2(x − [x]) + a² = 0
⇒ a² = 3.{x}² − 2{x}
= 3({x} − 1/3)² − 1/3 (∵ {x} ≠ 0 ⇒ a² ≠ 0)
⇒ a² ∈ (0,1)
⇒ a ∈ (−1,0) ∪ (0,1)
34. If [ā×b̄ b̄×c̄ c̄×ā] = λ [ā b̄ c̄]² then λ is equal to:
(1) 2
(2) 3
(3) 0
(4) 1
Ans. (4)
Sol. [ā×b̄ b̄×c̄ c̄×ā] = (ā×b̄).((b̄×c̄)×(c̄×ā))
= (ā×b̄).([b̄ c̄ ā]c̄ − [b̄ c̄ c̄]ā)
= [ā b̄ c̄]²
⇒ λ = 1.
35. The variance of first 50 even natural numbers is:-
(1) 833/4
(2) 833
(3) 437
(4) 437/4
Ans. (2)
Sol. We have to calculate variance of the data
2, 4, 6, 8, ……, 100
σ² = ∑xᵢ²/n − (∑xᵢ/n)²
Now, ∑xᵢ² = 2² + 4² + … + 100² = 4.50.51.101/6
Now, ∑xᵢ²/50 = 3434
Also, (∑xᵢ/n)² = (2.50.51/2.50)² = 2601
∴ σ² = 3434 − 2601 = 833.
36. A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is :
(1) 40(√2 − 1)
(2) 40(√3 − √2)
(3) 20√2
(4) 20(√3 − 1)
Ans. (4)
Sol. OT = 20 = BT
Now, In ΔOB'L
tan30° = 20/(20 + TL) = 1/√3
∴ TL = 20(√3 − 1)
∴ speed is 20(√3 − 1) m/sec.
37. The integral ∫₀^{π/2} √(1 + 4sin²(x/2) − 4sin(x/2)) dx equals :
(1) π − 4
(2) 2π/3 − 4 − 4√3
(3) 4√3 − 4
(4) 4√3 − 4 − π/3
Ans. (4)
Sol. I = ∫₀^{π/2} |2sin(x/2) − 1| dx = 2∫₀^{π/2} |2sin x − 1| dx
= 2(∫₀^{π/6} (1 − 2sin x)dx + ∫{π/6}^{π/2} (2sin x − 1)dx)
= 2(x + 2cos x)₀^{π/6} + (−2cos x − x){π/6}^{π/2}
= 4√3 − 4 − π/3
38. The statement ~(p ↔ ~q) is :
(1) equivalent to p ↔ q
(2) equivalent to ~p ↔ q
(3) a tautology
(4) a fallacy
Ans. (1)
Sol. Given statement is ~(p ↔ ~q)
As we know ~(p ↔ q) ≡ ~p ↔ q or p ↔ ~q
∴ ~(p ↔ ~q) ≡ p ↔ q.
39. If A is an 3×3 non-singular matrix such that AA' = A'A and B = A⁻¹A', the BB' equals :
(1) I + B
(2) I
(3) B⁻¹
(4) (B⁻¹)'
Ans. (2)
Sol. B = A⁻¹Aᵀ
40. The integral ∫(1 + x − 1/x)e^{x + 1/x}dx is equal to :
(1) (x − 1)e^{x + 1/x} + c
(2) xe^{x + 1/x} + c
(3) (x + 1)e^{x + 1/x} + c
(4) −xe^{x + 1/x} + c
Ans. (2)
Sol. ∫(1 + x − 1/x)e^{x + 1/x}dx
= ∫(e^{x + 1/x} + xe^{x + 1/x}(1 − 1/x²))dx
= ∫(f(x) + xf'(x))dx (where f(x) = e^{x + 1/x})
= x·e^{x + 1/x} + c.
