Q1.
Match List I with List II.
List I (Spectral Lines of Hydrogen for transitions from):
A. $n_2 = 3$ to $n_1 = 2$
B. $n_2 = 4$ to $n_1 = 2$
C. $n_2 = 5$ to $n_1 = 2$
D. $n_2 = 6$ to $n_1 = 2$
List II (Wavelengths (nm)):
I. 410.2
II. 434.1
III. 656.3
IV. 486.1
Choose the correct answer from the options given below:
- A. A-I, B-II, C-III, D-IV
- B. A-II, B-I, C-IV, D-III
- C. A-III, B-IV, C-II, D-I ✓
- D. A-IV, B-III, C-I, D-II
Solution: Energy difference $\Delta E = \dfrac{hc}{\lambda}$
$$\therefore\ \lambda \propto \frac{1}{\Delta E}$$
The further the upper level, the larger the energy gap:
$$(\Delta E)_{6-2} > (\Delta E)_{5-2} > (\Delta E)_{4-2} > (\Delta E)_{3-2}$$
so the wavelengths run the other way:
$$\lambda_{6-2} < \lambda_{5-2} < \lambda_{4-2} < \lambda_{3-2}$$
Matching the four given wavelengths in increasing order gives A-III, B-IV, C-II, D-I.
Q2.
Match List-I with List-II.
List-I (Material):
A. Diamagnetic
B. Ferromagnetic
C. Paramagnetic
D. Non-magnetic
List-II (Susceptibility ($\chi$)):
I. $\chi = 0$
II. $0 > \chi \geq -1$
III. $\chi \gg 1$
IV. $0 < \chi < \varepsilon$ (a small positive number)
Choose the correct answer from the options given below
- A. A-IV, B-III, C-II, D-I
- B. A-II, B-III, C-IV, D-I ✓
- C. A-II, B-I, C-III, D-IV
- D. A-III, B-II, C-I, D-IV
Solution: Taking each class of material with its susceptibility:
Diamagnetic — $0 > \chi \geq -1$
Ferromagnetic — $\chi \gg 1$
Paramagnetic — $0 < \chi < \varepsilon$
Non-magnetic — $\chi = 0$
So A-II, B-III, C-IV, D-I.
Q3.
In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6}$ kg m$^{2}$. If the magnitude of magnetic moment of the needle is $x \times 10^{-5}$ Am$^{2}$, then the value of '$x$' is :
- A. $1280\pi^{2}$ ✓
- B. $5\pi^{2}$
- C. $128\pi^{2}$
- D. $50\pi^{2}$
Solution: Time period of oscillation, $T = 2\pi\sqrt{\dfrac{I}{MB}}$.
Twenty oscillations in 5 s gives $T = \dfrac{5}{20} = \dfrac{1}{4}$ s:
$$\Rightarrow\ \frac{1}{4} = 2\pi\sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}}$$
$$\Rightarrow\ \frac{1}{16} = 4\pi^{2} \times \frac{9.8 \times 10^{-6}}{M \times 49 \times 10^{-3}}$$
$$\Rightarrow\ M = \frac{4\pi^{2} \times 9.8 \times 10^{-6}}{49 \times 10^{-3}} \times 16$$
$$= \frac{4\pi^{2} \times 9.8 \times 16 \times 10^{-3}}{49}$$
$$= 12.8\pi^{2} \times 10^{-3} = 1280\pi^{2} \times 10^{-5}\ \text{Am}^{2}$$
So $x = 1280\pi^{2}$.
Q4.
An unpolarised light beam strikes a glass surface at Brewster's angle. Then
- A. The reflected light will be completely polarised but the refracted light will be partially polarised. ✓
- B. The reflected light will be partially polarised.
- C. The refracted light will be completely polarised.
- D. Both the reflected and refracted light will be completely polarised.
Solution: At Brewster's angle the reflected ray and the refracted ray are perpendicular to each other, and the reflected light is completely plane polarised with its vibrations perpendicular to the plane of incidence.
The refracted beam still carries both components, though unequally, so it is only partially polarised.
Q5.
Consider the following statements A and B and identify the correct answer.
The graph referred to has current $I$ on the vertical axis and voltage $V$ on the horizontal axis, with the quadrants labelled (I) top right, (II) top left, (III) bottom left and (IV) bottom right.
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased $pn$ junction diode, the current measured in ($\mu$A), is due to majority charge carriers.
- A. Both A and B are incorrect
- B. A is correct but B is incorrect ✓
- C. A is incorrect but B is correct
- D. Both A and B are correct
Solution: A: The solar cell characteristic runs from the open-circuit voltage $V_{OC}$ on the positive voltage axis down to the short-circuit current $I_{SC}$ on the negative current axis, so the curve lies in the fourth quadrant. Statement A is correct.
B: In a reverse biased $pn$ junction diode, the small current measured in $\mu$A is due to minority charge carriers, not majority ones. Statement B is incorrect.
Q6.
A light ray enters through a right angled prism at point $P$ with the angle of incidence $30^{\circ}$ as shown in figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC$. The refractive index of the prism is:
- A. $\dfrac{\sqrt{3}}{2}$
- B. $\dfrac{\sqrt{5}}{4}$
- C. $\dfrac{\sqrt{5}}{2}$ ✓
- D. $\dfrac{\sqrt{3}}{4}$
Solution: The prism has $A = 90^{\circ}$. The ray emerges along the face $AC$, which means it strikes $AC$ at the critical angle $c$.
In the prism, $r_1 + c = A$:
$$r_1 = 90^{\circ} - c \qquad \ldots(1)$$
$$\sin c = \frac{1}{\mu} \ \Rightarrow\ \cos c = \frac{\sqrt{\mu^{2} - 1}}{\mu}$$
Applying Snell's law on the incidence surface:
$$1 \cdot \sin 30^{\circ} = \mu\sin(r_1) \ \Rightarrow\ 1 \times \frac{1}{2} = \mu \times \sin(90^{\circ} - c)$$
$$\frac{1}{2} = \mu \times \frac{\sqrt{\mu^{2} - 1}}{\mu}$$
On squaring, $\dfrac{1}{4} = \mu^{2} - 1$:
$$\Rightarrow\ \mu^{2} = \frac{5}{4} \ \Rightarrow\ \mu = \frac{\sqrt{5}}{2}$$
Q7.
A particle moving with uniform speed in a circular path maintains:
- A. Varying velocity and varying acceleration ✓
- B. Constant velocity
- C. Constant acceleration
- D. Constant velocity but varying acceleration
Solution: A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration.
It is because the direction of both the velocity as well as the acceleration will change continuously, even though their magnitudes stay the same.
Q8.
The graph which shows the variation of $\left(\dfrac{1}{\lambda^{2}}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle):
- A. A straight line through the origin, rising with $E$ ✓
- B. A curve through the origin that bends upward (concave up), rising with $E$
- C. A curve falling from high values and decaying towards the $E$ axis
- D. A curve through the origin that bends over and saturates
Solution: de-Broglie wavelength
$$\lambda = \frac{h}{P} = \frac{h}{mv} = \frac{h}{\sqrt{2mE}} \qquad \text{where } E = \frac{1}{2}mv^{2}$$
Squaring both sides,
$$\lambda^{2} = \frac{h^{2}}{2mE}$$
$$\Rightarrow\ \frac{1}{\lambda^{2}} = \left(\frac{2m}{h^{2}}\right)E = (\text{constant})\,E$$
The graph is therefore a straight line passing through the origin with constant slope.
Q9.
A wire of length '$l$' and resistance 100 $\Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
- A. 60 $\Omega$
- B. 26 $\Omega$
- C. 52 $\Omega$ ✓
- D. 55 $\Omega$
Solution: $$R = \frac{\rho l}{A}$$
Each of the 10 parts has
$$R' = \frac{\rho l}{10A} = \frac{R}{10} = 10\ \Omega$$
Five in series:
$$R_S = 5 \times \frac{R}{10} = 50\ \Omega$$
Five in parallel:
$$R_P = \frac{R}{50} = 2\ \Omega$$
$$R_{\text{eq}} = R_S + R_P = 52\ \Omega$$
Q10.
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm$^{2}$. The length of the 400 g rod is nearly:
- A. 72.0 cm
- B. 8.5 cm ✓
- C. 17.5 cm
- D. 20.7 cm
Solution: Moment of inertia of a rod about a perpendicular axis through its midpoint:
$$I = \frac{m\ell^{2}}{12}$$
$$\Rightarrow\ 2400 = 400\frac{\ell^{2}}{12}$$
$$\Rightarrow\ 72 = \ell^{2}$$
$$\Rightarrow\ \ell = \sqrt{72} = 8.48\ \text{cm} \simeq 8.5\ \text{cm}$$
Q11.
The output ($Y$) of the given logic gate is similar to the output of an/a
- A. AND gate ✓
- B. NAND gate
- C. NOR gate
- D. OR gate
Solution: Input $A$ goes to both terminals of a NAND gate and input $B$ to both terminals of a NOR gate, so
$$Y_1 = \overline{A \cdot A} = \overline{A}$$
$$Y_2 = \overline{B + B} = \overline{B}$$
These two feed a NOR gate:
$$Y = \overline{Y_1 + Y_2} = \overline{\overline{A} + \overline{B}}$$
By De Morgan's law,
$$= \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$$
which is similar to the output of an AND gate.
Q12.
In a vernier callipers, ($N$ + 1) divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
- A. $10(N + 1)$
- B. $\dfrac{1}{10N}$
- C. $\dfrac{1}{100(N + 1)}$ ✓
- D. $100N$
Solution: $$V.C = \text{MSD} - \text{VSD} \qquad \ldots(1)$$
Given $(N + 1)\,\text{VSD} = N\,\text{MSD}$:
$$\text{VSD} = \left(\frac{N}{N + 1}\right)\text{MSD} \qquad \ldots(2)$$
From (1) and (2),
$$V.C = \text{MSD} - \frac{N}{N + 1}\text{MSD} = \text{MSD}\left(1 - \frac{N}{N + 1}\right) = \frac{\text{MSD}}{N + 1}$$
1 MSD = 0.1 mm = 0.01 cm, so
$$= \frac{0.01}{N + 1} = \frac{1}{100(N + 1)}\ \text{cm}$$
Q13.
At any instant of time $t$, the displacement of any particle is given by $2t - 1$ (SI unit) under the influence of force of 5 N. The value of instantaneous power is (in SI unit):
Solution: $$x = 2t - 1$$
$$v = \frac{dx}{dt} = 2\ \text{m s}^{-1}$$
$$P = F \cdot v = 2 \times 5 = 10\ \text{W}$$
Q14.
A thin spherical shell of radius $R = 3$ cm carries a charge $q = 1\ \mu$C. The potential difference between the centre $C$ of the shell and a point $P$ lying on the shell (in V) is:
(Take $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ SI units)
- A. Zero ✓
- B. $3 \times 10^{5}$
- C. $1 \times 10^{5}$
- D. $0.5 \times 10^{5}$
Solution: For a uniformly charged spherical shell, the potential everywhere inside and on the shell is the same:
$$V = \frac{kq}{R} \qquad (\text{for } r \leq R)$$
$$\therefore\ V_C = V_P$$
$$V_C - V_P = \text{Zero}$$
Q15.
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m$^{-1}$, then the excess force required to take it away from the surface is
- A. 99 N
- B. 19.8 mN ✓
- C. 198 N
- D. 1.98 mN
Solution: The surface tension acts along the circumference of the disc:
$$\text{Excess force} = T \times 2\pi R$$
$$= \frac{7}{100} \times 2 \times 3.14 \times \frac{4.5}{100}$$
$$= 197.82 \times 10^{-4} = 19.8 \times 10^{-3}\ \text{N} = 19.8\ \text{mN}$$
Q16.
A logic circuit provides the output $Y$ as per the following truth table :
$A = 0$, $B = 0$ gives $Y = 1$
$A = 0$, $B = 1$ gives $Y = 0$
$A = 1$, $B = 0$ gives $Y = 1$
$A = 1$, $B = 1$ gives $Y = 0$
The expression for the output $Y$ is :
- A. $B$
- B. $AB + \overline{A}$
- C. $A\overline{B} + \overline{A}$
- D. $\overline{B}$ ✓
Solution: Reading the table, whenever $B = 0$ the output is 1, and whenever $B = 1$ the output is 0, no matter what $A$ is.
According to the given truth table, the output is independent of the value of $A$.
$$\therefore\ \text{Output } Y = \overline{B}$$
Q17.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential ($V$) at any axial point, at 2 m distance ($r$) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6}$ C m, is $\pm 9 \times 10^{3}$ V.
(Take $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ SI units)
Reason R: $V = \pm \dfrac{2P}{4\pi\varepsilon_0 r^{2}}$, where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
- A. A is false but R is true.
- B. Both A and R are true and R is the correct explanation of A.
- C. Both A and R are true and R is NOT the correct explanation of A.
- D. A is true but R is false. ✓
Solution: The potential $V$ at any point at distance $r$ from the centre of a dipole is
$$V = \frac{KP\cos\theta}{r^{2}}$$
At an axial point where $\theta = 0^{\circ}$:
$$V = \frac{KP}{r^{2}} = \frac{9 \times 10^{9} \times 4 \times 10^{-6}}{2^{2}} = 9 \times 10^{3}\ \text{V}$$
At an axial point where $\theta = 180^{\circ}$:
$$V = \frac{-KP}{r^{2}} = -9 \times 10^{3}\ \text{V}$$
So the Assertion is true. But the correct expression carries no factor of 2 — it is $V = \pm\dfrac{P}{4\pi\varepsilon_0 r^{2}}$ — so the Reason as stated is false.
Q18.
The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 $\Omega$, when connected through an external resistance of 4 $\Omega$ as shown in the figure is:
- A. 10 V
- B. 4 V
- C. 6 V
- D. 8 V ✓
Solution: Current in circuit
$$i = \frac{10}{4 + 1} = 2\ \text{A}$$
Terminal voltage
$$= E - iR = 10 - 2 \times 1 = 8\ \text{V}$$
Q19.
In the given diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
- A. $BA$ and $DC$
- B. $AB$ and $DC$ ✓
- C. $BA$ and $CD$
- D. $AB$ and $CD$
Solution: The magnet's north pole is moving away from solenoid-1, so by Lenz's law solenoid-1 opposes the departure by attracting it: the end $B$ of solenoid 1 facing the magnet becomes South.
The magnet's south pole is approaching solenoid-2, so solenoid-2 opposes the approach by repelling it: the end $C$ of solenoid 2 facing the magnet becomes South.
Reading the sense of the winding needed to produce those poles gives the induced current through $AB$ in solenoid-1 and through $DC$ in solenoid-2.
Q20.
In an ideal transformer, the turns ratio is $\dfrac{N_P}{N_S} = \dfrac{1}{2}$. The ratio $V_S : V_P$ is equal to (the symbols carry their usual meaning) :
- A. 1 : 4
- B. 1 : 2
- C. 2 : 1 ✓
- D. 1 : 1
Solution: According to the transformer ratio,
$$\frac{V_S}{V_P} = \frac{N_S}{N_P} = 2 : 1$$
Q21.
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4\pi \times 10^{-7}$ SI units):
- A. 44 T
- B. 44 mT
- C. 4.4 T
- D. 4.4 mT ✓
Solution: The magnitude of magnetic field due to a circular coil of $N$ turns is given by
$$B_C = \frac{\mu_0 iN}{2R}$$
$$= \frac{4\pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1}$$
$$= 4.4 \times 10^{-3}\ \text{T} = 4.4\ \text{mT}$$
Q22.
Given below are two statements:
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below.
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are correct
- C. Both Statement I and Statement II are incorrect
- D. Statement I is correct but Statement II is incorrect ✓
Solution: Statement I is true, as atoms are electrically neutral because they contain equal numbers of positive and negative charges.
Statement II is wrong: atoms of most of the elements are stable and emit a characteristic spectrum, but this statement is not true for every atom — radioactive atoms are not stable.
Q23.
The quantities which have the same dimensions as those of solid angle are:
- A. angular speed and stress
- B. strain and angle ✓
- C. stress and angle
- D. strain and arc
Solution: Solid angle $d\Omega = \dfrac{dA}{r^{2}}$ has dimensions $[\text{M}^{0}\text{L}^{0}\text{T}^{0}]$.
Strain $= \dfrac{\Delta l}{l}$ has dimensions $[\text{M}^{0}\text{L}^{0}\text{T}^{0}]$.
Angle measured in radians is also dimensionless, $[\text{M}^{0}\text{L}^{0}\text{T}^{0}]$, since $\theta = \dfrac{l}{r}$.
So strain and angle share the dimensions of solid angle.
Q24.
If $c$ is the velocity of light in free space, the correct statements about photon among the following are:
A. The energy of a photon is $E = h\nu$.
B. The velocity of a photon is $c$.
C. The momentum of a photon, $p = \dfrac{h\nu}{c}$.
D. In a photon-electron collision, both total energy and total momentum are conserved.
E. Photon possesses positive charge.
Choose the correct answer from the options given below:
- A. A, B, D and E only
- B. A and B only
- C. A, B, C and D only ✓
- D. A, C and D only
Solution: (A) The energy of a photon is $E = h\nu$.
(B) The velocity of a photon is equal to the velocity of light, i.e. $c$.
(C) From $\lambda = \dfrac{h}{p}$, $p = \dfrac{h}{\lambda} = \dfrac{h\nu}{c}$.
(D) In a photon-electron collision both total energy and total momentum are conserved.
(E) A photon is electrically neutral, so it does not possess a positive charge.
Hence A, B, C and D only.
Q25.
