1. The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is
(1) 4v
(2) v
(3) 2v
(4) 3v
Answer (1)
Sol. Escape velocity from the Earth's surface
v_e = sqrt(2GM/R)
= sqrt(2G rho (4/3) pi R^3 / R)
v_e proportional to R (For same density)
v/v1 = R/4R
v1 = 4v
2. A cup of coffee cools from 90 degree C to 80 degree C in t minutes, when the room temperature is 20 degree C. The time taken by a similar cup of coffee to cool from 80 degree C to 60 degree C at a room temperature same at 20 degree C is
(1) 5/13 t
(2) 13/5 t
(3) 10/13 t
(4) 5/13 t
Answer (3)
Sol. From Average form of Newton's law of cooling
-((T1 + T2)/2 - Ts)K = (T1 - T2)/Delta t
T1 and T2 are initial and final temperature and Ts is surrounding temperature.
=> -K[(90 + 80)/2 - 20] = (90 - 80)/t
=> -K(65) = 10/t
=> K = -2/(13t)
In second case,
-K((80 + 60)/2 - 20) = (80 - 60)/t1
=> -K(50) = 20/t1
=> (2/(13t))(50) = 20/t1
=> t1 = 13t/5
3. A thick current carrying cable of radius R carries current I uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance r from the axis of the cable is represented by
Answer (4)
Sol. From Ampere's circuital law
B = (mu0 I / (2 pi R^2)) * r if r < R => B_inside proportional to r
B = mu0 I / (2 pi r) if r >= R => B_outside proportional to 1/r
Hence the correct plot of magnetic field B with distance r from axis of cable is given as
4. Polar molecules are the molecules
(1) Having a permanent electric dipole moment
(2) Having zero dipole moment
(3) Acquire a dipole moment only in the presence of electric field due to displacement of charges
(4) Acquire a dipole moment only when magnetic field is absent
Answer (1)
Sol. In polar molecules, the centre of positive charges does not coincide with the centre of negative charges.
Hence, these molecules have a permanent electric dipole moment of their own.
5. Find the value of the angle of emergence from the prism. Refractive index of the glass is sqrt(3).
(1) 90 degree
(2) 60 degree
(3) 30 degree
(4) 45 degree
Answer (2)
Sol. From the ray diagram shown in the figure.
At point P, from Snell's law
sin i / sin r = mu_air / mu_prism
=> sin 30 / sin e = 1 / sqrt(3) (angle r = angle e emergent angle)
=> sin e = sqrt(3) * 1/2
=> angle e = 60 degree
6. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A, the energy stored in the capacitor is
(epsilon0 = permittivity of free space)
(1) E^2 A d / epsilon0
(2) (1/2) epsilon0 E^2
(3) epsilon0 E A d
(4) (1/2) epsilon0 E^2 A d
Answer (4)
Sol. Energy density associated with electric field is given by
u = dU/dV = (1/2) epsilon0 E^2
=> dU = (1/2) epsilon0 E^2 dV
Total energy stored in the space between the capacitor will be
U = integral dU = integral (1/2) epsilon0 E^2 dV
= (1/2) epsilon0 E^2 integral dV [E is constant]
= (1/2) epsilon0 E^2 V = (1/2) epsilon0 E^2 A d [V = Ad]
7. A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively
(1) S/4, sqrt(3gS/2)
(2) S/4, 3gS/2
(3) S/4, sqrt(3gS/2)
(4) S/2, sqrt(3gS/2)
Answer (1)
Sol. Let required height of body is y.
When body from rest falls through height (S - y) Then under constant acceleration
v^2 = 0^2 + 2g(S - y)
v = sqrt(2g(S - y))
When body is at height y above ground. Potential energy of body of mass m
U = mgy
As per given condition kinetic energy, K = 3U
(1/2) m(v)^2 = 3 * mg(y)
(1/2) * m * 2g(S - y) = 3 * mgy (using (1))
S - y = 3y
therefore y = S/4 ...(2)
therefore v = sqrt(2 * g(S - S/4)) = sqrt(3gS/2) ...(3)
8. The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be
(1) 2Mg
(2) Mg/2
(3) Mg
(4) 3/2 Mg
Answer (2)
Sol. Let Fv be the viscous force and FB be the Bouyant force acting on the ball.
Then, when body moves with constant velocity
Mg = FB + Fv
Fv = Mg - FB
= dVg - (d/2)Vg (M = dVg) V = volume of ball.
= (d/2)Vg
Fv = (M/2)g
9. The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
(1) No current will flow in p-type, current will only flow in n-type
(2) Current in n-type = current in p-type
(3) Current in p-type > current in n-type
(4) Current in n-type > current in p-type.
Answer (4)
Sol. The current through a semiconductor is
I = neAvd
I = neA mu E
In/Ip = (ne e A mu_e E) / (nh e A mu_h E)
In/Ip = mu_e / mu_h
therefore mu_e > mu_h
=> In > Ip
10. An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10^5 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.
(1) 8 x 10^-20 N
(2) 4 x 10^-20 N
(3) 8 pi x 10^-20 N
(4) 4 pi x 10^-20 N
Answer (1)
Sol. Magnetic field produced due to current carrying wire at point 'A'
B = (mu0 / (4 pi)) * (2I / r)
B = (10^-7 * 2 * 5) / (20 * 10^-2) = (1/2) * 10^-5 Tesla, upward to the plane of paper
Now, force acting on electron due to this field
F = q(v x B)
|F| = 1.6 x 10^-19 * 10^5 * (1/2) x 10^-5
= 0.8 x 10^-19 N
|F| = 8 x 10^-20 N
11. An electromagnetic wave of wavelength lambda is incident on a photosensitive surface of negligible work function. If m mass is of photoelectron emitted from the surface has de-Broglie wavelength lambda_d, then
(1) lambda = (2h/mc) lambda_d^2
(2) lambda = (2m/hc) lambda_d^2
(3) lambda_d = (2mc/h) lambda^2
(4) lambda = (2mc/h) lambda_d^2
Answer (4)
Sol. As per Einstein's photoelectric equation
hc/lambda = phi0 + k
phi0 : work function
k = maximum kinetic energy of photoelectrons
As per question, phi -> 0
therefore hc/lambda = k = P^2/2m => P = sqrt(2mhc/lambda)
Now De-broglie wavelength,
lambda_d = h/P = h / sqrt(2mhc/lambda)
=> sqrt(lambda) = lambda_d sqrt(2mc/h)
=> lambda = (2mc/h) lambda_d^2
12. Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres (sigma1/sigma2) is
(1) R1^2/R2^2
(2) R1/R2
(3) R2/R1
(4) sqrt(R1/R2)
Answer (3)
Sol. When two conductors are connected by a conducting wire, then the two conductors should have same potential.
so, V1 = V2
therefore (1/(4 pi epsilon0)) (Q1/R1) = (1/(4 pi epsilon0)) (Q2/R2)
=> (1/(4 pi epsilon0)) (Q1/R1) * (R1/R1) = (1/(4 pi epsilon0)) (Q2/R2) * (R2/R2)
=> Q1 R1 / (4 pi R1^2 epsilon0) = Q2 R2 / (4 pi R2^2 epsilon0)
=> sigma1 R1 / epsilon0 = sigma2 R2 / epsilon0
=> sigma1/sigma2 = R2/R1
13. An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series to an ac source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10 sqrt(2) A. The impedance of the circuit is
(1) 5 ohm
(2) 4 sqrt(2) ohm
(3) 5/sqrt(2) ohm
(4) 4 ohm
Answer (1)
Sol. VL = 40 volt
VR = 40 volt
VC = 10 volt
Now, VRMS = sqrt(VR^2 + (VL - VC)^2)
= sqrt((40)^2 + (40 - 10)^2) = 50 V
IRMS = I0/sqrt(2) = 10 sqrt(2)/sqrt(2) = 10 A
therefore VRMS = IRMS * Z
therefore Z = VRMS/IRMS = 50/10 = 5 ohm
14. For a plane electromagnetic wave propagating in x-direction, which one of the following combination gives the correct possible directions for electric field (E) and magnetic field (B) respectively?
(1) -j + k, -j + k
(2) j + k, j + k
(3) -j + k, -j - k
(4) j + k, -j - k
Answer (3)
Sol. Direction of propagation of electromagnetic waves is along E x B
Given that direction of propagation is along x-axis
(1) (-j + k) x (-j + k) = 0
(2) (j + k) x (j + k) = 0
(3) (-j + k) x (-j - k) = 2j
(4) (j + k) x [-(j + k)] = 0
therefore Option (3) is correct.
15. Consider the following statements (A) and (B) and identify the correct answer.
(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.
(1) (A) is incorrect but (B) is correct.
(2) (A) and (B) both are correct.
(3) (A) and (B) both are incorrect.
(4) (A) is correct and (B) is incorrect.
Answer (4)
Sol. In reverse biased, after breakdown, voltage across the zener diode becomes constant. Therefore zener diode is connected in reverse biased when used as voltage regulator.
Potential barrier of silicon diode is nearly 0.7 V statement A is correct and statement B is incorrect.
16. The equivalent capacitance of the combination shown in the figure is
Answer (3)
Sol. Given circuit is
Points 1, 2, 3 are at same potential (as they are connected by conducting wire)
So the capacitor is short circuited. It does not store any charge.
The circuit can be redrawn as
CAB = C + C = 2C (Parallel combination)
17. In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
(1) 62 cm
(2) 60 cm
(3) 21.6 cm
(4) 64 cm
Answer (2)
Sol. From the application of potentiometer to compare two cells of emfs E1 and E2 by balancing lengths l1 and l2
E1/E2 = l1/l2
=> l2 = l1 (E2/E1) = (36 cm)(2.5 V/1.5 V)
= 60 cm
18. A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by
V = V0 sin omega t
The displacement current between the plates of the capacitor, would then be given by
(1) Id = V0 omega C sin omega t
(2) Id = V0 omega C cos omega t
(3) Id = V0/(omega C) cos omega t
(4) Id = V0/(omega C) sin omega t
Answer (2)
Sol. Given V = V0 sin omega t ...(1)
Now displacement current Id is given by
Id = C dV/dt
= C d/dt (V0 sin omega t) (using equation 1)
= C (V0 omega) cos omega t
Id = V0 omega C cos omega t
19. A dipole is placed in an electric field as shown. In which direction will it move?
(1) Towards the right as its potential energy will increase.
(2) Towards the left as its potential energy will increase.
(3) Towards the right as its potential energy will decrease.
(4) Towards the left as its potential energy will decrease.
Answer (3)
Sol. Potential energy of electric dipole in external electric field U = -P . E
Angle between electric field and electric dipole is 180 degree
U = -PE cos theta
U = -PE cos 180 degree
U = +PE
On moving towards right electric field strength decrease therefore potential energy decrease.
Net force on electric dipole is towards right and net torque acting on it is zero.
So, it will more towards right.
20. The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm when it delivers the power of 3.3 x 10^-3 watt will be (h = 6.6 x 10^-34 J s)
(1) 10^15
(2) 10^18
(3) 10^17
(4) 10^16
Answer (4)
Sol. The power of a source is given as
(Here n/t is number of photons emitted per second)
=> n/t = (3.3 x 10^-3 * 6 x 10^-7) / (6.6 x 10^-34 * 3 x 10^8) = 10^16 photons per second
21. A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval t = n - 1 to t = n. Then, the ratio Sn/S(n+1) is
(1) 2n/(2n - 1)
(2) (2n - 1)/2n
(3) (2n - 1)/(2n + 1)
(4) (2n + 1)/(2n - 1)
Answer (3)
Sol. Suppose theta is inclination of inclined plane acceleration along inclined plane a = g sin theta
Sn = distance travelled by object during nth second.
Initial speed u = 0
By equation of uniformly accelerated motion
Sn = u + (a/2)(2n - 1)
Sn = 0 + (g sin theta / 2)(2n - 1) = (g sin theta / 2)(2n - 1) ...(i)
Distance travelled during (n + 1)th second.