41. If z is a complex number such that |z| ≥ 2, then the minimum value of |z + 1/2| :
(1) is equal to 5/2
(2) lies in the interval (1, 2)
(3) is strictly greater than 5/2
(4) is strictly greater than 3/2 but less than 5/2
Ans. (2)
Sol. |z + 1/2| ≥ ||z| − 1/2|
Min. value of |z + 1/2| occurs at |z| = 2
∴ |z| ≥ 2
∴ |z + 1/2|_Min = |2 − 1/2| = 3/2
42. If g is the inverse of a function f and f'(x) = 1/(1 + x⁵), then g'(x) is equal to :
(1) 1 + x⁵
(2) 5x⁴
(3) 1/(1 + [g(x)]⁵)
(4) 1 + [g(x)]⁵
Ans. (4)
Sol. f(g(x)) = x
⇒ f'(g(x))·g'(x) = 1
∴ g'(x) = 1/f'(g(x)) = 1 + {g(x)}⁵
43. If α, β ≠ 0, and f(n) = αⁿ + βⁿ and
|3 1 + f(1) 1 + f(2)|
|1 + f(1) 1 + f(2) 1 + f(3)|
|1 + f(2) 1 + f(3) 1 + f(4)|
= K(1 − α)²(1 − β)²(α − β)², then K is equal to :
(1) αβ
(2) 1/αβ
(3) 1
(4) −1
Ans. (3)
Sol.
|3 1 + α + β 1 + α² + β²|
|1 + α + β 1 + α² + β² 1 + α³ + β³|
|1 + α² + β² 1 + α² + β³ 1 + α⁴ + β⁴|
= |1 1 1|
|1 α β|
|1 α² β²|² = (1 − α)²(1 − β)²(α − β)²
∴ K = 1
44. Let f_K(x) = 1/K(sin^K x + cos^K x) where x ∈ R and K ≥ 1. Then f₄(x) − f₆(x) equals:
(1) 1/6
(2) 1/3
(3) 1/4
(4) 1/12
Ans. (4)
Sol.
f_K(x) = 1/K(sin^K x + cos^K x)
f₄(x) − f₆(x) = (sin⁴x + cos⁴x)/4 − (sin⁶x + cos⁶x)/6
= (1 − 2sin²x·cos²x)/4 − (1 − 3sin²x·cos²x)/6
= 2/24 = 1/12
45. Let α and β be the roots of equation px² + qx + r = 0, p ≠ 0. If p, q, r are in A.P. and 1/α + 1/β = 4, then the value of |α − β| is:
(1) √61/9
(2) 2√17/9
(3) √34/9
(4) 2√13/9
Ans. (4)
Sol.
(α + β)/αβ = 4 = −q/r ⇒ q = −4r
∵ p, q & r are in A.P 2q = p + r
⇒ −8r = p + r ⇒ p = −9r
|α − β| = √((α + β)² − 4αβ)
= √((q² − 4pr)/p²) = √((16r² + 36r²)/p²)
= √(52r²/p²) = √(52/81) = 2√13/9
46. Let A and B be two events such that P(A̅ ∪ B̅) = 1/6, P(A ∩ B) = 1/4 and P(A̅) = 1/4, Where A̅ stands for the complement of the event A. Then the events A and B are :
(1) mutually exclusive and independent.
(2) equally likely but not independent.
(3) independent but not equally likely.
(4) independent and equally likely.
Ans. (3)
Sol.
P(A ∪ B) = 1 − P(A̅ ∪ B̅) = 5/6
P(A) + P(B) − P(A ∩ B) = P(A ∪ B)
3/4 + P(B) − 1/4 = 5/6
∴ P(B) = 1/3
∴ P(A) = 3/4 & P(B) = 1/3
∴ P(A)·P(B) = 1/4 = P(A ∩ B)
A & B are independent but not equally likely
47. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6 then for some c ∈ ]0,1[ ..
