NEET (UG) 2018 — Code ZZ — Answer Key & Solutions

Free, login-free worked solutions to every question in NEET (UG) 2018 — Code ZZ. Check your answers below, then predict your rank or practise with full mock tests.

Q1.

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27$^{\circ}$C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27$^{\circ}$C is

  • A. 330 m/s
  • B. 339 m/s ✓
  • C. 300 m/s
  • D. 350 m/s

Solution: Two successive resonance lengths differ by half a wavelength, so $$v = 2\nu(L_2 - L_1)$$ $$= 2 \times 320 \times (73 - 20) \times 10^{-2}$$ $$= 339.2\ \text{ms}^{-1} \approx 339\ \text{m/s}$$

Q2.

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

  • A. Smaller ✓
  • B. 5 times greater
  • C. Equal
  • D. 10 times greater

Solution: The electric force gives an acceleration $a = \dfrac{eE}{m}$, so $$h = \frac{1}{2}\frac{eE}{m}t^{2} \Rightarrow t = \sqrt{\frac{2hm}{eE}}$$ So $t \propto \sqrt{m}$, as 'e' is the same for the electron and the proton. The electron has a smaller mass, so it will take a smaller time.

Q3.

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s$^{2}$ at a distance of 5 m from the mean position. The time period of oscillation is

  • A. 2$\pi$ s
  • B. $\pi$ s ✓
  • C. 1 s
  • D. 2 s

Solution: $$|a| = \omega^{2}y$$ $$20 = \omega^{2}(5) \Rightarrow \omega = 2\ \text{rad/s}$$ $$T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi\ \text{s}$$

Q4.

The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

  • A. Independent of the distance between the plates ✓
  • B. Linearly proportional to the distance between the plates
  • C. Inversely proportional to the distance between the plates
  • D. Proportional to the square root of the distance between the plates

Solution: For an isolated capacitor the charge Q is constant, and the force on one plate due to the field of the other is $$F_{plate} = \frac{Q^{2}}{2A\epsilon_0}$$ This expression contains no separation term, so F is independent of the distance between the plates.

Q5.

Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

  • A. 40 $\Omega$
  • B. 25 $\Omega$
  • C. 500 $\Omega$
  • D. 250 $\Omega$ ✓

Solution: Current sensitivity $I_S = \dfrac{NBA}{C}$ and voltage sensitivity $V_S = \dfrac{NBA}{CR_G}$. So the resistance of the galvanometer is $$R_G = \frac{I_S}{V_S} = \frac{5 \times 1}{20 \times 10^{-3}} = \frac{5000}{20} = 250\ \Omega$$

Q6.

A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from

  • A. The current source ✓
  • B. The magnetic field
  • C. The induced electric field due to the changing magnetic field
  • D. The lattice structure of the material of the rod

Solution: Energy of the current source will be converted into the potential energy of the rod. The magnetic field itself does no work; the source maintaining the current supplies the energy.

Q7.

An inductor 20 mH, a capacitor 100 $\mu$F and a resistor 50 $\Omega$ are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

  • A. 0.79 W ✓
  • B. 0.43 W
  • C. 1.13 W
  • D. 2.74 W

Solution: $$Z = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}} = 56\ \Omega$$ $$P_{av} = \left(\frac{V_{RMS}}{Z}\right)^{2}R = \left(\frac{10}{\sqrt{2}\times 56}\right)^{2}\times 50 = 0.79\ \text{W}$$

Q8.

A metallic rod of mass per unit length 0.5 kg m$^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of 30$^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

  • A. 7.14 A
  • B. 5.98 A
  • C. 11.32 A ✓
  • D. 14.76 A

Solution: For equilibrium along the incline, $$mg\sin 30^{\circ} = IlB\cos 30^{\circ}$$ $$I = \frac{m}{l}\frac{g}{B}\tan 30^{\circ} = \frac{0.5 \times 9.8}{0.25 \times \sqrt{3}} = 11.32\ \text{A}$$

Q9.

A carbon resistor of (47 $\pm$ 4.7) k$\Omega$ is to be marked with rings of different colours for its identification. The colour code sequence will be

  • A. Violet – Yellow – Orange – Silver
  • B. Yellow – Violet – Orange – Silver ✓
  • C. Green – Orange – Violet – Gold
  • D. Yellow – Green – Violet – Gold

Solution: $(47 \pm 4.7)$ k$\Omega$ $= 47 \times 10^{3} \pm 10\%$ 4 is Yellow, 7 is Violet, the multiplier $10^{3}$ is Orange, and $\pm 10\%$ tolerance is Silver. So the sequence is Yellow – Violet – Orange – Silver.

Q10.

A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is

  • A. 10 ✓
  • B. 11
  • C. 9
  • D. 20

Solution: In series, $$I = \frac{E}{nR + R}$$ In parallel, $$10I = \frac{E}{\frac{R}{n} + R}$$ Dividing the second by the first, $$10 = \frac{(n + 1)R}{\left(\frac{1}{n} + 1\right)R}$$ Solving this equation gives $n = 10$.

Q11.

A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?

  • A. Graph (1) ✓
  • B. Graph (2)
  • C. Graph (3)
  • D. Graph (4)

Solution: When n identical cells are in series and the terminals are short-circuited, $$I = \frac{n\varepsilon}{nr} = \frac{\varepsilon}{r}$$ So I is independent of n and I is constant. The correct graph is a horizontal straight line.

Q12.

In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength $\lambda$ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20$^{\circ}$. To increase the fringe angular width to 0.21$^{\circ}$ (with same $\lambda$ and D) the separation between the slits needs to be changed to

  • A. 1.8 mm
  • B. 1.9 mm ✓
  • C. 1.7 mm
  • D. 2.1 mm

Solution: Angular width $= \dfrac{\lambda}{d}$ $$0.20^{\circ} = \frac{\lambda}{2\ \text{mm}}\qquad \ldots(i)$$ $$0.21^{\circ} = \frac{\lambda}{d}\qquad \ldots(ii)$$ Dividing we get $$\frac{0.20}{0.21} = \frac{d}{2\ \text{mm}}$$ $$d = 1.9\ \text{mm}$$

Q13.

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

  • A. Small focal length and large diameter
  • B. Large focal length and small diameter
  • C. Small focal length and small diameter
  • D. Large focal length and large diameter ✓

Solution: For a telescope, angular magnification $= \dfrac{f_0}{f_E}$, so the focal length of the objective lens should be large. Angular resolution $= \dfrac{D}{1.22\lambda}$ should be large, so the diameter should be large. So the objective should have a large focal length $f_0$ and a large diameter D.

Q14.