S(n+1) = 0 + (g sin theta / 2)[2(n + 1) - 1] = (g sin theta / 2)(2n + 1) ...(ii)
Dividing equations (i) and (ii)
Sn/S(n+1) = (2n - 1)/(2n + 1)
22. A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
(1) 0.052 cm
(2) 0.52 cm
(3) 0.026 cm
(4) 0.26 cm
Answer (1)
Sol. Here, pitch of the screw gauge, P = 1 mm Number of circular division, n = 100
Thus least count LC = P/n = 1/100 = 0.01 mm
= 0.001 cm
So, diameter of the wire = MSR + (CSR x LC)
= 0 + (52 x 0.001 cm)
= 0.052 cm
23. A radioactive nucleus A/Z X undergoes spontaneous decay in the sequence
A/Z X -> B(Z-1) -> C(Z-3) -> D(Z-2), where Z is the atomic number of element X. The possible decay particles in the sequence are
(1) beta^-, alpha, beta^+
(2) alpha, beta^-, beta^+
(3) alpha, beta^+, beta^-
(4) beta^+, alpha, beta^-
Answer (4)
Sol. On beta^+ decay atomic number decreases by 1
On beta^- decay atomic number increases by 1
On alpha decay atomic number decreases by 2
A/Z X --beta^+ decay--> B(Z-1) --alpha decay--> C(Z-3) --beta^- decay--> D(Z-2)
Hence correct order of decay are beta^+, alpha, beta^-
24. A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since
(1) A large aperture contributes to the quality and visibility of the images.
(2) A large area of the objective ensures better light gathering power.
(3) A large aperture provides a better resolution.
(4) All of the above
Answer (4)
Sol. With larger aperture of objective lens, the light gathering power in telescope is high.
Also, the resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective. Thus objective of large diameter is preferred.
Also, with large diameters fainter objects can be observed. Hence it also contributes to the better quality and visibility of images.
Hence, all options are correct.
25. A body is executing simple harmonic motion with frequency n, the frequency of its potential energy is
(1) 4n
(2) n
(3) 2n
(4) 3n
Answer (3)
Sol. Equation of displacement of particle executing SHM is given by x = A sin(omega t + phi) ...(I) Potential energy of particle executing SHM is given by
U = (1/2) kx^2
= (1/2) kA^2 sin^2(omega t + phi) ...(II)
From I and II, it is clear that
Time period of x = A sin(omega t + phi) is
T1 = 2 pi / omega => frequency n1 = omega / (2 pi)
while time period of x^2 = A^2 sin^2(omega t + phi) is
T2 = pi / omega => frequency n2 = omega / pi
Hence n2 = 2n1
26. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be
(1) 2/(3 sqrt(2))
(2) 1/2
(3) 1/(2 sqrt(2))
(4) 2/3
Answer (3)
Sol. The activity of a radioactive substance is given as
A = A0 (1/2)^(t/T1/2)
Now, A/A0 = (1/2)^(t/T1/2)
=> A/A0 = (1/2)^(150/100)
=> A/A0 = (1/2)^(3/2)
=> A/A0 = 1/(2 sqrt(2))
27. Match Column - I and Column - II and choose the correct match from the given choices.
Column - I Column - II
(A) Root mean square speed of gas molecules (P) (1/3) n m v^2
(B) Pressure exerted by ideal gas (Q) sqrt(3RT/M)
(C) Average kinetic energy of a molecule (R) (5/2) RT
(D) Total internal energy of 1 mole of a diatomic gas (S) (3/2) k_B T
(1) (A)- (R), (B)- (Q), (C)- (P), (D)- (S)
(2) (A)- (R), (B)- (P), (C)- (S), (D)- (Q)
(3) (A)- (Q), (B)- (R), (C)- (S), (D)- (P)
(4) (A)- (Q), (B)- (P), (C)- (S), (D)- (R)
Answer (4)
Sol. Root mean square speed of gas molecules = sqrt(3RT/M)
Pressure exerted by ideal gas = (1/3) n m v^2
Average kinetic energy of a molecule = (3/2) k_B T
Total internal energy of a gas is (U) = (1/2) nRT
Here, n = 1
f = 5
U = (5/2) RT
Hence, (A) - (Q), (B) - (P), (C) - (S), (D) - (R)
28. If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
(1) [F][A^-1][T]
(2) [F][A][T]
(3) [F][A][T^2]
(4) [F][A][T^-1]
Answer (3)
Sol. Energy, E proportional to F^a A^b T^c
[E] = [F^a][A^b][T^c]
=> [ML^2 T^-2] = [MLT^-2]^a [LT^-2]^b [T]^c
[ML^2 T^-2] = [M^a L^(a+b) T^(-2a - 2b + c)]
Comparing dimensions on both sides.
=> a = 1; a + b = 2 and -2 = -2a - 2b + c
=> b = 1 => -2 = -2 - 2 + c
=> c = 2
[E] = [F A T^2]
29. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?
(g = 10 m/s^2)
(1) 7.0 kW
(2) 10.2 kW
(3) 8.1 kW
(4) 12.3 kW
Answer (3)
Sol. Incident power on turbine = d(mgh)/dt
= gh dm/dt
= 10 x 60 x 15
= 9000 W
Now, losses are 10%
therefore power generated = (1 - 10/100) x 9000
= 8100 W
= 8.1 kW
30. The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 ohm. What will be the effective resistance if they are connected in series?
(1) 4 ohm
(2) 0.25 ohm
(3) 0.5 ohm
(4) 1 ohm
Answer (1)
Sol. All the wires are identical and of same material so they will have same value of resistance. Let it be R. When these are (four) connected in parallel.
Given Rp = 0.25 ohm
therefore 0.25 = R/4
therefore R = 1 ohm
Now these four resistances are arranged in series
RS = R + R + R + R = 4R
therefore RS = 4 x 1 = 4 ohm
31. Column-1 gives certain physical terms associated with flow of current through a metallic conductor. Column-1I gives some mathematical relations involving electrical quantities. Match Column-1 and Column-1I with appropriate relations.
Column-1 Column-1I
(A) Drift Velocity (P) m/(ne^2 rho)
(B) Electrical Resistivity (Q) n e v_d
(C) Relaxation Period (R) (eE/m) tau
(D) Current Density (S) E/J
(1) (A) - (R), (B) - (Q), (C) - (S), (D) - (P)
(2) (A) - (R), (B) - (S), (C) - (P), (D) - (Q)
(3) (A) - (R), (B) - (S), (C) - (Q), (D) - (P)
(4) (A) - (R), (B) - (P), (C) - (S), (D) - (Q)
Answer (2)
Sol. Drift velocity, v_d = eE tau / m
Electrical resistivity, rho = 1/sigma = E/J
Relaxation period, tau = m/(ne^2 rho)
Current density, J = I/A = n e v_d
(A) - (R), (B) - (S), (C) - (P), (D) - (Q)
32. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is
(1) 216 MeV
(2) 0.9 MeV
(3) 9.4 MeV
(4) 804 MeV
Answer (1)
Sol. Mass number of reactant = 240
BE per nucleon = 7.6 MeV
Mass number of products = 120
BE per nucleon of product = 8.5 MeV
Total gain in BE = (BE) of products - (BE) of reactants
= [120 + 120] x 8.5 - [240] x 7.6
= (240) x 8.5 - 240 x 7.6
= (2040 - 1824) MeV
Gain in BE = 216 MeV
33. A convex lens 'A' of focal length 20 cm and a concave lens 'B' of focal length 5 cm are kept along the same axis with a distance 'd' between them. If a parallel beam of light falling on 'A' leaves 'B' as a parallel beam, then the distance 'd' in cm will be
(1) 30
(2) 25
(3) 15
(4) 50
Answer (3)
Sol. Parallel beam of light after refraction from convex lens converge at the focus of convex lens. In question it is given light after refraction pass through concave lens becomes parallel. Therefore light refracted from convex lens virtually meet at focus of concave lens. According to above ray diagram d = fA - fB
= 20 - 5 = 15 cm
34. A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is
(1) 0.628 s
(2) 0.0628 s
(3) 6.28 s
(4) 3.14 s
Answer (1)
Sol. For a spring, kx = F
=> k(5 x 10^-2) = 10
=> k = 1000/5 = 200 N/m
Now, for spring-mass system undergoing SHM
T = 2 pi sqrt(m/k)
given, m = 2 kg
=> T = 2 pi sqrt(2/200) = 2 pi / 10 = 0.628 s
35. If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of
(1) [M^2][L^-2][T^-1]
(2) [M^2][L^-1][T^0]
(3) [M][L^-1][T^-1]
(4) [M][L^0][T^0]
Answer (2)
Sol. Dimensional formula of energy
[E] = [M^1 L^2 T^-2] ...(1)
Dimensional formula of gravitational constant
[G] = [M^-1 L^3 T^-2] ...(2)
From (1) & (2)
[E]/[G] = [M^1 L^2 T^-2] / [M^-1 L^3 T^-2] = [M^2 L^-1 T^0]
So, correct option is (2)
SECTION-B
36. Twenty seven drops of same size are charged at 200 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
(1) 1980 V
(2) 660 V
(3) 1320 V
(4) 1520 V
Answer (1)
Sol. Electric potential due to a charged sphere = kQ/R
k = 9 x 10^9 N-m^2/C^2
Q : charge on sphere
R : Radius of sphere
Let charge and radius of smaller drop is q and r respectively
For smaller drop, V = kq/r = 220 V
Let R be radius of bigger drop,
As volume remains the same
(4/3 pi r^3) x 27 = 4/3 pi R^3
=> R = cuberoot(27) r = 3r
Now, using charge conservation,
Q = 27q
V_bigdrop = kQ/R = k(27q)/(3r) = 9(kq/r)
= 9 x 220 = 1980 V
37. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle theta to the horizontal, the maximum height attained by it equals 4R. The angle of projection, theta, is then given by :
(1) theta = sin^-1 (2gT^2 / (pi^2 R))^(1/2)
(2) theta = cos^-1 (gT^2 / (pi^2 R))^(1/2)
(3) theta = cos^-1 (pi^2 R / (gT^2))^(1/2)
(4) theta = sin^-1 (pi^2 R / (gT^2))^(1/2)
Answer (1)
Sol. To complete a circular path of radius R, time period is T.
so speed of particle (U) = 2 pi R / T ...(1)
Now the particle is projected with same speed at angle theta to horizontal.
So Maximum Height (H) = U^2 sin^2 theta / (2g)
Given that : H = 4R
=> U^2 sin^2 theta / (2g) = 4R
=> sin^2 theta = 8gR / U^2 ...(2)
=> sin^2 theta = 8gRT^2 / (4 pi^2 R^2) = 2gT^2 / (pi^2 R) (using equation 1)
=> theta = sin^-1 (2gT^2 / (pi^2 R))^(1/2)
38. From a circular ring of mass M and radius R an arc corresponding to a 90 degree sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is K times MR^2. Then the value of K is
(1) 1/8
(2) 7/8
(3) 3/4
(4) 1/4
Answer (2)
Sol. Given that,
Mass of Ring = M Radius of Ring = R
Now 90 degree arc is removed from circular ring, then mass removed = M/4
Mass of remaining portion = 3M/4
Moment of inertia of remaining part = integral dm r^2
=> I = R^2 integral dm (because r = R)
=> I = 3MR^2/4. So the value of K is 3/4
39. A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of,
(i) an equilateral triangle of side a (ii) a square of side a
The magnetic dipole moments of the coil in each case respectively are
(1) 4Ia^2 and 3Ia^2
(2) sqrt(3) I a^2 and 3 I a^2
(3) 3 I a^2 and I a^2
(4) 3 I a^2 and 4 I a^2
Answer (2)
Sol. Current in the loop will be V/R = I which is same for both loops.
Now magnetic moment of Triangle loop = N I A
M1 = (12a/3a) * I * (sqrt(3)/4) a^2 = sqrt(3) I a^2
and magnetic moment of square loop = N I A'
= (12a/4a) * I * a^2
M2 = 3 I a^2
40. Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1 >> R2, the mutual inductance M between them will be directly proportional to
(1) R2^2/R1
(2) R1/R2
(3) R2/R1
(4) R1^2/R2
Answer (1)
Sol. Two concentric coils are of radius R1 and R2 as shown
Let current in outer loop be i
Magnetic field at centre = B = mu0 i / (2 R1)
Magnetic flux through inner coil = B x pi R2^2
phi = (mu0 i / (2 R1)) * pi R2^2
phi = (mu0 i / 2) * (pi R2^2 / R1)
as per definition, phi = M i
=> M = (mu0 pi / 2) * (R2^2 / R1)
therefore M proportional to R2^2 / R1
41. A particle of mass 'm' is projected with a velocity v = k Ve (k < 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is
(1) R k^2 / (1 - k^2)
(2) R (k/(1 - k))^2
(3) R (k/(1 + k))^2
(4) R^2 k / (1 + k)
Answer (1)
Sol. given v = k Ve where, k < 1
Thus, v < Ve
From conservation of mechanical energy,
(1/2) m v^2 - GmM/R = -GmM/(R + h)
=> v^2/2 = GM/R - GM/(R + h) = h GM / (R(R + h))
=> (1/2) k^2 Ve^2 = G M h / (R(R + h))
We know, Ve = sqrt(2GM/R)
=> (1/2) k^2 (2GM/R) = G M h / (R(R + h))
k^2 = h/(R + h)
R k^2 + h k^2 = h
R k^2 = h(1 - k^2)
therefore h = R k^2 / (1 - k^2)
42. A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
(1) 4 A
(2) 0.2 A
(3) 0.4 A
(4) 2 A
Answer (2)
Sol. In ideal transformer.