(1) 2f'(c) = g'(c)
(2) 2f'(c) = 3g'(c)
(3) f(c) = g'(c)
(4) f(c) = 2g'(c)
Ans. (4)
Sol. Consider a function
h(x) = f(x) − 2g(x)
∴ h(0) = 2, h(1) = 2
∴ h(x) is continuous and differentiable in [0, 1]
by Rolle's theorem, there exist at least one 'c such that
h'(c) = 0
∴ f'(c) = 2g'(c)
48. Let the population of rabbits surviving at a time t be governed by the differential equation dp(t)/dt = 1/2 p(t) − 200. If p(0) = 100, then p(t) equals :
(1) 400 − 300e^{t/2}
(2) 300 − 200e^{−t/2}
(3) 600 − 500e^{t/2}
(4) 400 − 300e^{−t/2}
Ans. (1)
Sol. dp(t)/dt = 1/2 p(t) − 200
∫_{100}^{p(t)} dp(t)/(p(t) − 400) = ∫₀ᵗ dt/2
⇒ log |(p(t) − 400)/−300| = t/2
|p(t) − 400| = 300 e^{t/2}
400 − p(t) = 300 e^{t/2}
∴ p(t) = 400 − 300e^{t/2}
49. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to :
(1) √3/√2
(2) √3/2
(3) 1/2
(4) 1/4
Ans. (4)
Sol. C₁C₂ = r₁ + r₂
√(1 + (1 − y)²) = 1 + y
∴ y = 1/4
∴ radius = 1/4
50. The area of the region described by A = {(x, y) : x² + y² ≤ 1 and y² ≤ 1 − x} is :
(1) π/2 + 4/3
(2) π/2 − 4/3
(3) π/2 − 2/3
(4) π/2 + 2/3
Ans. (1)
Sol. Shaded Area
= π/2 + 2∫_{y=0}^{1} x dy
= π/2 + 2∫₀¹ (1 − y²) dy
= π/2 + 2(1 − 1/3)
= π/2 + 4/3
51. Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then :
(1) 2bc − 3ad = 0
(2) 2bc + 3ad = 0
(3) 3bc − 2ad = 0
(4) 3bc + 2ad = 0
Ans. (3)
Sol. Put y = −x [because equidistant points in forth quadrant lies on y = −x]
4ax + 2ay + c = 0
⇒ 4ax − 2ax + c = 0
⇒ x = −c/2a ...(1)
5bx + 2by + d = 0
⇒ 5bx − 2bx + d = 0
⇒ 3bx + d = 0
⇒ x = −d/3b ...(2)
from (1) & (2)
−c/2a = −d/3b
⇒ 3bc − 2ad = 0
52. Let PS be the median of the triangle with vertices P(2, 2), Q(6, −1) and R(7, 3). The equation of the line passing through (1, −1) and parallel to PS is :
(1) 4x − 7y − 11 = 0
(2) 2x + 9y + 7 = 0
(3) 4x + 7y + 3 = 0
(4) 2x − 9y − 11 = 0
Ans. (2)
Sol. Slope of PS
M_PS = (2 − 1)/(2 − 13/2)
M_PS = −2/9
Equation of the line passing through (1, −1) and parallel to PS
y + 1 = −2/9 (x − 1)
⇒ 9y + 9 = −2x + 2
⇒ 2x + 9y + 7 = 0
53. lim_{x→0} sin(π cos²x)/x² is equal to:
(1) π/2
(2) 1
(3) −π
(4) π
Ans. (4)
Sol.
lim_{x→0} sin(π cos²x)/x²
⇒ lim_{x→0} sin(π − π sin²x)/x²
⇒ lim_{x→0} sin(π sin²x)/(π sin²x) × π sin²x/x²
⇒ lim_{x→0} sin(π sin²x)/(π sin²x) × π × lim_{x→0} (sin x/x)²
⇒ 1×π×1
⇒ π
54. If X = {4ⁿ − 3n − 1 : n ∈ N} and Y = {9(n − 1) : n ∈ N}, where N is the set of natural numbers, then X ∪ Y is equal to:
(1) N
(2) Y − X
(3) X
(4) Y
Ans. (4)
Sol. X = 4ⁿ − 3n − 1
= (1 + 3)ⁿ − 3n − 1
= [ⁿC₀3⁰ + ⁿC₁3¹ + ⁿC₂3² + ……… ⁿC_n3ⁿ] − 3n − 1
= ⁿC₂3² + ⁿC₃3³ + ⁿC₄3⁴ + … + ⁿC_n3ⁿ
= 3²[ⁿC₂ + ⁿC₃3¹ + ⁿC₄3² + … + ⁿC_n3^{n−2}]
= 9 × integer (non negative) where n ∈ N
= multiple of 9 (non negative)
Y = 9(n − 1) = all non-negative multiple of 9
so X ∪ Y = Y
55. The locus of the foot of perpendicular drawn from the centre of the ellipse x² + 3y² = 6 on any tangent to it is :
(1) (x² − y²)² = 6x² + 2y²
(2) (x² − y²)² = 6x² − 2y²
(3) (x² + y²)² = 6x² + 2y²
(4) (x² + y²)² = 6x² − 2y²
Ans. (3)
Let the foot of perpendicular be (h, k)
then m_op = k/h
equation of tangent is
y = mx ± √(a²m² + b²)
y = mx ± √(6m² + 2)
satisfied by (h, k) and m = −1/m_op = −h/k
(k + h²/k)² = 6h²/k² + 2
multiply by k²
(k² + h²)² = 6h² + 2k²
⇒ (x² + y²)² = 6x² + 2y²
56. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is :
(1) √2 + √3
(2) 3 + √2
(3) 2 − √3
(4) 2 + √3
Ans. (4)
Sol. Let a, ar, ar² are in G.P.