Unpolarised light is incident from air on a plane surface of a material of refractive index '$\mu$'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

  • A. Reflected light is polarised with its electric vector parallel to the plane of incidence
  • B. Reflected light is polarised with its electric vector perpendicular to the plane of incidence ✓
  • C. $i = \tan^{-1}\left(\dfrac{1}{\mu}\right)$
  • D. $i = \sin^{-1}\left(\dfrac{1}{\mu}\right)$

Solution: When the reflected and refracted rays are perpendicular, the incidence is at the Brewster angle and the reflected light is completely plane polarised with its electric field vector perpendicular to the plane of incidence. Also $\tan i = \mu$ at the Brewster angle, so options (3) and (4) are wrong.

Q15.

An em wave is propagating in a medium with a velocity $\vec{V} = V\hat{i}$. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along

  • A. –z direction
  • B. +z direction ✓
  • C. –x direction
  • D. –y direction

Solution: For an electromagnetic wave, $\vec{E} \times \vec{B}$ points along the direction of propagation. $$(E\hat{j}) \times \vec{B} = V\hat{i}$$ Since $\hat{j} \times \hat{k} = \hat{i}$, we get $\vec{B} = B\hat{k}$. So the magnetic field oscillates along the +z direction.

Q16.

The refractive index of the material of a prism is $\sqrt{2}$ and the angle of the prism is 30$^{\circ}$. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

  • A. 60$^{\circ}$
  • B. 45$^{\circ}$ ✓
  • C. Zero
  • D. 30$^{\circ}$

Solution: For retracing its path, the light ray should be normally incident on the silvered face. Then the angle of refraction at the first face is 30$^{\circ}$, equal to the prism angle. Applying Snell's law at the first face M, $$\frac{\sin i}{\sin 30^{\circ}} = \frac{\sqrt{2}}{1}$$ $$\sin i = \sqrt{2}\times\frac{1}{2} = \frac{1}{\sqrt{2}}$$ $$i = 45^{\circ}$$

Q17.

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

  • A. 30 cm away from the mirror
  • B. 36 cm away from the mirror ✓
  • C. 36 cm towards the mirror
  • D. 30 cm towards the mirror

Solution: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ For $u_1 = -40$ cm and $f = -15$ cm, $$\frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} \Rightarrow v_1 = -24\ \text{cm}$$ When the object is displaced by 20 cm towards the mirror, $u_2 = -20$ cm, $$\frac{1}{v_2} = \frac{1}{-15} + \frac{1}{20} \Rightarrow v_2 = -60\ \text{cm}$$ So the image shifts away from the mirror by $60 - 24 = 36$ cm.

Q18.

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance

  • A. 0.138 H
  • B. 138.88 H
  • C. 13.89 H ✓
  • D. 1.389 H

Solution: Energy stored in an inductor is $$U = \frac{1}{2}LI^{2}$$ $$25 \times 10^{-3} = \frac{1}{2}\times L \times (60 \times 10^{-3})^{2}$$ $$L = \frac{25 \times 2 \times 10^{6} \times 10^{-3}}{3600} = \frac{500}{36} = 13.89\ \text{H}$$

Q19.

For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is

  • A. 20 ✓
  • B. 10
  • C. 15
  • D. 30

Solution: Number of nuclei remaining $= 600 - 450 = 150$ $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{n}$$ $$\frac{150}{600} = \left(\frac{1}{2}\right)^{t/t_{1/2}}$$ $$\left(\frac{1}{2}\right)^{2} = \left(\frac{1}{2}\right)^{t/t_{1/2}}$$ $$t = 2t_{1/2} = 2 \times 10 = 20\ \text{minutes}$$

Q20.

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is

  • A. 1 : 1
  • B. 1 : –1 ✓
  • C. 1 : –2
  • D. 2 : –1

Solution: In a Bohr orbit, KE $= -$(total energy). So kinetic energy : total energy $= 1 : -1$.

Q21.

An electron of mass m with an initial velocity $\vec{V} = V_0\hat{i}$ (V$_0$ > 0) enters an electric field $\vec{E} = -E_0\hat{i}$ (E$_0$ = constant > 0) at t = 0. If $\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

  • A. $\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0}{mV_0}t\right)}$ ✓
  • B. $\lambda_0\left(1 + \dfrac{eE_0}{mV_0}t\right)$
  • C. $\lambda_0$
  • D. $\lambda_0 t$

Solution: Initial de-Broglie wavelength $\lambda_0 = \dfrac{h}{mV_0}$. The electron carries a negative charge, so in the field $-E_0\hat{i}$ the force on it is along $+\hat{i}$ and the acceleration is $$a = \frac{eE_0}{m}$$ Velocity after time t is $$V = V_0 + \frac{eE_0}{m}t$$ $$\lambda = \frac{h}{mV} = \frac{h}{mV_0\left[1 + \dfrac{eE_0}{mV_0}t\right]} = \frac{\lambda_0}{\left[1 + \dfrac{eE_0}{mV_0}t\right]}$$

Q22.

When the light of frequency 2$\nu_0$ (where $\nu_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v$_1$. When the frequency of the incident radiation is increased to 5$\nu_0$, the maximum velocity of electrons emitted from the same plate is v$_2$. The ratio of v$_1$ to v$_2$ is

  • A. 1 : 2 ✓
  • B. 1 : 4
  • C. 2 : 1
  • D. 4 : 1

Solution: $$E = W_0 + \frac{1}{2}mv^{2}$$ $$h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^{2} \Rightarrow h\nu_0 = \frac{1}{2}mv_1^{2}\qquad \ldots(i)$$ $$h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^{2} \Rightarrow 4h\nu_0 = \frac{1}{2}mv_2^{2}\qquad \ldots(ii)$$ Dividing (i) by (ii), $$\frac{1}{4} = \frac{v_1^{2}}{v_2^{2}} \Rightarrow \frac{v_1}{v_2} = \frac{1}{2}$$

Q23.

In the combination of the following gates the output Y can be written in terms of inputs A and B as

  • A. $\overline{A \cdot B}$
  • B. $A \cdot \overline{B} + \overline{A} \cdot B$ ✓
  • C. $\overline{A + B}$
  • D. $\overline{A \cdot B} + A \cdot B$

Solution: The upper AND gate receives A and $\overline{B}$, so its output is $A\cdot\overline{B}$. The lower AND gate receives $\overline{A}$ and B, so its output is $\overline{A}\cdot B$. The two outputs feed an OR gate, so $$Y = (A\cdot\overline{B} + \overline{A}\cdot B)$$

Q24.

