NEET Previous Year Question Paper 2017 with Solutions Code-D| Physics, Chemistry & Biology | Free PDF Download
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The Answer Sheet is inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars on Side-1 and Side-2 carefully with blue / black ball point pen only.
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The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be deducted from the total scores. The maximum marks are 720.
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Page 2
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The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?
(1) 30 Hz (2) 40 Hz (3) 10 Hz (4) 20 Hz
Answer (4)
Sol. Two successive frequencies of closed pipe
nv/4l = 220 ...(i)
(n + 2)v/4l = 260 ...(ii)
Dividing (ii) by (i), we get
(n + 2)/n = 260/220 = 13/11
11n + 22 = 13n
n = 11
So, 11v/4l = 220
v/4l = 20
So fundamental frequency is 20 Hz.
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A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is
(1) 800 kg m⁻³
(2) 928 kg m⁻³
(3) 650 kg m⁻³
(4) 425 kg m⁻³
Answer (2)
Sol. h_oil ρ_oil g = h_water ρ_water g
140 × ρ_oil = 130 × ρ_water
ρ_oil = (13/14) × 1000 kg/m³
ρ_oil = 928 kg m⁻³
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A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
(1) 4π/√5 (2) 2π/√3
(3) √5/π (4) √5/2π
Answer (1)
Sol. v = ω√(A² - x²)
a = xω²
v = a
ω√(A² - x²) = xω²
√(3² - 2²) = 2(2π/T)
√5 = 4π/T
T = 4π/√5
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Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
(1) Move away from each other
(2) Will become stationary
(3) Keep floating at the same distance between them
(4) Move towards each other
Answer (4)
Sol. Both the astronauts are in the condition of weightless. Gravitational force between them pulls towards each other.
Page 3
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The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is
(1) 2h/√(3mkT) (2) 2h/√(mkT) (3) h/√(mkT) (4) h/√(3mkT)
Answer (4)
Sol. de-Broglie wavelength
λ = h/mv
λ = h/√(2m(3kT))
λ = h/√(3mkT)
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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
(1) Remains the same
(2) Increases by a factor of 2
(3) Increases by a factor of 4
(4) Decreases by a factor of 2
Answer (4)
Sol. Charge on capacitor
q = CV
when it is connected with another uncharged capacitor.
V_c = (q₁ + q₂)/(C₁ + C₂) = (q + 0)/(C + C)
V_c = V/2
Initial energy
U_i = (1/2)CV²
Final energy
U_f = (1/2)C(V/2)² + (1/2)C(V/2)²
= CV²/4
Loss of energy = U_i - U_f
= CV²/4
i.e. decreases by a factor (2)
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Which one of the following represents forward bias diode?
Answer (3)
Sol. In forward bias, p-type semiconductor is at higher potential w.r.t. n-type semiconductor.
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The photoelectric threshold wavelength of silver is 3250 × 10⁻¹⁰ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10⁻¹⁰ m is
(Given h = 4.14 × 10⁻¹⁵ eV s and c = 3 × 10⁸ ms⁻¹)
(1) ≈ 61 × 10³ ms⁻¹ (2) ≈ 0.3 × 10⁶ ms⁻¹
(3) ≈ 6 × 10⁵ ms⁻¹ (4) ≈ 0.6 × 10⁶ ms⁻¹
Answer (3 & 4) * Both answers are correct.
Sol. λ₀ = 3250 × 10⁻¹⁰ m
λ = 2536 × 10⁻¹⁰ m
φ = 1242 eV-nm / 325 nm = 3.82 eV
hν = 1242 eV-nm / 253.6 nm = 4.89 eV
KE_max = (4.89 - 3.82) eV = 1.077 eV
Page 4
(1/2)mv² = 1.077 × 1.6 × 10⁻¹⁹
v = √(2 × 1.077 × 1.6 × 10⁻¹⁹ / 9.1 × 10⁻³¹)
v = 0.6 × 10⁶ m/s
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Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]
(1) 411 Hz (2) 448 Hz (3) 350 Hz (4) 361 Hz
Answer (2)
Sol. f_A = f[(v + v₀)/(v - v_s)]
= 400[(340 + 16.5)/(340 - 22)]
f_A = 448 Hz
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Figure shows a circuit contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf ε = 18 V. The current i through the battery just after the switch closed is
(1) 2A (2) 0 ampere (3) 2 mA (4) 0.2A
Answer (1*)
Sol. At t = 0 no current flows through R₁ and R₃
i = ε/R₂
= 18/9
= 2A
Note : Not correctly framed but the best option out of given is (1).
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A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is given by
(1) x/2y (2) x/y
(3) y/2x (4) y/x
Answer (3)
Sol. When mirror is rotated by θ angle reflected ray will be rotated by 2θ.
y/x = 2θ
θ = y/2x
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An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current i along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire B is given by
(1) √2μ₀i²/πd (2) μ₀i²/√2πd
(3) μ₀i²/2πd (4) 2μ₀i²/πd
Page 5
Answer (2)
Sol. Force between BC and AB will be same in magnitude.
F_BC = F_BA = μ₀i²/2πd
F = √2F_BC
= √2 μ₀i²/2πd
F = μ₀i²/√2πd
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Young's double slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in air. The refractive index of the medium is nearly
(1) 1.69
(2) 1.78
(3) 1.25
(4) 1.59
Answer (2)
Sol. X₁ = X_5th dark = (2 × 5 - 1)λD/2d
X₂ = X_8th bright = 8λD/μd
X₁ = X₂
(9/2)(λD/d) = 8λD/μd
μ = 16/9 = 1.78
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A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is
(1) 1 : 11 (2) 1 : 14
(3) 1 : 6 (4) 1 : 9
Answer (1)
Sol. Spring constant ∝ 1/length
k ∝ 1/l
i.e, k₁ = 6k
k₂ = 3k
k₃ = 2k
In series
1/k' = 1/6k + 1/3k + 1/2k
1/k' = 6/6k
k' = k
k'' = 6k + 3k + 2k
k'' = 11k
k'/k'' = 1/11 i.e. k' : k'' = 1 : 11
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A thin prism having refracting angle 10° is made of glass of refractive index 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be
(1) 8° (2) 10°
(3) 4° (4) 6°
Answer (4)
Sol. (μ - 1)A + (μ' - 1)A' = 0
|(μ - 1)A| = |(μ' - 1)A'|
(1.42 - 1) × 10° = (1.7 - 1)A'
4.2 = 0.7A'
A' = 6°
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A gas mixture consists of 2 moles of O₂ and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is
(1) 9 RT (2) 11 RT
(3) 4 RT (4) 15 RT
Page 6
Sol. U = n₁(f₁/2)RT + n₂(f₂/2)RT
= 2 × (5/2)RT + 4 × (3/2)RT
= 5RT + 6RT
U = 11RT
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Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g constant with a value 10 m/s². The work done by the (i) gravitational force and the (ii) resistive force of air is
(1) (i) 100 J (ii) 8.75 J
(2) (i) 10 J (ii) -8.75 J
(3) (i) -10 J (ii) -8.25 J
(4) (i) 1.25 J (ii) -8.25 J
Answer (2)
Sol. w_g + w_a = K_f - K_i
mgh + w_a = (1/2)mv² - 0
10⁻³ × 10 × 10³ + w_a = (1/2) × 10⁻³ × (50)²
w_a = -8.75 J i.e. work done due to air resistance and work done due to gravity = 10 J
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The x and y coordinates of the particle at any time are x = 5t - 2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is
(1) -4 m/s² (2) -8 m/s²
(3) 0 (4) 5 m/s²
Answer (1)
Sol. x = 5t - 2t², y = 10t
dx/dt = 5 - 4t, dy/dt = 10
v_x = 5 - 4t, v_y = 10
dv_x/dt = -4, dv_y/dt = 0
a_x = -4, a_y = 0
Acceleration of particle at t = 2 s is = -4 m/s²
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Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω₁ and ω₂. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
(1) I(ω₁ - ω₂)²
(2) (I/8)(ω₁ - ω₂)²
(3) (1/2)I(ω₁ + ω₂)²
(4) (1/4)I(ω₁ - ω₂)²
Answer (4)
Sol. ΔKE = (1/2)(I₁I₂/(I₁ + I₂))(ω₁ - ω₂)²
= (1/2)(I²/(2I))(ω₁ - ω₂)²
= (1/4)I(ω₁ - ω₂)²
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Thermodynamic processes are indicated in the following diagram.
Match the following
Column-1 Column-2
P. Process I a. Adiabatic
Q. Process II b. Isobaric
R. Process III c. Isochoric
S. Process IV d. Isothermal
(1) P → c, Q → d, R → b, S → a
(2) P → d, Q → b, R → a, S → c
(3) P → a, Q → c, R → d, S → b
(4) P → c, Q → a, R → d, S → b
Page 7
Answer (4)
Sol. Process I = Isochoric
II = Adiabatic
III = Isothermal
IV = Isobaric
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The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is
(1) 3p/B (2) p/3B (3) p/B (4) B/3p
Answer (2)
Sol. B = p/(ΔV/V)
ΔV/V = p/B
3Δr/r = p/B
Δr/r = p/3B
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The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then
(1) d = 3/2 km (2) d = 2 km
(3) d = 1/2 km (4) d = 1 km
Answer (2)
Sol. Above earth surface
g' = g(1 - 2h/R_e)
Δg' = g(2h/R_e) ...(1)
From (1) & (2)
d = 2h
d = 2 × 1 km
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The diagrams below show regions of equipotentials.