A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension in the string is $T$. If speed becomes $2\omega$ while keeping the same radius, the tension in the string becomes:
- A. $\sqrt{2}T$
- B. $T$
- C. $4T$ ✓
- D. $\dfrac{T}{4}$
Solution: The tension supplies the centripetal force:
$$T = m\ell\omega^{2}$$
With the angular speed doubled at the same radius,
$$T' = m\ell(2\omega)^{2} = 4m\ell\omega^{2}$$
$$T' = 4T$$
Q26.
If $x = 5\sin\left(\pi t + \dfrac{\pi}{3}\right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
- A. 5 m, 1 s
- B. 5 cm, 2 s
- C. 5 m, 2 s ✓
- D. 5 cm, 1 s
Solution: $$x = 5\sin\left(\pi t + \frac{\pi}{3}\right)\ \text{m}$$
Comparing with $x = A\sin(\omega t + \phi)$:
Amplitude $= 5$ m
$$\omega = \pi = \frac{2\pi}{T}$$
$$T = \frac{2\pi}{\pi} = 2\ \text{s}$$
Q27.
A thermodynamic system is taken through the cycle $abcda$, whose $P$-$V$ diagram has $a$ at (100 cm$^{3}$, 100 kPa), $b$ at (400 cm$^{3}$, 100 kPa), $c$ at (400 cm$^{3}$, 300 kPa) and $d$ at (100 cm$^{3}$, 300 kPa). The work done by the gas along the path $bc$ is:
- A. $-60$ J
- B. Zero ✓
- C. 30 J
- D. $-90$ J
Solution: Along $bc$ the volume stays at 400 cm$^{3}$ while the pressure rises from 100 kPa to 300 kPa, so path $bc$ is an isochoric process.
$$W = \int P\,dV = 0 \qquad (dV = 0)$$
$$\therefore\ \text{Work done by gas along path } bc \text{ is zero.}$$
Q28.
$$^{290}_{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^{+}\ } Z \xrightarrow{\ \beta^{-}\ } P \xrightarrow{\ e^{-}\ } Q$$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are
- A. 286, 81 ✓
- B. 280, 81
- C. 286, 80
- D. 288, 82
Solution: Following each emission in turn:
$$^{290}_{82}X \xrightarrow{\ \alpha\ } {}^{286}_{80}Y \xrightarrow{\ e^{+}\ } {}^{286}_{79}Z \xrightarrow{\ \beta^{-}\ } {}^{286}_{80}P \xrightarrow{\ e^{-}\ } {}^{286}_{81}Q$$
An $\alpha$ emission drops the mass number by 4 and the atomic number by 2; a positron ($e^{+}$) emission drops the atomic number by 1; a $\beta^{-}$ emission raises it by 1; and electron capture written here as $e^{-}$ raises it by 1 again.
$$A \rightarrow 286$$
$$Z = 81$$
Q29.
Two bodies $A$ and $B$ of same mass undergo completely inelastic one dimensional collision. The body $A$ moves with velocity $v_1$ while body $B$ is at rest before collision. The velocity of the system after collision is $v_2$. The ratio $v_1 : v_2$ is
- A. 1 : 4
- B. 1 : 2
- C. 2 : 1 ✓
- D. 4 : 1
Solution: Before collision, $A$ moves with $v_1$ and $B$ is at rest. It undergoes a completely inelastic collision, so the two move together afterwards.
Using conservation of linear momentum, initial momentum = final momentum:
$$\Rightarrow\ mv_1 = mv_2 + mv_2$$
$$\Rightarrow\ mv_1 = 2mv_2$$
$$\Rightarrow\ \frac{v_1}{v_2} = \frac{2}{1}$$
Q30.
In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is :
The network is a bridge of five 2 $\mu$F capacitors: two in series along the upper arm from $A$ to $B$, two in series along the lower arm from $A$ to $B$, and the fifth bridging the two midpoints.
- A. 4 $\mu$F
- B. 2 $\mu$F ✓
- C. 1 $\mu$F
- D. 0.5 $\mu$F
Solution: The given circuit is a balanced Wheatstone bridge: the two arms have equal ratios, so the bridging capacitor carries no charge and can be removed.
What is left is two arms in parallel, each of two 2 $\mu$F capacitors in series:
$$\text{each arm} = \frac{2 \times 2}{2 + 2} = 1\ \mu\text{F}$$
$$C_{AB} = 1 + 1 = 2\ \mu\text{F}$$
Q31.
A horizontal force 10 N is applied to a block $A$ as shown in figure. The mass of blocks $A$ and $B$ are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block $A$ on block $B$ is :
- A. 10 N
- B. Zero
- C. 4 N
- D. 6 N ✓
Solution: The two blocks move together, so
$$F = (M_1 + M_2)a$$
$$a = \frac{10}{2 + 3} = 2\ \text{ms}^{-2}$$
The only horizontal force on $B$ is the contact force from $A$:
$$F' = M_2 a = 3 \times 2 = 6\ \text{N}$$
Q32.
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ($P$ and $Q$ are the highest and lowest points on the wheel, respectively)?
- A. Point $P$ has zero speed
- B. Point $P$ moves slower than point $Q$
- C. Point $P$ moves faster than point $Q$ ✓
- D. Both the points $P$ and $Q$ move with equal speed
Solution: In the case of pure rolling, the contact point is instantaneously at rest and the speed of any point is proportional to its distance from that contact point.
The topmost point will have velocity $2v$, while point $Q$, i.e. the lowest point, will have zero velocity.
Hence point $P$ moves faster than point $Q$.
Q33.
The mass of a planet is $\dfrac{1}{10}^{\text{th}}$ that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
- A. 3.92 m s$^{-2}$ ✓
- B. 19.6 m s$^{-2}$
- C. 9.8 m s$^{-2}$
- D. 4.9 m s$^{-2}$
Solution: $$g = \frac{GM}{R^{2}}$$
$$\frac{g_p}{g_e} = \frac{M_p}{M_e}\left(\frac{R_e}{R_p}\right)^{2}$$
$$= \frac{1}{10} \times (2)^{2} = \frac{4}{10}$$
$$g_p = 0.4 \times 9.8 = 3.92\ \text{m s}^{-2}$$
Q34.
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^{8}$ N m$^{-2}$ and $2 \times 10^{11}$ N m$^{-2}$, is:
- A. 8 mm
- B. 4 mm ✓
- C. 0.4 mm
- D. 40 mm
Solution: In the case of maximum elongation, stress = elastic limit.
$$\delta_{\max} = \frac{\sigma_{\text{elastic}} \times L}{\text{Young's modulus}} = \frac{8 \times 10^{8} \times 1}{2 \times 10^{11}} = 4 \times 10^{-3}\ \text{m}$$
$$= 4\ \text{mm}$$
Q35.
If the monochromatic source in Young's double slit experiment is replaced by white light, then
- A. All bright fringes will be of equal width
- B. Interference pattern will disappear
- C. There will be a central dark fringe surrounded by a few coloured fringes
- D. There will be a central bright white fringe surrounded by a few coloured fringes ✓
Solution: At the central point on the screen, the path difference is zero for all wavelengths. So the central bright fringe is white.
The other fringes depend on wavelength through
$$\beta = \frac{\lambda D}{d}$$
Therefore the other fringes will be coloured.
Q36.
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
Solution: $f_0 = 140$ cm and $f_e = 5$ cm.
For a distant object,
$$m = \frac{f_0}{f_e} = \frac{140}{5} = 28$$
Q37.
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
- A. A, B and C only
- B. A, B and E only
- C. A, C and E only ✓
- D. B, D and E only
Solution: The capacitor stays connected to the battery, so $V' = V$ = constant.
(i) $C' = \dfrac{\varepsilon_0 A}{d'}$ and $C = \dfrac{\varepsilon_0 A}{d}$ with $d' < d$, so $C' > C$: the final capacitance is greater than the initial capacitance.
(ii) $U' = \dfrac{1}{2}C'V^{2}$ and $U = \dfrac{1}{2}CV^{2}$, so $U' > U$: the final energy is greater than the initial energy, so B is wrong.
(iii) $\dfrac{Q'}{V'} = C'$ and $\dfrac{Q}{V} = C$, and since $C' \neq C$, the ratio of charge to potential does not remain the same, so D is wrong.
(iv) Product of charge and voltage: $X' = Q'V = C'V^{2}$ and $X = QV = CV^{2}$, so $X' > X$.
Charge $Q = CV$ also rises with $C$, so A is right. Hence A, C and E only.
Q38.
A 10 $\mu$F capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly ($\pi$ = 3.14):
- A. 0.35 A
- B. 0.58 A
- C. 0.93 A ✓
- D. 1.20 A
Solution: Capacitive reactance
$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1000}{3.14}\ \Omega$$
$V_{\text{rms}} = 210$ V
$$i_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{210}{X_C}$$
$$\text{Peak current} = \sqrt{2}\,i_{\text{rms}} = \sqrt{2} \times \frac{210}{1000} \times 3.14 = 0.932 \simeq 0.93\ \text{A}$$
Q39.
Choose the correct circuit which can achieve the bridge balance.
- A. Circuit (1)
- B. Circuit (2) ✓
- C. Circuit (3)
- D. Circuit (4)
Solution: In option (2), the balance condition reads
$$\frac{10}{15} = \frac{10}{5 + R_D}$$
The diode can conduct and have resistance $R_D = 10\ \Omega$, because a diode has a dynamic resistance. In that case the bridge will be balanced.
Q40.
Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
- A. 2 : 3
- B. 1 : 1
- C. 2 : 9 ✓
- D. 1 : 2
Solution: Power consumed $P = \dfrac{V^{2}}{R}$, so
$$\frac{P_A}{P_B} = \frac{R_B}{R_A} \quad \Rightarrow \quad R_A = 2R_B$$
For the series combination,
$$P_S = \frac{V^{2}}{R_A + R_B} = \frac{V^{2}}{3R_B}$$
For the parallel combination,
$$P_P = \frac{V^{2}}{R_A} + \frac{V^{2}}{R_B} = \frac{V^{2}}{2R_B} + \frac{V^{2}}{R_B} = \frac{3V^{2}}{2R_B}$$
$$\frac{P_S}{P_P} = \frac{\dfrac{1}{3}}{\dfrac{3}{2}} = \frac{2}{9}$$
Q41.
The velocity ($v$) – time ($t$) plot of the motion of a body is a trapezium: the velocity rises linearly from zero, then stays constant, then falls linearly back to zero.
The acceleration ($a$) – time ($t$) graph that best suits this motion is :
- A. A negative rectangle followed by a line rising through the axis
- B. A single positive trapezium above the axis
- C. A line rising from the origin, then levelling off at a constant positive value
- D. A positive rectangle, then zero, then a negative rectangle ✓
Solution: Initially, the body has zero velocity and zero slope; hence the acceleration would be zero initially.
After that, the slope of the $v$-$t$ curve is constant and positive, so the acceleration is a constant positive value.
After some time, velocity becomes constant and acceleration is zero.
After that, the slope of the $v$-$t$ curve is constant and negative, so the acceleration is a constant negative value.
The $a$-$t$ graph is therefore a positive rectangle, followed by zero, followed by a negative rectangle.
Q42.
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\dfrac{x}{2}$ times its original time period. Then the value of $x$ is:
- A. 4
- B. $\sqrt{3}$
- C. $\sqrt{2}$ ✓
- D. $2\sqrt{3}$
Solution: The time period of a simple pendulum does not depend on the mass of the bob:
$$T' = 2\pi\sqrt{\frac{\ell'}{g}} \qquad \text{where } \ell' = \frac{\ell}{2}$$
$$T = 2\pi\sqrt{\frac{\ell}{g}}$$
Given $T' = \dfrac{x}{2}T$:
$$2\pi\sqrt{\frac{\ell}{2g}} = \frac{x}{2} \cdot 2\pi\sqrt{\frac{\ell}{g}}$$
$$\frac{1}{\sqrt{2}} = \frac{x}{2} \ \Rightarrow\ x = \sqrt{2}$$
Q43.
A metallic bar of Young's modulus, $0.5 \times 10^{11}$ N m$^{-2}$ and coefficient of linear thermal expansion $10^{-5}\ ^{\circ}$C$^{-1}$, length 1 m and area of cross-section $10^{-3}$ m$^{2}$ is heated from 0$^{\circ}$C to 100$^{\circ}$C without expansion or bending. The compressive force developed in it is :
- A. $2 \times 10^{3}$ N
- B. $5 \times 10^{3}$ N
- C. $50 \times 10^{3}$ N ✓
- D. $100 \times 10^{3}$ N
Solution: Thermal strain = longitudinal strain = $\alpha\Delta T$
$$\Rightarrow\ \text{Longitudinal strain}, \delta = 10^{-5} \times 10^{2} = 10^{-3}$$
$$\Rightarrow\ \text{Compressive stress} = \delta \times \text{Young's Modulus}$$
$$= 10^{-3} \times 0.5 \times 10^{11} = 0.5 \times 10^{8}\ \text{N m}^{-2}$$
$$\Rightarrow\ \text{Compressive force} = 0.5 \times 10^{8} \times 10^{-3} = 0.5 \times 10^{5}$$
$$= 50 \times 10^{3}\ \text{N}$$
Q44.
A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement(s) from the options given below:
- A. C only
- B. B and D only
- C. A and C only ✓
- D. A, C and D only
Solution: A. A magnetic pole will repel or attract a magnetic sheet, so a force is needed to hold it there.
B. If the sheet is non-magnetic, no force is needed.
C. If it is conducting, then there will be an eddy current in the sheet which opposes the motion, so a force is needed to move the sheet with uniform speed.
D. The non-conducting and non-polar sheet does not interact with the magnetic field of the magnet, so no force is needed.
Hence A and C only.
Q45.
The property which is not of an electromagnetic wave travelling in free space is that:
- A. They originate from charges moving with uniform speed ✓
- B. They are transverse in nature
- C. The energy density in electric field is equal to energy density in magnetic field
- D. They travel with a speed equal to $\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$
Solution: The EM waves originate from an accelerating charge. A charge moving with uniform velocity produces a steady state magnetic field and radiates no electromagnetic wave, so that statement is not a property of an electromagnetic wave.
Q46.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:
- A. Displacement current of magnitude greater than $I$ flows but can be in any direction
- B. There is no current
- C. Displacement current of magnitude equal to $I$ flows in the same direction as $I$ ✓
- D. Displacement current of magnitude equal to $I$ flows in a direction opposite to that of $I$
Solution: According to modified Ampere's law,
$$\oint \vec{B} \cdot d\vec{l} = \mu_0\left(I_C + I_D\right)$$
For a loop $L_1$ enclosing the wire: $I_C \neq 0$ and $I_D = 0$.
For a loop $L_2$ enclosing the gap between the plates: $I_C = 0$ and $I_D \neq 0$.
Since the same $\oint \vec{B} \cdot d\vec{l}$ must result, and by KCL,
$$I_C = I_D$$
so the displacement current has the same magnitude and the same direction as the conduction current.
Q47.
An iron bar of length $L$ has magnetic moment $M$. It is bent at the middle of its length such that the two arms make an angle $60^{\circ}$ with each other. The magnetic moment of this new magnet is :
- A. $\dfrac{M}{\sqrt{3}}$
- B. $M$
- C. $\dfrac{M}{2}$ ✓
- D. $2M$
Solution: The bar is bent at its midpoint, so each arm has length $\dfrac{L}{2}$ and magnetic moment $\dfrac{M}{2}$.
The two arm moments are vectors with $60^{\circ}$ between them, so the resultant is
$$M' = \sqrt{\left(\frac{M}{2}\right)^{2} + \left(\frac{M}{2}\right)^{2} + 2\left(\frac{M}{2}\right)\left(\frac{M}{2}\right)\cos 120^{\circ}}$$
The angle between the moment vectors is $120^{\circ}$ when the arms make $60^{\circ}$, giving
$$M' = \frac{M}{2}$$
Q48.
The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$ from the surface of the earth is:
- A. $\dfrac{GmM}{3R}$
- B. $\dfrac{5GmM}{6R}$ ✓
- C. $\dfrac{2GmM}{3R}$
- D. $\dfrac{GmM}{2R}$
Solution: The orbit radius is $R + 2R = 3R$. Apply energy conservation,
$$U_i + K_i = U_f + K_f$$
$$\Rightarrow\ -\frac{GMm}{R} + K_i = -\frac{GMm}{3R} + \frac{1}{2}mv^{2}$$
For a circular orbit of radius $3R$, $v^{2} = \dfrac{GM}{3R}$:
$$\Rightarrow\ -\frac{GMm}{R} + K_i = -\frac{GMm}{3R} + \frac{1}{2} \times m \times \frac{GM}{3R}$$
$$\Rightarrow\ K_i = -\frac{1}{6}\frac{GMm}{R} + \frac{GMm}{R}$$
$$K_i = \frac{5}{6}\frac{GMm}{R}$$
Q49.
The following graph represents the $T$-$V$ curves of an ideal gas (where $T$ is the temperature and $V$ the volume) at three pressures $P_1$, $P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. The curve labelled $P_1$ lies furthest to the left, $P_2$ in the middle and $P_3$ furthest to the right.
Then the correct relation is:
- A. $P_1 > P_2 > P_3$ ✓
- B. $P_3 > P_2 > P_1$
- C. $P_1 > P_3 > P_2$
- D. $P_2 > P_1 > P_3$
Solution: Draw a straight line parallel to the volume axis, i.e. take the three curves at the same temperature.
At the same temperature, the curve with the higher volume corresponds to lower pressure, since $PV = nRT$.