In the circuit shown in the figure, the input voltage V$_i$ is 20 V, V$_{BE}$ = 0 and V$_{CE}$ = 0. The values of I$_B$, I$_C$ and $\beta$ are given by

  • A. I$_B$ = 40 $\mu$A, I$_C$ = 10 mA, $\beta$ = 250
  • B. I$_B$ = 25 $\mu$A, I$_C$ = 5 mA, $\beta$ = 200
  • C. I$_B$ = 40 $\mu$A, I$_C$ = 5 mA, $\beta$ = 125 ✓
  • D. I$_B$ = 20 $\mu$A, I$_C$ = 5 mA, $\beta$ = 250

Solution: With $V_{CE} = 0$, the whole 20 V appears across R$_C$: $$I_C = \frac{20 - 0}{4 \times 10^{3}} = 5 \times 10^{-3} = 5\ \text{mA}$$ With $V_{BE} = 0$, $$V_i = V_{BE} + I_BR_B \Rightarrow 20 = I_B \times 500 \times 10^{3}$$ $$I_B = \frac{20}{500 \times 10^{3}} = 40\ \mu\text{A}$$ $$\beta = \frac{I_C}{I_B} = \frac{5 \times 10^{-3}}{40 \times 10^{-6}} = 125$$

Q25.

In a p-n junction diode, change in temperature due to heating

  • A. Affects only reverse resistance
  • B. Affects only forward resistance
  • C. Affects the overall V - I characteristics of p-n junction ✓
  • D. Does not affect resistance of p-n junction

Solution: Due to heating, the number of electron-hole pairs will increase, so the overall resistance of the diode will change. Because of this, both forward biasing and reverse biasing characteristics are changed, i.e. the overall V-I characteristic is affected.

Q26.

A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

  • A. Angular velocity
  • B. Moment of inertia
  • C. Angular momentum ✓
  • D. Rotational kinetic energy

Solution: The sphere is rotating freely in free space, so no external torque acts on it. Hence the angular momentum of the sphere remains constant. Since the moment of inertia increases with radius, the angular velocity and the rotational kinetic energy both decrease.

Q27.

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K$_A$, K$_B$ and K$_C$, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

  • A. K$_A$ < K$_B$ < K$_C$
  • B. K$_A$ > K$_B$ > K$_C$ ✓
  • C. K$_B$ > K$_A$ > K$_C$
  • D. K$_B$ < K$_A$ < K$_C$

Solution: Point A is the perihelion and C is the aphelion, and B lies on the perpendicular to AC through the Sun. By conservation of angular momentum the planet moves fastest at the perihelion and slowest at the aphelion, so $$V_A > V_B > V_C$$ Hence $$K_A > K_B > K_C$$

Q28.

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

  • A. Raindrops will fall faster
  • B. Walking on the ground would become more difficult
  • C. 'g' on the Earth will not change ✓
  • D. Time period of a simple pendulum on the Earth would decrease

Solution: If the universal gravitational constant becomes ten times larger, then $G' = 10G$. Since $g = \dfrac{GM_e}{R_e^{2}}$ depends on G and on the mass of the Earth (not of the Sun), the acceleration due to gravity increases. So the statement that 'g' on the Earth will not change is the wrong option.

Q29.

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K$_t$) as well as rotational kinetic energy (K$_r$) simultaneously. The ratio K$_t$ : (K$_t$ + K$_r$) for the sphere is

  • A. 7 : 10
  • B. 5 : 7 ✓
  • C. 2 : 5
  • D. 10 : 7

Solution: $$K_t = \frac{1}{2}mv^{2}$$ $$K_t + K_r = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} = \frac{1}{2}mv^{2} + \frac{1}{2}\left(\frac{2}{5}mr^{2}\right)\left(\frac{v}{r}\right)^{2} = \frac{7}{10}mv^{2}$$ $$\frac{K_t}{K_t + K_r} = \frac{\frac{1}{2}mv^{2}}{\frac{7}{10}mv^{2}} = \frac{5}{7}$$

Q30.

A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

  • A. r$^{3}$
  • B. r$^{2}$
  • C. r$^{4}$
  • D. r$^{5}$ ✓

Solution: Rate of production of heat is the power dissipated against the viscous force: $$\text{Power} = 6\pi\eta r V_T \cdot V_T = 6\pi\eta r V_T^{2}$$ The terminal velocity satisfies $V_T \propto r^{2}$, so $$\text{Power} \propto r \times (r^{2})^{2} = r^{5}$$

Q31.

A sample of 0.1 g of water at 100$^{\circ}$C and normal pressure ($1.013 \times 10^{5}$ Nm$^{-2}$) requires 54 cal of heat energy to convert to steam at 100$^{\circ}$C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

  • A. 104.3 J
  • B. 208.7 J ✓
  • C. 84.5 J
  • D. 42.2 J

Solution: $$\Delta Q = \Delta U + \Delta W$$ $$54 \times 4.18 = \Delta U + 1.013 \times 10^{5}(167.1 \times 10^{-6} - 0)$$ $$\Delta U = 208.7\ \text{J}$$

Q32.

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by $\Delta l$ on applying a force F, how much force is needed to stretch the second wire by the same amount?

  • A. 9 F ✓
  • B. 6 F
  • C. F
  • D. 4 F

Solution: Equal volumes mean wire 1 has area A and length $3l$, while wire 2 has area $3A$ and length $l$. For wire 1, $$\Delta l = \left(\frac{F}{AY}\right)3l\qquad \ldots(i)$$ For wire 2, $$\frac{F'}{3A} = Y\frac{\Delta l}{l} \Rightarrow \Delta l = \left(\frac{F'}{3AY}\right)l\qquad \ldots(ii)$$ From (i) and (ii), $$\left(\frac{F}{AY}\right)3l = \left(\frac{F'}{3AY}\right)l$$ $$F' = 9F$$

Q33.

The power radiated by a black body is P and it radiates maximum energy at wavelength, $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\dfrac{3}{4}\lambda_0$, the power radiated by it becomes nP. The value of n is

  • A. $\dfrac{3}{4}$
  • B. $\dfrac{4}{3}$
  • C. $\dfrac{81}{256}$
  • D. $\dfrac{256}{81}$ ✓

Solution: By Wien's law, $\lambda_{max}T = $ constant, so $$\lambda_0 T = \frac{3\lambda_0}{4}T' \Rightarrow T' = \frac{4}{3}T$$ By Stefan's law $P \propto T^{4}$, so $$\frac{P_2}{P_1} = \left(\frac{T'}{T}\right)^{4} = \left(\frac{4}{3}\right)^{4} = \frac{256}{81}$$

Q34.