A positive charge is moved from A to B in each diagram.
(1) Minimum work is required to move q in figure (a).
(2) Maximum work is required to move q in figure (b).
(3) Maximum work is required to move q in figure (c).
(4) In all the four cases the work done is the same.
Answer (4)
Sol. Work done w = qΔV
ΔV is same in all the cases so work is done will be same in all the cases.
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Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be
(1) t₁t₂/(t₂ + t₁) (2) t₁ - t₂
(3) (t₁ + t₂)/2 (4) t₁t₂/(t₂ - t₁)
Answer (1)
Sol. Velocity of girl w.r.t. elevator = d/t₁ = v_gc
Velocity of elevator w.r.t. ground v_eg = d/t₂
then velocity of girl w.r.t. ground
v_gG = v_gc + v_eg
i.e. v_gG = v_gc + v_eg
d/t = d/t₁ + d/t₂
1/t = 1/t₁ + 1/t₂
t = t₁t₂/(t₁ + t₂)
Page 8
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Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively
(1) g, g (2) g/3, g/3 (3) g, g/3 (4) g/3, g
Answer (4)
Sol. Before the string is cut
kx = T + 3mg ...(1)
T = mg ...(2)
⇒ kx = 4mg
After the string is cut, T = 0
a = (kx - 3mg)/3m
a = (4mg - 3mg)/3m
a = g/3 ↑
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A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
(1) 1000 (2) 1800
(3) 225 (4) 450
Answer (2)
Sol. Rate of power loss
r ∝ R²T⁴
r₁/r₂ = R₁²T₁⁴/R₂²T₂⁴
= 4 × 1/16
450/r₂ = 1/4
r₂ = 1800 wt
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Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K₁ and K₂. The thermal conductivity of the composite rod will be
(1) K₁ + K₂ (2) 2(K₁ + K₂)
(3) (K₁ + K₂)/2 (4) 3(K₁ + K₂)/2
Answer (3)
Sol. Thermal current
H = H₁ + H₂
= K₁A(T₁ - T₂)/d + K₂A(T₁ - T₂)/d
K_EQ 2A(T₁ - T₂)/d = A(T₁ - T₂)/d [K₁ + K₂]
K_EQ/2 = (K₁ + K₂)/2
Page 9
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One end of string of length l is connected to a particle of mass m and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed v, the net force on the particle (directed towards center) will be (T represents the tension in the string)
(1) T - mv²/l (2) Zero
(3) T (4) T + mv²/l
Answer (3)
Sol. Centripetal force (mv²/l) is provided by tension so the net force will be equal to tension i.e., T.
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In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kΩ. If current gain is 100 and the base resistance is 2 kΩ, the voltage and power gain of the amplifier is
(1) 150 and 15000 (2) 20 and 2000
(3) 200 and 1000 (4) 15 and 200
Answer (1)
Sol. Current gain (β) = 100
Voltage gain (A_v) = β(R_c/R_b)
= 100(3/2)
= 150
Power gain = A_v β
= 150(100)
= 15000
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Which of the following statements are correct?
(a) Centre of mass of a body always coincides with the centre of gravity of the body.
(b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero
(c) A couple on a body produce both translational and rotational motion in a body.
(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
(1) (b) and (c)
(2) (c) and (d)
(3) (b) and (d)
(4) (a) and (b)
Answer (3)
Sol. Centre of mass may or may not coincide with centre of gravity.
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A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?
(1) 25 rad/s² (2) 5 m/s²
(3) 25 m/s² (4) 0.25 rad/s²
Answer (1)
Sol. τ = Iα
F × R = MR²α
30 × 0.4 = 3 × (0.4)²α
12 = 3 × 0.16 α
400 = 16 α
α = 25 rad/s²
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A long solenoid of diameter 0.1 m has 2 × 10⁴ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π² Ω, the total charge flowing through the coil during this time is
(1) 32 μC (2) 16π μC
(3) 32π μC (4) 16 μC
Answer (1)
Sol. ε = -N(dφ/dt)
|ε/R| = (N/R)(dφ/dt)
dq = (N/R)dφ
ΔQ = N(Δφ)/R
ΔQ = Δφ_total/R
= (NBA)/R
= μ₀niπr²/R
Page 10
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The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
(1) 4 (2) 0.5 (3) 2 (4) 1
Answer (1)
Sol. For last Balmer series
1/λ_b = R[1/2² - 1/∞²]
λ_b = 4/R
For last Lyman series
1/λ_l = R[1/1² - 1/∞²]
λ_l = 1/R
λ_b/λ_l = 4/R × R/1
λ_b/λ_l = 4
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Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen m_h = 1.67 × 10⁻²⁷ kg]
(1) 10⁻³⁷ C (2) 10⁻⁴⁷ C (3) 10⁻²⁰ C (4) 10⁻²³ C
Answer (1)
Sol. F_e = F_g
(1/4πε₀)(Δe²/d²) = Gm²/d²
9 × 10⁹(Δe²) = 6.67 × 10⁻¹¹ × 1.67 × 10⁻²⁷ × 1.67 × 10⁻²⁷
Δe² = (6.67 × 1.67 × 1.67/9) × 10⁻⁷⁴
Δe = 10⁻³⁷
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The given electrical network is equivalent to
(1) NOR gate (2) NOT gate (3) AND gate (4) OR gate
Answer (1)
Sol. Y = A + B (bar)
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A Carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
(1) 99 J (2) 100 J (3) 1 J (4) 90 J
Answer (4)
Sol. β = (1 - η)/η
β = (1 - 1/10)/(1/10) = 9
β = 9
β = Q₂/W
Q₂ = 9 × 10 = 90 J
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In an electromagnetic wave in free space the root mean square value of the electric field is E_rms = 6 V/m. The peak value of the magnetic field is
(1) 0.70 × 10⁻⁸ T
(2) 4.23 × 10⁻⁸ T
(3) 1.41 × 10⁻⁸ T
(4) 2.83 × 10⁻⁸ T
Page 11
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If θ₁ and θ₂ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by
(1) cot²θ = cot²θ₁ - cot²θ₂
(2) tan²θ = tan²θ₁ - tan²θ₂
(3) cot²θ = cot²θ₁ + cot²θ₂
(4) tan²θ = tan²θ₁ + tan²θ₂
Answer (3)
Sol. cot²θ = cot²θ₁ + cot²θ₂
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A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is
(1) 2.3 μJ (2) 1.15 μJ
(3) 9.1 μJ (4) 4.55 μJ
Answer (3)
Sol. W = MB(cos θ₁ - cos θ₂)
When it is rotated by angle 180° then
W = 2MB
W = 2(NIA)B
= 2 × 250 × 85 × 10⁻⁶[1.25 × 2.1 × 10⁻⁴] × 85 × 10⁻²
= 9.1 μJ
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The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be
(1) n²R (2) R/n²
(3) nR (4) R/n
Answer (1)
Sol. R₂/R₁ = l₂²/l₁²
= n²l₁²/l₁²
R₂/R₁ = n²
R₂ = n²R₁
Page 12
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A physical quantity of the dimensions of length that can be formed out of c, G and e²/4πε₀ is [c is velocity of light, G is universal constant of gravitation and e is charge]
(1) (1/c²)[e²/(G4πε₀)]^(1/2)
(2) (1/c)G(e²/4πε₀)
(3) (1/c²)[G(e²/4πε₀)]^(1/2)
(4) c²[G(e²/4πε₀)]^(1/2)
Answer (3)
Sol. Let e²/4πε₀ = A = ML³T⁻²
l = C^x G^y (A)^z
L = [LT⁻¹]^x [M⁻¹L³T⁻²]^y [ML³T⁻²]^z
-y + z = 0 ⇒ y = z ...(i)
x + 3y + 3z = 1 ...(ii)
-x - 4z = 0 ...(iii)
From (i), (ii) & (iii)
z = y = 1/2, x = -2
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Radioactive material 'A' has decay constant '8λ' and material 'B' has decay constant 'λ'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be 1/e?
(1) 1/8λ (2) 1/9λ
(3) 1/λ (4) 1/7λ
Answer (4)
Sol. No option is correct
If we take N_A/N_B = 1/e
Then
N_A/N_B = e^(-8λt)/e^(-λt)
1/e = e^(-7λt)
-1 = -7λt
t = 1/7λ
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A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves:
(1) A condition of no current flow through the galvanometer
(2) A combination of cells, galvanometer and resistances
(3) Cells
(4) Potential gradients
Answer (1)
Sol. Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit.
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The ratio of resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is
(1) 3 : 2 (2) 16 : 81
(3) 8 : 27 (4) 9 : 4
Answer (1)
Sol. Resolving power ∝ 1/λ
R₁/R₂ = λ₂/λ₁
= 6000 Å/4000 Å
= 3/2
Page 13
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A disease caused by an autosomal primary non-disjunction is
(1) Turner's syndrome
(2) Sickle cell anemia
(3) Down's syndrome
(4) Klinefelter's syndrome
Answer (3)
Sol. Down's syndrome is caused by non-disjunction of 21st chromosome.