a, 2ar, ar² are in AP
⇒ 4ar = a + ar²
⇒ r² − 4r + 1 = 0
⇒ r = 2 + √3, 2 − √3
Since GP is an increasing G.P
⇒ r = 2 + √3
57. If (10)⁹ + 2(11)¹(10)⁸ + 3(11)²(10)⁷ + … + 10(11)⁹ = k(10)⁹, then k is equal to :
(1) 121/10
(2) 441/100
(3) 100
(4) 110
Ans. (3)
Sol. S = 10⁹ + 2(11)¹(10)⁸ + 3(11)²(10)⁷ + … + 10(11)⁹
(11/10)S = (11)(10)⁸ + 2(11)²(10)⁷ + … 11¹⁰
−S/10 = 10⁹ + (11·10⁸ + 11²·10⁷ + … + 11⁹) − 11¹⁰
−S/10 = 10⁹ + 11·10⁸((1 − (11/10)⁹)/(1 − 11/10)) − 11¹⁰
= 10⁹ + 10⁸·11(10⁹ − 11⁹)/(10⁹(−1))·10 − 11¹⁰
= 10⁹ + 11(11⁹ − 10⁹) − 11¹⁰
= 10⁹(1 − 11)
S = 10¹¹ = K10⁹
⇒ K = 100
58. The angle between the lines whose direction cosines satisfy the equations ℓ + m + n = 0 and ℓ² = m² + n² is :
(1) π/3
(2) π/4
(3) π/6
(4) π/2
Ans. (1)
Sol. ℓ² = m² + n² (1)
ℓ + m + n = 0 ⇒ ℓ = −(m + n) …(2)
(m + n)² = m² + n²
mn = 0
Either m = 0 or n = 0
for m = 0, ℓ = −n ⇒ ℓ/1 = m/0 = n/−1
⇒ cos θ = |(1 + 0 + 0)/(√2√2)| = 1/2
⇒ θ = π/3
59. The slope of the line touching both, the parabolas y² = 4x and x² = −32y is :
(1) 1/2
(2) 3/2
(3) 1/8
(4) 2/3
Ans. (1)
Sol. Tangent to curve y² = 4x is y = mx + 1/m
Tangent to curve x² = −32y is y = mx + 8m²
On comparing (1) and (2)
m = 1/2
60. If x = −1 and x = 2 are extreme points of f(x) = α log |x| + βx² + x then :
(1) α = −6, β = 1/2
(2) α = −6, β = −1/2
(3) α = 2, β = −1/2
(4) α = 2, β = 1/2
Ans. (3)
Sol. f(x) = α log |x| + βx² + x
⇒ f'(x) = α/x + 2βx + 1
put x = −1 ⇒ −α − 2β + 1 = 0 …(1)
put x = 2 ⇒ α/2 + 4β + 1 = 0 …(2)
On solving (1) and (2)
α = 2, β = −1/2
61. Which one of the following properties is not shown by NO ?
(1) It combines with oxygen to form nitrogen dioxide
(2) It's bond order is 2.5
(3) It is diamagnetic in gaseous state
(4) It is a neutral oxide
Ans. (3)
Sol.
(1) NO + 1/2 O₂ → NO₂ (Correct statement)
(Colourless) (Air) (Brown)
(2) NO is odd e- molecule which have B.O. = 2.5 (correct statement)
(3) NO has an unpaired e- in its antibonding M.O. : it is paramagnetic compound (given statement in Q. is incorrect)
(4) It is neutral towards litmus (correct statement)
62. If Z is a compressibility factor, van der Waals equation at low pressure can be written as :
(1) Z = 1 − Pb/RT
(2) Z = 1 + Pb/RT
(3) Z = 1 + RT/Pb
(4) Z = 1 − a/RT
Ans. (4)
Sol. At low pressure, V_m → large
(V_m − b) ≈ V_m
⇒ (P + a/V_m²)V_m = RT
⇒ PV_m + a/V_m = RT
⇒ Z = PV_m/RT = 1 − a/V_mRT.
63. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is:
(1) Cu
(2) Cr
(3) Ag
(4) Ca
Ans. (4)
Sol. Ca is a reactive metal and extraction of reactive metal is not possible by electrolysis of aq. salt solution.
64. Resistance of 0.2M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.4 Sm⁻¹. The resistance of 0.5M solution of the same electrolyte is 280Ω. The molar conductivity of 0.5M solution of the electrolyte in S m² mol⁻¹ is :
(1) 5×10³
(2) 5×10²
(3) 5×10⁻⁴
(4) 5×10⁻³
Ans. (3)
Sol. K = 1/R × ℓ/A
K₁/K₂ = R₂/R₁
1.4/K₂ = 280/50
K₂ = 7/28 = 1/4 Sm⁻¹
λ_m = K/(M×1000) = 1/4 × 1/(0.5×1000)
= 5×10⁻⁴ Sm²mol⁻¹
65. CsCl crystallises in body centred cubic lattice. if 'a' is its edge length then which of the following expression is correct :
(1) r_Cs⁺ + r_Cl⁻ = √3/2 a
(2) r_Cs⁺ + r_Cl⁻ = √3a
(3) r_Cs⁺ + r_Cl⁻ = 3a
(4) r_Cs⁺ + r_Cl⁻ = 3a/2
Ans. (1)
Sol. In CsCl crystal Cs⁺ ion is present in cubic void therefore
r_Cs⁺ + r_Cl⁻ = √3a/2
66. Consider separate solution of 0.500 M C₂H₅OH(aq), 0.100 M Mg₃(PO₄)₂(aq), 0.250 M KBr(aq) and 0.125 M Na₃PO₄(aq) at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes ?
(1) 0.125 M Na₃PO₄(aq) has the highest osmotic pressure.
(2) 0.500 M C₂H₅OH(aq) has the highest osmotic pressure.
(3) They all have the same osmotic pressure.
(4) 0.100 M Mg₃(PO₄)₂(aq) has the highest osmotic pressure.
Ans. (3)
Sol. π = i C R T
S.No. Electrolyte i C i×C
1 C₂H₅OH 1 0.5M 0.5
2 Mg₃(PO₄)₂ 5 0.1M 0.5
3 KBr 2 0.25M 0.5
4 Na₃PO₄ 4 0.125M 0.5
∴ all have same osmotic pressure.
67. In which of the following reaction H₂O₂ acts as a reducing agent ?
(a) H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O
(b) H₂O₂ − 2e⁻ → O₂ + 2H⁺
(c) H₂O₂ + 2e⁻ → 2OH⁻
(d) H₂O₂ + 2OH⁻ − 2e⁻ → O₂ + 2H₂O
(1) (a), (c)
(2) (b), (d)
(3) (a), (b)
(4) (c), (d)
Ans. (2)
Sol. A chemical species will act as reducing agent when it looses electron therefore correct options are (b) and (d).
68. In S_N2 reactions, the correct order of reactivity for the following compounds : CH₃Cl, CH₃CH₂Cl, (CH₃)₂CHCl and (CH₃)₃CCl is :
(1) CH₃CH₂Cl > CH₃Cl > (CH₃)₂CHCl > (CH₃)₃CCl
(2) (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl > (CH₃)₃CCl
(3) CH₃Cl > (CH₃)₂CHCl > CH₃CH₂Cl > (CH₃)₃CCl
(4) CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl
Ans. (4)
Sol. In SN² reaction reactivity order for R-X with respect to 'R' follows 1°>2°>3° which is based on less steric crowding which helps for case of transition state formation.
rate for SN² ∝ 1/Steric hindrance on C of C−LG
then correct order will be
CH₃−Cl > CH₃−CH₂−Cl > (CH₃)₂CH−Cl > (CH₃)₃C−Cl
69. The octahedral complex of a metal ion M³⁺ with four monodentate ligands L₁, L₂, L₃ and L₄ absorb wavelength in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is:
(1) L₃ < L₂ < L₄ < L₁
(2) L₁ < L₂ < L₄ < L₃
(3) L₄ < L₃ < L₂ < L₁
(4) L₁ < L₃ < L₂ < L₄
Ans. (4)
Sol. The frequency order of given absorbed light is:
Blue > Green > Yellow > Red.