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given : Mass of oxygen molecule (m) = $2.76 \times 10^{-26}$ kg, Boltzmann's constant k$_B$ = $1.38 \times 10^{-23}$ JK$^{-1}$)

  • A. $2.508 \times 10^{4}$ K
  • B. $8.360 \times 10^{4}$ K ✓
  • C. $1.254 \times 10^{4}$ K
  • D. $5.016 \times 10^{4}$ K

Solution: $V_{escape} = 11200$ m/s. Say at temperature T the rms speed attains $V_{escape}$: $$\sqrt{\frac{3k_BT}{m_{O_2}}} = 11200\ \text{m/s}$$ On solving, $$T = 8.360 \times 10^{4}\ \text{K}$$

Q35.

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

  • A. $\dfrac{2}{5}$ ✓
  • B. $\dfrac{2}{3}$
  • C. $\dfrac{2}{7}$
  • D. $\dfrac{1}{3}$

Solution: The graph of V against T is a straight line through the origin, so $\dfrac{V}{T}$ is constant and the process is isobaric. For an isobaric process, $$W = nR\Delta T$$ $$Q = nC_P\Delta T = n\left(\frac{5}{2}R\right)\Delta T\quad\text{for a monatomic gas}$$ $$\frac{W}{Q} = \frac{nR\Delta T}{\frac{5}{2}nR\Delta T} = \frac{2}{5}$$

Q36.

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

  • A. 13.2 cm ✓
  • B. 8 cm
  • C. 16 cm
  • D. 12.5 cm

Solution: For a closed organ pipe, the third harmonic is $\dfrac{3v}{4l}$. For an open organ pipe, the fundamental frequency is $\dfrac{v}{2l'}$. Given that these are equal, $$\frac{3v}{4l} = \frac{v}{2l'}$$ $$l' = \frac{4l}{3 \times 2} = \frac{2l}{3} = \frac{2 \times 20}{3} \approx 13.3\ \text{cm}$$ which matches the option 13.2 cm.

Q37.

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

  • A. 26.8% ✓
  • B. 20%
  • C. 12.5%
  • D. 6.25%

Solution: Efficiency of an ideal heat engine is $$\eta = \left(1 - \frac{T_2}{T_1}\right)$$ where T$_2$ is the sink temperature and T$_1$ the source temperature. $$\%\eta = \left(1 - \frac{273}{373}\right)\times 100 = \left(\frac{100}{373}\right)\times 100 = 26.8\%$$

Q38.

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

  • A. $\dfrac{3}{2}D$
  • B. $D$
  • C. $\dfrac{5}{4}D$ ✓
  • D. $\dfrac{7}{5}D$

Solution: As the track is frictionless, the total mechanical energy remains constant. $$0 + mgh = \frac{1}{2}mv_L^{2} + 0 \Rightarrow h = \frac{v_L^{2}}{2g}$$ For completing the vertical circle, $v_L \geq \sqrt{5gR}$, so $$h = \frac{5gR}{2g} = \frac{5}{2}R = \frac{5}{4}D$$

Q39.

Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

  • A. W$_C$ > W$_B$ > W$_A$ ✓
  • B. W$_A$ > W$_B$ > W$_C$
  • C. W$_A$ > W$_C$ > W$_B$
  • D. W$_B$ > W$_A$ > W$_C$

Solution: Work done required to bring them to rest is $$\Delta W = \Delta KE = \frac{1}{2}I\omega^{2}$$ So $\Delta W \propto I$ for the same $\omega$. $$W_A : W_B : W_C = \frac{2}{5}MR^{2} : \frac{1}{2}MR^{2} : MR^{2} = \frac{2}{5} : \frac{1}{2} : 1 = 4 : 5 : 10$$ $$W_C > W_B > W_A$$

Q40.

Which one of the following statements is incorrect?

  • A. Rolling friction is smaller than sliding friction.
  • B. Limiting value of static friction is directly proportional to normal reaction.
  • C. Coefficient of sliding friction has dimensions of length. ✓
  • D. Frictional force opposes the relative motion.

Solution: Coefficient of sliding friction has no dimensions. $$f = \mu_s N \Rightarrow \mu_s = \frac{f}{N}$$ It is the ratio of two forces, so it is dimensionless — statement (3) is incorrect.

Q41.

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

  • A. 0.5
  • B. 0.25 ✓
  • C. 0.4
  • D. 0.8

Solution: According to the law of conservation of linear momentum, $$mv + 4m \times 0 = 4mv' + 0 \Rightarrow v' = \frac{v}{4}$$ $$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v/4}{v}$$ $$e = \frac{1}{4} = 0.25$$

Q42.

A block of mass m is placed on a smooth inclined wedge ABC of inclination $\theta$ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and $\theta$ for the block to remain stationary on the wedge is

  • A. $a = \dfrac{g}{\text{cosec}\,\theta}$
  • B. $a = \dfrac{g}{\sin\theta}$
  • C. $a = g\tan\theta$ ✓
  • D. $a = g\cos\theta$

Solution: Working in the frame of the wedge, a pseudo force $ma$ acts on the block towards the left. Resolving along and perpendicular to the incline, for the block to stay put: $$N\sin\theta = ma$$ $$N\cos\theta = mg$$ Dividing, $$\tan\theta = \frac{a}{g} \Rightarrow a = g\tan\theta$$

Q43.

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field $\vec{E}$. Due to the force $q\vec{E}$, its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

  • A. 2 m/s, 4 m/s
  • B. 1 m/s, 3 m/s ✓
  • C. 1.5 m/s, 3 m/s
  • D. 1 m/s, 3.5 m/s

Solution: Acceleration $a = \dfrac{6 - 0}{1} = 6$ ms$^{-2}$ For $t = 0$ to $t = 1$ s, $$S_1 = \frac{1}{2}\times 6(1)^{2} = 3\ \text{m}$$ For $t = 1$ s to $t = 2$ s (field reversed, car decelerates to rest), $$S_2 = 6(1) - \frac{1}{2}\times 6(1)^{2} = 3\ \text{m}$$ For $t = 2$ s to $t = 3$ s (car moves back), $$S_3 = 0 - \frac{1}{2}\times 6(1)^{2} = -3\ \text{m}$$ Total displacement $S = S_1 + S_2 + S_3 = 3$ m, so $$\text{Average velocity} = \frac{3}{3} = 1\ \text{ms}^{-1}$$ Total distance travelled $= 3 + 3 + 3 = 9$ m, so $$\text{Average speed} = \frac{9}{3} = 3\ \text{ms}^{-1}$$

Q44.