Input power = Output power
=> Vp Ip = Vs Is = Given power
=> 220 x Ip = 44
=> Ip = 0.2 A
43. For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal y ?
Answer (3)
Sol. Output of combination of logic gates is given as y = A . B + B(bar) . C
Input Signals Output Signal
Time duration A B C A B A.B B(bar).C y = A.B + B(bar).C
0 - t1 0 0 1 0 1 0 1 1
t1 - t2 1 0 1 0 1 0 1 1
So the output y is high (1) that is v0 = 5 V
44. A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s^2) nearly
(1) 1.4 kg m/s
(2) 0 kg m/s
(3) 4.2 kg m/s
(4) 2.1 kg m/s
Answer (3)
Sol. Given that :
Mass of ball = 0.15 kg
Height from which ball is dropped = 10 m
Impulse, I = Change in linear momentum = Delta P = Pf - Pi
Velocity of ball at ground (v) = sqrt(2gh)
= sqrt(2 x 10 x 10) = 10 sqrt(2) m/s
I = 0.15 x 10 sqrt(2) (-j) - 0.15 x 10 sqrt(2) (j)
I = 2 x 0.15 x 10 sqrt(2) (-j) = 4.2 (-j)
=> magnitude of impulse = 4.2 kg m/s
45. A car starts from rest and accelerates at 5 m/s^2. At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s ?
(Take g = 10 m/s^2)
(1) 20 sqrt(2) m/s, 10 m/s^2
(2) 20 m/s, 5 m/s^2
(3) 20 m/s, 0
(4) 20 sqrt(2) m/s, 0
Answer (1)
Sol. Initial velocity of car = 0
Acceleration of car = 5 m/s^2
Velocity of car at t = 4 s ; v = u + at
=> v = 0 + 5 x 4 = 20 ms^-1
At t = 4 s, A ball is dropped out of a window so velocity of ball at this instant is 20 ms^-1 along horizontal.
After 2 seconds of motion :
Horizontal velocity of ball = 20 ms^-1 (therefore ax = 0)
Vertical velocity of ball (vy) = uy + ay t
vy = 0 + 10 x 2 = 20 ms^-1 (therefore ay = g = 10 ms^-2)
So magnitude of velocity of ball
(v) = sqrt(vx^2 + vy^2) = 20 sqrt(2) m/s
Acceleration of ball at t = 6 s is g = 10 m/s^2
As ball is under free fall.
46. A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of
(1) 20 cm from the plane mirror, it would be a virtual image
(2) 20 cm from the lens, it would be a real image
(3) 30 cm from the lens, it would be a real image
(4) 30 cm from the plane mirror, it would be a virtual image
Answer (1)
47. The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be
(1) 2Mg
(2) Mg/2
(3) Mg
(4) 3/2 Mg
Answer (2)
48. A particle of mass m is projected with a velocity v = k Ve (k < 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is
(1) R k^2 / (1 - k^2)
(2) R (k/(1 - k))^2
(3) R (k/(1 + k))^2
(4) R^2 k / (1 + k)
Answer (1)
49. Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio i3/i1 of currents in terms of resistances used in the circuit is
(1) r2/(r1 + r3)
(2) r1/(r2 + r3)
(3) r2/(r2 + r3)
(4) r1/(r1 + r2)
Answer (3)
Sol. In parallel combination of resistances r2 and r3 potential difference will be equal across both resistance.
So, i2 r2 = i3 r3 => i2 = i3 r3 / r2 ...(1)
As per Kirchhoff's first law
=> i1 = i2 + i3
=> i1 = (r3/r2 + 1) i3 (from equation 1)
=> i3/i1 = r2/(r2 + r3)
50. A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass m is suspended from the rod at 160 cm mark as shown in the figure. Find the value of m such that the rod is in equilibrium. (g = 10 m/s^2)
(1) 1/12 kg
(2) 1/2 kg
(3) 1/3 kg
(4) 1/6 kg
Answer (1)
Sol. Given that
Mass of rod = 500 g Length of rod = 200 cm
Rod will be in equilibrium, when net torque about point O will be zero.
Torque at point O due to 2 kg mass
tau = r x F = r F sin theta (n)
tau1 = 20 x 20 x 10^-2 x sin 90 degree (k) = 4 N m (k)
Torque due to mass of rod :
tau2 = 5 x 60 x 10^-2 x sin 90 degree (-k) = 3 N m (-k)
Torque due to mass m
tau3 = mg x 120 x 10^-2 x sin 90 degree (-k) = 12m N m (-k)
Net torque about point O will be zero
So tau1 + tau2 + tau3 = 0
=> 4 - 3 - 12m = 0
=> 12m = 1
m = 1/12 kg
SECTION-A
51. Dihedral angle of least stable conformer of ethane is :
(1) 0 degree
(2) 120 degree
(3) 180 degree
(4) 60 degree
Answer (1)
Sol. Ethane has two conformers (i) Eclipsed
(ii) Staggered
Eclipsed conformer is least stable while staggered conformer is most stable. In eclipsed conformer the dihedral angle is 0 degree
52. The structures of beryllium chloride in solid state and vapour phase, are :
(1) Chain in both
(2) Chain and dimer, respectively
(3) Linear in both
(4) Dimer and Linear, respectively
Answer (2)
Sol. Beryllium chloride has a chain structure in the solid state as shown below
In vapour phase Beryllium chloride tends to form a chloro-bridged dimer.
53. Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.
Answer (4)
Sol. Benzene sulphonyl chloride (C6H5SO2Cl) is also known as Hinsberg's reagent.
The reaction of Hinsberg's reagent (C6H5SO2Cl) with primary amine (CH3CH2NH2) yields N-ethylbenzene sulphonamide.
The reaction of Hinsberg's reagent (C6H5SO2Cl) with secondary amine (C2H5NHCH3) gives, N-Ethyl-N-Methyl benzene sulphonamide
3 degree amine do not react with Hinsberg reagent
54. The molar conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm^2 mol^-1 respectively. The molar conductance of CH3COOH at infinite dilution is. Choose the right option for your answer.
(1) 540.48 S cm^2 mol^-1
(2) 201.28 S cm^2 mol^-1
(3) 390.71 S cm^2 mol^-1
(4) 698.28 S cm^2 mol^-1
Answer (3)
Sol. Lambda_m^0(CH3COOH) = Lambda_m^0(CH3COONa) + Lambda_m^0(HCl) - Lambda_m^0(NaCl)
= 91.0 S cm^2 mol^-1 + 426.16 S cm^2 mol^-1 - 126.45 S cm^2 mol^-1
= 390.71 S cm^2 mol^-1
55. Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are :
(1) 12, 6
(2) 8, 4
(3) 6, 12
(4) 2, 1
Answer (1)
Sol. Number of octahedral and tetrahedral voids formed by N closed packed atoms are N and 2N respectively. Each hexagonal unit cell contains 6 atoms therefore, number of tetrahedral and octahedral voids are 12 and 6 respectively.
56. Tritium, a radioactive isotope of hydrogen, emits which of the following particles?
(1) Neutron
(2) Beta (beta^-)
(3) Alpha (alpha)
(4) Gamma (gamma)
Answer (2)
Sol. Hydrogen has three isotopes : protium, 1H deuterium, 2H or D and tritium 3H or T. Of these isotopes, only tritium is radioactive and emits low energy beta^- particles (t1/2, 12.33 years).
57. An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1]
(1) CH4
(2) CH
(3) CH2
(4) CH3
Answer (4)
Sol. Element Mass percentage No. of mole Mole ratio
Based on above calculation, possible empirical formula is CH3
58. What is the IUPAC name of the organic compound formed in the following chemical reaction?
Acetone (i) C2H5MgBr dry Ether Product (ii) H2O H+
(1) 2-methylbutan-2-ol
(2) 2-methylpropan-2-ol
(3) pentan-2-ol
(4) pentan-3-ol
Answer (1)
Sol. Product: 2-methylbutan-2-ol
59. Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperatures :
Answer (1)
Sol. According to Boyle's law
P proportional to 1/V => P = k/V => PV = k
where k is proportionality constant and equal to nRT.
Graph between P vs. V should be rectangular hyperbola and product of PV increases with increase in temperature.
60. The correct structure of 2, 6-Dimethyl-dec-4-ene is
Answer (2)
Sol. 2,6-Dimethyldec-4-ene
61. Match List-1 with List-1I
List-1 List-1I
(a) PCl5 (i) Square pyramidal
(b) SF6 (ii) Trigonal planar
(c) BrF5 (iii) Octahedral
(d) BF3 (iv) Trigonal bipyramidal
Choose the correct answer from the options given below.
(1) (a)-iv), (b)-iii), (c)-ii), (d)-i)
(2) (a)-iv), (b)-iii), (c)-i), (d)-ii)
(3) (a)-ii), (b)-iii), (c)-iv), (d)-i)
(4) (a)-iii), (b)-i), (c)-iv), (d)-ii)
Answer (2)
Sol.
(a) PCl5: sp^3d hybridised and trigonal bipyramidal in shape
(b) SF6: sp^3d^2 hybridised and octahedral in shape
(c) BrF5: sp^3d^2 hybridised and square pyramidal in shape
(d) BF3: sp^2 hybridised and trigonal planar in shape
62. The maximum temperature that can be achieved in blast furnace is :
(1) Upto 5000 K
(2) Upto 1200 K
(3) Upto 2200 K
(4) Upto 1900 K
Answer (3)
Sol. Maximum temperature that can be achieved in blast furnace is upto 2200 K.
(As per NCERT text: 2170 K maximum temperature is given in the figure of blast furnace)
63. Which one among the following is the correct option for right relationship between Cp and Cv for one mole of ideal gas?
(1) Cv = R Cp
(2) Cp + Cv = R
(3) Cp - Cv = R
(4) Cp = R Cv
Answer (3)
Sol. At constant volume, qV = Cv Delta T = Delta U
At constant pressure, qP = Cp Delta T = Delta H
For a mole of an ideal gas,
Delta H = Delta U + Delta(PV)
= Delta U + Delta(RT)
= Delta U + R Delta T
On putting the values of Delta H and Delta U, we have
Cp Delta T = Cv Delta T + R Delta T
Cp = Cv + R
Cp - Cv = R
64. Statement I : Acid strength increases in the order given as HF << HCl << HBr << HI.
Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases.
In the light of the above statements, choose the correct answer from the options given below.
(1) Statement I is incorrect but Statement II is true
(2) Both statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is correct but statement II is false
Answer (2)
Sol. In the modern periodic table, moving down the group as the size of halogen atom increases, the H-X bond length also increases as a result the bond enthalpy decreases. Hence, The acidic strength also increases.
So, the correct order of acidic strength is
HI > HBr > HCl > HF
65. The right option for the statement "Tyndall effect is exhibited by", is :
(1) Urea solution
(2) NaCl solution
(3) Glucose solution
(4) Starch solution
Answer (4)
Sol. Tyndall effect is exhibited by colloidal solution only.
Among the given options, Urea, NaCl and Glucose solutions are true solutions, so cannot show Tyndall effect. Starch solution is a colloidal solution therefore can show Tyndall effect.
66. The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is :
(1) 3
(2) 7
(3) 5
(4) 2
Answer (1)
Sol. In 14 types of Bravais lattices, body centred unit cell is present in cubic, tetragonal and orthorhombic crystal systems. Hence, body centred possible variation is present in three crystal systems.
67. Which of the following reactions is the metal displacement reaction? Choose the right option.
(1) 2Pb(NO3)2 -> 2PbO + 4NO2 + O2
(2) 2KClO3 -> Delta -> 2KCl + 3O2
(3) Cr2O3 + 2Al -> Delta -> Al2O3 + 2Cr
(4) Fe + 2HCl -> FeCl2 + H2
Answer (3)
Sol. Both reactions (1) and (2) are examples of decomposition reactions. Reactions (3) and (4), both are examples of displacement reactions, while reaction (3) is an example of metal displacement reaction.
68. For a reaction A -> B enthalpy of reaction is -4.2 kJ mol^-1 and enthalpy of activation is 9.6 kJ mol^-1. The correct potential energy profile for the reaction is shown in option.
Answer (3)
Sol. Delta Hrxn = (Ea)f - (Ea)b
-4.2 = (Ea)f - (Ea)b
-4.2 = 9.6 - (Ea)b
(Ea)b = 9.6 + 4.2 = 13.8 kJ mol^-1
Since reaction is exothermic, so possible graph is (3) only.