Respective ligands L₄ L₂ L₃ L₁
Hence the ligand strength order is L₄ > L₂ > L₃ > L₁. Because strong field ligand causes higher splitting gap for octahedral-complex and absorbs high frequency light for d-d transition. Hence the answer is (4).
70. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is :
(1) 3%
(2) 5%
(3) 6%
(4) 10%
Ans. (4)
Sol.: mass of organic compound = 1.4 gm
⇒ 2NH₃ + H₂SO₄ → (NH₄)₂SO₄ ....(i)
excess M/10, 60 ml
⇒ 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O …(ii)
M/10, 20 ml
remaining ⇒ m moles of remaining H₂SO₄ = 1
m moles of NH₃ = 2 × m moles of H₂SO₄ used in reaction (i)
= 2 × (6 − 1)
= 10
⇒ moles of N = moles of NH₃ = 10 × 10⁻³
⇒ mass of N = 1/100 × 14 = 0.14 gm
∴ % by mass of nitrogen = 0.14/1.4 × 100
= 10%
71. The equivalent conductance of NaCl at concentration C and at infinite dilution are λ_C and λ_∞ respectively. The correct relationship between λ_C and λ_∞ is given as : (where the constant B is positive)
(1) λ_C = λ_∞ − (B)√C
(2) λ_C = λ_∞ + (B)√C
(3) λ_C = λ_∞ + (B)C
(4) λ_C = λ_∞ − (B)C
Ans. (1)
λ_C = λ_∞ − B√C ; Debye-Huckle equation
72. For the reaction SO₂(g) + 1/2 O₂(g) ⇌ SO₃(g), if K_p = K_C (RT)ˣ where the symbols have usual meaning then the value of x is : (assuming ideality)
(1) 1/2
(2) 1
(3) −1
(4) −1/2
Ans. (4)
SO₂(g) + 1/2 O₂(g) ⇌ SO₃(g)
K_p = K_C (RT)ˣ
x = 1 − 1 − 1/2 = −1/2
73. In the reaction, CH₃COOH —LiAlH₄→ A —PCl₅→ B —alc KOH→ C, the product C is :-
(1) Ethylene
(2) Acetyl chloride
(3) Acetaldehyde
(4) Acetylene
Ans. (1)
Sol.
CH₃−COOH —LiAlH₄→ CH₃−CH₂−OH (A) —PCl₅→ CH₃CH₂−Cl (B) —alc KOH→ CH₂=CH₂ (C)
Reduction of CH₃COOH will produce CH₃−CH₂−OH(A)
Now reaction of CH₃−CH₂−OH(A) with PCl₅ will produce CH₃−CH₂−Cl(B).
Now alcoholic KOH with Et-Cl will produce CH₂=CH₂(C) Ethylene by E₂ elemination.
Thus Ans. is (1) Ethylene.
74. Sodium phenoxide when heated with CO₂ under pressure at 125°C yields a product which on acetylation produces C. The major product C would be :
(1)
(2)
(3)
(4)
Ans. (3)
Sol. First step is carboxylation (Kolbe schmidt reaction) & second step is acetylation of sodium salt of aspirin (B) :-
75. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is :-
(1) an alkyl cyanide
(2) an alkyl isocyanide
(3) an alkanol
(4) an alkanediol
Ans. (2)
Sol. It is a carbyl amine reaction used for identification of primary amine also known as isocyanide test because of offensive smell of isocyanide.
R−NH₂ + CHCl₃ —Ethanol,KOH,Δ→ R−NC (Organic)
76. The correct statement for the molecule, CsI₃ is:
(1) it contains Cs³⁺ and I⁻ ions
(2) it contains Cs⁺ I⁻ and lattice I₂ molecule
(3) it is a covalent molecule
(4) it contains Cs⁺ and I₃⁻ ions
Ans. (4)
Sol. CsI₃ is an ionic compound and consisting of Cs⁺ and I₃⁻
77. The equation which is balanced and represents the correct product (s) is :
(1) [Mg(H₂O)₆]²⁺ + (EDTA)⁴⁻ —excess NaOH→ [Mg(EDTA)]²⁺ + 6H₂O
(2) [Mg(EDTA)]²⁺ + 6H₂O
(3) CuSO₄ + 4KCN → K₂[Cu(CN)₄] + K₂SO₄
(4) Li₂O + 2KCl → 2LiCl + K₂O
(5) [CoCl(NH₃)₅]⁺ + 5H⁺ → Co²⁺ + 5NH₄⁺ + Cl⁻
Ans. (4)
Sol.