The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$ at (2, 0, –3), about the point (2, –2, –2), is given by

  • A. $-8\hat{i} - 4\hat{j} - 7\hat{k}$
  • B. $-4\hat{i} - \hat{j} - 8\hat{k}$
  • C. $-7\hat{i} - 4\hat{j} - 8\hat{k}$ ✓
  • D. $-7\hat{i} - 8\hat{j} - 4\hat{k}$

Solution: $$\vec{\tau} = (\vec{r} - \vec{r}_0)\times\vec{F}$$ $$\vec{r} - \vec{r}_0 = (2\hat{i} + 0\hat{j} - 3\hat{k}) - (2\hat{i} - 2\hat{j} - 2\hat{k}) = 0\hat{i} + 2\hat{j} - \hat{k}$$ Taking the cross product with $\vec{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$, $$\vec{\tau} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 0 & 2 & -1\\ 4 & 5 & -6\end{vmatrix} = -7\hat{i} - 4\hat{j} - 8\hat{k}$$

Q45.

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is

  • A. 0.521 cm
  • B. 0.525 cm
  • C. 0.529 cm ✓
  • D. 0.053 cm

Solution: Diameter of the ball $=$ MSR $+$ CSR $\times$ (Least count) $-$ Zero error $$= 0.5\ \text{cm} + 25 \times 0.001 - (-0.004)$$ $$= 0.5 + 0.025 + 0.004 = 0.529\ \text{cm}$$

Q46.

The difference between spermiogenesis and spermiation is

  • A. In spermiogenesis spermatids are formed, while in spermiation spermatozoa are formed.
  • B. In spermiogenesis spermatozoa are formed, while in spermiation spermatids are formed.
  • C. In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules. ✓
  • D. In spermiogenesis spermatozoa from sertoli cells are released into the cavity of seminiferous tubules, while in spermiation spermatozoa are formed.

Solution: Spermiogenesis is the transformation of spermatids into spermatozoa. Spermiation is the release of the sperms from Sertoli cells into the lumen of the seminiferous tubule.

Q47.

The amnion of mammalian embryo is derived from

  • A. ectoderm and mesoderm ✓
  • B. endoderm and mesoderm
  • C. ectoderm and endoderm
  • D. mesoderm and trophoblast

Solution: The extraembryonic or foetal membranes are amnion, chorion, allantois and yolk sac. Amnion is formed from mesoderm on the outer side and ectoderm on the inner side. Chorion is formed from trophoectoderm and mesoderm, whereas the allantois and yolk sac membrane have mesoderm on the outer side and endoderm on the inner side.

Q48.

The contraceptive 'SAHELI'

  • A. blocks estrogen receptors in the uterus, preventing eggs from getting implanted. ✓
  • B. increases the concentration of estrogen and prevents ovulation in females.
  • C. is a post-coital contraceptive.
  • D. is an IUD.

Solution: Saheli is the first non-steroidal, once a week pill. It contains centchroman and its functioning is based upon selective estrogen receptor modulation, i.e. it blocks estrogen receptors in the uterus so that eggs cannot be implanted.

Q49.

Hormones secreted by the placenta to maintain pregnancy are

  • A. hCG, hPL, progestogens, prolactin
  • B. hCG, hPL, estrogens, relaxin, oxytocin
  • C. hCG, progestogens, estrogens, glucocorticoids
  • D. hCG, hPL, progestogens, estrogens ✓

Solution: The placenta releases human chorionic gonadotropic hormone (hCG), which stimulates the corpus luteum during pregnancy to release estrogen and progesterone and also rescues the corpus luteum from regression. Human placental lactogen (hPL) is involved in growth of the body of the mother and the breast. Progesterone maintains pregnancy and keeps the uterus silent by increasing the uterine threshold to contractile stimuli.

Q50.

Match the items given in Column I with those in Column II and select the correct option given below : Column I: (a) Proliferative Phase (b) Secretory Phase (c) Menstruation Column II: (i) Breakdown of endometrial lining (ii) Follicular Phase (iii) Luteal Phase

  • A. (a)-(iii), (b)-(ii), (c)-(i)
  • B. (a)-(i), (b)-(iii), (c)-(ii)
  • C. (a)-(iii), (b)-(i), (c)-(ii)
  • D. (a)-(ii), (b)-(iii), (c)-(i) ✓

Solution: During the proliferative phase the follicles start developing, hence it is called the follicular phase. The secretory phase is also called the luteal phase, mainly controlled by progesterone secreted by the corpus luteum. Estrogen further thickens the endometrium maintained by progesterone. Menstruation occurs due to a decline in progesterone level and involves breakdown of the overgrown endometrial lining.

Q51.

All of the following are part of an operon except

  • A. an operator
  • B. structural genes
  • C. a promoter
  • D. an enhancer ✓

Solution: The operon concept is for prokaryotes, and an operon consists of a promoter, an operator and structural genes. Enhancer sequences are present in eukaryotes, so an enhancer is not part of an operon.

Q52.

A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by

  • A. Only daughters
  • B. Only sons
  • C. Both sons and daughters ✓
  • D. Only grandchildren

Solution: The woman is a carrier, so either of her two X chromosomes can pass to any child. Both son and daughter can inherit that X-chromosome, although only the son will be diseased if the condition is recessive.

Q53.

According to Hugo de Vries, the mechanism of evolution is

  • A. Multiple step mutations
  • B. Saltation ✓
  • C. Minor mutations
  • D. Phenotypic variations

Solution: As per the mutation theory given by Hugo de Vries, evolution is a discontinuous phenomenon or saltatory phenomenon, i.e. saltation (single step large mutation).

Q54.

AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?

  • A. AGGUAUCGCAU ✓
  • B. UGGTUTCGCAT
  • C. UCCAUAGCGUA
  • D. ACCUAUGCGAU

Solution: The coding strand and the mRNA have the same nucleotide sequence, except that 'T' – thymine is replaced by 'U' – uracil in mRNA. So AGGTATCGCAT gives AGGUAUCGCAU.

Q55.

Among the following sets of examples for divergent evolution, select the incorrect option :

  • A. Forelimbs of man, bat and cheetah
  • B. Heart of bat, man and cheetah
  • C. Eye of octopus, bat and man ✓
  • D. Brain of bat, man and cheetah

Solution: Divergent evolution occurs in the same structure, for example the forelimbs, heart and brain of vertebrates, which have developed along different directions due to adaptation to different needs. Whereas the eye of octopus, bat and man are examples of analogous organs showing convergent evolution, so option (3) is the incorrect one.

Q56.

Conversion of milk to curd improves its nutritional value by increasing the amount of

  • A. Vitamin D
  • B. Vitamin A
  • C. Vitamin E
  • D. Vitamin B$_{12}$ ✓

Solution: Curd is more nourishing than milk. It has an enriched presence of vitamins, especially vitamin B$_{12}$.

Q57.

Which of the following is not an autoimmune disease?