Also (Ea)f < (Ea)b so answer is option (3).
69. The pKb of dimethylamine and pKa of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate solution is :
(1) 6.25
(2) 8.50
(3) 5.50
(4) 7.75
Answer (4)
Sol. Dimethylammonium acetate is a salt of weak acid and weak base whose pH can be calculated as
pH = 7 + (1/2)(pKa - pKb)
= 7 + (1/2)(4.77 - 3.27)
= 7.75
70. The major product formed in dehydrohalogenation reaction of 2-Bromo pentane is Pent-2-ene. This product formation is based on?
(1) Huckel's Rule
(2) Saytzeff's Rule
(3) Hund's Rule
(4) Hofmann Rule
Answer (2)
Sol. Major product formed in dehydrohalogenation reaction of 2-bromopentane is pent-2-ene because according to Saytzeff's rule, in dehydrohalogenation reactions, the preferred product is that alkene which has greater number of alkyl group(s) attached to the doubly bonded carbon atoms.
CH3-CH2-CH2-CH(Br)-CH3 -> CH3-CH2-CH=CH-CH3 (Pent-2-ene, 81%) + CH3-CH2-CH2-CH=CH2 (Pent-1-ene, 19%)
71. Which one of the following polymers is prepared by addition polymerisation?
(1) Dacron
(2) Teflon
(3) Nylon-66
(4) Novolac
Answer (2)
Sol. Dacron, Nylon-66 and Novolac are prepared by condensation polymerisation.
Teflon is an addition polymer. Monomer of teflon is tetrafluoroethene.
n CF2=CF2 --Catalyst--> -(CF2-CF2)n-
Tetrafluoroethene
72. The RBC deficiency is deficiency disease of :
(1) Vitamin B2
(2) Vitamin B12
(3) Vitamin B6
(4) Vitamin B1
Answer (2)
Sol. Deficiency of vitamin B2 (Riboflavin) causes cheilosis, digestive disorders and burning sensation of the skin. Deficiency of vitamin B12 causes Pernicious anaemia which is RBC deficiency in haemoglobin. Deficiency of vitamin B6 (Pyridoxine) causes Convulsions. Deficiency of vitamin B1 (Thiamine) causes Beri-Beri (loss of appetite and retarded growth).
73. The following solutions were prepared by dissolving 10 g of glucose (C6H12O6) in 250 ml of water (P1), 10 g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose (C12H22O11) in 250 ml of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is :
(1) P3 > P1 > P2
(2) P2 > P1 > P3
(3) P1 > P2 > P3
(4) P2 > P3 > P1
Answer (2)
Sol. Osmotic pressure (pi) = iCRT where C is molar concentration of the solution With increase in molar concentration of solution osmotic pressure increases. Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure. Order of molar mass of solute decreases as Sucrose > Glucose > Urea So, correct order of osmotic pressure of solution is P3 < P1 < P2
74. A particular station of All India Radio, New Delhi broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light c = 3.0 x 10^8 ms^-1]
(1) 21.92 cm
(2) 219.3 m
(3) 219.2 m
Answer (2)
Sol. Energy of electromagnetic radiation (E)
= hc/lambda = h gamma
So, c/lambda = gamma => lambda = c/gamma
lambda = (3 x 10^8) / (1368 x 10^3) = 219.3 m
75. Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.
(1) Noble gases have large positive values of electron gain enthalpy
(2) Noble gases are sparingly soluble in water
(3) Noble gases have very high melting and boiling points
(4) Noble gases have weak dispersion forces
Answer (3)
Sol. Noble gases have weak dispersion forces hence they have low melting and boiling points.
76. Given below are two statements :
Statement I :
Aspirin and Paracetamol belong to the class of narcotic analgesics.
Statement II :
Morphine and Heroin are non-narcotic analgesics. In the light of the above statements, choose the correct answer from the options given below.
(1) Statement I is incorrect but Statement II is true.
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is correct but Statement II is false
Answer (3)
Sol. Aspirin and paracetamol belong to the class of non-narcotic analgesics Morphine and Heroin are Narcotic analgesics Both statement I and statement II are false
77. Which one of the following methods can be used to obtain highly pure metal which is liquid at room temperature?
(1) Zone refining
(2) Electrolysis
(3) Chromatography
(4) Distillation
Answer (4)
Sol. Distillation method is generally used for the purification of metals having low boiling point such as Hg, Zn etc.
78. The correct sequence of bond enthalpy of C-X bond is :
(1) CH3-Cl > CH3-F > CH3-Br > CH3-I
(2) CH3-F < CH3-Cl < CH3-Br < CH3-I
(3) CH3-F > CH3-Cl > CH3-Br > CH3-I
(4) CH3-F < CH3-Cl > CH3-Br > CH3-I
Answer (3)
Sol. The size of halogen atom increases from F to I hence bond length from C-F to C-I increases : Bond enthalpy from CH3-F to CH3-I decreases
C-X Bond Bond dissociation enthalpies/kJ mol^-1
CH3-F 452
CH3-Cl 351
CH3-Br 293
CH3-I 234
79. The compound which shows metamerism is :
(1) C4H10O
(2) C5H12
(3) C3H6O
(4) C3H6O
Answer (1)
Sol. Compounds with formula C4H10O can be ethers which may exhibit metamerism. For example
CH3-CH2-O-CH2-CH3, CH3-O-CH(CH3)-CH3
and CH3-O-CH2-CH2-CH3 are metamers as structure of alkyl chains are different around the functional group.
80. Ethylene diaminetetraacetate (EDTA) ion is :
(1) Tridentate ligand with three "N" donor atoms
(2) Hexadentate ligand with four "O" and two "N" donor atoms
(3) Unidentate ligand
(4) Bidentate ligand with two "N" donor atoms
Answer (2)
81. Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of :
(1) Having similar chemical properties
(2) Belonging to same group
(3) Diagonal relationship
(4) Lanthanoid contraction
Answer (4)
Sol. The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr (160 pm) and Hf (159 pm) is a consequence of the lanthanoid contraction.
82. BF3 is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :
(1) sp^2 and 8
(2) sp^3 and 4
(3) sp^3 and 6
(4) sp^2 and 6
Answer (4)
Sol. F B F
Number of electrons around boron atom is 6. Hybridization of B is sp^2 Shape is trigonal planar.
83. The major product of the following chemical reaction is :
Answer (2)
Sol. Mechanism : Peroxide effect proceeds via free radical chain mechanism.
(i) C6H5-C(=O)-O-O-C(=O)-C6H5 --Homolysis--> 2 C6H5-C(=O)-O* -> 2 C6H5* + CO2
(ii) C6H5* + H-Br --Homolysis--> C6H6 + Br*
(iii) CH3-CH(CH3)-CH=CH2 + Br* -> CH3-CH(CH3)-CH*-CH2-Br (More Stable secondary free radical)
(iv) CH3-CH(CH3)-CH*-CH2-Br + H-Br --Homolysis--> CH3-CH(CH3)-CH2-CH2-Br (Major product)
84. The incorrect statement among the following is :
(1) Actinoids are highly reactive metals, especially when finely divided.
(2) Actinoid contraction is greater for element to element than lanthanoid contraction
(3) Most of the trivalent Lanthanoid ions are colorless in the solid state
(4) Lanthanoids are good conductors of heat and electricity
Answer (3)
Sol. Actinoids are highly reactive metals, especially when finely divided Actinoid contraction is greater from element to element than lanthanoid contraction resulting from poor shielding by 5f electrons Many trivalent lanthanoids ions are coloured both in the solid state and in aqueous solutions. Lanthanoids have typical metallic structure and are good conductors of heat and electricity
85. Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is :
(1) Beryllium chloride
(2) Calcium chloride
(3) Strontium chloride
(4) Magnesium chloride
Answer (1)
Sol. Except for beryllium chloride all other chloride of alkaline earth metals are ionic in nature. Due to small size of Be, Beryllium chloride is essentially covalent and soluble in organic solvents.
SECTION-B
86. The correct option for the value of vapour pressure of a solution at 45 degree C with benzene to octane in molar ratio 3 : 2 is :
[At 45 degree C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
(1) 350 mm of Hg
(2) 160 mm of Hg
(3) 168 mm of Hg
(4) 336 mm of Hg
Answer (4)
Sol. Given: n_C6H6 : n_C8H18 = 3 : 2
So, chi_C6H6 = 3/5, chi_C8H18 = 2/5
ps = p_C6H6^0 chi_C6H6 + p_C8H18^0 chi_C8H18
= 280 x 3/5 + 420 x 2/5
= 168 + 168
= 336 mm of Hg
87. For irreversible expansion of an ideal gas under isothermal condition, the correct option is:
(1) Delta U != 0, Delta S_total = 0
(2) Delta U = 0, Delta S_total = 0
(3) Delta U != 0, Delta S_total != 0
(4) Delta U = 0, Delta S_total != 0
Answer (4)
Sol. For a spontaneous process, Delta S_total > 0 and since irreversible process is always spontaneous therefore Delta S_total > 0. Since Delta U = n Cv Delta T and Delta T = 0 for isothermal process therefore Delta U = 0.
88. The intermediate compound 'X' in the following chemical reaction is:
Answer (2)
Sol. Etard's reaction
Toluene + CrO2Cl2 --CS2--> X --H3O+--> Benzaldehyde
X is benzal chloride? Actually intermediate is complex.
89. Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O2 and 2 g H2 confined in a total volume of one litre at 0 degree C is :
[Given R = 0.082 L atm mol^-1 K^-1 T = 273 K]
(1) 26.02
(2) 2.518
(3) 2.602
(4) 25.18
Answer (4)
Sol. n_O2 = 4/32 = 1/8
n_H2 = 2/2 = 1
n_t = 1/8 + 1 = 9/8
P_total = n_total RT/V = (9/8) * 0.082 * 273 / 1 = 25.18 atm
90. From the following pairs of ions which one is not an iso-electronic pair?
(1) Fe^2+, Mn^2+
(2) O^2-, F^-
(3) Na^+, Mg^2+
(4) Mn^2+, Fe^3+
Answer (1)
Sol. Isoelectronic species have same number of electrons.
Species Number of electrons
Fe^2+ 26 - 2 = 24
Mn^2+ 25 - 2 = 23
O^2- 8 + 2 = 10
F^- 9 + 1 = 10
Na^+ 11 - 1 = 10
Mg^2+ 12 - 2 = 10
Fe^3+ 26 - 3 = 23
91. CH3CH2COO^-Na^+ --NaOH, +?--> CH3CH3 + Na2CO3
Consider the above reaction and identify the missing reagent/chemical.
(1) DIBAL-H
(2) B2H6
(3) Red Phosphorus
(4) CaO
Answer (4)
Sol. Alkane is produced by heating sodium salt of carboxylic acid with sodalime (NaOH and CaO in the ratio of 3 : 1)
CH3CH2COO^-Na^+ --NaOH + CaO--> CH3CH3 + Na2CO3
92. Which of the following molecules is non-polar in nature?
(1) NO2
(2) POCl3
(3) CH2O
(4) SbCl5
Answer (4)
Sol. SbCl5 : Net vector summation of bond moments will be zero so SbCl5 is a non-polar molecule.
NO2 : polar molecule
POCl3 : polar molecule
CH2O : polar molecule
93. The slope of Arrhenius plot (ln k v/s 1/T) of first order reaction is -5 x 10^3 K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R = 8.314 J K^-1 mol^-1]
(1) -83 kJ mol^-1
(2) 41.5 kJ mol^-1
(3) 83.0 kJ mol^-1
(4) 166 kJ mol^-1
Answer (2)
Sol. Arrhenius equation
k = A e^(-Ea/RT)
ln k = ln A - Ea/R (1/T) ...(1)
Slope of ln k vs 1/T curve,
m = -Ea/R
-5 x 10^3 = -Ea/R
Ea = 5 x 10^3 x 8.314 J/mol
= 41.57 x 10^3 J/mol
= 41.5 kJ/mol
94. Match List-1 with List-1.
List-1 List-1
(a) 2SO2(g) + O2(g) -> 2SO3(g) (i) Acid rain
(b) HOCl(g) --hv--> OH + Cl (ii) Smog
(c) CaCO3 + H2SO4 -> CaSO4 + H2O + CO2 (iii) Ozone depletion
(d) NO2(g) --hv--> NO(g) + O(g) (iv) Tropospheric pollution
Choose the correct answer from the options given below.
(1) (a)- (iii), (b)- (ii), (c)- (iv), (d)- (i)
(2) (a)- (i), (b)- (ii), (c)- (iii), (d)- (iv)
(3) (a)- (ii), (b)- (iii), (c)- (iv), (d)- (i)
(4) (a)- (iv), (b)- (iii), (c)- (i), (d)- (ii)
Answer (4)
Sol. Tropospheric pollution: In the presence of pollutant, SO2 converts into SO3.