-
A dioecious flowering plant prevents both:
(1) Geitonogamy and xenogamy
(2) Cleistogamy and xenogamy
(3) Autogamy and xenogamy
(4) Autogamy and geitonogamy
Answer (4)
Sol. When unisexual male and female flowers are present on different plants the condition is called dioecious and it prevents both autogamy and geitonogamy.
-
Attractants and rewards are required for
(1) Hydrophily
(2) Cleistogamy
(3) Anemophily
(4) Entomophily
Answer (4)
Sol. Insect pollinated plants provide rewards as edible pollen grain and nectar as usual rewards. While some plants also provide safe place for deposition of eggs.
-
Alexander Von Humboldt described for the first time
(1) Species area relationships
(2) Population Growth equation
(3) Ecological Biodiversity
(4) Laws of limiting factor
Answer (1)
Sol. Alexander Von Humboldt observed that within a region species richness increases with the increases in area.
-
Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP?
(1) Chloroplast
(2) Mitochondrion
(3) Lysosome
(4) Ribosome
Answer (2)
Sol. Mitochondria are the site of aerobic oxidation of carbohydrates to generate ATP.
-
Zygotic meiosis is characteristic of
(1) Funaria
(2) Chlamydomonas
(3) Marchantia
(4) Fucus
Answer (2)
Sol. Chlamydomonas has haplontic life cycle hence showing zygotic meiosis or initial meiosis.
-
Good vision depends on adequate intake of carotene rich food
Select the best option from the following statements
(a) Vitamin A derivatives are formed from carotene
(b) The photopigments are embedded in the membrane discs of the inner segment
(c) Retinal is a derivative of vitamin A
(d) Retinal is a light absorbing part of all the visual photopigments
(1) a and (2) (b), (c) and
(3) (a) and (b) (4) (a), (c) and (d)
Answer (4)
Sol. Carotene is the source of retinal which is involved in formation of rhodopsin of rod cells. Retinal, a derivative of vitamin A, is the light-absorbing part of all visual photopigments.
-
Among the following characters, which one was not considered by Mendel in his experiments on pea?
(1) Seed - Green or Yellow
(2) Pod - Inflated or Constricted
(3) Stem - Tall or Dwarf
(4) Trichomes - Glandular or non-glandular
Answer (4)
Sol. During his experiments Mendel studied seven characters.
Nature of trichomes i.e., glandular or non-glandular was not considered by Mendel.
Page 14
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The association of histone H1 with a nucleosome indicates:
(1) The DNA is condensed into a Chromatin Fibre
(2) The DNA double helix is exposed
(3) Transcription is occurring
(4) DNA replication is occurring
Answer (1)
Sol. The association of H1 protein indicates the complete formation of nucleosome.
Therefore the DNA is in condensed form.
-
The pivot joint between atlas and axis is a type of
(1) Synovial joint
(2) Saddle joint
(3) Fibrous joint
(4) Cartilaginous joint
Answer (1)
Sol. Synovial joints are freely movable joint which allow considerable movements. Pivot joint is a type of synovial joint which provide rotational movement as in between atlas and axis vertebrae of vertebral column.
-
Receptor sites for neurotransmitters are present on
(1) Tips of axons
(2) Post-synaptic membrane
(3) Membranes of synaptic vesicles
(4) Pre-synaptic membrane
Answer (2)
Sol. Pre-synaptic membrane is involved in the release of neurotransmitter in the chemical synapse. The receptors sites for neurotransmitters are present on post-synaptic membrane.
-
GnRH, a hypothalamic hormone, needed in reproduction, acts on
(1) Posterior pituitary gland and stimulates secretion of oxytocin and FSH
(2) Posterior pituitary gland and stimulates secretion of LH and relaxin
(3) Anterior pituitary gland and stimulates secretion of LH and oxytocin
(4) Anterior pituitary gland and stimulates secretion of LH and FSH
Answer (4)
Sol. Hypothalamus secretes GnRH which stimulates anterior pituitary gland for the secretion of gonadotropins (FSH and LH).
-
Hypersecretion of Growth Hormone in adults does not cause further increase in height, because
(1) Bones loose their sensitivity to Growth Hormone in adults
(2) Muscle fibres do not grow in size after birth
(3) Growth Hormone becomes inactive in adults
(4) Epiphyseal plates close after adolescence
Answer (4)
Sol. Epiphyseal plate is responsible for the growth of bone which close after adolescence so hypersecretion of growth hormone in adults does not cause further increase in height.
-
Select the mismatch:
(1) Anabaena - Nitrogen fixer
(2) Rhizobium - Alfalfa
(3) Frankia - Alnus
(4) Rhodospirillum - Mycorrhiza
Answer (4)
Sol. Rhodospirillum is anaerobic, free living nitrogen fixer. Mycorrhiza is a symbiotic relationship between fungi and roots of higher plants.
-
Which one of the following statements is not valid for aerosols?
(1) They cause increased agricultural productivity
(2) They have negative impact on agricultural land
(3) They are harmful to human health
(4) They alter rainfall and monsoon patterns
Answer (1)
Sol. Aerosols can cause various problems to agriculture through its direct or indirect effects on plants. However continually increasing air pollution may represent a persistent and largely irreversible threat to agriculture in the future.
-
Which one of the following is related to Ex-situ conservation of threatened animals and plants?
(1) Amazon rainforest
(2) Himalayan region
(3) Wildlife Safari parks
(4) Biodiversity hot spots
Answer (3)
Page 15
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Which of the following facilitates opening of stomatal aperture?
(1) Radial orientation of cellulose microfibrils in the cell wall of guard cells
(2) Longitudinal orientation of cellulose microfibrils in the cell wall of guard cells
(3) Contraction of outer wall of guard cells
(4) Decrease in turgidity of guard cells
Answer (1)
Sol. Cellulose microfibrils are oriented radially rather than longitudinally which makes easy for the stoma to open.
-
Select the mismatch:
(1) Salvinia - Heterosporous
(2) Equisetum - Homosporous
(3) Pinus - Dioecious
(4) Cycas - Dioecious
Answer (3)
Sol. Pinus is monoecious plant having both male and female cones on same plant.
-
Asymptote in a logistic growth curve is obtained when
(1) K > N (2) K < N
(3) The value of 'r' approaches zero
(4) K = N
Answer (4)
Sol. A population growing in a habitat with limited resources shows logistic growth curve.
For logistic growth
dN/dt = rN((K - N)/K)
If K = N then (K - N)/K = 0
∴ the dN/dt = 0, the population reaches asymptote.
-
The process of separation and purification of expressed protein before marketing is called
(1) Bioprocessing
(2) Postproduction processing
(3) Upstream processing
(4) Downstream processing
Answer (4)
Sol. Biosynthetic stage for synthesis of product in recombinant DNA technology is called upstreaming process while after completion of biosynthetic stage, the product has to be subjected through a series of processes which include separation and purification are collectively referred to as downstreaming processing.
-
The water potential of pure water is
(1) More than zero but less than one
(2) More than one
(3) Zero
(4) Less than zero
Answer (3)
Sol. By convention, the water potential of pure water at standard temperature, which is not under any pressure, is taken to be zero.
-
The function of copper ions in copper releasing IUD's is:
(1) They make uterus unsuitable for implantation
(2) They inhibit ovulation
(3) They suppress sperm motility and fertilising capacity of sperms
(4) They inhibit gametogenesis
Answer (3)
Sol. Cu²⁺ interfere in the sperm movement, hence suppress the sperm motility and fertilising capacity of sperms.
-
Double fertilization is exhibited by
(1) Fungi
(2) Angiosperms
(3) Gymnosperms
(4) Algae
Answer (2)
Sol. Double fertilization is a characteristic feature exhibited by angiosperms. It involves syngamy and triple fusion.
Page 16
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Presence of plants arranged into well defined vertical layers depending on their height can be seen best in:
(1) Grassland
(2) Temperate Forest
(3) Tropical Savannah
(4) Tropical Rain Forest
Answer (4)
Sol. The tropical rain forest have five vertical strata on the basis of height of plants. i.e., ground vegetation, shrubs, short canopy trees, tall canopy trees and tall emergent trees.
-
Which ecosystem has the maximum biomass?
(1) Pond ecosystem
(2) Lake ecosystem
(3) Forest ecosystem
(4) Grassland ecosystem
Answer (3)
Sol. High productive ecosystem are-
Tropical rain forest
Coral reef
Estuaries
Sugarcane fields
-
Root hairs develop from the region of
(1) Root cap
(2) Meristematic activity
(3) Maturation
(4) Elongation
Answer (3)
Sol. In roots, the root hairs arise from zone of maturation. This zone is differentiated zone thus bearing root hairs.
-
DNA replication in bacteria occurs
(1) Prior to fission
(2) Just before transcription
(3) During S-phase
(4) Within nucleolus
Answer (1)
Sol. DNA replication in bacteria occurs prior to fission. Prokaryotes do not show well marked S-phase due to their primitive nature.
-
Homozygous purelines in cattle can be obtained by
(1) mating of individuals of different breed
(2) mating of individuals of different species
(3) mating of related individuals of same breed
(4) mating of unrelated individuals of same breed
Answer (3)
Sol. Inbreeding results in increase in the homozygosity. Therefore, mating of the related individuals of same breed will increase homozygosity.