(1) [Mg(H₂O)₆]²⁺ + (EDTA)⁻⁴ —excess NaOH→ [Mg(EDTA)]²⁺ + 6H₂O
(2) CuSO₄ + 5KCN → K₃[Cu(CN)₄] + K₂SO₄ + 1/2(CN)₂
(3) 2LiCl + K₂O → Li₂O + 2KCl
backward reaction of given equation is correct
(4) [CoCl(NH₃)₅]⁺ + 5H⁺ → Co²⁺ + 5NH₄⁺ + Cl⁻
Hence NH₃ gets protonated and makes Co²⁺ ion free in solution.
78. For which of the following molecule significant μ ≠ 0
(a)
(b)
(c)
(d)
(1) Only (c)
(2) (c) and (d)
(3) Only (a)
(4) (a) and (b)
Ans. (2)
Sol. Due to presence of ℓ.p.(s) on oxygen and sulphur atom which are out of plane hence and are polar and μ ≠ 0.
79. For the non-stoichiometre reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.
Initial Concentration (A) Initial Concentration (B) Initial rate of formation of C (mol L⁻¹ S⁻¹)
0.1M 0.1M 1.2×10⁻³
0.1M 0.2M 1.2×10⁻³
0.2M 0.1M 2.4×10⁻³
(1) dc/dt = k[A][B]²
(2) dc/dt = k[A]
(3) dc/dt = k[A][B]
(4) dc/dt = k[A]²[B]
Ans. (2)
Sol. Rate law -
r = k[A]^α [B]^β
from table
1.2 × 10⁻³ = k [0.1]^α [0.1]^β ……(i)
1.2 × 10⁻³ = k [0.1]^α [0.2]^β ……(ii)
2.4 × 10⁻³ = k [0.2]^α [0.1]^β ……(iii)
from (i) and (ii) ⇒ β = 0
from (i) and (iii) ⇒ α = 1
∴ rate law would be -
r = k[A]
80. Which series of reactions correctly represents chemical relations related to iron and its compound?
(1) Fe —Cl₂,heat→ FeCl₃ —heat,air→ FeCl₂ —Zn→ Fe
(2) Fe —O₂,heat→ Fe₃O₄ —CO,600°C→ FeO —CO,700°C→ Fe
(3) Fe —dil H₂SO₄→ FeSO₄ —H₂SO₄,O₂→ Fe₂(SO₄)₃ —heat→ Fe
(4) Fe —O₂,heat→ FeO —dil H₂SO₄→ FeSO₄ —heat→ Fe
Ans. (2)
Sol. Ans. is (2) because all steps are correct as per information.
(1) is wrong because FeCl₃ —Δ→ FeOCl + Cl₂↑
(in presence of air)
(3) is wrong because Fe₂(SO₄)₃ —Δ→ Fe₂O₃(s) + 3SO₃↑
(4) is wrong because 2FeSO₄ —Δ→ Fe₂O₃(s) + SO₂↑ + SO₃↑
81. Considering the basic strength of amines in aqueous solution, which one has the smallest pK_b value ?
(1) (CH₃)₃N
(2) C₆H₅NH₂
(3) (CH₃)₂NH
(4) CH₃NH₂
Ans. (3)
Sol. Basicity order : (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > C₆H₅NH₂ (in aqu medium)
pK_b order : (CH₃)₂NH < CH₃NH₂ < (CH₃)₃N < C₆H₅NH₂
smallest pK_b : (CH₃)₂NH
82. Which one of the following bases is not present in DNA ?
(1) Cytosine
(2) Thymine
(3) Quinoline
(4) Adenine
Ans. (3)
Sol. All cytosine, thymine and adenine are present in DNA. Only quindine is not present in DNA.
83. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is:
(1) 5,1,1,+1/2
(2) 5,0,1,+1/2
(3) 5,0,0,+1/2
(4) 5,1,0,+1/2
Ans. (3)
Sol. Z = 37 ; [Kr]5s¹
n = 5, ℓ = 0, m = 0, s = +1/2 or −1/2
84. The major organic compound formed by the reaction of 1, 1, 1- trichloroethane with silver powder is :-
(1) 2-Butyne
(2) 2-Butene
(3) Acetylene
(4) Ethene
Ans. (1)
Sol.