2SO2 + O2 -> 2SO3
In spring season, sunlight breaks HOCl and Cl2 to give chlorine radicals.
HOCl --hv--> OH(g) + Cl(g)
These chlorine radicals deplete ozone layer
High level of sulphur causes acid rain which reacts with marble and causes discolouring and disfiguring
CaCO3 + H2SO4 -> CaSO4 + H2O + CO2
A chain reaction occurs from interaction of NO with sunlight in which NO is converted to NO2 which absorb energy from sunlight and breaks into NO and O, which causes photochemical smog.
NO2(g) --hv--> NO(g) + O(g)
95. Match List-1 with List-1
List-1 List-1
(a) Benzene + CO, HCl Anhyd. AlCl3/CuCl (i) Hell-Volhard-Zelinsky reaction
(b) R-CO-CH3 + NaOX (ii) Gattermann-Koch reaction
(c) R-CH2-OH + R'COOH Conc. H2SO4 (iii) Haloform reaction
(d) R-CH2COOH (i) X2/Red P (ii) H2O (iv) Esterification
Choose the correct answer from the options given below.
(1) (a)- (ii), (b)- (iii), (c)- (iv), (d)- (i)
(2) (a)- (iv), (b)- (i), (c)- (ii), (d)- (iii)
(3) (a)- (iii), (b)- (ii), (c)- (i), (d)- (iv)
(4) (a)- (i), (b)- (iv), (c)- (iii), (d)- (ii)
Answer (1)
Sol. Gattermann-Koch reaction: Benzene + CO, HCl Anhyd. AlCl3/CuCl -> Benzaldehyde
Haloform reaction: R-CO-CH3 + NaOX -> R-COONa + CHX3
Esterification: R-CH2-OH + R'COOH Conc. H2SO4 -> R'COOCH2R
Hell-Volhard-Zelinsky reaction: R-CH2COOH (i) X2/Red P (ii) H2O -> R-CH(X)-COOH
96. Match List-1 with List-1.
List-1 List-1
[Fe(CN)6]^3- (i) 5.92 BM
[Fe(H2O)6]^3+ (ii) 0 BM
[Fe(CN)6]^4- (iii) 4.90 BM
[Fe(H2O)6]^2+ (iv) 1.73 BM
Choose the correct answer from the options given below.
(1) (a)- (iv), (b)- (i), (c)- (ii), (d)- (iii)
(2) (a)- (iv), (b)- (ii), (c)- (i), (d)- (iii)
(3) (a)- (ii), (b)- (iv), (c)- (iii), (d)- (i)
(4) (a)- (i), (b)- (iii), (c)- (iv), (d)- (ii)
Answer (1)
Sol. Magnetic moment, mu = sqrt(n(n + 2)) BM (where n = number of unpaired electrons)
Complex No. of unpaired electron(s) mu (BM)
(a) [Fe(CN)6]^3- 1 1.73
(b) [Fe(H2O)6]^3+ 5 5.92
(c) [Fe(CN)6]^4- 0 0
(d) [Fe(H2O)6]^2+ 4 4.90
97. The product formed in the following chemical reaction is:
Answer (1)
Sol. NaBH4 is a reducing agent. It reduces carbonyl group into alcohols but does not reduce esters.
Reaction: cyclohexanone with ester side chain -> NaBH4/C2H5OH -> alcohol (carbonyl reduced) ester remains.
98. In which one of the following arrangements the given sequence is not strictly according to the properties indicated against it?
(1) CO2 < SiO2 < SnO2 < PbO2 : Increasing oxidizing power
(2) HF < HCl < HBr < HI : Increasing acidic strength
(3) H2O < H2S < H2Se < H2Te : Increasing pKa values
(4) NH3 < PH3 < AsH3 < SbH3 : Increasing acidic character
Answer (3)
Sol. Stronger is the acid, lower is the value of pKa. On moving down the group, bond dissociation enthalpy of hydrides of group 16 elements decreases hence acidity increases and pKa value decreases. Correct order of pKa value will be
H2O > H2S > H2Se > H2Te
99. The molar conductivity of 0.007 M acetic acid is 20 S cm^2 mol^-1. What is the dissociation constant of acetic acid? Choose the correct option.
[Lambda_H+^0 = 350 S cm^2 mol^-1]
[Lambda_CH3COO-^0 = 50 S cm^2 mol^-1]
(1) 2.50 x 10^-5 mol L^-1
(2) 1.75 x 10^-4 mol L^-1
(3) 2.50 x 10^-4 mol L^-1
(4) 1.75 x 10^-5 mol L^-1
Answer (4)
Sol. Lambda_m^0 = 20 S cm^2 mol^-1
Lambda_m^0 CH3COOH = Lambda_m^0 CH3COO^- + Lambda_m^0 H+
= 50 + 350 = 400 S cm^2 mol^-1
alpha = Lambda_m / Lambda_m^0 = 20/400 = 1/20
Ka = C alpha^2 / (1 - alpha) = C alpha^2 = 7 x 10^-3 x (1/20)^2
= 7 x 10^-3 x 1/4 x 10^-2
= 1.75 x 10^-5 mol L^-1
100. The reagent R in the given sequence of chemical reaction is:
(1) CuCN/KCN
(2) H2O
(3) CH3CH2OH
(4) HI
Answer (3)
Sol. Reagent R is C2H5OH with diazonium salt.
101. Match List-1 with List-1.
List - I List - II
(a) Cells with active cell division capacity (i) Vascular tissues
(b) Tissue having all cells similar in structure and function (ii) Meristematic tissue
(c) Tissue having different types of cells (iii) Sclereids
(d) Dead cells with highly thickened walls and narrow lumen (iv) Simple tissue
Select the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iii) (ii) (iv) (i)
(2) (ii) (iv) (i) (iii)
(3) (iv) (iii) (ii) (i)
(4) (i) (ii) (iii) (iv)
Answer (2)
Sol. (a) Meristematic tissues are those tissues which have cells with active cell division capacity.
(b) Simple tissues are those tissues which have all the cells similar in structure and function.
(c) Vascular tissues are complex permanent tissues hence they have different types of cells.
(d) Sclereids are sclerenchymatous cells which are dead with highly thickened walls and narrow lumen.
102. Which of the following is an incorrect statement?
(1) Nuclear pores act as passages for proteins and RNA molecules in both directions between nucleus and cytoplasm
(2) Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles
(3) Microbodies are present both in plant and animal cells
(4) The perinuclear space forms a barrier between the materials present inside the nucleus and that of the cytoplasm
Answer (2)
Sol. A mature sieve tube elements possess a peripheral cytoplasm and a large central vacuole but lacks a nucleus.
Rest of other statements are correct.
103. When gene targeting involving gene amplification is attempted in an individual's tissue to treat disease, it is known as :
(1) Safety testing
(2) Biopiracy
(3) Gene therapy
(4) Molecular diagnosis
Answer (3)
Sol. The correct option is (3)
Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child/embryo. Biopiracy is the term used to refer to the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment. Molecular diagnosis refers to the act or process of determining the nature and cause of a disease.
104. During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out :
(1) Polysaccharides
(2) RNA
(3) DNA
(4) Histones
Answer (3)
Sol. Various enzymes like protease, RNase, etc. are added to break down substances like proteins, RNA, etc. Once all these substances are broken down, DNA is left which is precipitated out by adding chilled ethanol.
Histones are basic proteins that help condense DNA in a cell.
105. Match List-I with List-II
List-I List-II
(a) Protoplast fusion (i) Totipotency
(b) Plant tissue culture (ii) Pomato
(c) Meristem culture (iii) Somaclones
(d) Micropropagation (iv) Virus free plants
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (iii) (iv) (ii) (i)
(3) (ii) (i) (iv) (iii)
(4) (iii) (iv) (i) (ii)
Answer (3)
Sol. Pomato is obtained as a result of protoplast fusion. Totipotency is a property of explant to develop into whole plant body during plant tissue culture. Virus free plants can be obtained through meristem culture. Somaclones are obtained by the process of micropropagation.
106. When the centromere is situated in the middle of two equal arms of chromosomes, the chromosome is referred as :
(1) Acrocentric
(2) Metacentric
(3) Telocentric
(4) Sub-metacentric
Answer (2)
Sol. When the centromere is situated in the middle of two equal arms of chromosomes, the chromosome is referred as Metacentric.
When the centromere is present slightly away from the middle, it is called sub-metacentric chromosome.
When the centromere is present very close to one end of the chromosome, it is called acrocentric chromosome.
When the centromere is present at terminal position, the chromosome is called telocentric.
107. The factor that leads to Founder effect in a population is :
(1) Genetic drift
(2) Natural selection
(3) Genetic recombination
(4) Mutation
Answer (1)
Sol. Change in gene frequency in a small population by chance is known as genetic drift. Genetic drift has two ramifications, one is bottle neck effect and another is founder's effect.
When accidentally a few individuals are dispersed and act as founders of a new isolated population, founder's effect is said to be observed. Crossing over which occurs during gamete formation results in genetic recombination. Mutations are random and directionless.
108. Which of the following is a correct sequence of steps in a PCR (Polymerase Chain Reaction)?
(1) Annealing, Denaturation, Extension
(2) Denaturation, Annealing, Extension
(3) Denaturation, Extension, Annealing
(4) Extension, Denaturation, Annealing
Answer (2)
Sol. The first step in the polymerase chain reaction is denaturation during which strands of dsDNA separate. This requires temperature around 94 degree C.
This is followed by annealing in which primers anneal to 3' end of template DNA strand.
Annealing is followed by extension in which Taq polymerase adds nucleotides to 3'OH end of primers.
109. Inspite of interspecific competition in nature, which mechanism the competing species might have evolved for their survival?
(1) Predation
(2) Resource partitioning
(3) Competitive release
(4) Mutualism
Answer (2)
Sol. Inspite of interspecific competition the competing species may co-exist by doing resource partitioning.
In mutualism two organisms are equally benefitted. In predation one organism (Predator) eats the another one (Prey). In competition release there occurs dramatical increase in population of a less distributed species when its superior competitor is removed.
110. A typical angiosperm embryo sac at maturity is:
(1) 8-nucleate and 8-celled
(2) 8-nucleate and 7-celled
(3) 7-nucleate and 8-celled
(4) 7-nucleate and 7-celled
Answer (2)
Sol. A typical angiospermic embryo sac has seven cells that are three antipodals, one central cell, one egg cell and two synergids.
The central cell has two polar nuclei, hence the embryo sac is eight nucleated.
111. The site of perception of light in plants during photoperiodism is
(1) Leaf
(2) Shoot apex
(3) Stem
(4) Axillary bud
Answer (1)
Sol. The site of perception of light in plants during photoperiodism is leaf. The site of perception of low temperature stimulus during vernalisation is shoot apex and embryo. Axillary bud are not sites of perception of photoperiod.
112. Which of the following plants is monoecious?
(1) Cycas cirinalis
(2) Carica papaya
(3) Chara
(4) Marchantia polymorpha
Answer (3)
Sol. When male and female sex organs are present on same plant body, such plants are said to be monoecious. Most of the species of Chara are monoecious. Cycas cirinalis, Carica papaya and Marchantia polymorpha are dioecious.
113. Match List-1 with List-1
List-1 List-1I
(a) Cristae (i) Primary constriction in chromosome
(b) Thylakoids (ii) Disc-shaped sacs in Golgi apparatus
(c) Centromere (iii) Infoldings in mitochondria
(d) Cisternae (iv) Flattened membranous sacs in stroma of plastids
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (iii) (iv) (i)
(2) (iv) (iii) (ii) (i)
(3) (i) (iv) (iii) (ii)
(4) (iii) (iv) (i) (ii)
Answer (4)
Sol. The inner membrane of mitochondria forms infoldings called cristae.
Thylakoids are flattened membranous sacs in stroma of plastids. Cisternae are disc shaped sacs in Golgi apparatus. Primary constriction in chromosome that holds two chromatids together is called centromere.
Hence correct option is (4)- a(iii), b(iv), c(i), d(ii)
114. Match List-1 with List-1l.
List-1 List-1l
(a) Cohesion (i) More attraction in liquid phase
(b) Adhesion (ii) Mutual attraction among water molecules
(c) Surface tension (iii) Water loss in liquid phase
(d) Guttation (iv) Attraction towards polar surfaces
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (i) (iv) (iii)
(2) (ii) (iv) (i) (iii)
(3) (iv) (iii) (ii) (i)
(4) (iii) (i) (iv) (ii)
Answer (2)
Sol. (a) Cohesion is mutual attraction among water molecules.