-
In Bougainvillea thorns are the modifications of
(1) Stem
(2) Leaf
(3) Stipules
(4) Adventitious root
Answer (1)
Sol. Thorns are hard, pointed straight structures for protection. These are modified stem.
-
A decrease in blood pressure/volume will not cause the release of
(1) Aldosterone
(2) ADH
(3) Renin
(4) Atrial Natriuretic Factor
Answer (4)
Sol. A decrease in blood pressure / volume stimulates the release of renin, aldosterone, and ADH while increase in blood pressure / volume stimulates the release of Atrial Natriuretic Factor (ANF) which cause vasodilation and also inhibits RAAS (Renin Angiotensin Aldosterone System) mechanism that decreases the blood volume/pressure.
-
Which statement is wrong for Krebs' cycle?
(1) During conversion of succinyl CoA to succinic acid, a molecule of GTP is synthesised
(2) The cycle starts with condensation of acetyl group (acetyl CoA) with pyruvic acid to yield citric acid
(3) There are three points in the cycle where NAD⁺ is reduced to NADH + H⁺
(4) There is one point in the cycle where FAD⁺ is reduced to FADH₂
Answer (2)
Sol. Krebs cycle starts with condensation of acetyl CoA (2C) with oxaloacetic acid (4C) to form citric acid (6C).
Page 17
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Anaphase promoting complex (APC) is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in a human cell, which of the following is expected to occur?
(1) Chromosomes will not segregate
(2) Recombination of chromosome arms will occur
(3) Chromosomes will not condense
(4) Chromosomes will be fragmented
Answer (1)
Sol. Anaphase Promoting Complex (APC) is a protein necessary for separation of daughter chromosomes during anaphase. If APC is defective then the chromosomes will fail to segregate during anaphase.
-
Which of the following options best represents the enzyme composition of pancreatic juice?
(1) Peptidase, amylase, pepsin, rennin
(2) Lipase, amylase, trypsinogen, procarboxypeptidase
(3) Amylase, peptidase, trypsinogen, rennin
(4) Amylase, pepsin, trypsinogen, maltase
Answer (2)
Sol. Rennin and Pepsin enzymes are present in the gastric juice. Maltase is present in the intestinal juice.
-
Life cycle of Ectocarpus and Fucus respectively are
(1) Haplodiplontic, Diplontic
(2) Haplodiplontic, Haplontic
(3) Haplontic, Diplontic
(4) Diplontic, Haplodiplontic
Answer (1)
Sol. Ectocarpus has haplodiplontic life cycle and Fucus has diplontic life cycle.
-
Which of the following is made up of dead cells?
(1) Phellen
(2) Phloem
(3) Xylem parenchyma
(4) Collenchyma
Answer (1)
Sol. Cork cambium undergoes periclinal division and cuts off thick walled suberised dead cork cells towards outside and it cuts off thin walled living cells i.e., phelloderm on inner side.
-
Which of the following is correctly matched for the product produced by them?
(1) Penicillium notatum : Acetic acid
(2) Saccharomyces cerevisiae : Ethanol
(3) Acetobacter aceti : Antibiotics
(4) Methanobacterium : Lactic acid
Answer (2)
Sol. Saccharomyces cerevisiae is commonly called Brewer's yeast. It causes fermentation of carbohydrates producing ethanol.
-
Fruit and leaf drop at early stages can be prevented by the application of
(1) Auxins
(2) Gibberellic acid
(3) Cytokinins
(4) Ethylene
Answer (1)
Sol. Auxins prevent premature leaf and fruit fall. NAA prevents fruit drop in tomato; 2,4-D prevents fruit drop in Citrus.
-
Viroids differ from viruses in having:
(1) RNA molecules with protein coat
(2) RNA molecules without protein coat
(3) DNA molecules with protein coat
(4) DNA molecules without protein coat
Answer (2)
Sol. Viroids are sub-viral agents as infectious RNA particles, without protein coat.
-
Which of the following are not polymeric?
(1) Polysaccharides
(2) Lipids
(3) Nucleic acids
(4) Proteins
Answer (2)
Sol. Nucleic acids are polymers of nucleotides
Proteins are polymers of amino acids
Polysaccharides are polymers of monosaccharides
Lipids are the esters of fatty acids and alcohol
Page 18
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A temporary endocrine gland in the human body is
(1) Corpus luteum
(2) Corpus allatum
(3) Pineal gland
(4) Corpus cardiacum
Answer (1)
Sol. Corpus luteum is the temporary endocrine structure formed in the ovary after ovulation. It is responsible for the release of the hormones like progesterone, oestrogen etc.
-
Phosphonol pyruvate (PEP) is the primary CO₂ acceptor in:
(1) C₂ plants
(2) C₃ and C₄ plants
(3) C₃ plants
(4) C₄ plants
Answer (4)
Sol. PEP is 3C compound which serves as primary CO₂ acceptor in the mesophyll cell cytoplasm of C₄ plants like maize, sugarcane, Sorghum etc.
-
Plants which produce characteristic pneumatophores and show vivipary belong to
(1) Psammophytes
(2) Hydrophytes
(3) Mesophytes
(4) Halophytes
Answer (4)
Sol. Halophytes growing in saline soils show
(i) Vivipary which is in-situ seed germination
(ii) Pneumatophores for gaseous exchange
-
Mycorrhizae are the example of
(1) Antibiosis
(2) Mutualism
(3) Fungistasis
(4) Amensalism
Answer (2)
Sol. Mycorrhizae is a symbiotic association of fungi with roots of higher plants.
-
If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?
(1) 33
(2) 333
(3) 1
(4) 11
Answer (1)
Sol. If deletion occurs at 901st position the remaining 98 bases specifying for 33 codons of amino acids will be altered.
-
A gene whose expression helps to identify transformed cell is known as
(1) Plasmid
(2) Structural gene
(3) Selectable marker
(4) Vector
Answer (3)
Sol. In recombinant DNA technology, selectable markers helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
-
Which of the following are found in extreme saline conditions?
(1) Cyanobacteria
(2) Mycobacterium
(3) Archaebacteria
(4) Eubacteria
Answer (3)
Sol. Archaebacteria are able to survive in harsh conditions because of branched lipid chain in cell membrane which reduces fluidity of cell membrane.
Halophiles are exclusively found in saline habitats.
-
Out of 'X' pairs of ribs in humans only 'Y' pairs are true ribs. Select the option that correctly represents values of X and Y and provides their explanation:
(1) X = 24, Y = 7 True ribs are dorsally attached to vertebral column but are free on ventral side
(2) X = 24, Y = 12 True ribs are dorsally attached to vertebral column but are free on ventral side
(3) X = 12, Y = 7 True ribs are attached dorsally to vertebral column and ventrally to the sternum
(4) X = 12, Y = 5 True ribs are attached dorsally to vertebral column and sternum on the two ends
Answer (3)
Sol. In human, 12 pairs of ribs are present in which 7 pairs of ribs (1st to 7th pairs) are attached dorsally to vertebral column and ventrally to the sternum.
Page 19
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MALT constitutes about percent of the lymphoid tissue in human body
(1) 70% (2) 10% (3) 50% (4) 20%
Answer (3)
Sol. MALT is Mucosa Associated Lymphoid Tissue and it constitutes about 50 percent of the lymphoid tissue in human body.
-
Which one from those given below is the period for Mendel's hybridization experiments?
(1) 1857-1869
(2) 1870-1877
(3) 1856-1863
(4) 1840-1850
Answer (3)
Sol. Mendel conducted hybridization experiments on Pea plant for 7 years between 1856 to 1863 and his data was published in 1865 (according to NCERT).
-
Adult human RBCs are enucleate. Which of the following statement(s) is/are most appropriate explanation for this feature?
(a) They do not need to reproduce
(b) They are somatic cells
(c) They do not metabolize
(d) All their internal space is available for oxygen transport
(1) (a), (c) and (d)
(2) (b) and (c)
(3) Only (d)
(4) Only (a)
Answer (3)
Sol. In Human RBCs, nucleus degenerates during maturation which provide more space for oxygen carrying pigment (Haemoglobin). It lacks most of the cell organelles including mitochondria so respires anaerobically.
-
Myelin sheath is produced by
(1) Oligodendrocytes and Osteoclasts
(2) Osteoclasts and Astrocytes
(3) Schwann Cells and Oligodendrocytes
(4) Astrocytes and Schwann Cells
Answer (3)
Sol. Oligodendrocytes are neuroglial cells which produce myelin sheath in central nervous system while Schwann cell produces myelin sheath in peripheral nervous system.
-
Which of the following statements is correct?
(1) The ascending limb of loop of Henle is permeable to water
(2) The descending limb of loop of Henle is permeable to electrolytes
(3) The ascending limb of loop of Henle is impermeable to water
(4) The descending limb of loop of Henle is impermeable to water
Answer (3)
Sol. Descending limb of loop of Henle is permeable to water but impermeable to electrolytes while ascending limb is impermeable to water but permeable to electrolytes.
-
During DNA replication, Okazaki fragments are used to elongate
(1) The leading strand away from replication fork
(2) The lagging strand away from the replication fork
(3) The leading strand towards replication fork
(4) The lagging strand towards replication fork
Answer (2)
Sol. Two DNA polymerase molecules work simultaneously at the DNA fork, one on the leading strand and the other on the lagging strand.