  • A. Psoriasis
  • B. Rheumatoid arthritis
  • C. Vitiligo
  • D. Alzheimer's disease ✓

Solution: Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage. Vitiligo causes white patches on skin and is also characterised as an autoimmune disorder. Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune. Whereas Alzheimer's disease is due to deficiency of the neurotransmitter acetylcholine, so it is not autoimmune.

Q58.

The similarity of bone structure in the forelimbs of many vertebrates is an example of

  • A. Homology ✓
  • B. Analogy
  • C. Adaptive radiation
  • D. Convergent evolution

Solution: The forelimbs of many vertebrates have the same fundamental bone structure but perform different functions. Such structures are homologous organs, so this is an example of homology, resulting from divergent evolution.

Q59.

Which of the following characteristics represent 'Inheritance of blood groups' in humans? (a) Dominance (b) Co-dominance (c) Multiple allele (d) Incomplete dominance (e) Polygenic inheritance

  • A. (b), (c) and (e)
  • B. (a), (b) and (c) ✓
  • C. (a), (c) and (e)
  • D. (b), (d) and (e)

Solution: I$^{A}$I$^{O}$ and I$^{B}$I$^{O}$ show a dominant-recessive relationship, i.e. dominance. I$^{A}$I$^{B}$ shows codominance. I$^{A}$, I$^{B}$ and I$^{O}$ are three different allelic forms of one gene, i.e. multiple allelism. So (a), (b) and (c) are the characteristics shown.

Q60.

In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?

  • A. Elephantiasis ✓
  • B. Ascariasis
  • C. Amoebiasis
  • D. Ringworm disease

Solution: Elephantiasis is caused by the roundworm $Wuchereria$ $bancrofti$ and it is transmitted by the $Culex$ mosquito. The worms cause chronic inflammation of the lymphatic vessels.

Q61.

All of the following are included in '$ex$-$situ$ conservation' except

  • A. Wildlife safari parks
  • B. Sacred groves ✓
  • C. Seed banks
  • D. Botanical gardens

Solution: Sacred groves are a method of in-situ conservation. They represent pristine forest patches protected by tribal groups.

Q62.

Which part of poppy plant is used to obtain the drug "Smack"?

  • A. Flowers
  • B. Latex ✓
  • C. Leaves
  • D. Roots

Solution: 'Smack', also called brown sugar or heroin, is formed by acetylation of morphine. It is obtained from the latex of the unripe capsule of the poppy plant.

Q63.

In a growing population of a country,

  • A. pre-reproductive individuals are more than the reproductive individuals. ✓
  • B. reproductive individuals are less than the post-reproductive individuals.
  • C. pre-reproductive individuals are less than the reproductive individuals.
  • D. reproductive and pre-reproductive individuals are equal in number.

Solution: Whenever the pre-reproductive individuals, i.e. the younger population size, is larger than the reproductive group, the population will be an increasing population. The age pyramid for such a population is broad at the base.

Q64.

Which one of the following population interactions is widely used in medical science for the production of antibiotics?

  • A. Commensalism
  • B. Mutualism
  • C. Amensalism ✓
  • D. Parasitism

Solution: Amensalism or antibiosis is a (0, –) interaction. Antibiotics are chemicals secreted by one microbial group, e.g. $Penicillium$, which harm other microbes, e.g. $Staphylococcus$. It has no effect on $Penicillium$ or the organism which produces it.

Q65.

Match the items given in Column I with those in Column II and select the correct option given below : Column-I: (a) Eutrophication (b) Sanitary landfill (c) Snow blindness (d) Jhum cultivation Column-II: (i) UV-B radiation (ii) Deforestation (iii) Nutrient enrichment (iv) Waste disposal

  • A. (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
  • B. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
  • C. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
  • D. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) ✓

Solution: Eutrophication is the natural ageing of a lake by nutrient enrichment of its water. Sanitary landfills are used for waste disposal. Snow blindness is caused by inflammation of the cornea due to high doses of UV-B radiation. Jhum cultivation, i.e. slash and burn agriculture, is a cause of deforestation.

Q66.

Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively?

  • A. Inflammation of bronchioles; Decreased respiratory surface ✓
  • B. Increased number of bronchioles; Increased respiratory surface
  • C. Decreased respiratory surface; Inflammation of bronchioles
  • D. Increased respiratory surface; Inflammation of bronchioles

Solution: Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. Emphysema is a chronic disorder in which alveolar walls are damaged, due to which the respiratory surface is decreased.

Q67.

Match the items given in Column I with those in Column II and select the correct option given below : Column I: (a) Tricuspid valve (b) Bicuspid valve (c) Semilunar valve Column II: (i) Between left atrium and left ventricle (ii) Between right ventricle and pulmonary artery (iii) Between right atrium and right ventricle

  • A. (a)-(iii), (b)-(i), (c)-(ii) ✓
  • B. (a)-(i), (b)-(iii), (c)-(ii)
  • C. (a)-(ii), (b)-(i), (c)-(iii)
  • D. (a)-(i), (b)-(ii), (c)-(iii)

Solution: Tricuspid valve is the AV valve present between the right atrium and right ventricle. Bicuspid valve is the AV valve present between the left atrium and left ventricle. Semilunar valves are present at the openings of the aorta and pulmonary artery.

Q68.

Match the items given in Column I with those in Column II and select the correct option given below: Column I: (a) Tidal volume (b) Inspiratory Reserve volume (c) Expiratory Reserve volume (d) Residual volume Column II: (i) 2500 – 3000 mL (ii) 1100 – 1200 mL (iii) 500 – 550 mL (iv) 1000 – 1100 mL

  • A. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
  • B. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) ✓
  • C. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
  • D. (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)

Solution: Tidal volume is the volume of air inspired or expired during normal respiration. It is approximately 500 mL. Inspiratory reserve volume is the additional volume of air a person can inspire by a forceful inspiration. It is around 2500 – 3000 mL. Expiratory reserve volume is the additional volume of air a person can expire by a forceful expiration. This averages 1000 – 1100 mL. Residual volume is the volume of air remaining in the lungs even after forceful expiration. This averages 1100 – 1200 mL.

Q69.

Which of the following is an amino acid derived hormone?

  • A. Epinephrine ✓
  • B. Ecdysone
  • C. Estriol
  • D. Estradiol

Solution: Epinephrine is derived from the amino acid tyrosine by removal of the carboxyl group. It is a catecholamine. Ecdysone, estriol and estradiol are steroid hormones.

Q70.

Which of the following structures or regions is incorrectly paired with its functions?