(b) Adhesion is attraction towards polar surfaces.
(c) Surface tension explains water molecules are more attracted in liquid phase than gaseous phase.
(d) Guttation is loss of water is liquid form from the leaf margins.
115. Match List-1 with List-1l.
List-1 List-1l
(a) Lenticels (i) Phellogen
(b) Cork cambium (ii) Suberin deposition
(c) Secondary cortex (iii) Exchange of gases
(d) Cork (iv) Phelloderm
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (iv) (i) (iii) (ii)
(3) (iii) (i) (iv) (ii)
(4) (ii) (iii) (iv) (i)
Answer (3)
Sol. Lenticels are meant for exchange of gases.
Phellogen is also known as cork cambium. Phelloderm is also called secondary cortex because it is the cortex that develops during secondary growth. Cork has deposition of suberin in their cell walls when they get mature.
116. The term used for transfer of pollen grains from anthers of one plant to stigma of a different plant which, during pollination, brings genetically different types of pollen grains to stigma, is :
(1) Cleistogamy
(2) Xenogamy
(3) Geitonogamy
(4) Chasmogamy
Answer (2)
Sol. Xenogamy refers to the transfer to pollen grains from anthers of one plant to stigma of a different plant which during pollination, brings genetically different types of pollen grains to stigma. Cleistogamy is a condition is which flower does not open. Geitonogamy refers to the transfer of pollen grain from anther to stigma of another flower of the same plant. Chasmogamy is a condition in which flowers remain open.
117. Which of the following is not an application of PCR (Polymerase Chain Reaction)?
(1) Detection of gene mutation
(2) Molecular diagnosis
(3) Gene amplification
(4) Purification of isolated protein
Answer (4)
Sol. PCR is Polymerase Chain Reaction. It is used for making multiple copies of the gene. Hence PCR is used for Gene amplification. PCR-based assays have been developed that detect the presence of gene sequences of the infectious agents. It is also used in detecting mutations. Protein is not the target of PCR. Hence, plays no role in its purification.
118. The production of gametes by the parents, formation of zygotes, the F1 and F2 plants, can be understood from a diagram called :
(1) Net square
(2) Bullet square
(3) Punch square
(4) Punnett square
Answer (4)
Sol. The production of gametes (n) by the parents (2n), the formation of the zygote (2n), the F1 and F2 plants can be understood from a diagram called Punnett square.
119. Which of the following algae produce Carrageen?
(1) Blue-green algae
(2) Green algae
(3) Brown algae
(4) Red algae
Answer (4)
Sol. The cell wall of red algae is composed of agar, carrageen and funori along with cellulose. In brown algae cell wall contains align while in green algae it is composed of cellulose and pectin. In blue green algae cell wall is composed of mucopeptides.
120. Amensalism can be represented as:
(1) Species A (+) Species B 0
(2) Species A (-) Species B 0
(3) Species A (+) Species B (+)
(4) Species A (-) Species B (-)
Answer (2)
Sol. Amensalism is an interaction between two organisms of different species in which one species inhibits the growth of other species by secreting certain chemicals. The first species is neither get benefited nor harmed.
(+):(0) interaction is observed in commensalism (+):(+) interaction is observed in mutualism. (-):(-) interaction is seen in competition
121. Which of the following stages of meiosis involves division of centromere?
(1) Telophase II
(2) Metaphase I
(3) Metaphase II
(4) Anaphase II
Answer (4)
Sol. Division of centromere occurs in anaphase II.
Telophase II is the last stage of meiosis II. During this phase, the chromatids reach the poles and start uncoiling. Chromosomes form two parallel plates in metaphase I and one plate in metaphase II.
122. Complete the flow chart on central dogma.
(a) DNA --(b)--> mRNA --(c)--> (d)
(1) (a)-Transduction; (b)-Translation; (c)-Replication; (d)-Protein
(2) (a)-Replication; (b)-Transcription; (c)-Transduction; (d)-Protein
(3) (a)-Translation; (b)-Replication; (c)-Transcription; (d)-Transduction
(4) (a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein
Answer (4)
Sol. Formation of DNA from DNA is replication.
Formation of mRNA from DNA is called Transcription.
Formation of protein from mRNA is called Translation.
So, (a) is Replication
(b) is Transcription
(c) is Translation
(d) is Protein
Transduction is transfer of genetic material from one bacterium to another with the help of virus or a bacteriophage.
123. Mutations in plant cells can be induced by:
(1) Zeatin
(2) Kinetin
(3) Infrared rays
(4) Gamma rays
Answer (4)
Sol. Several kinds of radiation like gamma rays, X-rays, UV-rays cause mutation. These are physical mutagens. Such induced mutation in plants is done to develop improved varieties. The first natural cytokinin was isolated from unripe maize grain known as zeatin. The cytokinin that was obtained from degraded product of autoclaved herring sperm DNA was kinetin (N6-furfuryl aminopurine). Infrared rays cause heating effect.
124. In the equation GPP - R = NPP
R represents :
(1) Respiration losses
(2) Radiant energy
(3) Retardation factor
(4) Environmental factor
Answer (1)
Sol. In the equation, GPP - R = NPP R refers to respiratory loss GPP is gross primary productivity NPP is net primary productivity
125. Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called
(1) Maturity
(2) Elasticity
(3) Flexibility
(4) Plasticity
Answer (4)
Sol. Plants show plasticity which means the ability of plant to follow different pathways and produce different structures in response to environment.
126. The plant hormone used to destroy weeds in a field
(1) IBA
(2) IAA
(3) NAA
(4) 2,4-D
Answer (4)
Sol. Some synthetic auxins are used as weedicides. 2,4-D is widely used to remove broad leaved weeds or dicotyledonous weeds in cereal crops or monocotyledonous plants.
IAA and IBA are natural auxins.
NAA is a synthetic auxin.
127. Which of the following statements is not correct?
(1) Pyramid of numbers in a grassland ecosystem is upright.
(2) Pyramid of biomass in sea is generally inverted.
(3) Pyramid of biomass in sea is generally upright.
(4) Pyramid of energy is always upright.
Answer (3)
Sol. Pyramid of biomass in sea is inverted. For example, biomass of zooplankton is higher than that of phytoplankton as life span of former is longer and the latter multiply much faster though having shorter life span.
Small standing crop of phytoplankton supports large standing crop of zooplankton
128. Gemmae are present in
(1) Some Liverworts
(2) Mosses
(3) Pteridophytes
(4) Some Gymnosperms
Answer (1)
Sol. Gemmae are green, multicellular asexual buds that are produced by some liverworts like Marchantia. Mosses reproduce vegetatively by fragmentation and budding of protonema. Pteridophytes and Gymnosperms normally do not reproduce asexually
129. The first stable product of CO2 fixation in Sorghum is
(1) Phosphoglyceric acid
(2) Pyruvic acid
(3) Oxaloacetic acid
(4) Succinic acid
Answer (3)
Sol. Sorghum is a C4 plant. The first stable product of CO2 fixation in Sorghum is oxaloacetic acid. The first stable product in C3 cycle is 3-phosphoglyceric acid. Pyruvic acid is the end product of glycolysis. Succinic acid is an intermediate product in krebs cycle.
130. Which of the following algae contains mannitol as reserve food material?
(1) Ulothrix
(2) Ectocarpus
(3) Gracilaria
(4) Volvox
Answer (2)
Sol. Ectocarpus is a brown alga belongs to the class Phaeophyceae. Members of this class have mannitol and laminarin as stored food material.
Ulothrix and Volvox belong to Chlorophyceae (green algae). Members of this class have starch as reserve food material. Gracilaria is a member of red algae (Rhodophyceae). This class is characterised by having floridean starch as stored food material.
131. DNA strands on a gel stained with ethidium bromide when viewed under UV radiation, appear as
(1) Bright blue bands
(2) Yellow bands
(3) Bright orange bands
(4) Dark red bands
Answer (3)
Sol. After the bands are stained, they are viewed in UV light. The bands appear bright orange in colour. Ethidium bromide is the intercalating agent that stacks in between the nitrogenous bases.
132. Dialephous stamens are found in
(1) China rose and citrus
(2) China rose
(3) Citrus
(4) Pea
Answer (4)
Sol. Stamens are said to be diadelphous when these are united in two bundles e.g. Pea. China rose has monoadelphous stamens while, Citrus has polyadelp hous stamens. Monoadelphous stamens are grouped in single bundle whereas polyadelphous stamens occur in more than two bundles.
133. Which of the following are not secondary metabolites in plants?
(1) Rubber, gums
(2) Morphine, codeine
(3) Amino acids, glucose
(4) Vinblastin, curcumin
Answer (3)
Sol. The correct option is (3)
Amino acids and glucose are included under the category of primary metabolites as they have identifiable functions and play known roles in normal physiological processes. Rubber, gums, morphine, codeine, vinblastin and curcumin are included under the category of secondary metabolites as their role or functions in host organisms is not known yet. However, many of them are useful to human welfare.
134. The amount of nutrients, such as carbon, nitrogen, phosphorus and calcium present in the soil at any given time, is referred as :
(1) Standing crop
(2) Climax
(3) Climax community
(4) Standing state
Answer (4)
Sol. Amount of all the inorganic substances or nutrients, such as carbon, nitrogen, phosphorus and calcium present in soil at any given time, is referred as standing state. Amount of living material present in different trophic levels at a given time, is referred as standing crop. Climax community is the last community in biotic succession which is relatively stable and is in near equilibrium with the environment of that area.
135. Genera like Selaginella and Salvinia produce two kinds of spores. Such plants are known as:
(1) Heterosporous
(2) Homosporus
(3) Heterosporus
(4) Homosporous
Answer (1)
Sol. Plants like Selaginella and Salvinia produce two kinds of spore i.e., microspores and macrospores. They are known as heterosporous.
Most of the pteridophytes produce single type of spores and are called homosporous
Sorus are brownish or yellowish cluster of spore-producing structures located on the lower surface of fern leaves.
SECTION-B
136. Plasmid pBR322 has PstI restriction enzyme site within gene ampR that confers ampicillin resistance. If this enzyme is used for inserting a gene for beta-galactoside production and the recombinant plasmid is inserted in an E.coli strain
(1) It will be able to produce a novel protein with dual ability
(2) It will not be able to confer ampicillin resistance to the host cell
(3) The transformed cells will have the ability to resist ampicillin as well as produce beta-galactoside
(4) It will lead to lysis of host cell
Answer (2)
Sol. pBR322 is a commonly used cloning vector. When the gene for beta-galactoside is inserted in the ampicillin resistance gene by using Pst I, the recombinant E.coli will lose ampicillin resistance due to insertional inactivation of the antibiotic resistance gene.
The host (recombinant) cell will produce beta-galactoside which is not a novel protein nor does it have dual ability.
The transformed cells cannot resist ampicillin as they have lost ampicillin resistance.
A recombinant E. coli is produced and the host cell will not undergo lysis due to insertion of beta-galactoside gene.
137. In some members of which of the following pairs of families, pollen grains retain their viability for months after release?
(1) Rosaceae; Leguminosae
(2) Poaceae; Rosaceae
(3) Poaceae; Leguminosae
(4) Poaceae; Solanaceae
Answer (1)
Sol. In members of some plant families like Solanaceae, Rosaceae and Leguminosae the pollen grains retain their viability for several months. In cereals (Poaceae) pollen grains retain viability for around 30 minutes.
138. Identify the correct statement.
(1) Split gene arrangement is characteristic of prokaryotes
(2) In capping, methyl guanosine triphosphate is added to the 3' end of hnRNA
(3) RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria
(4) The coding strand in a transcription unit is copied to an mRNA
Answer (3)
Sol. Split gene arrangement is characteristic of eukaryotes.
In capping 5-methyl guanosine triphosphate is added at 5' end of hnRNA. At 3' end poly-A tail is added. The non coding or template strand is copied to an mRNA. RNA polymerase associate with rho factor (Rho factor) and it alters the specificity of the RNA polymerase to terminate the processes.
139. Which of the following statements is correct?
(1) Some of the organisms can fix atmospheric nitrogen in specialized cells called sheath cells
(2) Fusion of two cells is called Karyogamy
(3) Fusion of protoplasm between two motile on non-motile gametes is called plasmogamy
(4) Organisms that depend on living plants are called saprophytes
Answer (3)
Sol. In some blue-green algae specialised cells called heterocyst fixes atmospheric nitrogen into ammonia. Fusion of two nuclei is called Karyogamy. Organisms that depend on living plants are parasites, saprophytes grow on dead material. Fusion of protoplasts of two cells is called plasmogamy.
140. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as
(1) Polymorphic DNA
(2) Satellite DNA
(3) Repetitive DNA
(4) Single nucleotides
Answer (3)
Sol. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA. The basis of DNA fingerprinting is VNTR (a satellite DNA as probe that show very high degree of polymorphism) Polymorphism is the variation at genetic level. Allelic sequence variation has traditionally been described as a DNA polymorphism.
141. Select the correct pair.
(1) Loose parenchyma cells - Spongy rupturing the epidermis parenchyma and forming a lens shaped opening in bark
(2) Large colorless empty - Subsidiary cells cells in the epidermis of grass leaves
(3) In dicot leaves, vascular - Conjunctive bundles are surrounded tissue by large thick-walled cells
(4) Cells of medullary rays - Interfascicular that form part of cambium ring
Answer (4)
Sol. When the cells of medullary rays differentiated, they give rise to the new cambium called interfascicular cambium. Loose parenchyma cells rupturing the epidermis and forming a lens-shaped opening in bark are called complementary cells. Large colourless empty cells in the epidermis of grass leaves are called bulliform cells. In dicot leave, vascular bundles are surrounded by large thick walled cells called bundle sheath cells.
142. Which of the following statements is incorrect?
(1) Oxidation-reduction reactions produce proton gradient in respiration
(2) During aerobic respiration, role of oxygen is limited to the terminal stage
(3) In ETC (Electron Transport Chain), one molecule of NADH + H+ gives rise to 2 ATP molecules, and one FADH2 gives rise to 3 ATP molecules
(4) ATP is synthesized through complex V
Answer (3)
Sol. During respiration, process of ATP synthesis is explained by chemiosmotic model. It says that a proton gradient is required for ATP synthesis that is established by oxidation-reduction reactions. In ETC, one NADH + H+ produces 3 ATP while one FADH2 produces 2 ATP molecules. ATP is synthesised via complex V. In ETS, oxygen acts as terminal electron acceptor.
143. Which of the following statements is incorrect?
(1) Cyclic photophosphorylation involves both PS I and PS II
(2) Both ATP and NADPH + H+ are synthesized during non-cyclic photophosphorylation
(3) Stroma lamellae have PS I only and lack NADP reductase
(4) Grana lamellae have both PS I and PS II
Answer (1)
Sol. Cyclic photophosphorylation involves only PS I. Both PS I and PS II are involved in non-cyclic photophosphorylation where both ATP and NADPH + H+ are synthesized. Both PS I and PS II are found on grana lamellae whereas stroma lamellae have PS I only and lack NADP reductase.
144. Match Column-I with Column-II.
Column-I Column-II
(a) Nitrococcus (i) Denitrification
(b) Rhizobium (ii) Conversion of ammonia to nitrite
(c) Thiobacillus (iii) Conversion of nitrite to nitrate
(d) Nitrobacter (iv) Conversion of atmospheric nitrogen to ammonia
Choose the correct answer from options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (ii) (i)
(2) (ii) (iv) (i) (iii)
(3) (i) (ii) (iii) (iv)
(4) (iii) (i) (iv) (ii)
Answer (2)
Sol. Nitrogen fixation is conversion of atmospheric N2 to NH3 (ammonia). It is carried out by N2 fixers such as Rhizobium. NH3 is converted to NO2 (nitrite) by nitrifying bacteria such as Nitrococcus. Then NO2 is converted to NO3 (nitrate) by nitrifying bacteria called Nitrobacter. Thiobacillus carries out denitrification, a process where NO2^-/NO3^- is converted to N2.
145. What is the role of RNA polymerase III in the process of transcription in eukaryotes?
(1) Transcribes only snRNAs
(2) Transcribes rRNAs (28S, 18S and 5.8S)
(3) Transcribes tRNA, 5s rRNA and snRNA
(4) Transcribes precursor of mRNA
Answer (3)
Sol. RNA polymerase III transcribes tRNA, ScRNA, 5S rRNA and SnRNA. RNA polymerase I transcribes 5.8S, 18S and 28S rRNA. RNA polymerase II transcribes hnRNA which is precursor of mRNA
146. In the exponential growth equation Nt = N0 e^(rt), e represents
(1) The base of geometric logarithms
(2) The base of number logarithms
(3) The base of exponential logarithms
(4) The base of natural logarithms
Answer (4)
Sol. In the exponential growth equation Nt = N0 e^(rt), e represents the base of natural logarithms Nt = Population density after time t N0 = Population density at time zero r = Intrinsic rate of natural increase called biotic potential.
147. Now a days it is possible to detect the mutated gene causing cancer by allowing radioactive probe to hybridise its complimentary DNA in a clone of cells, followed by its detection using autoradiography because :
(1) Mutated gene does not appear on photographic film as the probe has complementarity with it
(2) Mutated gene partially appears on a photographic film
(3) Mutated gene completely and clearly appears on a photographic film
(4) Mutated gene does not appear on a photographic film as the probe has no complementarity with it
Answer (4)
Sol. Autoradiography allows the detection/localisation of radioactive isotope within a biological sample. Probe is a radiolabelled ss DNA or ss RNA depending on the technique. To identify the mutated gene probe is allowed to hybridise to its complementary DNA in a clone of cells followed by detection using autoradiography. The mutated gene will not appear on the photographic film, because the probe does not have complementarity with the mutated gene.
148. Match List-1 with List-1L.
List-1 List-1I
(a) Protein (i) C = C double bonds
(b) Unsaturated fatty acid (ii) Phosphodiester bonds
(c) Nucleic acid (iii) Glycosidic bonds
(d) Polysaccharide (iv) Peptide bonds
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (iii) (i) (ii)
(2) (iv) (i) (ii) (iii)
(3) (i) (iv) (iii) (ii)
(4) (ii) (i) (iv) (iii)
Answer (2)
Sol. In a polypeptide or a protein, amino acids are linked by a peptide bond which is formed when the carboxyl (-COOH) group of one amino acid reacts with amino (-NH2) group of the next amino acid with the elimination of a water moiety.
Unsaturated fatty acids are with one or more C=C double bonds.
In nucleic acids, a phosphate moiety links the 3'-carbon of one sugar of one nucleotide to the 5'-carbon of the sugar of the succeeding nucleotide. The bond between the phosphate and hydroxyl group is an ester bond. As there is one such ester bond on either side, it is called phosphodiester bond.
In a polysaccharide, the individual monosaccharides are linked by a glycosidic bond.
149. Match List-1 with List-1I.
List-1 List-1I
(a) S phase (i) Proteins are synthesized
(b) G2 phase (ii) Inactive phase
(c) Quiescent stage (iii) Interval between mitosis and initiation of DNA replication
(d) G1 phase (iv) DNA replication
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (ii) (iv) (iii) (i)
(2) (iii) (ii) (i) (iv)
(3) (iv) (iii) (ii) (i)
(4) (iv) (i) (ii) (iii)
Answer (4)
Sol. In S phase DNA replication takes place. In G2 phase there is synthesis of proteins, RNA etc. Quiescent stage is inactive stage of cell cycle but cells remain metabolically active in this stage. G1 phase is the interval between mitosis and initiation of DNA replication.
150. Match Column-1 with Column-1I
Column-1 Column-1I
(a) K(5)C(1+2+(2))A(5)+G(1) (i) Brassicaceae
(b) K(5)C(5)A(5)G(2) (ii) Liliaceae
(c) P(3+3)A(3+3)G(3) (iii) Fabaceae
(d) K(2+2)C(4)A(2-4)G(2) (iv) Solanaceae
Select the correct answer from the options given below.
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (iii) (iv) (ii) (i)
(3) (i) (ii) (iii) (iv)
(4) (ii) (iii) (iv) (i)
Answer (2)
Sol. The floral formula of
Brassicaceae family - K(2+2)C(4)A(2+4)G(2) Solanaceae family - K(5)C(5)A(5)G(2) Fabaceae family - K(5)C(1+2+(2))A(9)+1G(1) Liliaceae family - P(3+3)A(3+3)G(3) So a(iii), b(iv), c(ii), d(i) is correct matching.
151. Receptors for sperm binding in mammals are present on :
(1) Zona pellucida
(2) Corona radiata
(3) Vitelline membrane
(4) Perivitelline space
Answer (1)
Sol. Option (1) is correct because zona pellucida has receptors for sperm binding (ZP3 receptors) in mammals. Corona radiata is a layer of radially arranged cells of membrana granulosa. Perivitelline space is present in between vitelline membrane and zona pellucida.
152. Which stage of meiotic prophase shows terminalisation of chiasmata as its distinctive feature?
(1) Pachytene
(2) Leptotene
(3) Zygotene
(4) Diakinesis
Answer (4)
Sol. In meiosis I, chiasmata (X shaped structure) is formed in diplotene stage while it terminalise in diakinesis stage. Bivalents are formed in zygotene stage and crossing over takes place in pachytene stage. Compaction of chromosomal material occurs in leptotene stage.
153. The organelles that are included in the endomembrane system are
(1) Golgi complex, Endoplasmic reticulum, Mitochondria and Lysosomes
(2) Endoplasmic reticulum, Mitochondria, Ribosomes and Lysosomes
(3) Endoplasmic reticulum, Golgi complex, Lysosomes and Vacuoles
(4) Golgi complex, Mitochondria, Ribosomes and Lysosomes
Answer (3)
Sol. Endomembrane system consist of endoplasmic reticulum, Golgi complex, vacuoles and lysosomes. Mitochondria is semi-autonomous cell organelle. Ribosome is non-membranous cell organelle.
154. A specific recognition sequence identified by endonucleases to make cuts at specific positions within the DNA is:
(1) Poly(A) tail sequences
(2) Degenerate primer sequence
(3) Okazaki sequences
(4) Palindromic Nucleotide sequences
Answer (4)
Sol. Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA. Once it finds its specific recognition sequence it bind to DNA and cuts each of the two strands of DNA. During post transcriptional modification in eukaryotes, poly(A) tail (200-300 adenylate residues) are added at 3' end of hnRNA. During DNA replication Okazaki fragments are synthesized discontinuously and joined by DNA ligase. A PCR primer sequence is termed degenerate if some of its position have several possible bases.
155. Select the favourable conditions required for the formation of oxyhaemoglobin at the alveoli.
(1) Low pO2, low pCO2, more H+, higher temperature
(2) High pO2, low pCO2, less H+, lower temperature
(3) Low pO2, high pCO2, more H+, higher temperature
(4) High pO2, high pCO2, less H+, higher temperature
Answer (2)
Sol. The factors favourable for the formation of oxyhaemoglobin at the alveolar level are; high pO2, low pCO2, less H+ concentration and lower temperature. The conditions favourable for the dissociation of oxygen from oxyhaemoglobin at the tissue level are; low pO2, high pCO2, high H+ concentration and high temperature.
156. Match the following:
List-I List-II
(a) Physalia (i) Pearl oyster
(b) Limulus (ii) Portuguese Man of War
(c) Ancylostoma (iii) Living fossil
(d) Pinctada (iv) Hookworm
Choose the correct answer from the options given below.
(a) (b) (c) (d)
(1) (i) (iv) (iii) (ii)
(2) (ii) (iii) (i) (iv)
(3) (iv) (i) (iii) (ii)
(4) (ii) (iii) (iv) (i)
Answer (4)
Sol. Option (4) is correct because Physalia is commonly known as Portuguese man of war. Limulus is considered as a living fossil and commonly known as king crab. Ancylostoma is a roundworm and commonly known as hookworm. Pinctada is commonly known known as pearl oyster, included in phylum Mollusca.
157. Dobson units are used to measure thickness of:
(1) Troposphere
(2) CFCs
(3) Stratosphere
(4) Ozone
Answer (4)
Sol. The thickness of the ozone in a column of air from the ground to the top of atmosphere is measured in term of Dobson unit (1 DU = 1ppb).
The lowermost layer of atmosphere is called troposphere.
CFCs are ozone depleting substances. Ozone found in upper part of atmosphere (the stratosphere) is called good ozone.
158. Which one of the following belongs to the family Muscidae?
(1) House fly
(2) Fire fly
(3) Grasshopper
(4) Cockroach
Answer (1)
Sol. Option (1) is correct because housefly belongs to the family Muscidae, class Insecta and phylum Arthropoda.
Fire flies are placed in family Lampyridae of class insecta. Grasshopper is also an insect placed in family Acrididae. Cockroach is also an insect placed in family Blattidae.
159. Veneral diseases can spread through :
(a) Using sterile needles
(b) Transfusion of blood from infected person
(c) Infected mother to foetus
(d) Kissing
(e) Inheritance
Choose the correct answer from the options given below.