Each Okazaki fragment is synthesized by DNA polymerase at lagging strand in 5'→3' direction. New Okazaki fragments appear as the replication fork opens further.
As the first Okazaki fragment appears away from the replication fork, the direction of elongation would be away from replication fork.
-
Which one of the following statements is correct, with reference to enzymes?
(1) Coenzyme = Apoenzyme + Holoenzyme
(2) Holoenzyme = Coenzyme + Cofactor
(3) Apoenzyme = Holoenzyme + Coenzyme
(4) Holoenzyme = Apoenzyme + Coenzyme
Answer (4)
Sol. Holoenzyme is conjugated enzyme in which protein part is apoenzyme while non-protein is cofactor.
Coenzyme are also organic compounds but their association with apoenzyme is only transient and serve as cofactors.
Page 20
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DNA fragments are
(1) Neutral
(2) Either positively or negatively charged depending on their size
(3) Positively charged
(4) Negatively charged
Answer (4)
Sol. DNA fragments are negatively charged because of phosphate group.
-
The DNA fragments separated on an agarose gel can be visualised after staining with
(1) Aniline blue
(2) Ethidium bromide
(3) Bromophenol blue
(4) Acetocarmine
Answer (2)
Sol. Ethidium bromide is used to stain the DNA fragments and will appear as orange coloured bands under UV light.
-
Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen?
(1) Mycoplasma
(2) Nostoc
(3) Bacillus
(4) Pseudomonas
Answer (1)
Sol. Mycoplasmas are smallest, wall-less prokaryotes, pleomorphic in nature. These are pathogenic on both plants and animals.
-
The morphological nature of the edible part of coconut is
(1) Endosperm
(2) Pericarp
(3) Perisperm
(4) Cotyledon
Answer (1)
Sol. Coconut has double endosperm with liquid endosperm and cellular endosperm.
-
Select the correct route for the passage of sperms in male frogs:
(1) Testes → Vasa efferentia → Bidder's canal → Ureter → Cloaca
(2) Testes → Vasa efferentia → Kidney → Bidder's canal → Urinogenital duct → Cloaca
(3) Testes → Bidder's canal → Kidney → Vasa efferentia → Urinogenital duct → Cloaca
(4) Testes → Vasa efferentia → Kidney → Seminal Vesicle → Urinogenital duct → Cloaca
Answer (2)
Sol. In male frog the sperms will move from
Testes → Vasa efferentia → Kidney → Bidder's canal → Urinogenital duct → Cloaca.
-
Identify the wrong statement in context of heartwood.
(1) It conducts water and minerals efficiently
(2) It comprises dead elements with highly lightened walls
(3) Organic compounds are deposited in it
(4) It is highly durable
Answer (1)
Sol. Heartwood is physiologically inactive due to deposition of organic compounds and tyloses formation, so this will not conduct water and minerals.
-
Transplantation of tissues/organs fails often due to non-acceptance by the patient's body. Which type of immune-response is responsible for such rejections?
(1) Hormonal immune response
(2) Physiological immune response
(3) Autoimmune response
(4) Cell-mediated immune response
Answer (4)
Sol. Non-acceptance or rejection of graft or transplanted tissues/organs is due to cell mediated immune response.
-
The region of Biosphere Reserve which is legally protected and where no human activity is allowed is known as
(1) Transition zone
(2) Restoration zone
(3) Core zone
(4) Buffer zone
Answer (3)
Page 21
Sol. Biosphere reserve is protected area with multipurpose activities.
It has three zones
(a) Core zone - without any human interference
(b) Buffer zone - with limited human activity
(c) Transition zone - human settlement, grazing cultivation etc., are allowed.
-
Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the correct statement.
(1) Thalassemia is due to less synthesis of globin molecules
(2) Sickle cell anemia is due to a quantitative problem of globin molecules
(3) Both are due to a qualitative defect in globin chain synthesis
(4) Both are due to a quantitative defect in globin chain synthesis
Answer (1)
Sol. Thalassemia differs from sickle-cell anaemia in that the former is a quantitative problem of synthesising too few globin molecules while the latter is a qualitative problem of synthesising an incorrectly functioning globin.
-
Flowers which have single ovule in the ovary and are packed into inflorescence are usually pollinated by
(1) Wind
(2) Bat
(3) Water
(4) Bee
Answer (1)
Sol. Wind pollination or anemophily is favoured by flowers having a single ovule in each ovary, and numerous flowers packed in an inflorescence. Wind pollination is a non-directional pollination.
-
An important characteristic that Hemichordates share with Chordates is
(1) Pharynx with gill slits
(2) Pharynx without gill slits
(3) Absence of notochord
(4) Ventral tubular nerve cord
Answer (1)
Sol. Pharyngeal gill slits are present in hemichordates as well as in chordates. Notochord is present in chordates only. Ventral tubular nerve cord is characteristic feature of non-chordates.
-
Which of the following options gives the correct sequence of events during mitosis?
(1) Condensation → crossing over → nuclear membrane disassembly → segregation → telophase
(2) Condensation → arrangement at equator → centromere division → segregation → telophase
(3) Condensation → nuclear membrane disassembly → crossing over → segregation → telophase
(4) Condensation → nuclear membrane disassembly → arrangement at equator → centromere division → segregation → telophase
Answer (4)
Sol. The correct sequence of events during mitosis would be as follows
(i) Condensation of DNA so that chromosomes become visible occurs during early to mid-prophase.
(ii) Nuclear membrane disassembly begins at late prophase or transition to metaphase.
(iii) Arrangement of chromosomes at equator occurs during metaphase, called congression.
(iv) Centromere division or splitting occurs during anaphase forming daughter chromosomes.
(v) Segregation also occurs during anaphase as daughter chromosomes separate and move to opposite poles.
(vi) Telophase leads to formation of two daughter nuclei.
-
The final proof for DNA as the genetic material came from the experiments of
(1) Avery, Mcleod and McCarty
(2) Hargobind Khorana
(3) Griffith
(4) Hershey and Chase
Answer (4)
Sol. Hershey and Chase gave unequivocal proof which ended the debate between protein and DNA as genetic material.
Page 22
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What is the criterion for DNA fragments movement on agarose gel during gel electrophoresis?
(1) Positively charged fragments move to farther end
(2) Negatively charged fragments do not move
(3) The larger the fragment size, the farther it moves
(4) The smaller the fragment size, the farther it moves
Answer (4)
Sol. During gel electrophoresis, DNA fragments separate (resolve) according to their size through sieving effect provided by agarose gel.
-
With reference to factors affecting the rate of photosynthesis, which of the following statements is not correct?
(1) C₃ plants responds to higher temperatures with enhanced photosynthesis while C₄ plants have much lower temperature optimum
(2) Tomato is a greenhouse crop which can be grown in CO₂-enriched atmosphere for higher yield
(3) Light saturation for CO₂ fixation occurs at 10% of full sunlight
(4) Increasing atmospheric CO₂ concentration upto 0.05% can enhance CO₂ fixation rate
Answer (1)
Sol. In C₃ plants photosynthesis is decreased at higher temperature due to increased photorespiration. C₄ plants have higher temperature optimum because of the presence of pyruvate phosphate dikinase enzyme, which is sensitive to low temperature.
-
Artificial selection to obtain cows yielding higher milk output represents
(1) Disruptive as it splits the population into two one yielding higher output and the other lower output
(2) Stabilizing followed by disruptive as it stabilizes the population to produce higher yielding cows
(3) Stabilizing selection as it stabilizes this character in the population
(4) Directional as it pushes the mean of the character in one direction
Answer (4)
Sol. Artificial selection to obtain cow yielding higher milk output will shift the peak to one direction, hence, will be an example of Directional selection. In stabilizing selection, the organisms with the mean value of the trait are selected. In disruptive selection, both extremes get selected.
suspended solids?
(1) Primary treatment
(2) Sludge treatment
(3) Tertiary treatment
(4) Secondary treatment
Answer (1)
Sol. Primary treatment is a physical process which involves sequential filtration and sedimentation.
-
Spliceosomes are not found in cells of
(1) Animals
(2) Bacteria
(3) Plants
(4) Fungi
Answer (2)
Sol. Spliceosomes are used in removal of introns during post-transcriptional processing of hnRNA in eukaryotes only as split genes are absent as prokaryotes.
-
Functional megaspore in an angiosperm develops into
(1) Embryo sac
(2) Embryo
(3) Ovule
(4) Endosperm
Answer (1)
Sol. Megaspore is the first cell of female gametophytic generation in angiosperm. It undergoes three successive generations of free nuclear mitosis to form 8-nucleated and 7-celled embryo sac.
-
Which of the following components provides sticky character to the bacterial cell?
(1) Plasma membrane
(2) Glycocalyx
(3) Cell wall
(4) Nuclear membrane
Answer (2)
Sol. Sticky character of the bacterial wall is due to glycocalyx or slime layer. This layer is rich in glycoproteins.
-
Which among these is the correct combination of aquatic mammals?
(1) Whales, Dolphins, Seals
(2) Trygon, Whales, Seals
(3) Seals, Dolphins, Sharks
(4) Dolphins, Seals, Trygon
Answer (1)
Page 23
Sol. Sharks and Trygon (sting ray) are the members of chondrichthyes (cartilaginous fish) while whale, Dolphin and Seals are aquatic mammals belong to class mammalia.