  • A. Medulla oblongata : controls respiration and cardiovascular reflexes.
  • B. Limbic system : consists of fibre tracts that interconnect different regions of brain; controls movement. ✓
  • C. Corpus callosum : band of fibers connecting left and right cerebral hemispheres.
  • D. Hypothalamus : production of releasing hormones and regulation of temperature, hunger and thirst.

Solution: The limbic system is the emotional brain. It controls all emotions in our body but not movements. So the pairing in option (2) is incorrect; the other three pairings are correct.

Q71.

The transparent lens in the human eye is held in its place by

  • A. ligaments attached to the ciliary body ✓
  • B. ligaments attached to the iris
  • C. smooth muscles attached to the ciliary body
  • D. smooth muscles attached to the iris

Solution: The lens in the human eye is held in its place by suspensory ligaments attached to the ciliary body.

Q72.

Which of the following hormones can play a significant role in osteoporosis?

  • A. Aldosterone and Prolactin
  • B. Progesterone and Aldosterone
  • C. Parathyroid hormone and Prolactin
  • D. Estrogen and Parathyroid hormone ✓

Solution: Estrogen promotes the activity of osteoblasts and inhibits osteoclasts. In an ageing female, osteoporosis occurs due to deficiency of estrogen. Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.

Q73.

Which of the following gastric cells indirectly help in erythropoiesis?

  • A. Chief cells
  • B. Mucous cells
  • C. Parietal cells ✓
  • D. Goblet cells

Solution: Parietal or oxyntic cells are the source of HCl and intrinsic factor. HCl converts iron present in the diet from the ferric to the ferrous form so that it can be absorbed easily and used during erythropoiesis. Intrinsic factor is essential for the absorption of vitamin B$_{12}$ and its deficiency causes pernicious anaemia.

Q74.

Match the items given in Column I with those in Column II and select the correct option given below : Column I: (a) Fibrinogen (b) Globulin (c) Albumin Column II: (i) Osmotic balance (ii) Blood clotting (iii) Defence mechanism

  • A. (a)-(iii), (b)-(ii), (c)-(i)
  • B. (a)-(i), (b)-(ii), (c)-(iii)
  • C. (a)-(ii), (b)-(iii), (c)-(i) ✓
  • D. (a)-(i), (b)-(iii), (c)-(ii)

Solution: Fibrinogen forms fibrin strands during coagulation. These strands form a network whose meshes are occupied by blood cells, and this structure finally forms a clot. Antibodies are derived from the $\gamma$-globulin fraction of plasma proteins, which means globulins are involved in defence mechanisms. Albumin is a plasma protein mainly responsible for blood colloidal osmotic pressure, i.e. osmotic balance.

Q75.

Which of the following is an occupational respiratory disorder?

  • A. Anthracis
  • B. Silicosis ✓
  • C. Emphysema
  • D. Botulism

Solution: Silicosis is an occupational respiratory disorder found in workers of the grinding and stone-breaking industries, caused by long exposure to silica dust. It causes inflammation leading to fibrosis of the upper part of the lung.

Q76.

Calcium is important in skeletal muscle contraction because it

  • A. Binds to troponin to remove the masking of active sites on actin for myosin. ✓
  • B. Activates the myosin ATPase by binding to it.
  • C. Prevents the formation of bonds between the myosin cross bridges and the actin filament.
  • D. Detaches the myosin head from the actin filament.

Solution: The signal for contraction increases the Ca$^{++}$ level many fold in the sarcoplasm. Ca$^{++}$ binds with the sub-unit of troponin (troponin "C") which is masking the active site on the actin filament, and displaces that sub-unit of troponin. Once the active site is exposed, the head of the myosin attaches and initiates contraction by sliding the actin over myosin.

Q77.

Select the incorrect match :

  • A. Lampbrush chromosomes – Diplotene bivalents
  • B. Allosomes – Sex chromosomes
  • C. Polytene chromosomes – Oocytes of amphibians ✓
  • D. Submetacentric chromosomes – L-shaped chromosomes

Solution: Polytene chromosomes are found in the salivary glands of insects of order Diptera, not in oocytes of amphibians. Lampbrush chromosomes are the ones found in the diplotene bivalents of amphibian oocytes, so option (3) is the incorrect match.

Q78.

Nissl bodies are mainly composed of

  • A. Proteins and lipids
  • B. DNA and RNA
  • C. Free ribosomes and RER ✓
  • D. Nucleic acids and SER

Solution: Nissl granules are present in the cyton and even extend into the dendrite, but are absent in the axon and the rest of the neuron. Nissl granules are in fact composed of free ribosomes and RER. They are responsible for protein synthesis.

Q79.

Which of these statements is incorrect?

  • A. Enzymes of TCA cycle are present in mitochondrial matrix
  • B. Glycolysis occurs in cytosol
  • C. Oxidative phosphorylation takes place in outer mitochondrial membrane ✓
  • D. Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms

Solution: Oxidative phosphorylation takes place in the INNER mitochondrial membrane, where the electron transport chain and ATP synthase are located. So statement (3) is incorrect; the other three are correct.

Q80.

Which of the following events does not occur in rough endoplasmic reticulum?

  • A. Protein folding
  • B. Protein glycosylation
  • C. Phospholipid synthesis ✓
  • D. Cleavage of signal peptide

Solution: Phospholipid synthesis does not take place in RER. Smooth endoplasmic reticulum is involved in lipid synthesis.

Q81.

Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as

  • A. Polysome ✓
  • B. Polyhedral bodies
  • C. Nucleosome
  • D. Plastidome

Solution: The phenomenon of association of many ribosomes with a single mRNA leads to the formation of polyribosomes, or polysomes, or ergasomes.

Q82.

Which of the following terms describe human dentition?

  • A. Thecodont, Diphyodont, Homodont
  • B. Thecodont, Diphyodont, Heterodont ✓
  • C. Pleurodont, Diphyodont, Heterodont
  • D. Pleurodont, Monophyodont, Homodont

Solution: In humans, dentition is Thecodont : Teeth are present in the sockets of the jaw bone called alveoli. Diphyodont : Teeth erupt twice; temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth. Heterodont : Dentition consists of different types of teeth, namely incisors, canine, premolars and molars.

Q83.

Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system

  • A. Amphibia
  • B. Reptilia
  • C. Osteichthyes
  • D. Aves ✓

Solution: The digestive tract of Aves has additional chambers, the crop and gizzard. Crop is concerned with storage of food grains. Gizzard is a masticatory organ in birds used to crush food grain.

Q84.

Which one of these animals is not a homeotherm?

  • A. $Macropus$
  • B. $Chelone$ ✓
  • C. $Psittacula$
  • D. $Camelus$

Solution: Homeotherms are animals that maintain a constant body temperature, irrespective of surrounding temperature. Birds and mammals are homeotherms. $Chelone$ (turtle) belongs to class Reptilia, which is poikilotherm or cold blooded.