Reading off the graph at fixed $T$,
$$V_3 > V_2 > V_1$$
$$\Rightarrow\ P_1 > P_2 > P_3$$
Q50.
A force defined by $F = \alpha t^{2} + \beta t$ acts on a particle at a given time $t$. The factor which is dimensionless, if $\alpha$ and $\beta$ are constants, is:
- A. $\dfrac{\alpha\beta}{t}$
- B. $\dfrac{\beta t}{\alpha}$
- C. $\dfrac{\alpha t}{\beta}$ ✓
- D. $\alpha\beta t$
Solution: From the principle of homogeneity, every term must have the dimensions of force:
$$[F] = [\alpha t^{2}] = [\beta t]$$
$$[\alpha] = \frac{[F]}{[t^{2}]} \quad \text{and} \quad [\beta] = \frac{[F]}{[t]}$$
$$\therefore\ [\alpha][t] = [\beta]$$
$$\therefore\ \frac{\alpha t}{\beta} = \text{dimensionless}$$
Q51.
Given below are two statements:
Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II : Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is incorrect but Statement II is true
- B. Both statement I and Statement II are true ✓
- C. Both Statement I and Statement II are false
- D. Statement I is correct but Statement II is false
Solution: Aniline does not undergo Friedel-Crafts alkylation reaction due to salt formation with aluminium chloride, the Lewis acid which is used as a catalyst.
Aniline (an aromatic primary amine) cannot be prepared by Gabriel phthalimide synthesis, because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Both statements are therefore true.
Q52.
Match List I with List II.
List I (Compound):
A. NH$_3$
B. BrF$_5$
C. XeF$_4$
D. SF$_6$
List II (Shape/geometry):
I. Trigonal Pyramidal
II. Square Planar
III. Octahedral
IV. Square Pyramidal
Choose the correct answer from the options given below:
- A. A-II, B-III, C-IV, D-I
- B. A-I, B-IV, C-II, D-III ✓
- C. A-II, B-IV, C-III, D-I
- D. A-III, B-IV, C-I, D-II
Solution: NH$_3$ is $sp^{3}$ hybridised with 1 lone pair, so the structure will be trigonal pyramidal.
BrF$_5$ is $sp^{3}d^{2}$ hybridised with 1 lone pair, so the structure will be square pyramidal.
XeF$_4$ is $sp^{3}d^{2}$ with two lone pairs, so the structure will be square planar.
SF$_6$ is $sp^{3}d^{2}$ with no lone pair, so the structure will be octahedral.
So A-I, B-IV, C-II, D-III.
Q53.
Match List I with List II.
List I (Molecule):
A. ethane
B. ethene
C. carbon molecule, C$_2$
D. ethyne
List II (Number and types of bond/s between two carbon atoms):
I. one $\sigma$-bond and two $\pi$-bonds
II. two $\pi$-bonds
III. one $\sigma$-bond
IV. one $\sigma$-bond and one $\pi$-bond
Choose the correct answer from the options given below:
- A. A-III, B-IV, C-I, D-II
- B. A-I, B-IV, C-II, D-III
- C. A-IV, B-III, C-II, D-I
- D. A-III, B-IV, C-II, D-I ✓
Solution: (A) Ethane, H$_3$C–CH$_3$ — one (C–C) $\sigma$ bond
(B) Ethene, H$_2$C=CH$_2$ — one (C–C) $\sigma$ and one (C–C) $\pi$ bond
(C) C$_2$ — two (C–C) $\pi$ bonds, as the molecular orbital picture of dicarbon has no net $\sigma$ bond
(D) Ethyne, H–C$\equiv$C–H — two (C–C) $\pi$ bonds and one (C–C) $\sigma$ bond
So A-III, B-IV, C-II, D-I.
Q54.
For the reaction 2A $\rightleftharpoons$ B + C, K$_C$ = $4 \times 10^{-3}$. At a given time, the composition of reaction mixture is: [A] = [B] = [C] = $2 \times 10^{-3}$ M.
Then, which of the following is correct?
- A. Reaction has gone to completion in forward direction.
- B. Reaction is at equilibrium.
- C. Reaction has a tendency to go in forward direction.
- D. Reaction has a tendency to go in backward direction. ✓
Solution: At a given time $t$, $Q_C$ is to be calculated and compared with $K_C$.
$$Q_C = \frac{[B][C]}{[A]^{2}} = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^{2}}$$
$$Q_C = 1$$
As $Q_C > K_C$, the reaction has a tendency to move backward.
Q55.
Match List I with List II.
List-I (Process):
A. Isothermal process
B. Isochoric process
C. Isobaric process
D. Adiabatic process
List-II (Conditions):
I. No heat exchange
II. Carried out at constant temperature
III. Carried out at constant volume
IV. Carried out at constant pressure
Choose the correct answer from the options given below:
- A. A-II, B-III, C-IV, D-I ✓
- B. A-IV, B-III, C-II, D-I
- C. A-IV, B-II, C-III, D-I
- D. A-I, B-II, C-III, D-IV
Solution: (A) Isothermal process — temperature is constant throughout the process
(B) Isochoric process — volume is constant throughout the process
(C) Isobaric process — pressure is constant throughout the process
(D) Adiabatic process — no exchange of heat ($q$) between system and surroundings
So A-II, B-III, C-IV, D-I.
Q56.
Match List I with List II
List I (Quantum Number):
A. $m_l$
B. $m_s$
C. $l$
D. $n$
List II (Information provided):
I. Shape of orbital
II. Size of orbital
III. Orientation of orbital
IV. Orientation of spin of electron
Choose the correct answer from the options given below :
- A. A-II, B-I, C-IV, D-III
- B. A-I, B-III, C-II, D-IV
- C. A-III, B-IV, C-I, D-II ✓
- D. A-III, B-IV, C-II, D-I
Solution: Magnetic quantum number $m_l$ informs about the orientation of the orbital.
Spin quantum number $m_s$ informs about the orientation of the spin of the electron.
Azimuthal quantum number ($l$) informs about the shape of the orbital.
Principal quantum number ($n$) informs about the size of the orbital.
So A-III, B-IV, C-I, D-II.
Q57.
Arrange the following elements in increasing order of first ionization enthalpy:
Li, Be, B, C, N
Choose the correct answer from the options given below:
- A. Li < Be < N < B < C
- B. Li < Be < B < C < N
- C. Li < B < Be < C < N ✓
- D. Li < Be < C < B < N
Solution: The first ionization enthalpies ($\Delta_i$H / kJ mol$^{-1}$) are
Li — 520, Be — 899, B — 801, C — 1086, N — 1402
Boron sits below beryllium because removing boron's 2p electron is easier than removing an electron from beryllium's filled 2s subshell.
So the increasing order of first ionization enthalpy is Li < B < Be < C < N.
Q58.
Among Group 16 elements, which one does NOT show $-2$ oxidation state?
Solution: Oxygen shows $-2$, $-1$, $+1$ and $+2$ oxidation states.
Selenium shows $-2$, $+2$, $+4$ and $+6$ oxidation states.
Tellurium shows $-2$, $+2$, $+4$ and $+6$ oxidation states.
Polonium shows $+2$ and $+4$ oxidation states only, so it does not show the $-2$ state.
Q59.
In which of the following equilibria, K$_p$ and K$_c$ are NOT equal?
- A. $2\text{BrCl}_{(g)} \rightleftharpoons \text{Br}_{2(g)} + \text{Cl}_{2(g)}$
- B. $\text{PCl}_{5(g)} \rightleftharpoons \text{PCl}_{3(g)} + \text{Cl}_{2(g)}$ ✓
- C. $\text{H}_{2(g)} + \text{I}_{2(g)} \rightleftharpoons 2\text{HI}_{(g)}$
- D. $\text{CO}_{(g)} + \text{H}_2\text{O}_{(g)} \rightleftharpoons \text{CO}_{2(g)} + \text{H}_{2(g)}$
Solution: $$K_p = K_c(RT)^{\Delta n_g}$$
For $K_p \neq K_c$ we need $\Delta n_g \neq 0$, where $\Delta n_g = n_p - n_r$.
(1) $\Delta n_g = 2 - 2 = 0$
(2) $\Delta n_g = 2 - 1 = 1$
(3) $\Delta n_g = 2 - 2 = 0$
(4) $\Delta n_g = 2 - 2 = 0$
Only the PCl$_5$ dissociation has $\Delta n_g \neq 0$.
Q60.
The reagents with which glucose does not react to give the corresponding tests/products are
A. Tollen's reagent
B. Schiff's reagent
C. HCN
D. NH$_2$OH
E. NaHSO$_3$
Choose the correct options from the given below:
- A. E and D
- B. B and C
- C. A and D
- D. B and E ✓
Solution: Despite having the aldehyde group, glucose does not give Schiff's test and it does not form the hydrogen sulphite addition product with NaHSO$_3$.
These two facts are part of the evidence that the aldehyde group in glucose is not free but is tied up in the cyclic hemiacetal form.
Glucose does react with Tollen's reagent, with HCN to give the cyanohydrin, and with NH$_2$OH to give the oxime.
Q61.
Identify the correct reagents that would bring about the following transformation.
Allylbenzene, C$_6$H$_5$CH$_2$–CH=CH$_2$, is converted to 3-phenylpropanal, C$_6$H$_5$CH$_2$–CH$_2$–CHO.
- A. (i) H$_2$O/H$^{+}$, (ii) PCC
- B. (i) H$_2$O/H$^{+}$, (ii) CrO$_3$
- C. (i) BH$_3$, (ii) H$_2$O$_2$/$\overset{\ominus}{\text{O}}$H, (iii) PCC ✓
- D. (i) BH$_3$, (ii) H$_2$O$_2$/$\overset{\ominus}{\text{O}}$H, (iii) alk. KMnO$_4$, (iv) H$_3$O$^{\oplus}$
Solution: The aldehyde carbon ends up on the terminal carbon of the chain, so the water must add anti-Markovnikov. That is hydroboration-oxidation, not acid-catalysed hydration.
BH$_3$ adds across the double bond with boron on the less substituted carbon; H$_2$O$_2$/OH$^{-}$ then replaces boron by OH, giving the primary alcohol C$_6$H$_5$CH$_2$CH$_2$CH$_2$OH.
PCC oxidises a primary alcohol only as far as the aldehyde, giving C$_6$H$_5$CH$_2$CH$_2$CHO.
Alkaline KMnO$_4$ in option (4) would carry the oxidation on to the carboxylic acid, and H$_2$O/H$^{+}$ in options (1) and (2) would put the OH on the middle carbon.
Q62.
The E$^{\circ}$ value for the Mn$^{3+}$/Mn$^{2+}$ couple is more positive than that of Cr$^{3+}$/Cr$^{2+}$ or Fe$^{3+}$/Fe$^{2+}$ due to change of
- A. $d^{3}$ to $d^{5}$ configuration
- B. $d^{5}$ to $d^{4}$ configuration
- C. $d^{5}$ to $d^{2}$ configuration
- D. $d^{4}$ to $d^{5}$ configuration ✓
Solution: $$E^{\circ}_{\text{Mn}^{3+}/\text{Mn}^{2+}} > E^{\circ}_{\text{Cr}^{3+}/\text{Cr}^{2+}} \ \text{or}\ E^{\circ}_{\text{Fe}^{3+}/\text{Fe}^{2+}}$$
Electronic configuration of Mn$^{3+}$ = [Ar]3$d^{4}$
Electronic configuration of Mn$^{2+}$ = [Ar]3$d^{5}$
Electronic configuration of Cr$^{3+}$ = [Ar]3$d^{3}$
Electronic configuration of Cr$^{2+}$ = [Ar]3$d^{4}$
As Mn$^{3+}$ goes from the $d^{4}$ configuration to the more stable $d^{5}$ configuration (half filled), due to the greater exchange energy in the $d^{5}$ configuration, the reduction is easier and E$^{\circ}$ is more positive.
Q63.
Given below are two statements :
Statement I: Both [Co(NH$_3$)$_6$]$^{3+}$ and [CoF$_6$]$^{3-}$ complexes are octahedral but differ in their magnetic behaviour.
Statement II: [Co(NH$_3$)$_6$]$^{3+}$ is diamagnetic whereas [CoF$_6$]$^{3-}$ is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true ✓
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false
Solution: In [Co(NH$_3$)$_6$]$^{3+}$, the Co$^{3+}$ ion has a 3$d^{6}$ configuration. In the presence of the strong field NH$_3$ ligand, pairing of electrons takes place and it becomes a diamagnetic complex ion.
$$\therefore\ [\text{Co(NH}_3)_6]^{3+} \text{ is octahedral with } d^{2}sp^{3} \text{ hybridisation and is diamagnetic in nature.}$$
In the case of [CoF$_6$]$^{3-}$, Co is in the $+3$ oxidation state and also has a 3$d^{6}$ configuration. In the presence of the weak field F$^{-}$ ligand, pairing does not take place.
$$\therefore\ \text{In } [\text{CoF}_6]^{3-}, \text{Co}^{3+} \text{ is } sp^{3}d^{2} \text{ hybridised with four unpaired electrons, so it is paramagnetic in nature.}$$
Both statements are therefore true.
Q64.
Fehling's solution 'A' is
- A. aqueous sodium citrate
- B. aqueous copper sulphate ✓
- C. alkaline copper sulphate
- D. alkaline solution of sodium potassium tartrate (Rochelle's salt)
Solution: Fehling solution 'A' = aqueous copper sulphate
Fehling solution 'B' = alkaline sodium potassium tartrate (Rochelle salt)
The two are mixed in equal volumes immediately before use.
Q65.
In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C. $2\text{NaHCO}_{3(s)} \rightarrow \text{Na}_2\text{CO}_{3(s)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(g)}$
D. $\text{Cl}_{2(g)} \rightarrow 2\text{Cl}_{(g)}$
Choose the correct answer from the options given below:
- A. C and D
- B. A and C
- C. A, B and D
- D. A, C and D ✓
Solution: When a liquid evaporates to vapour, entropy increases.
Lowering the temperature of a crystalline solid from 130 K to 0 K decreases the entropy, so B is wrong.
$$2\text{NaHCO}_{3(s)} \rightarrow \text{Na}_2\text{CO}_{3(s)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(g)}$$
The number of gaseous product molecules increases, so entropy increases.
$$\text{Cl}_{2(g)} \rightarrow 2\text{Cl}_{(g)}$$
1 mole of Cl$_{2(g)}$ forms 2 mol Cl$_{(g)}$, so entropy increases.
Hence A, C and D.
Q66.
The energy of an electron in the ground state ($n$ = 1) for He$^{+}$ ion is $-x$ J, then that for an electron in $n$ = 2 state for Be$^{3+}$ ion in J is
- A. $-\dfrac{4}{9}x$
- B. $-x$ ✓
- C. $-\dfrac{x}{9}$
- D. $-4x$
Solution: $$E_n = -R_H\left(\frac{Z^{2}}{n^{2}}\right)\ \text{J}$$
For He$^{+}$ ($n$ = 1, $Z$ = 2),
$$E_n = -x = -R_H\left(\frac{2^{2}}{1^{2}}\right) = -4R_H$$
$$\therefore\ R_H = \frac{x}{4}$$
For Be$^{3+}$ ($n$ = 2, $Z$ = 4),
$$E_n = -R_H\left(\frac{Z^{2}}{n^{2}}\right) = -\frac{x}{4} \times \left(\frac{4 \times 4}{2 \times 2}\right) = -x\ \text{J}$$
Q67.
Match List I with List II.
List I (Reaction):
A. Bicyclohexylidene (two cyclohexane rings joined by a C=C) gives 2 molecules of cyclohexanone
B. Benzene gives benzophenone (diphenyl ketone)
C. Cyclohexanol gives cyclohexanone
D. Ethylbenzene gives potassium benzoate
List II (Reagents/Condition):
I. C$_6$H$_5$COCl / Anhyd. AlCl$_3$
II. CrO$_3$
III. KMnO$_4$/KOH, $\Delta$
IV. (i) O$_3$ (ii) Zn-H$_2$O
Choose the correct answer from the options given below:
- A. A-I, B-IV, C-II, D-III
- B. A-IV, B-I, C-III, D-II
- C. A-III, B-I, C-II, D-IV
- D. A-IV, B-I, C-II, D-III ✓
Solution: A. Reductive ozonolysis, (i) O$_3$ then (ii) Zn-H$_2$O, cleaves the C=C of bicyclohexylidene into two carbonyl groups, giving 2 molecules of cyclohexanone.
B. Friedel-Crafts acylation of benzene with benzoyl chloride and anhydrous AlCl$_3$ gives benzophenone.
C. CrO$_3$ oxidises the secondary alcohol cyclohexanol to the ketone cyclohexanone.
D. Hot alkaline KMnO$_4$ oxidises the whole side chain of ethylbenzene to a carboxylate, giving potassium benzoate.
So A-IV, B-I, C-II, D-III.
Q68.
Activation energy of any chemical reaction can be calculated if one knows the value of
- A. rate constant at two different temperatures ✓
- B. rate constant at standard temperature
- C. probability of collision
- D. orientation of reactant molecules during collision
Solution: To calculate the value of $E_a$, the equation used is
$$\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
Hence $E_a$ can be calculated if the value of the rate constant $k$ is known at two different temperatures $T_1$ and $T_2$.
Q69.
The most stable carbocation among the following is :
- A. The 1-methylcyclohexyl cation (a cyclohexane ring carrying a methyl group and the positive charge on the same carbon) ✓
- B. CH$_3$CH$_2$–$\overset{\oplus}{\text{C}}$H–CH(CH$_3$)$_2$
- C. CH$_3$–$\overset{\oplus}{\text{C}}$H–CH$_2$–CH(CH$_3$)$_2$
- D. Cyclopentyl–$\overset{\oplus}{\text{C}}$H$_2$
Solution: The stability of a carbocation can be described by hyperconjugation. The greater the extent of hyperconjugation, the more is the stability of the carbocation.