-
Which of the following represents order of 'Horse'?
(1) Caballus
(2) Ferus
(3) Equidae
(4) Perissodactyla
Answer (4)
Sol. Horse belongs to order perissodactyla of class mammalia. Perissodactyla includes odd-toed mammals.
-
Lungs are made up of air-filled sacs the alveoli. They do not collapse even after forceful expiration, because of:
(1) Tidal Volume
(2) Expiratory Reserve Volume
(3) Residual Volume
(4) Inspiratory Reserve Volume
Answer (3)
Sol. Volume of air present in lungs after forceful expiration as residual volume which prevents the collapsing of alveoli even after forceful expiration.
-
Capacitation occurs in
(1) Vas deferens
(2) Female Reproductive tract
(3) Rete testis
(4) Epididymis
Answer (2)
Sol. Capacitation is increase in fertilising capacity of sperms which occurs in female reproductive tract.
-
Which of the following RNAs should be most abundant in animal cell?
(1) m-RNA
(2) mi-RNA
(3) r-RNA
(4) t-RNA
Answer (3)
Sol. rRNA is most abundant in animal cell. It constitutes 80% of total RNA of the cell.
-
Which cells of 'Crypts of Lieberkuhn' secrete antibacterial lysozyme?
(1) Zymogen cells
(2) Kupffer cells
(3) Argentaffin cells
(4) Paneth cells
Answer (4)
Sol. Kupffer-cells are phagocytic cells of liver.
Zymogen cells are enzyme producing cells.
Paneth cell secretes lysozyme which acts as anti-bacterial agent.
Argentaffin cells are hormone producing cells.
-
In case of a couple where the male is having a very low sperm count, which technique will be suitable for fertilisation?
(1) Artificial Insemination
(2) Intracytoplasmic sperm injection
(3) Intrauterine transfer
(4) Gamete intracytoplasmic fallopian transfer
Answer (1)
Sol. Infertility cases due to inability of the male partner to inseminate the female or due to very low sperm count in the ejaculates, could be corrected by artificial insemination (AI).
-
Frog's heart when taken out of the body continues to beat for some time
Select the best option from the following statements
(a) Frog is a poikilotherm
(b) Frog does not have any coronary circulation
(c) Heart is "myogenic" in nature
(d) Heart is autoexcitable
Options
(1) (a) & (b)
(2) (c) & (d)
(3) Only (c)
(4) Only (d)
Answer (2)
Sol. Frog or the vertebrates have myogenic heart having self contractile system or are autoexcitable; because of this condition, it will keep on working outside the body for some time.
Page 24
-
Match the following sexually transmitted diseases (Column - I) with their causative agent (Column - II) and select the correct option.
Column - I Column - II
(a) Gonorrhea (i) HIV
(b) Syphilis (ii) Neisseria
(c) Genital Warts (iii) Treponema
(d) AIDS (iv) Human Papilloma virus
Options:
(a) (b) (c) (d)
(1) (iv) (ii) (iii) (i)
(2) (iv) (iii) (ii) (i)
(3) (ii) (iii) (iv) (i)
(4) (iii) (iv) (i) (ii)
Answer (3)
Sol. Gonorrhoea - Neisseria (Bacteria)
Syphilis - Treponema (Bacteria)
Genital Warts - Human papilloma virus (Virus)
AIDS - HIV (Virus)
-
The genotypes of a Husband and Wife are I^A I^B and I^A i.
Among the blood types of their children, how many different genotypes and phenotypes are possible?
(1) 4 genotypes; 3 phenotypes
(2) 4 genotypes; 4 phenotypes
(3) 3 genotypes; 3 phenotypes
(4) 3 genotypes; 4 phenotypes
Answer (1)
Sol. Husband × Wife
I^A I^B × I^A i
Number of genotypes = 4
Number of phenotypes = 3
I^A I^A and I^A i = A
I^A I^B = AB
I^B i = B
-
The hepatic portal vein drains blood to liver from
(1) Kidneys
(2) Intestine
(3) Heart
(4) Stomach
Answer (2)
Sol. In hepatic portal system, hepatic portal vein carries maximum amount of nutrients from intestine to liver.
-
Coconut fruit is a
(1) Nut
(2) Capsule
(3) Drupe
(4) Berry
Answer (3)
Sol. Coconut fruit is a drupe. A drupe develops from monocarpellary superior ovary and are one seeded.
-
The vascular cambium normally gives rise to
(1) Secondary xylem
(2) Periderm
(3) Phelloderm
(4) Primary phloem
Answer (1)
Sol. During secondary growth, vascular cambium gives rise to secondary xylem and secondary phloem. Phelloderm is formed by cork cambium.
-
In case of poriferans the spongocoel is lined with flagellated cells called:
(1) Choanocytes
(2) Mesenchymal cells
(3) Ostia
(4) Oscula
Answer (1)
Sol. Choanocytes (collar cells) form lining of spongocoel in poriferans (sponges). Flagella in collar cells provide circulation to water in water canal system.
-
A baby boy aged two years is admitted to play school and passes through a dental check-up. The dentist observed that the boy had twenty teeth. Which teeth were absent?
(1) Pre-molars
(2) Molars
(3) Incisors
(4) Canines
Answer (1)
Sol. Total number of teeth in human child = 20. Premolars are absent in primary dentition.
-
An example of colonial alga is
(1) Ulothrix
(2) Spirogyra
(3) Chlorella
(4) Volvox
Answer (4)
Sol. Volvox is motile colonial fresh water alga with definite number of vegetative cells.
Page 25
-
An example of a sigma bonded organometallic compound is
(1) Ferrocene
(2) Cobaltocene
(3) Ruthenocene
(4) Grignard's reagent
Answer (4)
Sol. Grignard's reagent i.e., RMgX is σ-bonded organometallic compound.
-
Which one is the correct order of acidity?
(1) CH ≡ CH > CH₂ = CH₂ > CH₃ - C ≡ CH > CH₃ - CH₃
(2) CH₃ - CH₃ > CH₂ = CH₂ > CH₃ - C ≡ CH > CH ≡ CH
(3) CH₂ = CH₂ > CH₃ - CH = CH₂ > CH₃ - C ≡ CH > CH ≡ CH
(4) CH ≡ CH > CH₃ - C ≡ CH > CH₂ = CH₂ > CH₃ - CH₃
Answer (4)
Sol. Correct order is
H-C≡C-H > H₃C-C≡C-H > H₂C=CH₂ > CH₃-CH₃
(Two acidic hydrogens) (One acidic hydrogen)
-
Predict the correct intermediate and product in the following reaction
H₃C-C≡CH → (H₂O, H₂SO₄, HgSO₄) → intermediate (A) → product (B)
(1) A : H₃C-C(=O)-CH₃, B : H₃C-C≡CH
(2) A : H₃C-C(OH)=CH₂, B : H₃C-C(=O)-CH₃
(3) A : H₃C-C(=CH₂)-OH, B : H₃C-C(=O)-CH₃
(4) A : H₃C-C(=CH₂)-OH, B : H₃C-C(=O)-CH₂
Answer (2)
Sol. H₃C-C≡CH → H₃C-C(OH)=CH₂ (A) → Tautomerism → H₃C-C(=O)-CH₃ (B)
-
It is because of inability of ns² electrons of the valence shell to participate in bonding that
(1) Sn²⁺ and Pb²⁺ are both oxidising and reducing
(2) Sn⁴⁺ is reducing while Pb⁴⁺ is oxidising
(3) Sn²⁺ is reducing while Pb⁴⁺ is oxidising
(4) Sn²⁺ is oxidising while Pb⁴⁺ is reducing
Answer (3)
Sol. Inability of ns² electrons of the valence shell to participate in bonding on moving down the group in heavier p-block elements is called inert pair effect
As a result, Pb(II) is more stable than Pb(IV)
Sn(IV) is more stable than Sn(II)
∴ Pb(IV) is easily reduced to Pb(II)
∴ Pb(IV) is oxidising agent
Sn(II) is easily oxidised to Sn(IV)
∴ Sn(II) is reducing agent
-
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
(1) Rb
(2) Li
(3) Na
(4) K
Answer (2)
Sol. Li⁺ being smallest, has maximum charge density
∴ Li⁺ is most heavily hydrated among all alkali metal ions. Effective size of Li⁺ in aq solution is therefore, largest.
∴ Moves slowest under electric field.
-
Match the interhalogen compounds of column I with the geometry in column II and assign the correct code
Column I Column II
(a) XX' (i) T-shape
(b) XX'₃ (ii) Pentagonal bipyramidal
(c) XX'₅ (iii) Linear
(d) XX'₇ (iv) Square-pyramidal
(v) Tetrahedral
Code:
(a) (b) (c) (d)
(1) (v) (iv) (iii) (ii)
(2) (iv) (iii) (ii) (i)
(3) (iii) (iv) (i) (ii)
(4) (iii) (i) (iv) (ii)
Page 26
Answer (4)
Sol. XX' → Linear
XX'₃ → Example: ClF₃ → T-shape
XX'₅ → Example: BrF₅ → Square pyramidal
XX'₇ → Example: IF₇ → Pentagonal bipyramidal
-
Which is the incorrect statement?