CH₃−C(Cl)₃ + 6Ag + Cl−C(Cl)₂−CH₃ —Δ→ CH₃−C≡C−CH₃
But-2-yne
85. Given below are the half-cell reactions :-
Mn²⁺ + 2e⁻ → Mn; E° = −1.18V
2(Mn³⁺ + e⁻ → Mn²⁺); E° = +1.51V
The E° for 3Mn²⁺ → Mn + 2Mn³⁺ will be :
(1) −0.33 V; the reaction will not occur
(2) −0.33 V; the reaction will occur
(3) −2.69 V; the reaction will not occur
(4) −2.69 V; the reaction will occur
Ans. (3)
Sol. Mn²⁺ + 2e⁻ → Mn; E° = −1.18V
(Mn²⁺ → Mn³⁺ + e⁻)×2; E° = −1.51V
3Mn²⁺ → Mn + 2Mn³⁺
E°_cell = −1.18 − 1.51
E°_cell = −2.69 volt
the reaction will not occur.
86. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecule is :
(1) 1:8
(2) 3:16
(3) 1:4
(4) 7:32
Ans. (4)
Sol. Given W_O₂/W_N₂ = 1/4
⇒ n_O₂/n_N₂ = (W_O₂×M_N₂)/(W_N₂×M_O₂)
= 1/4 × 28/32 = 7/32.
87. Which one is classified as a condensation polymer?
(1) Teflon
(2) Acrylonitrile
(3) Dacron
(4) Neoprene
Ans. (3)
Sol. F₂C=CF₂ —polymerisation→ (−F₂C−CF₂−)
Teflon addition polymer
Acrylonitrile in monomeric unit.
Dacron is condensation polymer (Polyester).
HO−C(=O)−C₆H₄−C(=O)−OH + HO−CH₂−CH₂−OH
Terphthalic acid Ethylene glycol
(−C(=O)−C₆H₄−C(=O)−O−CH₂−CH₂−O−)ₙ
Condensation polymer
CH₂=C(Cl)−CH=CH₂ —polymerisation→ (−CH₂−C(Cl)=CH−CH₂−)ₙ
Neoprene addition polymer
88. Among the following oxoacids, the correct decreasing order of acid strength is :-
(1) HClO₄ > HClO₃ > HClO₂ > HOCl
(2) HClO₂ > HClO₄ > HClO₃ > HOCl
(3) HOCl > HClO₂ > HClO₃ > HClO₄
(4) HClO₄ > HOCl > HClO₂ > HClO₃
Ans. (1)
Sol.
HClO₄ HClO₃ HClO₂ HOCl
−H⁺ ↓ −H⁺ ↓ −H⁺ ↓ −H⁺ ↓
O O O O
Cl−O⁻ Cl−O⁻ Cl−O⁻ Cl⁻
O O O
Stability of conjugate base is decreasing due to decreasing no. of resonating structures.
HClO₄ > HClO₃ > HClO₂ > HOCl
89. For complete combustion of ethanol, C₂H₅OH(ℓ) + 3O₂(g) → 2CO₂(g) + 3H₂O(ℓ), the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol⁻¹ at 25°C. Assuming ideality the Enthalpy of combustion, Δ_cH, for the reaction will be :- (R = 8.314 kJ mol⁻¹)
(1) −1460.50 kJ mol⁻¹
(2) −1350.50 kJ mol⁻¹
(3) −1366.95 kJ mol⁻¹
(4) −1361.95 kJ mol⁻¹
Ans. (3)
Sol. C₂H₅OH(ℓ) + 3O₂(g) ⟶ 2CO₂(g) + 3H₂O(ℓ)
q_V = ΔE = −1364.67 kJmol⁻¹
(Since calorimeter is Bomb calorimeter)
ΔH = ΔE + Δn_gRT
= −1364.67 + (−1)(8.314)(298)/1000
= −1364.67 + (−2.477)
≈ −1366.95 kJ/mol
90. The most suitable reagent for the conversion of R−CH₂−OH → R−CHO is :-
(1) CrO₃
(2) PCC (Pyridinium chlorochromate)
(3) KMnO₄
(4) K₂Cr₂O₇
Ans. (2)
Sol.
R−CH₂−OH —PCC→ R−C(=O)−H
[Pyridinium chloro chromate]
PCC is milder oxidising agent, oxidises only into aldehyde.
refer page number 82 to 104 from given PDF
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