Counting the $\alpha$-hydrogens available for hyperconjugation:
(1) 7 $\alpha$-H
(2) 3 $\alpha$-H
(3) 5 $\alpha$-H
(4) 1 $\alpha$-H
Stability order of carbocations = (1) > (3) > (2) > (4)
Option (1) is also tertiary while (4) is primary, which reinforces the same order.
Q70.
Given below are two statements:
Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane
Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are correct ✓
- C. Both Statement I and Statement II are incorrect
- D. Statement I is correct but Statement II is incorrect
Solution: Both statement I and statement II are correct.
Boiling point of n-pentane = 309 K, isopentane = 301 K, neopentane = 282.5 K.
As branching increases, molecules attain the shape of a sphere, which results in a smaller area of contact and thus weak intermolecular forces between the spherical molecules. These are overcome at relatively lower temperature, leading to a decrease in boiling point.
Q71.
Which one of the following alcohols reacts instantaneously with Lucas reagent?
- A. (CH$_3$)$_3$C–OH ✓
- B. CH$_3$–CH$_2$–CH$_2$–CH$_2$OH
- C. CH$_3$–CH$_2$–CH(OH)–CH$_3$
- D. CH$_3$–CH(CH$_3$)–CH$_2$OH
Solution: Tertiary alcohols react instantaneously with Lucas reagent and give immediate turbidity.
In the case of tertiary alcohols, they form halides easily with Lucas reagent (conc. HCl and ZnCl$_2$), because the reaction goes through a stable tertiary carbocation.
Of the four, only (CH$_3$)$_3$C–OH is tertiary; the others are primary or secondary.
Q72.
The Henry's law constant (K$_H$) values of three gases (A, B, C) in water are 145, $2 \times 10^{-5}$ and 35 kbar, respectively. The solubility of these gases in water follow the order:
- A. A > B > C
- B. B > A > C
- C. B > C > A ✓
- D. A > C > B
Solution: Value of Henry's law constant $\propto \dfrac{1}{\text{Solubility of gas}}$
The higher the value of K$_H$ at a given pressure, the lower is the solubility of the gas in the liquid.
K$_H$ value of gases (given): A > C > B
$$\therefore\ \text{Order of solubility of gases in water : B > C > A}$$
Q73.
Which plot of $\ln k$ vs $\dfrac{1}{T}$ is consistent with Arrhenius equation?
- A. A straight line of negative slope, starting high on the $\ln k$ axis and falling ✓
- B. A straight line of positive slope starting from the $\ln k$ axis above the origin
- C. A straight line of positive slope starting at a lower intercept
- D. A straight line of positive slope passing near the origin
Solution: The Arrhenius equation is given as
$$k = Ae^{-\frac{E_a}{RT}}$$
$$\therefore\ \ln k = \ln A - \frac{E_a}{RT}$$
$\ln k$ v/s $\dfrac{1}{T}$ gives a straight line graph with slope $= -\dfrac{E_a}{R}$ and intercept $= \ln A$.
Since $E_a$ is positive, the slope is negative: the line falls as $\dfrac{1}{T}$ increases.
Q74.
On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as
- A. Chromatography
- B. Crystallization
- C. Sublimation ✓
- D. Distillation
Solution: The direct conversion of a solid into vapour without passing through the liquid state is called sublimation.
The technique that separates a sublimable volatile compound from a non-sublimable impurity on this principle is known as sublimation.
Q75.
Intramolecular hydrogen bonding is present in
- A. HF
- B. o-Nitrophenol ✓
- C. p-Nitrophenol
- D. m-Nitrophenol
Solution: In o-nitrophenol, intramolecular H-bonding is present.
The OH group and the NO$_2$ group sit on adjacent ring carbons, so the hydroxyl hydrogen can reach an oxygen of the nitro group and close a six-membered ring within the same molecule.
In the para and meta isomers the two groups are too far apart, so the hydrogen bonding there is intermolecular. HF forms intermolecular hydrogen bonds only.
Q76.
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to
- A. 200 mg
- B. 750 mg
- C. 250 mg ✓
- D. Zero mg
Solution: $$M = \frac{W \times 1000}{M_2 \times V\ (\text{in mL})}$$
$$W = \frac{M \times M_2 \times V\ (\text{in mL})}{1000} = \frac{0.75 \times 36.5 \times 25}{1000}$$
$$= 0.684\ \text{g (Mass of HCl)}$$
$$\text{HCl} + \text{NaOH} \longrightarrow \text{NaCl} + \text{H}_2\text{O}$$
36.5 g HCl reacts with NaOH = 40 g
$$0.684\ \text{g HCl reacts with NaOH} = \frac{40}{36.5} \times 0.684 = 0.750\ \text{g}$$
Amount of NaOH left = 1 g $-$ 0.750 g = 0.250 g = 250 mg
Q77.
Which reaction is NOT a redox reaction?
- A. BaCl$_2$ + Na$_2$SO$_4$ $\rightarrow$ BaSO$_4$ + 2NaCl ✓
- B. Zn + CuSO$_4$ $\rightarrow$ ZnSO$_4$ + Cu
- C. 2KClO$_3$ + I$_2$ $\rightarrow$ 2KIO$_3$ + Cl$_2$
- D. H$_2$ + Cl$_2$ $\rightarrow$ 2HCl
Solution: In BaCl$_2$ + Na$_2$SO$_4$ $\rightarrow$ BaSO$_4$ + 2NaCl, barium stays $+2$, chlorine stays $-1$, sodium stays $+1$ and the sulphate is unchanged. This is not a redox reaction, as there is no change in oxidation state.
In Zn + CuSO$_4$ $\rightarrow$ ZnSO$_4$ + Cu, Zn goes from 0 to $+2$ (oxidation) and Cu from $+2$ to 0 (reduction).
In 2KClO$_3$ + I$_2$ $\rightarrow$ 2KIO$_3$ + Cl$_2$, iodine goes 0 to $+5$ and chlorine $+5$ to 0.
In H$_2$ + Cl$_2$ $\rightarrow$ 2HCl, hydrogen goes 0 to $+1$ and chlorine 0 to $-1$.
Q78.
'Spin only' magnetic moment is same for which of the following ions?
A. Ti$^{3+}$
B. Cr$^{2+}$
C. Mn$^{2+}$
D. Fe$^{2+}$
E. Sc$^{3+}$
Choose the most appropriate answer from the options given below.
- A. A and D only
- B. B and D only ✓
- C. A and E only
- D. B and C only
Solution: Counting unpaired electrons:
Ti$^{3+}$ — 3$d^{1}$ — 1 unpaired electron
Cr$^{2+}$ — 3$d^{4}$ — 4 unpaired electrons
Mn$^{2+}$ — 3$d^{5}$ — 5 unpaired electrons
Fe$^{2+}$ — 3$d^{6}$ — 4 unpaired electrons
Sc$^{3+}$ — 3$d^{0}$ — 0 unpaired electrons
Spin only magnetic moment is given by $\sqrt{n(n + 2)}$ BM.
$$\therefore\ \text{Cr}^{2+} \text{ and Fe}^{2+} \text{ will have the same spin only magnetic moment.}$$
Q79.
The highest number of helium atoms is in
- A. 2.271098 L of helium at STP
- B. 4 mol of helium ✓
- C. 4 u of helium
- D. 4 g of helium
Solution: (1) 2.271098 L of He at STP $= \dfrac{2.271}{22.710982}$ mole $= 0.1$ mole $= 0.1\,N_A$ He atoms
(2) 4 mol of He $= 4N_A$ He atoms
(3) 4 u of He $= \dfrac{4\ \text{u}}{4\ \text{u}} = 1$ He atom
(4) 4 g of Helium $= \dfrac{4\ \text{g}}{4\ \text{g}}$ mole $= 1$ mole $= N_A$ He atoms
The largest of these is 4 mol of helium.
Q80.
Match List I with List II.
List I (Complex):
A. [Co(NH$_3$)$_5$(NO$_2$)]Cl$_2$
B. [Co(NH$_3$)$_5$(SO$_4$)]Br
C. [Co(NH$_3$)$_6$][Cr(CN)$_6$]
D. [Co(H$_2$O)$_6$]Cl$_3$
List II (Type of isomerism):
I. Solvate isomerism
II. Linkage isomerism
III. Ionization isomerism
IV. Coordination isomerism
Choose the correct answer from the options given below:
- A. A-II, B-IV, C-III, D-I
- B. A-II, B-III, C-IV, D-I ✓
- C. A-I, B-III, C-IV, D-II
- D. A-I, B-IV, C-III, D-II
Solution: A. [Co(NH$_3$)$_5$(NO$_2$)]Cl$_2$ — linkage isomerism, due to 'N' and 'O' linkage by NO$_2$
B. [Co(NH$_3$)$_5$(SO$_4$)]Br — ionization isomerism, as sulphate and bromide can exchange between the sphere and the counter ion
C. [Co(NH$_3$)$_6$][Cr(CN)$_6$] — coordination isomerism, as the ligands can be swapped between the two metal centres
D. [Co(H$_2$O)$_6$]Cl$_3$ — solvate isomerism, according to how many water molecules sit inside the coordination sphere
So A-II, B-III, C-IV, D-I.
Q81.
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follow the order H$_2$O > H$_2$Te > H$_2$Se > H$_2$S.
Statement II: On the basis of molecular mass, H$_2$O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H$_2$O, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true ✓
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false
Solution: Statement I is correct, because the boiling point of hydrides of group 16 follows the order H$_2$O > H$_2$Te > H$_2$Se > H$_2$S.
Statement II is correct too: due to intermolecular H-bonding, H$_2$O shows a higher boiling point than the respective hydrides of group 16.
The order from H$_2$Te down to H$_2$S is due to decreasing molar mass, and hence weaker van der Waals forces.
Q82.
A compound with a molecular formula of C$_6$H$_{14}$ has two tertiary carbons. Its IUPAC name is :
- A. 2,2-dimethylbutane
- B. n-hexane
- C. 2-methylpentane
- D. 2,3-dimethylbutane ✓
Solution: A tertiary carbon is one attached to three other carbon atoms.
In 2,3-dimethylbutane, (CH$_3$)$_2$CH–CH(CH$_3$)$_2$, both C-2 and C-3 carry a methyl branch and are each bonded to three carbons, so there are two tertiary carbons.
n-Hexane has none, 2-methylpentane has one, and 2,2-dimethylbutane has a quaternary carbon rather than two tertiary ones.
Q83.
Match List I with List II.
List I (Conversion):
A. 1 mol of H$_2$O to O$_2$
B. 1 mol of MnO$_4^{-}$ to Mn$^{2+}$
C. 1.5 mol of Ca from molten CaCl$_2$
D. 1 mol of FeO to Fe$_2$O$_3$
List II (Number of Faraday required):
I. 3F
II. 2F
III. 1F
IV. 5F
Choose the correct answer from the options given below:
- A. A-III, B-IV, C-II, D-I
- B. A-II, B-IV, C-I, D-III ✓
- C. A-III, B-IV, C-I, D-II
- D. A-II, B-III, C-I, D-IV
Solution: $$4\text{OH}^{-} \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^{-}$$
For 2 mole of H$_2$O, 4F charge is required, so for 1 mole of H$_2$O $= \dfrac{4F}{2} = 2F$ required.
$$\overset{+7}{\text{Mn}}\text{O}_4^{-} \rightarrow \overset{+2}{\text{Mn}}^{2+}$$
For 1 mole MnO$_4^{-}$, 5F charge is required.
$$\text{Ca}^{2+} \xrightarrow{\ +2e^{-}\ } \text{Ca}$$
For 1 mole Ca$^{2+}$ ion required = 2F, so 1.5 mole Ca$^{2+}$ requires $\dfrac{2}{1} \times 1.5 = 3F$.
$$\overset{+2}{\text{Fe}}\text{O} \rightarrow \overset{+3}{\text{Fe}_2}\text{O}_3$$
For 1 mole FeO, 1F charge is required.
So A-II, B-IV, C-I, D-III.
Q84.
The compound that will undergo S$_N$1 reaction with the fastest rate is
- A. (1-Bromoethyl)benzene, C$_6$H$_5$CH(Br)CH$_3$ ✓
- B. (Bromomethyl)cyclohexane
- C. Bromocyclohexane
- D. Bromobenzene
Solution: Reactivity towards S$_N$1 depends upon the stability of the carbocation formed.
The order of stability of the corresponding cations is
benzylic (C$_6$H$_5$$\overset{\oplus}{\text{C}}$H–CH$_3$) > cyclohexyl (secondary) > cyclohexylmethyl (primary) > phenyl
The benzylic cation is resonance-stabilised by the ring, and the phenyl cation cannot form at all.
Hence (1-bromoethyl)benzene is most reactive.
Q85.
Arrange the following elements in increasing order of electronegativity:
N, O, F, C, Si
Choose the correct answer from the options given below:
- A. F < O < N < C < Si
- B. Si < C < N < O < F ✓
- C. Si < C < O < N < F
- D. O < F < N < C < Si
Solution: Electronegativity increases across the period on moving left to right. It decreases on moving down the group.
Si lies below C in group 14, so it is the least electronegative. Across period 2, C < N < O < F.
The correct option is Si < C < N < O < F.
Q86.
Major products A and B formed in the following reaction sequence, are
2-Methylcyclohexan-1-ol is treated with PBr$_3$ to give A (major), which on treatment with alc. KOH and heat gives B (major).
- A. A = a bromo-alcohol (OH retained, Br added); B = 2-methylcyclohexanone
- B. A = 1-bromo-2-methylcyclohexane; B = 1-methylcyclohexene ✓
- C. A = 1-bromo-2-methylcyclohexane; B = 3-methylcyclohexene
- D. A = a bromo-alcohol (OH retained, Br added); B = a methylcyclohexenol
Solution: PBr$_3$ replaces the hydroxyl group of the alcohol by bromine, giving A = 1-bromo-2-methylcyclohexane. The OH is not retained.
Alcoholic KOH on heating brings about dehydrohalogenation: the base removes a $\beta$-hydrogen and bromide leaves, forming an alkene.
By Saytzeff's rule the more substituted alkene is the major product, so B = 1-methylcyclohexene, in which the double bond carries the methyl-bearing carbon.
Q87.
Identify the correct answer.
- A. Three canonical forms can be drawn for CO$_3^{2-}$ ion ✓
- B. Three resonance structures can be drawn for ozone
- C. BF$_3$ has non-zero dipole moment
- D. Dipole moment of NF$_3$ is greater than that of NH$_3$
Solution: The carbonate ion has one C=O and two C–O$^{-}$ bonds, and the double bond can be placed on any of the three oxygens, so three equivalent canonical forms can be drawn for CO$_3^{2-}$.
Ozone has only two resonance structures, not three.
BF$_3$ is trigonal planar and symmetric, so its bond dipoles cancel and the dipole moment is zero.
In NH$_3$ the bond dipoles and the lone pair moment point the same way, while in NF$_3$ they oppose, so the dipole moment of NF$_3$ is smaller than that of NH$_3$.
Q88.
The work done during reversible isothermal expansion of one mole of hydrogen gas at 25$^{\circ}$C from pressure of 20 atmosphere to 10 atmosphere is
(Given R = 2.0 cal K$^{-1}$ mol$^{-1}$)
- A. 100 calories
- B. 0 calorie
- C. $-413.14$ calories ✓
- D. 413.14 calories
Solution: $$W_{\text{rev, iso}} = -2.303\,nRT\log\frac{P_i}{P_f}$$
$$= -2.303 \times 1 \times 2 \times 298 \times \log 2$$
$$= -2.303 \times 1 \times 2 \times 298 \times 0.3$$
$$= -413.14\ \text{calories}$$
The sign is negative because the gas expands and therefore does work on the surroundings.
Q89.
Consider the following reaction in a sealed vessel at equilibrium with concentrations of N$_2$ = $3.0 \times 10^{-3}$ M, O$_2$ = $4.2 \times 10^{-3}$ M and NO = $2.8 \times 10^{-3}$ M.
$$2\text{NO}_{(g)} \rightleftharpoons \text{N}_{2(g)} + \text{O}_{2(g)}$$
If 0.1 mol L$^{-1}$ of NO$_{(g)}$ is taken in a closed vessel, what will be degree of dissociation ($\alpha$) of NO$_{(g)}$ at equilibrium?
- A. 0.717 ✓
- B. 0.00889
- C. 0.0889
- D. 0.8889
Solution: $$K_c = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^{2}} = \frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} = 1.607$$
Starting with 0.1 M NO:
at equilibrium [NO] $= 0.1 - 0.1\alpha$, [N$_2$] $= 0.05\alpha$, [O$_2$] $= 0.05\alpha$
$$K_c = \frac{0.05\alpha \times 0.05\alpha}{(0.1 - 0.1\alpha)^{2}} = \frac{(0.05)^{2}\alpha^{2}}{0.01(1 - \alpha)^{2}}$$
$$1.607 = \frac{(0.05)^{2}\alpha^{2}}{0.01(1 - \alpha)^{2}}$$
$$\frac{\alpha^{2}}{(1 - \alpha)^{2}} = \frac{1.607 \times (0.1)^{2}}{(0.05)^{2}}$$
$$\frac{\alpha}{1 - \alpha} = \frac{1.27 \times 0.1}{0.05} = 2.54$$
$$\alpha = 2.54 - 2.54\alpha \ \Rightarrow\ 3.54\alpha = 2.54$$
$$\alpha = \frac{2.54}{3.54} = 0.717$$
Q90.