(1) NaCl(s) is insulator, silicon is semiconductor, silver is conductor, quartz is piezo electric crystal
(2) Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal
(3) FeO₀.₉₈ has non stoichiometric metal deficiency defect
(4) Density decreases in case of crystals with Schottky's defect
Answer (2 & 3)
Sol. Frenkel defect occurs in those ionic compounds in which size of cation and anion is largely different.
Non-stoichiometric ferrous oxide is Fe₀.₉₃₋₀.₉₆O₁.₀₀ and it is due to metal deficiency defect.
-
Which one of the following statements is not correct?
(1) Enzymes catalyse mainly bio-chemical reactions
(2) Coenzymes increase the catalytic activity of enzyme
(3) Catalyst does not initiate any reaction
(4) The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium
Answer (4)
Sol. A catalyst decreases activation energies of both the forward and backward reaction by same amount, therefore, it speeds up both forward and backward reaction by same rate.
Equilibrium constant is therefore not affected by catalyst at a given temperature.
-
In the electrochemical cell
Zn|ZnSO₄(0.01M)||CuSO₄(1.0 M)|Cu, the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0M and that of CuSO₄ changed to 0.01M the emf changes to E₂. From the following, which one is the relationship between E₁ and E₂? (Given, RT/F = 0.059)
(1) E₁ > E₂ (2) E₂ = 0 ≠ E₁
(3) E₁ = E₂ (4) E₁ < E₂
Answer (1)
Sol. Zn|ZnSO₄(0.01M)||CuSO₄(1.0M)|Cu
∴ E₁ = E°_cell - (2.303RT/2F) × log(0.01/1)
When concentrations are changed
∴ E₂ = E°_cell - (2.303RT/2F) × log(1/0.01)
i.e., E₁ > E₂
-
The correct statement regarding electrophile is
(1) Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
(2) Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile
(3) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile
(4) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
Answer (2)
Sol. Fact.
-
The correct order of the stoichiometries of AgCl formed when AgNO₃ in excess is treated with the complexes: CoCl₃·6NH₃, CoCl₃·5NH₃, CoCl₃·4NH₃ respectively is
(1) 3 AgCl, 2 AgCl, 1 AgCl
(2) 2 AgCl, 3 AgCl, 1 AgCl
(3) 1 AgCl, 3 AgCl, 2 AgCl
(4) 3 AgCl, 1 AgCl, 2 AgCl
Answer (1)
Sol. Complexes are respectively [Co(NH₃)₆]Cl₃, [Co(NH₃)₅Cl]Cl₂ and [Co(NH₃)₄Cl₂]Cl
-
The IUPAC name of the compound
(1) 5-methyl-4-oxohex-2-en-5-al
(2) 3-keto-2-methylhex-5-enal
(3) 3-keto-2-methylhex-4-enal
(4) 5-formylhex-2-en-3-one
Page 27
Aldehydes get higher priority over ketone and alkene in numbering of principal C-chain.
3-keto-2-methylhex-4-enal
-
The species, having bond angles of 120° is
(1) NCl₃ (2) BCl₃ (3) PH₃ (4) ClF₃
Answer (2)
Sol. BCl₃ has bond angle of 120°.
-
The equilibrium constants of the following are
N₂ + 3H₂ ⇌ 2NH₃ K₁
N₂ + O₂ ⇌ 2NO K₂
H₂ + (1/2)O₂ → H₂O K₃
The equilibrium constant (K) of the reaction
2NH₃ + (5/2)O₂ ⇌ 2NO + 3H₂O, will be
(1) K₂K₃/K₁ (2) K₂³K₃/K₁
(3) K₁K₃³/K₂ (4) K₂K₃³/K₁
Answer (4)
Sol. (I) N₂ + 3H₂ ⇌ 2NH₃; K₁ = [NH₃]²/([N₂][H₂]³)
(II) N₂ + O₂ ⇌ 2NO; K₂ = [NO]²/([N₂][O₂])
(III) H₂ + (1/2)O₂ → H₂O; K₃ = [H₂O]/([H₂][O₂]^(1/2))
(II + 3 × III - I) will give
2NH₃ + (5/2)O₂ ⇌ 2NO + 3H₂O;
∴ K = K₂ × K₃³/K₁
-
Name the gas that can readily decolourises acidified KMnO₄ solution:
(1) NO₂ (2) P₂O₅
(3) CO₂ (4) SO₂
Answer (4)
Sol. SO₂ is readily decolourises acidified KMnO₄.
-
The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenols is
(1) Crystallisation
(2) Steam distillation
(3) Sublimation
(4) Chromatography
Answer (2)
Sol. Steam distillation is the most suitable method of separation of 1:1 mixture of ortho and para nitrophenols as there is intramolecular H-bonds in ortho nitro phenol.
-
The reason for greater range of oxidation states in actinoids is attributed to
(1) 5f, 6d and 7s levels having comparable energies
(2) 4f and 5d levels being close in energies
(3) The radioactive nature of actinoids
(4) Actinoid contraction
Answer (1)
Sol. It is a fact.
-
The element Z = 114 has been discovered recently. It will belong to which of the following family group and electronic configuration?
(1) Oxygen family, [Rn] 5f¹⁴6d¹⁰7s²7p⁴
(2) Nitrogen family, [Rn] 5f¹⁴6d¹⁰7s²7p⁶
(3) Halogen family, [Rn] 5f¹⁴6d¹⁰7s²7p⁵
(4) Carbon family, [Rn] 5f¹⁴6d¹⁰7s²7p²
Answer (4)
Sol. Z = 114 belong to Group 14, carbon family
Electronic configuration = [Rn] 5f¹⁴6d¹⁰7s²7p²
-
Mechanism of a hypothetical reaction X₂ + Y₂ → 2XY is given below:
(i) X₂ → X + X (fast)
(ii) X + Y₂ ⇌ XY + Y (slow)
(iii) X + Y → XY (fast)
The overall order of the reaction will be
(1) 0 (2) 1.5 (3) 1 (4) 2
Page 28
Sol. The solution of this question is given by assuming step (i) to be reversible which is not given in question
Overall rate = Rate of slowest step (ii)
k = k[X][Y₂] ...(1)
k = rate constant of step (ii)
Assuming step (i) to be reversible, its equilibrium constant,
k_eq = [X]²/[X₂] ⇒ [X] = k_eq^(1/2)[X₂]^(1/2) ...(2)
Put (2) in (1)
Rate = k k_eq^(1/2)[X₂]^(1/2)[Y₂]
Overall order = 1/2 + 1 = 3/2
-
If molality of the dilute solution is doubled, the value of molal depression constant (K_f) will be
(1) Tripled
(2) Unchanged
(3) Doubled
(4) Halved
Answer (2)
Sol. K_f (molal depression constant) is a characteristic of solvent and is independent of molality.
-
With respect to the conformers of ethane, which of the following statements is true?
(1) Both bond angle and bond length change
(2) Both bond angles and bond length remains same
(3) Bond angle remains same but bond length changes
(4) Bond angle changes but bond length remains same
Answer (2)
Sol. There is no change in bond angles and bond lengths in the conformations of ethane. There is only change in dihedral angle.
-
The heating of phenyl-methyl ethers with HI produces.
(1) Phenol
(2) Benzene
(3) Ethyl chlorides
(4) Iodobenzene
Answer (1)
Sol. C₆H₅-O-CH₃ + HI → C₆H₅-OH + CH₃I
-
The correct increasing order of basic strength for the following compounds is
(I) Aniline
(II) p-Nitroaniline
(III) p-Toluidine
(1) III < II < I (2) II < I < III
(3) II < III < I (4) III < I < II
Answer (2)
Sol. -NO₂ has strong -R effect and -CH₃ shows +R effect.
Order of basic strength is
p-Nitroaniline < Aniline < p-Toluidine
-
In which pair of ions both the species contain S-S bond?
(1) S₂O₇²⁻, S₂O₆²⁻
(2) S₂O₇²⁻, S₂O₃²⁻
(3) S₄O₆²⁻, S₂O₃²⁻
(4) S₄O₆²⁻, S₂O₃²⁻
Answer (4)
Sol. [S₄O₆]²⁻ and [S₂O₃]²⁻ both contain S-S bond.
-
Which of the following is dependent on temperature?
(1) Mole fraction
(2) Weight percentage
(3) Molality
(4) Molarity
Answer (4)
Sol. Molarity includes volume of solution which can change with change in temperature.
Page 29
-
Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating?
Answer (4)
Sol. Cyclohexanone + Cyclohexanone → (i) OH⁻ (ii) Δ → Product (4)
-
Mixture of chloroxylenol and terpineol acts as
(1) Antipyretic
(2) Antibiotic
(3) Analgesic
(4) Antiseptic
Answer (4)
Sol. Mixture of chloroxylenol and terpineol acts as antiseptic.