(1) (a) and (c) only
(2) (a), (b) and (c) only
(3) (b), (c) and (d) only
(4) (b) and (c) only
Answer (4)
Sol. Veneral diseases or sexually transmitted diseases or infections are transmitted by sharing of infected needles, surgical instruments with infected person, transfusion of blood or from an infected mother to foetus. Veneral diseases are not transmitted through kissing or inheritance.
160. Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes?
(1) DNase
(2) DNA dependent DNA polymerase
(3) DNA dependent RNA polymerase
(4) DNA Ligase
Answer (3)
Sol. In prokaryotes, the DNA dependent RNA polymerase is a holoenzyme that is made of polypeptides (alpha2 beta beta' omega) sigma. It is responsible for initiation, elongation and termination during transcription.
DNase degrades DNA.
DNA dependent DNA polymerase is involved in replication of DNA.
DNA ligase joins the discontinuously synthesised fragments of DNA.
161. Match List-1 with List-1I.
List-1 List-II
(a) Vaults (i) Entry of sperm through Cervix is blocked
(b) IUDs (ii) Removal of Vas deferens
(c) Vasectomy (iii) Phagocytosis of sperms within the Uterus
(d) Tubectomy (iv) Removal of fallopian tube
Choose the correct answer from the options given below
(a) (b) (c) (d)
(1) (iii) (i) (iv) (ii)
(2) (iv) (ii) (i) (iii)
(3) (i) (iii) (ii) (iv)
(4) (ii) (iv) (iii) (i)
Answer (3)
Sol. Diaphragms, cervical caps and vaults are barrier methods of contraception for female which works by blocking the entry of sperms through the cervix. IUDs increase phagocytosis of sperms within the uterus. Vasectomy is a surgical method of contraception in males in which a small part of the vas deferens is removed or tied up through a small incision on the scrotum. Tubectomy is a surgical method of contraception in females where a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.
162. Match List-1 with List-1I
List-I List-II
(a) Aspergillus niger (i) Acetic Acid
(b) Acetobacter aceti (ii) Lactic Acid
(c) Clostridium butylicum (iii) Citric Acid
(d) Lactobacillus (iv) Butyric Acid
Choose the correct answer from the options given below
(a) (b) (c) (d)
(1) (iv) (ii) (i) (iii)
(2) (iii) (i) (iv) (ii)
(3) (i) (iii) (ii) (iv)
(4) (ii) (iii) (i) (iv)
Answer (2)
Sol. Aspergillus niger is involved in production of citric acid. Acetobacter aceti is involved in production of acetic acid. Clostridium butylicum is involved in production of butyric acid whereas Lactobacillus is involved in the production of lactic acid.
So a(iii), b(i), c(iv), d(ii) is correct matching.
163. Identify the incorrect pair
(1) Drugs - Ricin
(2) Alkaloids - Codeine
(3) Toxin - Abrin
(4) Lectins - Concanavalin A
Answer (1)
Sol. Option (1) is incorrect because ricin is a toxin obtained from Ricinus plant. Vinblastin and curcumin are drugs. Morphine and codeine are alkaloids. Abrin is also a toxin obtained by plant Abrus. Concanavalin A is a lectin.
164. The partial pressures (in mm Hg) of oxygen (O2) and carbon dioxide (CO2) at alveoli (the site of diffusion) are:
(1) pO2 = 159 and pCO2 = 0.3
(2) pO2 = 104 and pCO2 = 40
(3) pO2 = 40 and pCO2 = 45
(4) pO2 = 95 and pCO2 = 40
Answer (2)
Sol. Option (2) is correct because pO2 in alveoli is 104 mm Hg and pCO2 in alveoli is 40 mm Hg In atmosphere, pO2 is 159 mm Hg and pCO2 is 0.3 mm Hg In deoxygenated blood, pO2 is 40 mm Hg and pCO2 is 45 mm Hg In oxygenated blood, pO2 is 95 mm Hg and pCO2 is 40 mm Hg
165. Sphincter of oddi is present at:
(1) Junction of jejunum and duodenum
(2) lleo-caecal junction
(3) Junction of hepato-pancreatic duct and duodenum
(4) Gastro-oesophageal junction
Answer (3)
Sol. The bile duct and the pancreatic duct open together into the duodenum as the common hepato-pancreatic duct which is guarded by a sphincter called the sphincter of Oddi. lleo-caecal valve is present at the junction of ileum and caecum to prevent the backflow of faecal matter into the ileum in humans. Gastro-oesophageal sphincter regulates the opening of oesophagus into stomach.
166. Which of the following RNAs is not required for the synthesis of protein?
(1) siRNA
(2) mRNA
(3) tRNA
(4) rRNA
Answer (1)
Sol. siRNA are small interfering RNA also called silencing RNA. It is a class of double-stranded RNA, non-coding RNA molecules. mRNA is messenger RNA that carries genetic information provided by DNA. tRNA carries amino acids to the mRNA during translation. rRNA is structural RNA that forms ribosomes which are involved in translation.
167. Succus entericus is referred to as:
(1) Chyme
(2) Pancreatic juice
(3) Intestinal juice
(4) Gastric juice
Answer (3)
Sol. Option 3 is correct because succus entericus is referred to as intestinal juice. Chyme is name given to acidic food present in stomach.
Exocrine secretion of pancreatic acini is called pancreatic juice. Secretion of gastric glands present in stomach is called gastric juice.
168. Persons with 'AB' blood group are called as "Universal recipients". This is due to :
(1) Absence of antibodies, anti-A and anti-B, in plasma
(2) Absence of antigens A and B on the surface of RBCs
(3) Absence of antigens A and B in plasma
(4) Presence of antibodies, anti-A and anti-B, on RBCs
Answer (1)
Sol. Option 1 is correct because persons with 'AB' blood group contain antigens 'A' and 'B' but lack antibodies anti-A and anti-B in plasma. So, persons with 'AB' blood group can accept blood from persons with AB as well as the other groups of blood due to lack of antibodies in their blood. Therefore, such persons are called "Universal recipients".
169. Which of the following characteristics is incorrect with respect to cockroach?
(1) 10th abdominal segment in both sexes, bears a pair of anal cerci
(2) A ring of gastric caeca is present at the junction of midgut and hind gut
(3) Hypopharynx lies within the cavity enclosed by the mouth parts
(4) In females, 7th-9th sterna together form a genital pouch
Answer (2)
Sol. Option 2 is incorrect because a ring of gastric caeca is present at the junction of foregut and midgut. At the junction of midgut and hindgut, malpighian tubules are present. Hypopharynx lies within the cavity enclosed by mouthparts. In female cockroach, the 7th sternum is boat shaped and together with the 8th and 9th sterna forms a genital pouch. 10th abdominal segment in both sexes, bears a pair of anal cerci and 9th sternum only in male cockroach, bears a pair of chitinous anal style.
170. Which of the following statements wrongly represents the nature of smooth muscle?
(1) These muscles are present in the wall of blood vessels
(2) These muscle have no striations
(3) They are involuntary muscles
(4) Communication among the cells is performed by intercalated discs
Answer (4)
Sol. Option (4) is incorrect because intercalated discs are found only in cardiac muscle tissue. Smooth muscle fibres are non-striated and involuntary in nature and are present in the wall of blood vessels, uterus, gall bladder, alimentary canal etc.
171. Which one of the following organisms bears hollow and pneumatic long bones?
(1) Ornithorhynchus
(2) Neophron
(3) Hemidactylus
(4) Macropus
Answer (2)
Sol. Hollow and pneumatic long bones are present in animals that belong to class Aves e.g., Neophron (vulture). Ornithorhynchus (Platypus) and Macropus (Kangaroo) belong to class Mammalia. Hemidactylus (Wall lizard) is a member of class Reptilia.
172. If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it?
(1) T:20; G:25; C:25
(2) T:20; G:30; C:20
(3) T:20; G:20; C:30
(4) T:30; G:20; C:20
Answer (4)
Sol. According to Chargaff's rule, for a double stranded DNA,
[A] = [T],
therefore [A] = 30%, => [T] = 30%
173. Which enzyme is responsible for the conversion of inactive fibrinogens to fibrins?
(1) Thromboxane
(2) Thrombin
(3) Renin
(4) Epinephrine
Answer (2)
Sol. During coagulation of blood, an enzyme complex thromboxinase helps in the conversion of prothrombin (present in plasma) into thrombin.
Thrombin further helps in the conversion of inactive fibrinogens into fibrins which form network of threads.
Renin is secreted by JG cells in response to fall in glomerular blood flow, which converts angiotensinogen in blood to angiotensin-I
Epinephrine or adrenaline is secreted by adrenal medulla in response to stress of any kind and during emergency.
174. In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased?
(1) 100%
(2) 50%
(3) 75%
(4) 25%
Answer (4)
Sol. According to given question;
Parents: HbA HbS x HbA HbS
Progenies: HbS HbS, HbA HbS, HbA HbS, HbA HbA
Total number of affected progenies = 1
Percentage of diseased/affected progenies
= 1/4 x 100 = 25%
175. Which one of the following is an example of Hormone releasing IUD?
(1) Multiload 375
(2) CuT
(3) LNG 20
(4) Cu 7
Answer (3)
Sol. LNG-20 is a hormone releasing IUD which makes the uterus unsuitable for implantation and the cervix hostile to sperms. Multiload 375, CuT and Cu7 are copper releasing IUDs which suppress sperm motility and the fertilizing capacity of sperms.
176. The centriole undergoes duplication during:
(1) G2 phase
(2) S-phase
(3) Prophase
(4) Metaphase
Answer (2)
Sol. During S phase of cell cycle replication of DNA takes place. In animal cells during S phase, centriole duplicates in the cytoplasm.
In G2 phase there is duplication of mitochondria, chloroplast and Golgi bodies. Tubulin protein is also synthesized during this phase.
During prophase, condensation of chromatin starts.
During metaphase, chromosomes get aligned at equator to form metaphasic plate.
177. Chronic auto immune disorder affecting neuro muscular junction leading to fatigue, weakening and paralysis of skeletal muscle is called as:
(1) Gout
(2) Arthritis
(3) Muscular dystrophy
(4) Myasthenia gravis
Answer (4)
Sol. Option (4) is correct because myasthenia gravis is a chronic auto immune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle. Gout is caused due to deposition of uric acid crystals in joints leading to its inflammation. Inflammation of joints is commonly known as arthritis. Muscular dystrophy is a genetic disorder which results in progressive degeneration of skeletal muscle.
178. For effective treatment of the disease, early diagnosis and understanding its pathophysiology is very important. Which of the following molecular diagnostic techniques is very useful for early detection?
(1) Hybridization Technique
(2) Western Blotting Technique
(3) Southern Blotting Technique
(4) ELISA Technique
Answer (3/4)
Sol. ELISA can be used for early detection of an infection either by detecting the presence of pathogenic antigen or by detecting the antibodies synthesized against the pathogen. Option (3) Southern blotting is used to detect a specific DNA sequence in the given sample and can be detected prior to antibody formation. One can detect presence of pathogenic DNA/RNA. In hybridization technique a ssDNA/ssRNA tagged with a radioactive molecule (probe) is allowed to hybridize its complementary DNA in a clone of cells followed by detection using autoradiography. It is used to find a mutated gene. Western blotting technique is used to detect a specific protein molecule among a mixture of proteins.
179. Read the following statements
(a) Metagenesis is observed in Helminths.
(b) Echinoderms are triploblastic and coelomate animals.
(c) Round worms have organ-system level of body organization.
(d) Comb plates present in ctenophores help in digestion.
(e) Water vascular system is characteristic of Echinoderms.
Choose the correct answer from the options given below.
(1) (b), (c) and (e) are correct
(2) (c), (d) and (e) are correct
(3) (a), (b) and (c) are correct
(4) (a), (d) and (e) are correct
Answer (1)
Sol. Metagenesis (alternation of generation) is observed in members of phylum Coelenterata (Cnidaria). - Echinoderms are triploblastic and coelomate animals as true coelom is observed in them. - Roundworms (Aschelminthes) have organ system level of organization. - Comb plates present in ctenophores help in locomotion. - Water vascular system is seen in echinoderms, which helps in locomotion, capture and transport of food and respiration.
180. Erythropoietin hormone which stimulates R.B.C. formation is produced by:
(1) Juxtaglomerular cells of the kidney
(2) Alpha cells of pancreas
(3) The cells of rostral adenohypophysis
(4) The cells of bone marrow
Answer (1)
Sol. Option (1) is correct because Juxtaglomerular cells of kidney secrete erythropoietin hormone which stimulates RBC formation. Alpha cells of pancreas produce hormone glucagon. The cells of rostral adenohypophysis synthesizes hormones of anterior lobe of pituitary. The cells of bone marrow are responsible for formation of formed elements.
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