For the given reaction:
1,2-Dicyclohexylethene, in which each carbon of the C=C carries one cyclohexyl group and one hydrogen, is treated with KMnO$_4$/H$^{+}$ to give 'P' (major product).
'P' is
- A. A 1,2-diketone, cyclohexyl–CO–CO–cyclohexyl
- B. Cyclohexanecarbaldehyde, cyclohexyl–CHO
- C. Cyclohexanecarboxylic acid, cyclohexyl–COOH ✓
- D. A 1,2-diol, cyclohexyl–CH(OH)–CH(OH)–cyclohexyl
Solution: Acidified KMnO$_4$ brings about oxidative cleavage of the carbon-carbon double bond.
Each doubly-bonded carbon carries one hydrogen, so on cleavage each fragment is oxidised past the aldehyde stage to a carboxylic acid.
$$\text{C}_6\text{H}_{11}\text{CH}=\text{CHC}_6\text{H}_{11} \xrightarrow{\ \text{KMnO}_4/\text{H}^{+}\ } 2\ \text{C}_6\text{H}_{11}\text{COOH}$$
The major product is cyclohexanecarboxylic acid.
Q91.
The pair of lanthanoid ions which are diamagnetic is
- A. Pm$^{3+}$ and Sm$^{3+}$
- B. Ce$^{4+}$ and Yb$^{2+}$ ✓
- C. Ce$^{3+}$ and Eu$^{2+}$
- D. Gd$^{3+}$ and Eu$^{3+}$
Solution: Magnetic moment $\mu = \sqrt{n(n + 2)}$, where $n$ is the number of unpaired electrons. A species is diamagnetic only when $n = 0$.
Ce$^{4+}$ — (Xe)4$f^{0}$ — $\mu = 0$ — diamagnetic
Yb$^{2+}$ — (Xe)4$f^{14}$ — $\mu = 0$ — diamagnetic
Ce$^{3+}$ — (Xe)4$f^{1}$ — $\mu = \sqrt{3}$ — paramagnetic
Eu$^{2+}$ — (Xe)4$f^{7}$ — $\mu = \sqrt{63}$ — paramagnetic
Gd$^{3+}$ — (Xe)4$f^{7}$ — $\mu = \sqrt{63}$ — paramagnetic
Eu$^{3+}$ — (Xe)4$f^{6}$ — $\mu = \sqrt{48}$ — paramagnetic
Pm$^{3+}$ — (Xe)4$f^{4}$ — $\mu = \sqrt{24}$ — paramagnetic
Sm$^{3+}$ — (Xe)4$f^{5}$ — $\mu = \sqrt{35}$ — paramagnetic
Hence Ce$^{4+}$ and Yb$^{2+}$ are the only diamagnetic pair.
Q92.
The products A and B obtained in the following reactions, respectively, are
3ROH + PCl$_3$ $\rightarrow$ 3RCl + A
ROH + PCl$_5$ $\rightarrow$ RCl + HCl + B
- A. H$_3$PO$_3$ and POCl$_3$ ✓
- B. POCl$_3$ and H$_3$PO$_3$
- C. POCl$_3$ and H$_3$PO$_4$
- D. H$_3$PO$_4$ and POCl$_3$
Solution: These reactions are the preparation of haloalkanes from alcohols.
$$3\text{ROH} + \text{PCl}_3 \longrightarrow 3\text{RCl} + \text{H}_3\text{PO}_3$$
$$\text{ROH} + \text{PCl}_5 \longrightarrow \text{RCl} + \text{HCl} + \text{POCl}_3$$
A and B are H$_3$PO$_3$ and POCl$_3$ respectively.
Q93.
Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI.
A. Al$^{3+}$
B. Cu$^{2+}$
C. Ba$^{2+}$
D. Co$^{2+}$
E. Mg$^{2+}$
Choose the correct answer from the options given below.
- A. E, A, B, C, D
- B. B, A, D, C, E ✓
- C. B, C, A, D, E
- D. E, C, D, B, A
Solution: Placing each cation in its analytical group:
Group-II — Cu$^{2+}$
Group-III — Al$^{3+}$
Group-IV — Co$^{2+}$
Group-V — Ba$^{2+}$
Group-VI — Mg$^{2+}$
The correct order of group number of ions is Cu$^{2+}$ < Al$^{3+}$ < Co$^{2+}$ < Ba$^{2+}$ < Mg$^{2+}$
$$\therefore\ \text{The correct order is B, A, D, C, E}$$
Q94.
Given below are two statements :
Statement I : [Co(NH$_3$)$_6$]$^{3+}$ is a homoleptic complex whereas [Co(NH$_3$)$_4$Cl$_2$]$^{+}$ is a heteroleptic complex.
Statement II : Complex [Co(NH$_3$)$_6$]$^{3+}$ has only one kind of ligands but [Co(NH$_3$)$_4$Cl$_2$]$^{+}$ has more than one kind of ligands.
In the light of the above statements, choose the correct answer from the options given below.
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true ✓
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false
Solution: [Co(NH$_3$)$_6$]$^{3+}$ is a homoleptic complex, as only one type of ligand (NH$_3$) is coordinated with the Co$^{3+}$ ion.
While [Co(NH$_3$)$_4$Cl$_2$]$^{+}$ is a heteroleptic complex, in which the Co$^{3+}$ ion is ligated with more than one type of ligand, i.e. NH$_3$ and Cl$^{-}$.
Both statements are true, and the second is exactly the definition that makes the first correct.
Q95.
A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is :
(Given atomic masses of A = 64; B = 40; C = 32 u)
- A. ABC$_4$
- B. A$_2$BC$_2$
- C. ABC$_3$ ✓
- D. AB$_2$C$_2$
Solution: The percentage of C is $100 - 32 - 20 = 48\%$.
Moles of each element per 100 g:
A: $\dfrac{32}{64} = \dfrac{1}{2}$
B: $\dfrac{20}{40} = \dfrac{1}{2}$
C: $\dfrac{48}{32} = \dfrac{3}{2}$
Dividing by the smallest and clearing the halves by multiplying by 2:
A : B : C = 1 : 1 : 3
$$\therefore\ \text{The correct empirical formula of compound X is ABC}_3$$
Q96.
The plot of osmotic pressure ($\Pi$) vs concentration (mol L$^{-1}$) for a solution gives a straight line with slope 25.73 L bar mol$^{-1}$. The temperature at which the osmotic pressure measurement is done is
(Use R = 0.083 L bar mol$^{-1}$ K$^{-1}$)
- A. 12.05$^{\circ}$C
- B. 37$^{\circ}$C ✓
- C. 310$^{\circ}$C
- D. 25.73$^{\circ}$C
Solution: $$\Pi = CRT$$
So a plot of $\Pi$ against $C$ is a straight line of slope $RT$:
$$\text{Slope} = RT$$
$$25.73 = 0.083 \times T$$
$$T = \frac{25.73}{0.083} = 309.47 \approx 310\ \text{K}$$
$$\therefore\ \text{Temperature in }^{\circ}\text{C} = 310 - 273 = 37^{\circ}\text{C}$$
Q97.
Identify the major product C formed in the following reaction sequence:
CH$_3$–CH$_2$–CH$_2$–I reacts with NaCN to give A; A on partial hydrolysis with OH$^{-}$ gives B; B with NaOH/Br$_2$ gives C (major).
- A. $\alpha$-bromobutanoic acid
- B. propylamine ✓
- C. butylamine
- D. butanamide
Solution: $$\text{CH}_3\text{CH}_2\text{CH}_2\text{I} \xrightarrow{\ \text{NaCN}\ } \text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\ (A)$$
Step-I is an S$_N$ reaction with the $\overset{\ominus}{\text{C}}$N nucleophile.
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} \xrightarrow{\ \text{partial hydrolysis}\ } \text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2\ (B)$$
Step-II gives the amide.
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2 \xrightarrow{\ \text{NaOH}/\text{Br}_2\ } \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2\ (C)$$
Step-III is the Hoffmann bromamide degradation reaction, which shortens the chain by one carbon, so C is propylamine.
Q98.
During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe$^{2+}$ ion?
- A. dilute sulphuric acid ✓
- B. dilute hydrochloric acid
- C. concentrated sulphuric acid
- D. dilute nitric acid
Solution: During the preparation of Mohr's salt, dilute sulphuric acid is added to prevent the hydrolysis of the Fe$^{2+}$ ion.
Nitric acid would oxidise Fe$^{2+}$ to Fe$^{3+}$, and hydrochloric acid introduces chloride which interferes with the later titration.
Q99.
The rate of a reaction quadruples when temperature changes from 27$^{\circ}$C to 57$^{\circ}$C. Calculate the energy of activation.
Given R = 8.314 J K$^{-1}$ mol$^{-1}$, log 4 = 0.6021
- A. 3804 kJ/mol
- B. 38.04 kJ/mol ✓
- C. 380.4 kJ/mol
- D. 3.80 kJ/mol
Solution: $$\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
$$\log\left(\frac{4}{1}\right) = \frac{E_a}{2.303R}\left(\frac{1}{300} - \frac{1}{330}\right)$$
$$E_a = \frac{\left(\log 4\right) \times 2.303 \times 8.314 \times 300 \times 330}{30}$$
$$= 3.804 \times 10^{4}\ \text{J/mol} = 38.04\ \text{kJ/mol}$$
Q100.
Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol$^{-1}$, 1 F = 96487 C)
- A. 0.0315 g
- B. 3.15 g
- C. 0.315 g ✓
- D. 31.5 g
Solution: $$\text{Cu}^{2+}(aq) + 2e^{-} \rightarrow \text{Cu}(s)$$
Mass of Cu deposited
$$w = \frac{M \times i \times t}{nF}$$
$$= \frac{63 \times 9.6487 \times 100}{2 \times 96487}$$
$$= 0.315\ \text{g}$$
Q101.
The cofactor of the enzyme carboxypeptidase is:
- A. Haem
- B. Zinc ✓
- C. Niacin
- D. Flavin
Solution: The cofactor of the enzyme carboxypeptidase is zinc, a metal ion cofactor held by coordinate bonds to the side chains of the enzyme.
Niacin is associated with coenzyme NAD and NADP.
Haem is the prosthetic group in peroxidase and catalase, not in carboxypeptidase.
Q102.
Given below are two statements:
Statement I : Parenchyma is living but collenchyma is dead tissue.
Statement II : Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is false but Statement II is true ✓
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false
Solution: Collenchyma is also a living tissue — it is a simple permanent tissue with thickened corners that gives flexible support to growing organs. So Statement I is false.
Gymnosperms lack xylem vessels, but the presence of xylem vessels is the characteristic of angiosperms. So Statement II is true.
Q103.
Spindle fibers attach to kinetochores of chromosomes during
- A. Telophase
- B. Prophase
- C. Metaphase ✓
- D. Anaphase
Solution: Spindle fibers attach to kinetochores of chromosomes in the metaphase stage.
This is what allows the chromosomes to be aligned at the metaphase plate and then pulled apart in anaphase.
Q104.
In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it?
- A. BB/Bb
- B. BB
- C. bb ✓
- D. Bb
Solution: To determine the genotype of a black seed colour at F$_2$, the black seed from F$_2$ is crossed with the white seed colour. This is called a test cross.
$$\therefore\ \text{To determine the genotype of (BB/Bb) black seed we need to cross them with white seed, i.e. bb.}$$
If the black parent is BB all offspring are black; if it is Bb, half the offspring are white, which reveals the genotype at once.
Q105.
The equation of Verhulst-Pearl logistic growth is $\dfrac{dN}{dt} = rN\left[\dfrac{K - N}{K}\right]$.
From this equation, K indicates:
- A. Population density
- B. Intrinsic rate of natural increase
- C. Biotic potential
- D. Carrying capacity ✓
Solution: In the equation
$$\frac{dN}{dt} = rN\left(\frac{K - N}{K}\right)$$
K represents the carrying capacity — the maximum population size that the habitat's resources can sustain.
Here $N$ is the population density and $r$ the intrinsic rate of natural increase.
Q106.
How many molecules of ATP and NADPH are required for every molecule of CO$_2$ fixed in the Calvin cycle?
- A. 3 molecules of ATP and 2 molecules of NADPH ✓
- B. 2 molecules of ATP and 3 molecules of NADPH
- C. 2 molecules of ATP and 2 molecules of NADPH
- D. 3 molecules of ATP and 3 molecules of NADPH
Solution: For fixation of 1 molecule of CO$_2$ in the Calvin cycle, 3 ATP molecules and 2 NADPH molecules are required.
Over a full turn producing one molecule of glucose, six CO$_2$ are fixed and 18 ATP and 12 NADPH are consumed.
Q107.
Match List I with List II
List I:
A. Clostridium butylicum
B. Saccharomyces cerevisiae
C. Trichoderma polysporum
D. Streptococcus sp.
List II:
I. Ethanol
II. Streptokinase
III. Butyric acid
IV. Cyclosporin-A
Choose the correct answer from the options given below:
- A. A-IV, B-I, C-III, D-II
- B. A-III, B-I, C-II, D-IV
- C. A-II, B-IV, C-III, D-I
- D. A-III, B-I, C-IV, D-II ✓
Solution: Matching each microbe with the product it yields:
A. Clostridium butylicum — Butyric acid
B. Saccharomyces cerevisiae — Ethanol
C. Trichoderma polysporum — Cyclosporin-A
D. Streptococcus sp. — Streptokinase
So A-III, B-I, C-IV, D-II.
Q108.
These are regarded as major causes of biodiversity loss:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
- A. A, B and D only ✓
- B. A, C and D only
- C. A, B, C and D only
- D. A, B and E only
Solution: The major causes of biodiversity losses, often called the Evil Quartet, are
(1) Habitat loss and fragmentation
(2) Over-exploitation
(3) Alien species invasions
(4) Co-extinctions
Mutation and migration are not among them.
Hence the correct option is A, B and D only.
Q109.
Given below are two statements:
Statement I : Chromosomes become gradually visible under light microscope during leptotene stage.
Statement II : The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true ✓
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false
Solution: During the leptotene stage the chromosomes become gradually visible under the light microscope.
The beginning of diplotene is recognised by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other, except at the site of crossover.
Thus both Statement I and Statement II are correct.
Q110.
Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin
- A. can help in cell division in grasses, to produce growth.
- B. promotes apical dominance.
- C. promotes abscission of mature leaves only.
- D. does not affect mature monocotyledonous plants. ✓
Solution: Auxin does not affect mature monocot plants.
In monocots, especially grasses, there is limited translocation of the applied auxin and rapid degradation of the external auxin, so the grass is unharmed while the broad-leaved dicot weeds are killed.
Q111.
A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?
- A. Red, Pink as well as white flowered plants
- B. Only red flowered plants
- C. Red flowered as well as pink flowered plants ✓
- D. Only pink flowered plants
Solution: Pink colour flower in snapdragon has genotype Rr, and red flowered snapdragon has genotype RR.
Crossing Rr with RR gives the gametes R and r from the pink parent and R from the red parent:
RR, RR, Rr, Rr
Red : Pink : White = 2 : 2 : 0
So the progeny that we get are red and pink flowered plants only.
Q112.
Which of the following are required for the dark reaction of photosynthesis?
A. Light
B. Chlorophyll
C. CO$_2$
D. ATP
E. NADPH
Choose the correct answer from the options given below:
- A. D and E only
- B. A, B and C only
- C. B, C and D only
- D. C, D and E only ✓
Solution: For the dark reaction of photosynthesis there is the requirement of CO$_2$, ATP and NADPH.
The CO$_2$ is the carbon source that is fixed, while the ATP and NADPH are the assimilatory power carried over from the light reaction.
Light and chlorophyll are needed for the light reaction, not directly for the dark reaction.
Q113.
The lactose present in the growth medium of bacteria is transported to the cell by the action of
- A. Polymerase
- B. Beta-galactosidase
- C. Acetylase
- D. Permease ✓
Solution: The $y$ gene of the lac operon codes for the permease enzyme, which increases the permeability of the cell to $\beta$-galactosides.
So the lactose present in the growth medium of bacteria is transported into the cell by the action of permease.
$\beta$-galactosidase (the $z$ gene) then hydrolyses the lactose once it is inside.
Q114.
Match List I with List II
List-I:
A. Rhizopus
B. Ustilago
C. Puccinia
D. Agaricus
List-II:
I. Mushroom
II. Smut fungus
III. Bread mould
IV. Rust fungus
Choose the correct answer from the options given below:
- A. A-IV, B-III, C-II, D-I
- B. A-III, B-II, C-IV, D-I ✓
- C. A-I, B-III, C-II, D-IV
- D. A-III, B-II, C-I, D-IV
Solution: Rhizopus is a bread mould fungus.
Ustilago is a smut fungus.
Puccinia is known as rust fungus.
Agaricus is commonly called mushroom.
So A-III, B-II, C-IV, D-I.
Q115.
The type of conservation in which the threatened species are taken out from their natural habitat and placed in special setting where they can be protected and given special care is called
- A. Sustainable development
- B. in-situ conservation
- C. Biodiversity conservation ✓
- D. Semi-conservative method
Solution: The type of conservation in which threatened species are taken out from their natural habitat and placed in a special setting where they can be protected and given special care is called ex-situ conservation, which is a type of biodiversity conservation.
In-situ conservation, by contrast, protects the species where it already lives.
Q116.