-
For a given reaction, ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 JK⁻¹ mol⁻¹. The reaction is spontaneous at: (Assume that ΔH and ΔS do not vary with temperature)
(1) All temperatures
(2) T > 298 K
(3) T < 425 K
(4) T > 425 K
Answer (4)
Sol. ΔG = ΔH - TΔS
For a reaction to be spontaneous, ΔG = -ve
i.e., ΔH < TΔS
∴ T > ΔH/ΔS = 35.5 × 10³ J / 83.6 JK⁻¹
i.e., T > 425 K
-
HgCl₂ and I₂ both when dissolved in water containing I⁻ ions the pair of species formed is
(1) HgI₄²⁻, I₅⁻ (2) Hg₂I₂, I⁻
(3) HgI₂, I₃⁻ (4) HgI₂, I⁻
Answer (1)
Sol. In a solution containing HgCl₂, I₂ and I⁻ both HgCl₂ and I₂ compete for I⁻
Since formation constant of [HgI₄]²⁻ is 1.9 × 10³⁰ which is very large as compared with I₃⁻ K_f = 700
∴ I⁻ will preferentially combine with HgCl₂
HgCl₂ + 2I⁻ → HgI₂↓ + 2Cl⁻ (Red ppt)
HgI₂ + 2I⁻ → [HgI₄]²⁻ (soluble)
-
A first order reaction has a specific reaction rate of 10⁻² s⁻¹. How much time will it take for 20 g of the reactant to reduce to 5 g?
(1) 346.5 second
(2) 693.0 second
(3) 238.6 second
(4) 138.6 second
Answer (4)
Sol. t₁/₂ = 0.693/10⁻² second
For the reduction of 20 g of reactant to 5 g, two t₁/₂ is required.
∴ t = 2 × 0.693/10⁻² second
= 138.6 second
-
Which one is the most acidic compound?
(1) p-Nitrophenol
(2) 2,4,6-Trinitrophenol
(3) p-Cresol
(4) Phenol
Answer (2)
Sol. -NO₂ group has very strong -I & -R effects.
Page 30
-
Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co³⁺ is
(1) [Co(H₂O)₆]³⁺, [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺
(2) [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺, [Co(H₂O)₆]³⁺
(3) [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺, [Co(H₂O)₆]³⁺
(4) [Co(H₂O)₆]³⁺, [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺
Answer (3)
Sol. The order of the ligand in the spectrochemical series
H₂O < NH₃ < en
Hence, the wavelength of the light observed will be in the order
[Co(H₂O)₆]³⁺ < [Co(NH₃)₆]³⁺ < [Co(en)₃]³⁺
Thus, wavelength absorbed will be in the opposite order
i.e., [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺, [Co(H₂O)₆]³⁺
-
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10⁻⁴ mol L⁻¹. Solubility product of Ag₂C₂O₄ is
(1) 4.5 × 10⁻¹¹ (2) 5.3 × 10⁻¹²
(3) 2.42 × 10⁻⁸ (4) 2.66 × 10⁻¹²
Answer (2)
Sol. Ag₂C₂O₄(s) ⇌ 2Ag⁺(aq) + C₂O₄²⁻(aq)
2s s
K_sp = [Ag⁺]²[C₂O₄²⁻]
[Ag⁺] = 2.2 × 10⁻⁴ M
∴ [C₂O₄²⁻] = 2.2 × 10⁻⁴/2 M = 1.1 × 10⁻⁴ M
∴ K_sp = (2.2 × 10⁻⁴)²(1.1 × 10⁻⁴)
= 5.324 × 10⁻¹²
-
A 20 litre container at 400 K contains CO₂(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO₂ attains its maximum value, will be
(Given that: SrCO₃(s) ⇌ SrO(s) + CO₂(g), K_p = 1.6 atm)
(1) 4 litre (2) 2 litre
(3) 5 litre (4) 10 litre
Answer (3)
Sol. Max. pressure of CO₂ = Pressure of CO₂ at equilibrium
For reaction,
SrCO₃(s) ⇌ SrO(s) + CO₂
K_p = P_CO₂ = 1.6 atm = maximum pressure of CO₂
Volume of container at this stage,
V = nRT/P ...(i)
Since container is sealed and reaction was not earlier at equilibrium
∴ n = constant
n = PV/RT = 0.4 × 20/RT ...(ii)
Put equation (ii) in equation (i)
V = [0.4 × 20/RT] RT/1.6 = 5 L
-
Identify A and predict the type of reaction
Answer (3)
Page 31
Sol. More stable as -ve charge is close to electron withdrawing group
Incoming nucleophile ends on same C on which Br (Leaving group) was present NOT cine substitution.
-
Which of the following reactions is appropriate for converting acetamide to methanamine?
(1) Stephens reaction
(2) Gabriels phthalimide synthesis
(3) Carbylamine reaction
(4) Hoffmann hypobromide reaction
Answer (4)
Sol. CH₃-CONH₂ + Br₂ + 4NaOH → Δ → CH₃-NH₂ + 2NaBr + Na₂CO₃ + 3H₂O
This is Hoffmann Bromamide reaction.
-
Which of the following pairs of compounds is isoelectronic and isostructural?
(1) IBr₂⁻, XeF₂ (2) IF₃, XeF₂
(3) BeCl₂, XeF₂ (4) TeI₂, XeF₂
Answer (1)
Sol. IBr₂⁻, XeF₂
Total number of valence electrons are equal in both the species and both the species are linear also.
-
Which of the following is a sink for CO?
(1) Oceans
(2) Plants
(3) Haemoglobin
(4) Micro-organisms present in the soil
Answer (4)
Sol. Micro-organisms present in the soil is a sink for CO.
-
Which one of the following pairs of species have the same bond order?
(1) CN⁻, CO (2) N₂, O₂⁻
(3) CO, NO (4) O₂, NO⁺
Answer (1)
Sol. CN⁻ and CO have bond order 3 each.
-
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be
(1) -505 J (2) +505 J
(3) 1136.25 J (4) -500 J
Answer (1)
Sol. ΔU = q + w
For adiabatic process, q = 0
∴ ΔU = w
= -P·ΔV
= -2.5 atm × (4.5 - 2.5) L
= -2.5 × 2 L·atm
= -5 × 101.3 J
= -506.5 J
= -505 J
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Extraction of gold and silver involves leaching with CN⁻ ion. Silver is later recovered by
(1) Zone refining
(2) Displacement with Zn
(3) Liquation
(4) Distillation
Answer (2)
Sol. Zn being more reactive than Ag and Au, displaces them.
From Native ore,
4Ag + 8NaCN + 2H₂O + O₂ → Leaching → 4Na[Ag(CN)₂] + 4NaOH (Soluble Sodium dicyanoargentate(I))
2Na[Ag(CN)₂] + Zn → Displacement → Na₂[Zn(CN)₄] + 2Ag↓
Page 32
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Consider the reactions:
X (C₂H₆O) → Cu/573 K → A → [Ag(NH₃)₂]⁺ → Silver mirror observed
A → -OH, Δ → Y
A → NH₂-NH-C(=O)-NH₂ → Z
Identify A, X, Y and Z
(1) A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone
(2) A-Ethanol, X-Acetaldehyde, Y-Butanone, Z-Hydrozone
(3) A-Methoxymethane, X-Ethanoic acid, Y-Acetate ion, Z-Hydrozine
(4) A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazide
Answer (1)
Sol. Since 'A' gives positive silver mirror test therefore, it must be an aldehyde or α-Hydroxyketone.
Reaction with semicarbazide indicates that A can be an aldehyde or ketone.
Reaction with OH⁻ i.e., aldol condensation (by assuming alkali to be dilute) indicates that A is aldehyde as aldol reaction of ketones is reversible and carried out in special apparatus.
These indicates option (1).
CH₃-CH₂OH → Cu/573 K → CH₃-CHO (A) ethanal → [Ag(NH₃)₂]⁺, OH⁻/Δ → CH₃-COOH
CH₃-CHO + H₂N-NH-C(=O)-NH₂ → CH₃-CH=N-NH-C(=O)-NH₂ (Z)
CH₃-CHO → OH⁻ → CH₃-CH(OH)-CH₂-CHO (3-Hydroxybutanal) → Δ → CH₃-CH=CH-CHO (Y)
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Pick out the correct statement with respect [Mn(CN)₆]³⁻:
(1) It is d²sp³ hybridised and octahedral
(2) It is dsp² hybridised and square planar
(3) It is sp³d² hybridised and octahedral
(4) It is sp³d² hybridised and tetrahedral
Answer (1)
Sol. [Mn(CN)₆]³⁻ Mn(III) = [Ar]3d⁴
CN⁻ being strong field ligand forces pairing of electrons
This gives t₂g⁴ eg⁰
Mn(III) = [Ar] 3d⁴
Coordination number of Mn = 6
Structure = octahedral
[Mn(CN)₆]³⁻ = d²sp³
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Which one is the wrong statement?
(1) Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
(2) The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms
(3) de-Broglie's wavelength is given by λ = h/mv where m = mass of the particle, v = group velocity of the particle
(4) The uncertainty principle is ΔE × Δt ≥ h/4π
Answer (2)
Sol. Energy of 2s-orbital and 2p-orbital in case of hydrogen like atoms is equal.
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Which of the following statements is not correct?
(1) Blood proteins thrombin and fibrinogen are involved in blood clotting
(2) Denaturation makes the proteins more active
(3) Insulin maintains sugar level in the blood of a human body
(4) Ovalbumin is a simple food reserve in egg-white
Answer (2)
Sol. Due to denaturation of proteins, globules unfold and helix get uncoiled and protein loses its biological activity.
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