Match List I with List II
List-I:
A. Nucleolus
B. Centriole
C. Leucoplasts
D. Golgi apparatus
List-II:
I. Site of formation of glycolipid
II. Organization like the cartwheel
III. Site for active ribosomal RNA synthesis
IV. For storing nutrients
Choose the correct answer from the options given below:
- A. A-I, B-II, C-III, D-IV
- B. A-III, B-II, C-IV, D-I ✓
- C. A-II, B-III, C-I, D-IV
- D. A-III, B-IV, C-II, D-I
Solution: Nucleolus is a site for active ribosomal RNA synthesis.
Both the centrioles in a centrosome lie perpendicular to each other, in which each has an organisation like the cartwheel.
Leucoplasts are the colourless plastids of varied shapes and sizes with stored nutrients.
Golgi apparatus is the important site for formation of glycoproteins and glycolipids.
So A-III, B-II, C-IV, D-I.
Q117.
The capacity to generate a whole plant from any cell of the plant is called:
- A. Somatic hybridization
- B. Totipotency ✓
- C. Micropropagation
- D. Differentiation
Solution: Totipotency is defined as the capacity to generate a whole plant from any cell of the plant.
Micropropagation is the method of producing thousands of plants through tissue culture; somatic hybridization is the fusion of two protoplasts from different varieties.
Q118.
Formation of interfascicular cambium from fully developed parenchyma cells is an example for
- A. Maturation
- B. Differentiation
- C. Redifferentiation
- D. Dedifferentiation ✓
Solution: The phenomenon of formation of interfascicular cambium from fully differentiated parenchyma cells is called dedifferentiation.
Here living differentiated cells that had lost the capacity to divide regain it under certain conditions.
Q119.
Which one of the following can be explained on the basis of Mendel's Law of Dominance?
A. Out of one pair of factors one is dominant and the other is recessive.
B. Alleles do not show any expression and both the characters appear as such in F$_2$ generation.
C. Factors occur in pairs in normal diploid plants.
D. The discrete unit controlling a particular character is called factor.
E. The expression of only one of the parental characters is found in a monohybrid cross.
Choose the correct answer from the options given below:
- A. A, B, C, D and E
- B. A, B and C only
- C. A, C, D and E only ✓
- D. B, C and D only
Solution: According to the Law of Dominance:
(1) Characters are controlled by discrete units called factors
(2) Factors occur in pairs
(3) In a dissimilar pair of factors one member of the pair dominates (dominant) the other recessive
The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross.
Statement B belongs to the Law of Segregation, which is based on the fact that the alleles do not show any blending and both the characters are recovered as such in the F$_2$ generation.
Hence A, C, D and E only.
Q120.
What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism?
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integral part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below:
- A. A and E only
- B. A and B only
- C. D and E only
- D. B and C only ✓
Solution: A piece of DNA carrying only the gene of interest has no origin of replication of its own, so it cannot multiply by itself.
Its fate is that (B) it may get integrated into the genome of the recipient, and (C) it may then multiply and be inherited along with the host DNA.
So this piece of DNA would not be able to multiply itself in the progeny cells of the organism; but when it gets integrated into the genome of the recipient, it may multiply and be inherited along with the host DNA.
Q121.
Match List I with List II
List I:
A. Two or more alternative forms of a gene
B. Cross of F$_1$ progeny with homozygous recessive parent
C. Cross of F$_1$ progeny with any of the parents
D. Number of chromosome sets in plant
List II:
I. Back cross
II. Ploidy
III. Allele
IV. Test cross
Choose the correct answer from the options given below:
- A. A-IV, B-III, C-II, D-I
- B. A-I, B-II, C-III, D-IV
- C. A-II, B-I, C-III, D-IV
- D. A-III, B-IV, C-I, D-II ✓
Solution: A. Two or more alternative forms of a gene are called alleles.
B. Cross of F$_1$ progeny with homozygous recessive parent is a test cross.
C. Cross of F$_1$ progeny with any of the parents is a back cross.
D. Number of chromosome sets in plant is called ploidy.
So A-III, B-IV, C-I, D-II.
Q122.
Identify the set of correct statements:
A. The flowers of Vallisneria are colourful and produce nectar.
B. The flowers of water lily are not pollinated by water.
C. In most of water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
Choose the correct answer from the options given below.
- A. B, C, D and E only ✓
- B. C, D and E only
- C. A, B, C and D only
- D. A, C, D and E only
Solution: Flowers of Vallisneria are not colourful and do not produce nectar, so statement A is wrong.
Waterlily is pollinated by insect or wind, so statement B is correct.
In water-pollinated species, pollen grains are protected from wetting by a mucilaginous covering.
Pollen grains of some hydrophytes such as Zostera are long and ribbon like.
In some hydrophytes such as Vallisneria, pollen grains are carried passively by the water current.
Hence B, C, D and E only.
Q123.
Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
- A. Enzyme activation
- B. Cofactor inhibition
- C. Feedback inhibition
- D. Competitive inhibition ✓
Solution: Malonate shows close structural similarity with the substrate succinate, and it competes with the substrate for the substrate binding site of the enzyme succinic dehydrogenase.
That is the definition of competitive inhibition.
Enzyme activation, cofactor inhibition and feedback inhibition do not involve an inhibitor with structural similarity to the substrate binding at the active site.
Q124.
List of endangered species was released by
- A. IUCN ✓
- B. GEAC
- C. WWF
- D. FOAM
Solution: The list of endangered species, the IUCN Red List, was released by the International Union for Conservation of Nature (IUCN).
Q125.
Given below are two statements:
Statement I : Bt toxins are insect group specific and coded by a gene cry IAc.
Statement II : Bt toxin exists as inactive protoxin in B. thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false ✓
Solution: Specific Bt toxin genes were isolated from Bacillus thuringiensis and incorporated into several crop plants such as cotton. The choice of genes depends upon the crop and the targeted pest, as most Bt toxins are insect-group specific. The toxin is coded by a gene named cry. The proteins encoded by the genes cry IAc and cry IIAb control the cotton bollworms, and that of cry IAb controls corn borer. So Statement I is true.
Statement II is false because the inactive protoxin is converted into the active form due to the alkaline pH of the insect gut, not the acidic pH.
Q126.
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
- A. Carbohydrates
- B. Amino acids
- C. Phospholipids ✓
- D. Glycerides
Solution: Some lipids have phosphorus and a phosphorylated organic compound in them. These are phospholipids. They are found in the cell membrane. Lecithin is one example.
Glycerides are another group of lipids in which both glycerol and fatty acids are present, with no phosphate.
Carbohydrates and amino acids are separate groups of biomolecules.
Q127.
Which of the following is an example of actinomorphic flower?
- A. Sesbania
- B. Datura ✓
- C. Cassia
- D. Pisum
Solution: Datura shows an actinomorphic flower — it can be divided into two equal radial halves in any radial plane passing through the centre.
In Cassia, Pisum and Sesbania, zygomorphic flowers are seen, which can be divided into two similar halves only in one particular vertical plane.
Q128.
Tropical regions show greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below.
- A. A, B and D only
- B. A, C, D and E only ✓
- C. A and B only
- D. A, B and E only
Solution: Only statement B is incorrect, because tropical environments, unlike temperate ones, are less seasonal, relatively more constant and predictable.
Thus statements A, C, D and E are correct.
Q129.
Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
- A. 10 bp
- B. 8 bp
- C. 6 bp ✓
- D. 4 bp
Solution: The first restriction endonuclease, Hind II, whose functioning depends on a specific DNA nucleotide sequence, was isolated. It was found that Hind II always cuts DNA molecules at a particular point by recognising a sequence of six base pairs.
The other options are incorrect because they have either more than 6 or less than 6 bp.
Q130.
Bulliform cells are responsible for
- A. Providing large spaces for storage of sugars.
- B. Inward curling of leaves in monocots. ✓
- C. Protecting the plant from salt stress.
- D. Increased photosynthesis in monocots.
Solution: In grasses, certain adaxial epidermal cells along the veins modify themselves into large, empty, colourless cells. These are called bulliform cells.
When the bulliform cells in the leaves have absorbed water and are turgid, the leaf surface is exposed. When they are flaccid due to water stress, they make the leaves curl inwards to minimise water loss.
Q131.
A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end;
- A. Promotor, Structural gene, Terminator ✓
- B. Repressor, Operator gene, Structural gene
- C. Structural gene, Transposons, Operator gene
- D. Inducer, Repressor, Structural gene
Solution: A transcription unit of DNA is defined primarily by the three regions in the DNA:
(i) A promoter
(ii) The structural gene
(iii) A terminator
The promoter is said to be located towards the 5$'$-end (upstream) of the structural gene (the reference is made with respect to the polarity of the coding strand).
The terminator is located towards the 3$'$-end (downstream) of the coding strand.
Q132.
Which one of the following is not a criterion for classification of fungi?
- A. Fruiting body
- B. Morphology of mycelium
- C. Mode of nutrition ✓
- D. Mode of spore formation
Solution: The morphology of the mycelium, mode of spore formation and fruiting bodies form the basis for the division of the kingdom Fungi into various classes.
Mode of nutrition is not one of these criteria — all fungi are heterotrophic, so it does not distinguish the classes.
Q133.
Match List I with List II
List I (Types of Stamens):
A. Monoadelphous
B. Diadelphous
C. Polyadelphous
D. Epiphyllous
List II (Example):
I. Citrus
II. Pea
III. Lily
IV. China-rose
Choose the correct answer from the options given below:
- A. A-III, B-I, C-IV, D-II
- B. A-IV, B-II, C-I, D-III ✓
- C. A-IV, B-I, C-II, D-III
- D. A-I, B-II, C-IV, D-III
Solution: In China rose a monoadelphous androecium is present — all the filaments are united into a single bundle.
Diadelphous androecium is found in the pea plant, where the stamens are united into two bundles.
Polyadelphous androecium is found in citrus, where they are united into more than two bundles.
Epiphyllous androecium is found in lily, where the stamens are attached to the perianth.
So A-IV, B-II, C-I, D-III.
Q134.
The DNA present in chloroplast is:
- A. Circular, single stranded
- B. Linear, double stranded
- C. Circular, double stranded ✓
- D. Linear, single stranded
Solution: The DNA present in chloroplast is circular double stranded, like that of prokaryotes — which is part of the evidence for the endosymbiotic origin of the organelle.
Q135.
Read the following statements and choose the set of correct statements:
In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin.
Choose the correct answer from the options given below:
- A. A, B, C and E only
- B. A, B, C and D only
- C. B, C, D and E only
- D. A, C, D and E only ✓
Solution: In members of Phaeophyceae, sexual reproduction is by oogamous, isogamous or anisogamous methods — not by the oogamous method only. So statement B is incorrect.
Therefore the correct set of statements are A, C, D and E.
Q136.
Match List I with List II
List I:
A. Citric acid cycle
B. Glycolysis
C. Electron transport system
D. Proton gradient
List II:
I. Cytoplasm
II. Mitochondrial matrix
III. Intermembrane space of mitochondria
IV. Inner mitochondrial membrane
Choose the correct answer from the options given below:
- A. A-IV, B-III, C-II, D-I
- B. A-I, B-II, C-III, D-IV
- C. A-II, B-I, C-IV, D-III ✓
- D. A-III, B-IV, C-I, D-II
Solution: Citric acid cycle occurs in the mitochondrial matrix.
Glycolysis occurs in the cytosol in most organisms.
Electron transport system is present in the inner mitochondrial membrane.
Proton gradient is formed across the intermembrane space of mitochondria.
So A-II, B-I, C-IV, D-III.
Q137.
Given below are two statements:
Statement I: In C$_3$ plants, some O$_2$ binds to RuBisCO, hence CO$_2$ fixation is decreased.
Statement II: In C$_4$ plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false ✓
Solution: In a C$_3$ plant, some O$_2$ binds to RuBisCO, and hence CO$_2$ fixation is decreased. So Statement I is true.
Statement II is incorrect: photorespiration does not occur in C$_4$ plants at all, as they lack RuBisCO in the mesophyll — the mesophyll fixes CO$_2$ with PEP carboxylase instead.
Hence Statement I is the only correct one.
Q138.
Which of the following statement is correct regarding the process of replication in E. coli?
- A. The DNA dependent DNA polymerase catalyses polymerization in 5$'$ $\rightarrow$ 3$'$ direction ✓
- B. The DNA dependent DNA polymerase catalyses polymerization in one direction that is 3$'$ $\rightarrow$ 5$'$
- C. The DNA dependent RNA polymerase catalyses polymerization in one direction, that is 5$'$ $\rightarrow$ 3$'$
- D. The DNA dependent DNA polymerase catalyses polymerization in 5$'$ $\rightarrow$ 3$'$ as well as 3$'$ $\rightarrow$ 5$'$ direction
Solution: In prokaryotes like E. coli, during replication, the DNA dependent DNA polymerase catalyses polymerization only in one direction, that is 5$'$ $\rightarrow$ 3$'$.
This single-direction constraint is why one strand is made continuously and the other discontinuously as Okazaki fragments.
Q139.
Match List I with List II
List I:
A. Robert May
B. Alexander von Humboldt
C. Paul Ehrlich
D. David Tilman
List II:
I. Species-Area relationship
II. Long term ecosystem experiment using out door plots
III. Global species diversity at about 7 million
IV. Rivet popper hypothesis
Choose the correct answer from the options given below:
- A. A-III, B-IV, C-II, D-I
- B. A-II, B-III, C-I, D-IV
- C. A-III, B-I, C-IV, D-II ✓
- D. A-I, B-III, C-II, D-IV
Solution: Robert May places the global species diversity at about 7 million.
Alexander von Humboldt gave the species-area relationship.
Paul Ehrlich used the analogy "Rivet popper hypothesis" to explain the role of species in the ecosystem.
David Tilman performed long term ecosystem experiments using outdoor plots.
So A-III, B-I, C-IV, D-II.
Q140.
Match List I with List II
List I:
A. Frederick Griffith
B. Francois Jacob & Jacque Monod
C. Har Gobind Khorana
D. Meselson & Stahl
List II:
I. Genetic code
II. Semi-conservative mode of DNA replication
III. Transformation
IV. Lac operon
Choose the correct answer from the options given below:
- A. A-IV, B-I, C-II, D-III
- B. A-III, B-II, C-I, D-IV
- C. A-III, B-IV, C-I, D-II ✓
- D. A-II, B-III, C-IV, D-I
Solution: Frederick Griffith's series of experiments witnessed the miraculous transformation in the bacteria.
The elucidation of the Lac operon was a result of a close association between the geneticist Francois Jacob and the biochemist Jacques Monod.
Har Gobind Khorana developed a chemical method to define the combination of bases in the genetic code.
Meselson and Stahl gave the semi-conservative mode of DNA replication.
So A-III, B-IV, C-I, D-II.
Q141.
In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is $100x$ (kcal m$^{-2}$) yr$^{-1}$, what would be the GPP (Gross Primary Productivity) of the third trophic level of the same ecosystem?
- A. $\dfrac{100x}{3x}$ (kcal m$^{-2}$) yr$^{-1}$
- B. $\dfrac{x}{10}$ (kcal m$^{-2}$) yr$^{-1}$
- C. $x$ (kcal m$^{-2}$) yr$^{-1}$
- D. $10x$ (kcal m$^{-2}$) yr$^{-1}$ ✓
Solution: NPP at the first trophic level would be the GPP for the second trophic level, and NPP at the second trophic level would be the GPP for the third trophic level.
Therefore $100x$ (kcal/m$^{2}$/yr) would be GPP at the second trophic level.
Applying the 10 per cent law once more, $100x \times 10\% = 10x$ (kcal/m$^{2}$/yr) energy would be GPP at the third trophic level.
Q142.
Match List I with List II
List I:
A. Rose
B. Pea
C. Cotton
D. Mango
List II:
I. Twisted aestivation
II. Perigynous flower
III. Drupe
IV. Marginal placentation
Choose the correct answer from the options given below :
- A. A-II, B-III, C-IV, D-I
- B. A-II, B-IV, C-I, D-III ✓
- C. A-I, B-II, C-III, D-IV
- D. A-IV, B-III, C-II, D-I
Solution: Rose has a half-inferior ovary, thus it is known as a perigynous flower.
In Pea, the placenta forms a ridge along the ventral suture of the ovary and ovules are borne on this ridge forming two rows — that is marginal placentation.
In Cotton, twisted aestivation is present.
In Mango, the fruit is a drupe.
So A-II, B-IV, C-I, D-III.
Q143.
Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.
- A. Isocitrate $\rightarrow$ $\alpha$-ketoglutaric acid
- B. Malic acid $\rightarrow$ Oxaloacetic acid
- C. Succinic acid $\rightarrow$ Malic acid
- D. Succinyl-CoA $\rightarrow$ Succinic acid ✓
Solution: Oxidation involves the loss of electrons (often as part of hydrogen) from a molecule, leading to an increase in its oxidation state. This process is typically associated with the transfer of electrons to an electron acceptor which is reduced in the process.
The conversion of succinyl-CoA to succinic acid does not involve oxidation of substrate — it is a substrate-level phosphorylation that produces GTP/ATP, with no reduction of NAD$^{+}$ or FAD.
Q144.
Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield?
- A. Abscisic acid
- B. Auxin
- C. Gibberellin ✓
- D. Cytokinin
Solution: Sugarcanes store carbohydrate as sugar in their stems. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield — an increase of up to 20 tonnes per acre.
Q145.
Match List-I with List-II
List-I:
A. GLUT-4
B. Insulin
C. Trypsin
D. Collagen
List-II:
I. Hormone
II. Enzyme
III. Intercellular ground substance
IV. Enables glucose transport into cells
Choose the correct answer from the options given below.
- A. A-III, B-IV, C-I, D-II
- B. A-IV, B-I, C-II, D-III ✓
- C. A-I, B-II, C-III, D-IV
- D. A-II, B-III, C-IV, D-I
Solution: A. GLUT-4 — enables glucose transport into cells
B. Insulin — hormone
C. Trypsin — enzyme
D. Collagen — intercellular ground substance
So A-IV, B-I, C-II, D-III.
Q146.
Which of the following are fused in somatic hybridization involving two varieties of plants?
- A. Pollens
- B. Callus
- C. Somatic embryos
- D. Protoplasts ✓
Solution: Protoplasts of two varieties of plants are fused in somatic hybridization.
Isolated protoplasts, from which the cell wall has been digested away, can be fused to give a hybrid protoplast which is then grown into a new plant — the somatic hybrid.
Q147.
The flippers of the Penguins and Dolphins are the example of the
- A. Divergent evolution
- B. Adaptive radiation
- C. Natural selection
- D. Convergent evolution ✓
Solution: The flippers of the Penguins and Dolphins perform a similar function but they are not anatomically similar structures. This is an example of analogous structures, which arise by convergent evolution.
Adaptive radiation is the process of evolution of different species in a given geographical area starting from a point and radiating to other areas of geography.
Natural selection is a key mechanism of evolution, not a name for this pattern.
Divergent evolution results in the formation of homologous structures.
Q148.
Match List I with List II :
List I:
A. Pleurobrachia
B. Radula
C. Stomochord
D. Air bladder
List II:
I. Mollusca
II. Ctenophora
III. Osteichthyes
IV. Hemichordata
Choose the correct answer from the options given below :
- A. A-IV, B-III, C-II, D-I
- B. A-IV, B-II, C-III, D-I
- C. A-II, B-I, C-IV, D-III ✓
- D. A-II, B-IV, C-I, D-III
Solution: A. Pleurobrachia — is a member of phylum Ctenophora.
B. Radula — is a rasping feeding organ present in phylum Mollusca.
C. Stomochord — a rudimentary structure similar to notochord found in the collar region of members of phylum Hemichordata.
D. Air bladder — is found in Osteichthyes, which provides them buoyancy.
So A-II, B-I, C-IV, D-III.
Q149.
Which of the following is not a component of Fallopian tube?
- A. Ampulla
- B. Uterine fundus ✓
- C. Isthmus
- D. Infundibulum
Solution: The uterine fundus is the upper, dome-shaped part of the uterus, above the opening of the fallopian tubes — it belongs to the uterus, not the tube.
The isthmus is the last and narrow part of the oviduct that links to the uterus.
The infundibulum is the part of the oviduct which is closer to the ovary.
The ampulla is the wider part of the oviduct.
Q150.
Following are the stages of pathway for conduction of an action potential through the heart
A. AV bundle
B. Purkinje fibres
C. AV node
D. Bundle branches
E. SA node
Choose the correct sequence of pathway from the options given below
- A. E-A-D-B-C
- B. E-C-A-D-B ✓
- C. A-E-C-B-D
- D. B-D-E-C-A
Solution: The correct pathway of conduction of the action potential is
SA node $\rightarrow$ AV node $\rightarrow$ AV bundle $\rightarrow$ Bundle branches $\rightarrow$ Purkinje fibres
which is E-C-A-D-B.
Q151.
Given below are two statements:
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
In the light of the above statements, choose the correct answer from the options given below :
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false ✓
Solution: The presence or absence of hymen is not a reliable indicator of virginity, because the hymen can also be broken by a sudden jolt, insertion of a vaginal tampon, or active participation in some sports; and in some women the hymen persists even after coitus.
So Statement I is true, and Statement II — that it is torn during the first coitus only — is false.
Q152.
Match List I with List II :
List I:
A. $\alpha$-1 antitrypsin
B. Cry IAb
C. Cry IAc
D. Enzyme replacement therapy
List II:
I. Cotton bollworm
II. ADA deficiency
III. Emphysema
IV. Corn borer
Choose the correct answer from the options given below:
- A. A-II, B-IV, C-I, D-III
- B. A-II, B-I, C-IV, D-III
- C. A-III, B-I, C-II, D-IV
- D. A-III, B-IV, C-I, D-II ✓
Solution: $\alpha$-1 antitrypsin is used for treatment of emphysema.
Cry IAb gene controls corn borer.
Cry IAc gene controls cotton bollworms.
Enzyme replacement therapy can be used as a treatment option in ADA deficiency.
So A-III, B-IV, C-I, D-II.
Q153.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : FSH acts upon ovarian follicles in female and Leydig cells in male.
Reason R : Growing ovarian follicles secrete estrogen in female while interstitial cells secrete androgen in male human being.
In the light of the above statements, choose the correct answer from the options given below :
- A. A is false but R is true ✓
- B. Both A and R are true and R is the correct explanation of A
- C. Both A and R are true but R is NOT the correct explanation of A
- D. A is true but R is false
Solution: FSH is a gonadotropin that affects ovarian follicles in females and causes their growth, but in males it is LH that affects Leydig cells, leading to secretion of androgens. So the Assertion is false.
Growing ovarian follicles secrete estrogen in females while interstitial cells secrete androgen in male human beings, so the Reason is true.
Hence Assertion is false and Reason is true.
Q154.
Match List I with List II :
List I:
A. Cocaine
B. Heroin
C. Morphine
D. Marijuana
List II:
I. Effective sedative in surgery
II. Cannabis sativa
III. Erythroxylum
IV. Papaver somniferum
Choose the correct answer from the options given below:
- A. A-III, B-IV, C-I, D-II ✓
- B. A-IV, B-III, C-I, D-II
- C. A-I, B-III, C-II, D-IV
- D. A-II, B-I, C-III, D-IV
Solution: A. Cocaine — obtained from the plant Erythroxylum coca, with a stimulating action on the CNS.
B. Heroin — formed by the acetylation of morphine, which is obtained from the plant Papaver somniferum.
C. Morphine — obtained from Papaver somniferum, is an effective sedative in surgery.
D. Marijuana — obtained from Cannabis sativa, produces hallucinogenic effects and affects the cardiovascular system of the body.
So A-III, B-IV, C-I, D-II.
Q155.
Match List I with List II :
List-I:
A. Lipase
B. Nuclease
C. Protease
D. Amylase
List-II:
I. Peptide bond
II. Ester bond
III. Glycosidic bond
IV. Phosphodiester bond
Choose the correct answer from the options given below :
- A. A-IV, B-I, C-III, D-II
- B. A-IV, B-II, C-III, D-I
- C. A-III, B-II, C-I, D-IV
- D. A-II, B-IV, C-I, D-III ✓
Solution: A. Lipase — digests ester bonds found in lipids.
B. Nuclease — helps in digestion of phosphodiester bonds found in nucleic acids.
C. Protease — helps in digestion of peptide bonds found in proteins.
D. Amylase — digests the glycosidic bonds found in carbohydrates, i.e. digests starch into smaller molecules, ultimately yielding maltose, which in turn is cleaved into two glucose molecules by maltase.
So A-II, B-IV, C-I, D-III.
Q156.
Match List I with List II :
List I:
A. Pterophyllum
B. Myxine
C. Pristis
D. Exocoetus
List II:
I. Hag fish
II. Saw fish
III. Angel fish
IV. Flying fish
Choose the correct answer from the options given below :
- A. A-III, B-II, C-I, D-IV
- B. A-II, B-I, C-III, D-IV
- C. A-III, B-I, C-II, D-IV ✓
- D. A-IV, B-I, C-II, D-III
Solution: Pterophyllum is the scientific name for Angel fish.
Myxine is the scientific name for Hag fish.
Pristis is the scientific name for Saw fish.
Exocoetus is the scientific name for Flying fish.
So A-III, B-I, C-II, D-IV.
Q157.
Match List I with List II :
List I:
A. Fibrous joints
B. Cartilaginous joints
C. Hinge joints
D. Ball and socket joints
List II:
I. Adjacent vertebrae, limited movement
II. Humerus and Pectoral girdle, rotational movement
III. Skull, don't allow any movement
IV. Knee, help in locomotion
Choose the correct answer from the options given below :
- A. A-III, B-I, C-IV, D-II ✓
- B. A-IV, B-II, C-III, D-I
- C. A-I, B-III, C-II, D-IV
- D. A-II, B-III, C-I, D-IV
Solution: Fibrous joints do not allow any movement. This type of joint is shown by the flat skull bones which fuse end-to-end with the help of dense fibrous connective tissues in the form of sutures.
Cartilaginous joint is present between the adjacent vertebrae in the vertebral column and this permits limited movements.
Hinge joint is a type of synovial joint present in the knee, and helps in locomotion.
Ball and socket joint is also a type of synovial joint present between the humerus and pectoral girdle and allows rotational movement.
So A-III, B-I, C-IV, D-II.
Q158.
Following are the stages of cell division :
A. Gap 2 phase
B. Cytokinesis
C. Synthesis phase
D. Karyokinesis
E. Gap 1 phase
Choose the correct sequence of stages from the options given below :
- A. E-C-A-D-B ✓
- B. C-E-D-A-B
- C. E-B-D-A-C
- D. B-D-E-A-C
Solution: The correct sequence of stages of cell division is
Gap 1 phase (E) $\rightarrow$ Synthesis phase (C) $\rightarrow$ Gap 2 phase (A) $\rightarrow$ Karyokinesis (D) $\rightarrow$ Cytokinesis (B)
The correct sequence will be E $\rightarrow$ C $\rightarrow$ A $\rightarrow$ D $\rightarrow$ B.
Q159.
The "Ti plasmid" of Agrobacterium tumefaciens stands for
- A. Temperature independent plasmid
- B. Tumour inhibiting plasmid
- C. Tumor independent plasmid
- D. Tumor inducing plasmid ✓
Solution: The Ti plasmid of Agrobacterium tumefaciens is the tumor inducing plasmid, containing T-DNA which causes tumour in several dicot plants.
It has been modified into a cloning vector that no longer causes disease but is still able to deliver genes into plants.
Q160.
Match List I with List II :
List I:
A. Pons
B. Hypothalamus
C. Medulla
D. Cerebellum
List II:
I. Provides additional space for Neurons, regulates posture and balance.
II. Controls respiration and gastric secretions.
III. Connects different regions of the brain.
IV. Neuro secretory cells
Choose the correct answer from the options given below :
- A. A-II, B-I, C-III, D-IV
- B. A-II, B-III, C-I, D-IV
- C. A-III, B-IV, C-II, D-I ✓
- D. A-I, B-III, C-II, D-IV
Solution: Pons — part of hindbrain, it connects different regions of the brain.
Hypothalamus — also has neuro secretory cells which secrete hormones.
Medulla oblongata — part of hindbrain which controls respiration and gastric secretions.
Cerebellum — part of hindbrain with convoluted surface which provides additional space for neurons, and also regulates posture and balance.
So A-III, B-IV, C-II, D-I.
Q161.
Match List I with List II :
List I:
A. Axoneme
B. Cartwheel pattern
C. Crista
D. Satellite
List II:
I. Centriole
II. Cilia and flagella
III. Chromosome
IV. Mitochondria
Choose the correct answer from the options given below :
- A. A-II, B-I, C-IV, D-III ✓
- B. A-IV, B-III, C-II, D-I
- C. A-IV, B-II, C-III, D-I
- D. A-II, B-IV, C-I, D-III
Solution: Axoneme is seen in cilia and flagella.
Centriole shows cartwheel appearance.
Crista is found in mitochondria.
Satellite is present in chromosomes.
So A-II, B-I, C-IV, D-III.
Q162.
Consider the following statements :
A. Annelids are true coelomates
B. Poriferans are pseudocoelomates
C. Aschelminthes are acoelomates
D. Platyhelminthes are pseudocoelomates
Choose the correct answer from the options given below :
- A. D only
- B. B only
- C. A only ✓
- D. C only
Solution: Annelids are true coelomate animals, so statement A is correct.
The other statements are incorrect, because poriferans are acoelomates, aschelminths are pseudocoelomates and platyhelminthes are acoelomates.
Q163.
Which of the following is not a steroid hormone?
- A. Glucagon ✓
- B. Cortisol
- C. Testosterone
- D. Progesterone
Solution: Glucagon is a proteinaceous (peptide) hormone secreted from the $\alpha$-cells of the islets of Langerhans in the pancreas.
Cortisol, testosterone and progesterone are all steroid in nature, derived from cholesterol.
Q164.
Given below are two statements :
Statement I : In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II : The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
In the light of the above statements, choose the correct answer from the option given below :
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false ✓
- D. Statement I is true but Statement II is false
Solution: Statement I is false, as the descending limb of the loop of Henle is permeable to water and almost impermeable to electrolytes — it is the ascending limb that is impermeable to water.
Statement II is false as the proximal convoluted tubule is lined by simple cuboidal brush border epithelium, not columnar, which increases the surface area for reabsorption.
Q165.
Match List I with List II :
List I:
A. Common cold
B. Haemozoin
C. Widal test
D. Allergy
List II:
I. Plasmodium
II. Typhoid
III. Rhinoviruses
IV. Dust mites
Choose the correct answer from the options given below :
- A. A-IV, B-II, C-III, D-I
- B. A-II, B-IV, C-III, D-I
- C. A-I, B-III, C-II, D-IV
- D. A-III, B-I, C-II, D-IV ✓
Solution: Common cold is caused by Rhinoviruses.
Haemozoin is released in blood due to ruptured RBCs after Plasmodium infection.
Widal test is used to confirm typhoid fever.
Allergy is caused due to dust mites.
So A-III, B-I, C-II, D-IV.
Q166.
Match List I with List II :
List I:
A. Expiratory capacity
B. Functional residual capacity
C. Vital capacity
D. Inspiratory capacity
List II:
I. Expiratory reserve volume + Tidal volume + Inspiratory reserve volume
II. Tidal volume + Expiratory reserve volume
III. Tidal volume + Inspiratory reserve volume
IV. Expiratory reserve volume + Residual volume
Choose the correct answer from the options given below :
- A. A-I, B-III, C-II, D-IV
- B. A-II, B-IV, C-I, D-III ✓
- C. A-III, B-II, C-IV, D-I
- D. A-II, B-I, C-IV, D-III
Solution: Expiratory capacity = Tidal volume + Expiratory reserve volume
Functional residual capacity = Expiratory reserve volume + Residual volume
Vital capacity = Expiratory reserve volume + Tidal volume + Inspiratory reserve volume
Inspiratory capacity = Tidal volume + Inspiratory reserve volume
So A-II, B-IV, C-I, D-III.
Q167.
Match List I with List II :
List I (Sub Phases of Prophase I):
A. Diakinesis
B. Pachytene
C. Zygotene
D. Leptotene
List II (Specific Characters):
I. Synaptonemal complex formation
II. Completion of terminalisation of chiasmata
III. Chromosomes look like thin threads
IV. Appearance of recombination nodules
Choose the correct answer from the options given below
- A. A-IV, B-III, C-II, D-I
- B. A-IV, B-II, C-III, D-I
- C. A-I, B-II, C-IV, D-III
- D. A-II, B-IV, C-I, D-III ✓
Solution: (A) Diakinesis — completion of terminalisation of chiasmata
(B) Pachytene — appearance of recombination nodules
(C) Zygotene — synaptonemal complex formation
(D) Leptotene — chromosomes look like thin threads
So A-II, B-IV, C-I, D-III.
Q168.
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A : Breast-feeding during initial period of infant growth is recommended by doctors for bringing a healthy baby.
Reason R : Colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. A is not correct but R is correct
- B. Both A and R are correct and R is the correct explanation of A ✓
- C. Both A and R are correct but R is NOT the correct explanation of A
- D. A is correct but R is not correct
Solution: Breast-feeding during the initial period of infant growth is recommended by doctors for bringing up a healthy baby, precisely because colostrum contains several antibodies absolutely essential to develop resistance for the new born baby.
So both A and R are correct, and R is the correct explanation of A.
Q169.
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
- A. Constant gene pool ✓
- B. Genetic recombination
- C. Genetic drift
- D. Gene migration
Solution: A constant gene pool will not disturb the Hardy-Weinberg equilibrium — it is precisely the condition the equilibrium describes.
Genetic recombination, genetic drift and gene migration all change allele frequencies, affecting the equilibrium and leading to evolution.
Q170.
Match List I with List II :
List I:
A. Typhoid
B. Leishmaniasis
C. Ringworm
D. Filariasis
List II:
I. Fungus
II. Nematode
III. Protozoa
IV. Bacteria
Choose the correct answer from the options given below:
- A. A-II, B-IV, C-III, D-I
- B. A-I, B-III, C-II, D-IV
- C. A-IV, B-III, C-I, D-II ✓
- D. A-III, B-I, C-IV, D-II
Solution: Typhoid — caused by Salmonella typhi (Bacteria)
Leishmaniasis — caused by a protozoan, i.e. Leishmania donovani
Ringworm — caused by fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton
Filariasis — caused by Wuchereria bancrofti and Wuchereria malayi (Nematode)
So A-IV, B-III, C-I, D-II.
Q171.
Given below are some stages of human evolution.
Arrange them in correct sequence. (Past to Recent)
A. Homo habilis
B. Homo sapiens
C. Homo neanderthalensis
D. Homo erectus
Choose the correct sequence of human evolution from the options given below:
- A. A-D-C-B ✓
- B. D-A-C-B
- C. B-A-D-C
- D. C-B-D-A
Solution: The correct sequence of stages of human evolution from past to recent is
Homo habilis $\rightarrow$ Homo erectus $\rightarrow$ Homo neanderthalensis $\rightarrow$ Homo sapiens
which is A-D-C-B.
Q172.
Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
- A. Low pCO$_2$ and High temperature
- B. High pO$_2$ and High pCO$_2$
- C. High pO$_2$ and Lesser H$^{+}$ concentration ✓
- D. Low pCO$_2$ and High H$^{+}$ concentration
Solution: Conditions favourable for formation of oxyhaemoglobin in alveoli are high pO$_2$, less H$^{+}$ concentration, low pCO$_2$ and low temperature.
The other options each include at least one factor that shifts the oxygen dissociation curve to the right and so favours dissociation rather than formation.
Q173.
In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on
- A. 11$^{\text{th}}$ segment
- B. 5$^{\text{th}}$ segment
- C. 10$^{\text{th}}$ segment ✓
- D. 8$^{\text{th}}$ and 9$^{\text{th}}$ segment
Solution: In both sexes of cockroach, the 10$^{\text{th}}$ segment bears a pair of jointed filamentous structures called anal cerci.
The 5$^{\text{th}}$, 8$^{\text{th}}$ and 9$^{\text{th}}$ segments do not bear such structures. In adult cockroaches only 10 segments are present in the abdomen; the 11$^{\text{th}}$ abdominal segment is absent.
Q174.
Match List I with List II :
List I:
A. Down's syndrome
B. $\alpha$-Thalassemia
C. $\beta$-Thalassemia
D. Klinefelter's syndrome
List II:
I. 11$^{\text{th}}$ chromosome
II. 'X' chromosome
III. 21$^{\text{st}}$ chromosome
IV. 16$^{\text{th}}$ chromosome
Choose the correct answer from the options given below :
- A. A-IV, B-I, C-II, D-III
- B. A-I, B-II, C-III, D-IV
- C. A-II, B-III, C-IV, D-I
- D. A-III, B-IV, C-I, D-II ✓
Solution: Down's syndrome is due to the presence of an additional copy of chromosome number 21.
$\alpha$-Thalassemia is controlled by two closely linked genes on chromosome 16 of each parent.
$\beta$-Thalassemia is controlled by a single gene HBB on chromosome 11 of each parent.
Klinefelter's syndrome is caused due to the presence of an additional copy of the X-chromosome.
So A-III, B-IV, C-I, D-II.
Q175.
Which of the following is not a natural/traditional contraceptive method?
- A. Vaults ✓
- B. Coitus interruptus
- C. Periodic abstinence
- D. Lactational amenorrhea
Solution: A vault is a barrier method of contraception, made of rubber, that is inserted into the female reproductive tract to cover the cervix during coitus — so it is not a natural method.
Coitus interruptus is a natural method of contraception in which the male partner withdraws before ejaculation so as to avoid insemination.
Periodic abstinence is also a natural method in which couples avoid coitus during the fertile period.
Lactational amenorrhea is also a natural method, based on the fact that ovulation and therefore the cycle do not occur during the period of intense lactation following parturition.
Q176.
Which of the following are Autoimmune disorders?
A. Myasthenia gravis
B. Rheumatoid arthritis
C. Gout
D. Muscular dystrophy
E. Systemic Lupus Erythematosus (SLE)
Choose the most appropriate answer from the options given below:
- A. C, D & E only
- B. A, B & D only
- C. A, B & E only ✓
- D. B, C & E only
Solution: Myasthenia gravis, Rheumatoid arthritis and Systemic Lupus Erythematosus (SLE) are autoimmune disorders.
Muscular dystrophy is a genetic disorder which progressively affects the skeletal muscles.
Gout is the inflammation of joints due to deposition of uric acid crystals.
Hence A, B and E only.
Q177.
Which of the following statements is incorrect?
- A. Bio-reactors have an agitator system, an oxygen delivery system and foam control system
- B. A bio-reactor provides optimal growth conditions for achieving the desired product
- C. Most commonly used bio-reactors are of stirring type
- D. Bio-reactors are used to produce small scale bacterial cultures ✓
Solution: Bioreactors are vessels in which raw materials are biologically converted into specific products on a LARGE scale, not a small scale — so that statement is incorrect.
The other three are true: a bioreactor provides optimal growth conditions, the most commonly used are of the stirring type, and they have an agitator system, an oxygen delivery system and a foam control system, along with temperature and pH control systems and a sampling port.
Q178.
Match List I with List II
List I:
A. Non-medicated IUD
B. Copper releasing IUD
C. Hormone releasing IUD
D. Implants
List II:
I. Multiload 375
II. Progestogens
III. Lippes loop
IV. LNG-20
Choose the correct answer from the option given below:
- A. A-III, B-I, C-IV, D-II ✓
- B. A-III, B-I, C-II, D-IV
- C. A-I, B-III, C-IV, D-II
- D. A-IV, B-I, C-II, D-III
Solution: Lippes loop is a non-medicated IUD.
Multiload 375 is a copper releasing IUD.
LNG-20 is a hormone releasing IUD.
Progestogens are used as implants.
So A-III, B-I, C-IV, D-II.
Q179.
Which one is the correct product of DNA dependent RNA polymerase to the given template?
3$'$TACATGGCAAATATCCATTCA5$'$
- A. 5$'$ATGTACCGTTTATAGGTAAGT3$'$
- B. 5$'$AUGUACCGUUUAUAGGUAAGU3$'$ ✓
- C. 5$'$AUGUAAAGUUUAUAGGUAAGU3$'$
- D. 5$'$AUGUACCGUUUAUAGGGAAGU3$'$
Solution: The RNA is synthesised antiparallel and complementary to the template, with uracil in place of thymine.
Template DNA is
3$'$TACATGGCAAATATCCATTCA5$'$
Reading it 3$'$ to 5$'$ and pairing A with U, T with A, G with C and C with G gives
5$'$AUGUACCGUUUAUAGGUAAGU3$'$ m-RNA
Q180.
Given below are two statements:
Statement I: The cerebral hemispheres are connected by nerve tract known as corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but Statement II is correct.
- B. Both Statement I and Statement II are correct.
- C. Both Statement I and Statement II are incorrect.
- D. Statement I is correct but Statement II is incorrect. ✓
Solution: In the human brain, a deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The cerebral hemispheres are connected by a tract of nerve fibres called corpus callosum — so Statement I is correct.
Three major regions make up the brain stem, i.e. mid brain, pons and medulla oblongata. Cerebrum is a part of the forebrain which does not form the brain stem — so Statement II is incorrect.
Q181.
Given below are two statements:
Statement I: Gause's competitive exclusion principle states that two closely related species competing for different resources cannot exist indefinitely.
Statement II: According to Gause's principle, during competition, the inferior will be eliminated. This may be true if resources are limiting.
In the light of the above statements, choose the correct answer from the options given below :
- A. Statement I is false but Statement II is true. ✓
- B. Both Statement I and Statement II are true.
- C. Both Statement I and Statement II are false.
- D. Statement I is true but Statement II is false.
Solution: Gause's competitive exclusion principle states that two closely related species competing for the SAME resources cannot exist indefinitely — Statement I says "different resources", so it is false.
The competitively inferior one will be eliminated eventually. This may be true if resources are limiting — so Statement II is true.
Q182.
Identify the correct option (A), (B), (C), (D) with respect to spermatogenesis.
The flow chart runs: GnRH branches into two arms. The left arm is GnRH $\rightarrow$ LH $\rightarrow$ (B) $\rightarrow$ Androgens $\rightarrow$ Formation of spermatids. The right arm is GnRH $\rightarrow$ (A) $\rightarrow$ (C) $\rightarrow$ Factors $\rightarrow$ (D).
- A. ICSH, Leydig cells, Sertoli cells, spermatogenesis.
- B. FSH, Leydig cells, Sertoli cells, spermiogenesis. ✓
- C. ICSH, Interstitial cells, Leydig cells, spermiogenesis.
- D. FSH, Sertoli cells, Leydig cells, spermatogenesis.
Solution: (A) is FSH, which is a pituitary hormone.
(B) is Leydig cells, which are found in the interstitial space outside of the seminiferous tubules and secrete androgens under LH.
(C) is Sertoli cells, which are found inside the seminiferous tubules and respond to FSH by secreting factors.
(D) is Spermiogenesis, which is the process that helps in transformation of spermatids into spermatozoa.
Q183.
The following are the statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post anal tail is absent.
Choose the most appropriate answer from the options given below:
- A. B, C & D only
- B. A & C only
- C. A, B & D only
- D. B, D & E only ✓
Solution: In non-chordates the notochord is absent, the heart is dorsal if present, and a post-anal tail is absent — so B, D and E are correct.
A perforated pharynx with gill slits and a dorsal central nervous system are chordate features; in non-chordates the central nervous system is ventral. So A and C are incorrect.
Q184.
Match List I with List II:
List I:
A. Unicellular glandular epithelium
B. Compound epithelium
C. Multicellular glandular epithelium
D. Endocrine glandular epithelium
List II:
I. Salivary glands
II. Pancreas
III. Goblet cells of alimentary canal
IV. Moist surface of buccal cavity
Choose the correct answer from the options given below:
- A. A-II, B-I, C-IV, D-III
- B. A-II, B-I, C-III, D-IV
- C. A-IV, B-III, C-I, D-II
- D. A-III, B-IV, C-I, D-II ✓
Solution: A. Unicellular glandular epithelium — Goblet cells of alimentary canal
B. Compound epithelium — lines the moist surface of the buccal cavity
C. Multicellular glandular epithelium — Salivary glands
D. Endocrine glandular epithelium — Pancreas
So A-III, B-IV, C-I, D-II.
Q185.
Match List I with List II :
List I:
A. P wave
B. QRS complex
C. T wave
D. T-P gap
List II:
I. Heart muscles are electrically silent.
II. Depolarisation of ventricles.
III. Depolarisation of atria.
IV. Repolarisation of ventricles.
Choose the correct answer from the options given below :
- A. A-IV, B-II, C-I, D-III
- B. A-I, B-III, C-IV, D-II
- C. A-III, B-II, C-IV, D-I ✓
- D. A-II, B-III, C-I, D-IV
Solution: A. P wave — Depolarisation of atria.
B. QRS complex — Depolarisation of ventricles.
C. T wave — Repolarisation of ventricles.
D. T-P gap — Heart muscles are electrically silent.
So A-III, B-II, C-IV, D-I.
Q186.
Match List I with List II :
List I:
A. Exophthalmic goiter
B. Acromegaly
C. Cushing's syndrome
D. Cretinism
List II:
I. Excess secretion of cortisol, moon face & hyperglycemia.
II. Hypo-secretion of thyroid hormone and stunted growth.
III. Hyper secretion of thyroid hormone & protruding eye balls.
IV. Excessive secretion of growth hormone.
Choose the correct answer from the options given below :
- A. A-III, B-IV, C-I, D-II ✓
- B. A-I, B-III, C-II, D-IV
- C. A-IV, B-II, C-I, D-III
- D. A-III, B-IV, C-II, D-I
Solution: (A) Exophthalmic goiter — hyper secretion of thyroid hormone, characterized by protruding eye balls.
(B) Acromegaly — excessive secretion of growth hormone.
(C) Cushing's syndrome — excess secretion of cortisol, moon face and hyperglycaemia.
(D) Cretinism — hypo-secretion of thyroid hormone, characterized by stunted growth.
So A-III, B-IV, C-I, D-II.
Q187.
Match List I with List II:
List I:
A. Mesozoic Era
B. Proterozoic Era
C. Cenozoic Era
D. Paleozoic Era
List II:
I. Lower invertebrates
II. Fish & Amphibia
III. Birds & Reptiles
IV. Mammals
Choose the correct answer from the options given below :
- A. A-III, B-I, C-IV, D-II ✓
- B. A-II, B-I, C-III, D-IV
- C. A-III, B-I, C-II, D-IV
- D. A-I, B-II, C-IV, D-III
Solution: (A) Mesozoic Era — Birds & Reptiles
(B) Proterozoic Era — Lower invertebrates
(C) Cenozoic Era — Mammals
(D) Paleozoic Era — Fish & Amphibia
So A-III, B-I, C-IV, D-II.
Q188.
Match List I with List II related to digestive system of cockroach.
List I:
A. The structures used for storing of food
B. Ring of 6-8 blind tubules at junction of foregut and midgut.
C. Ring of 100-150 yellow coloured thin filaments at junction of midgut and hindgut.
D. The structures used for grinding the food.
List II:
I. Gizzard
II. Gastric Caeca
III. Malpighian tubules
IV. Crop
Choose the correct answer from the options given below:
- A. A-III, B-II, C-IV, D-I
- B. A-IV, B-II, C-III, D-I ✓
- C. A-I, B-II, C-III, D-IV
- D. A-IV, B-III, C-II, D-I
Solution: The structure used for storing of food — Crop
Ring of 6-8 blind tubules at the junction of foregut and midgut, which assists in secretion of digestive juices — Gastric Caeca
Ring of 100-150 yellow coloured thin filaments at the junction of midgut and hindgut, which assists in elimination of nitrogenous wastes — Malpighian tubules
The structure used for grinding the food particles — Gizzard
So A-IV, B-II, C-III, D-I.
Q189.
Given below are two statements :
Statement I : Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II : Both bone marrow and thymus provide micro environments for the development and maturation of T-lymphocytes.
In the light of above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but Statement II is correct.
- B. Both Statement I and Statement II are correct. ✓
- C. Both Statement I and Statement II are incorrect.
- D. Statement I is correct but Statement II is incorrect.
Solution: In humans, the bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced — so Statement I is correct.
Both bone marrow and thymus provide micro-environments for the development and maturation of T-lymphocytes — so Statement II is correct as well.
Q190.
Match List I with List II:
List I:
A. RNA polymerase III
B. Termination of transcription
C. Splicing of Exons
D. TATA box
List II:
I. snRNPs
II. Promotor
III. Rho factor
IV. SnRNAs, tRNA
Choose the correct answer from the options given below :
- A. A-IV, B-III, C-I, D-II ✓
- B. A-II, B-IV, C-I, D-III
- C. A-III, B-II, C-IV, D-I
- D. A-III, B-IV, C-I, D-II
Solution: In eukaryotes, RNA polymerase III codes for snRNAs, tRNA and 5s rRNA.
Rho factor is responsible for termination of transcription.
Splicing of exons is performed by snRNPs.
TATA box is present in the promoter region of the transcription unit.
So A-IV, B-III, C-I, D-II.
Q191.
Choose the correct statement given below regarding juxta medullary nephron.
- A. Juxta medullary nephrons outnumber the cortical nephrons.
- B. Juxta medullary nephrons are located in the columns of Bertini.
- C. Renal corpuscle of juxta medullary nephron lies in the outer portion of the renal medulla.
- D. Loop of Henle of juxta medullary nephron runs deep into medulla. ✓
Solution: The length of the loop of Henle of a juxta medullary nephron is longer than that of a cortical nephron and runs deep into the medulla.
Juxta medullary nephrons are fewer in number than cortical nephrons, so the first option is incorrect.
Juxta medullary nephrons are not present in the columns of Bertini.
The renal corpuscle of a juxta medullary nephron lies in the inner cortical region, not in the medulla.
Q192.
Given below are two statements:
Statement I: Mitochondria and chloroplasts both double membranes bound organelles.
Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but Statement II is correct.
- B. Both Statement I and Statement II are correct.
- C. Both Statement I and Statement II are incorrect.
- D. Statement I is correct but Statement II is incorrect. ✓
Solution: Both mitochondria and chloroplasts are double membrane bound cell organelles, so Statement I is correct.
Transport of ions occurs across the inner membrane of mitochondria, while the inner membrane of the chloroplast is impermeable to ions and metabolites. Therefore the inner membrane of mitochondria is relatively MORE permeable than that of the chloroplast, so Statement II is incorrect.
Q193.
Regarding catalytic cycle of an enzyme action, select the correct sequential steps :
A. Substrate enzyme complex formation.
B. Free enzyme ready to bind with another substrate.
C. Release of products.
D. Chemical bonds of the substrate broken.
E. Substrate binding to active site.
Choose the correct answer from the options given below :
- A. E, D, C, B, A
- B. E, A, D, C, B ✓
- C. A, E, B, D, C
- D. B, A, C, D, E
Solution: The catalytic cycle of an enzyme action runs in the following order:
First, the substrate binds to the active site of the enzyme, fitting into the active site (E).
The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate, forming the enzyme-substrate complex (A).
The active site of the enzyme, now in close proximity of the substrate, breaks the chemical bonds of the substrate and the new enzyme-product complex is formed (D).
The enzyme releases the products of the reaction (C), and the free enzyme is ready to bind to another molecule of substrate and run through the catalytic cycle once again (B).
So the correct sequence is E, A, D, C, B.
Q194.
As per ABO blood grouping system, the blood group of father is B$^{+}$, mother is A$^{+}$ and child is O$^{+}$. Their respective genotype can be
A. I$^{B}$i / I$^{A}$i / ii
B. I$^{B}$I$^{B}$ / I$^{A}$I$^{A}$ / ii
C. I$^{A}$I$^{B}$ / iI$^{A}$ / I$^{B}$i
D. I$^{A}$i / I$^{B}$i / I$^{A}$i
E. iI$^{B}$ / iI$^{A}$ / I$^{A}$I$^{B}$
Choose the most appropriate answer from the options given below :
- A. D & E only
- B. A only ✓
- C. B only
- D. C & B only
Solution: A child with blood group O must be ii, so each parent must carry an i allele to pass on.
Genotype of father with blood group B$^{+}$ = I$^{B}$i
Genotype of mother with blood group A$^{+}$ = I$^{A}$i
Genotype of child with blood group O$^{+}$ = ii
Hence only 'A' is correct.