Q85.

Which of the following features is used to identify a male cockroach from a female cockroach?

  • A. Presence of a boat shaped sternum on the 9$^{\text{th}}$ abdominal segment
  • B. Presence of caudal styles ✓
  • C. Presence of anal cerci
  • D. Forewings with darker tegmina

Solution: Males bear a pair of short, thread-like anal styles which are absent in females. Anal or caudal styles arise from the 9$^{\text{th}}$ abdominal segment in the male cockroach. Anal cerci are present in both sexes.

Q86.

Which of the following organisms are known as chief producers in the oceans?

  • A. Dinoflagellates
  • B. Diatoms ✓
  • C. Euglenoids
  • D. Cyanobacteria

Solution: Diatoms are the chief producers of the ocean. They are photosynthetic and form the base of most marine food chains.

Q87.

Ciliates differ from all other protozoans in

  • A. using flagella for locomotion
  • B. having a contractile vacuole for removing excess water
  • C. having two types of nuclei ✓
  • D. using pseudopodia for capturing prey

Solution: Ciliates differ from other protozoans in having two types of nuclei. For example, $Paramoecium$ has two types of nuclei, i.e. macronucleus and micronucleus.

Q88.

Which of the following animals does not undergo metamorphosis?

  • A. Earthworm ✓
  • B. Tunicate
  • C. Starfish
  • D. Moth

Solution: Metamorphosis refers to transformation of larva into adult. Animals that perform metamorphosis are said to have indirect development. In earthworm, development is direct, which means there is no larval stage and hence no metamorphosis.

Q89.

Match the items given in Column I with those in Column II and select the correct option given below: Column I (Function): (a) Ultrafiltration (b) Concentration of urine (c) Transport of urine (d) Storage of urine Column II (Part of Excretory system): (i) Henle's loop (ii) Ureter (iii) Urinary bladder (iv) Malpighian corpuscle (v) Proximal convoluted tubule

  • A. (a)-(iv), (b)-(v), (c)-(ii), (d)-(iii)
  • B. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓
  • C. (a)-(v), (b)-(iv), (c)-(i), (d)-(iii)
  • D. (a)-(v), (b)-(iv), (c)-(i), (d)-(ii)

Solution: Ultrafiltration refers to filtration of very fine particles having molecular weight less than 68,000 daltons, through the Malpighian corpuscle. Concentration of urine refers to water absorption from the glomerular filtrate as a result of hyperosmolarity in the medulla created by the counter-current mechanism in Henle's loop. Urine is carried from kidney to bladder through the ureter. Urinary bladder is concerned with storage of urine.

Q90.

Match the items given in Column I with those in Column II and select the correct option given below : Column I: (a) Glycosuria (b) Gout (c) Renal calculi (d) Glomerular nephritis Column II: (i) Accumulation of uric acid in joints (ii) Mass of crystallised salts within the kidney (iii) Inflammation in glomeruli (iv) Presence of glucose in urine

  • A. (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
  • B. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  • C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) ✓
  • D. (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)

Solution: Glycosuria denotes the presence of glucose in the urine. This is observed when blood glucose level rises above 180 mg/100 mL of blood, which is called the renal threshold value for glucose. Gout is due to deposition of uric acid crystals in the joint. Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney. Glomerular nephritis is the inflammatory condition of the glomerulus, characterised by proteinuria and haematuria.

Q91.

What is the role of NAD$^{+}$ in cellular respiration?

  • A. It functions as an enzyme.
  • B. It functions as an electron carrier. ✓
  • C. It is the final electron acceptor for anaerobic respiration.
  • D. It is a nucleotide source for ATP synthesis.

Solution: In cellular respiration, NAD$^{+}$ acts as an electron carrier. It picks up electrons and hydrogen from the substrate to form NADH + H$^{+}$ and passes them on to the electron transport chain.

Q92.

Which one of the following plants shows a very close relationship with a species of moth, where none of the two can complete its life cycle without the other?

  • A. $Hydrilla$
  • B. $Yucca$ ✓
  • C. $Viola$
  • D. Banana

Solution: $Yucca$ has an obligate mutualism with a species of moth, i.e. $Pronuba$. The moth pollinates the flower while it lays eggs in the ovary, and the larvae feed on some of the developing seeds, so neither can complete its life cycle without the other.

Q93.

Oxygen is not produced during photosynthesis by

  • A. Green sulphur bacteria ✓
  • B. $Nostoc$
  • C. $Chara$
  • D. $Cycas$

Solution: Green sulphur bacteria do not use H$_2$O as the source of protons, therefore they do not evolve O$_2$. They use H$_2$S instead, and sulphur rather than oxygen is released.

Q94.

In which of the following forms is iron absorbed by plants?

  • A. Ferric ✓
  • B. Ferrous
  • C. Both ferric and ferrous
  • D. Free element

Solution: Iron is absorbed by plants in the form of ferric ions, as stated in NCERT. Iron is required in larger amounts than the other micronutrients and is an important constituent of proteins involved in the transfer of electrons, like ferredoxin and cytochromes.

Q95.

Double fertilization is

  • A. Fusion of two male gametes of a pollen tube with two different eggs
  • B. Fusion of one male gamete with two polar nuclei
  • C. Syngamy and triple fusion ✓
  • D. Fusion of two male gametes with one egg

Solution: Double fertilization is a unique phenomenon that occurs in angiosperms only. Syngamy $+$ Triple fusion $=$ Double fertilization

Q96.

Which of the following elements is responsible for maintaining turgor in cells?

  • A. Magnesium
  • B. Sodium
  • C. Calcium
  • D. Potassium ✓

Solution: Potassium helps in maintaining the turgidity of cells. It is required in more abundant quantities by meristematic tissue, developing leaves and buds, and it also has a role in opening and closing of stomata.

Q97.

Pollen grains can be stored for several years in liquid nitrogen having a temperature of

  • A. –120$^{\circ}$C
  • B. –80$^{\circ}$C
  • C. –160$^{\circ}$C
  • D. –196$^{\circ}$C ✓

Solution: Pollen grains can be stored for several years in liquid nitrogen at $-196^{\circ}$C. This technique is called cryopreservation, and such stored pollen can be used as pollen banks in crop breeding programmes.

Q98.

Which among the following is not a prokaryote?

  • A. $Saccharomyces$ ✓
  • B. $Mycobacterium$
  • C. $Oscillatoria$
  • D. $Nostoc$

Solution: $Saccharomyces$, i.e. yeast, is a eukaryote (unicellular fungi). $Mycobacterium$ is a bacterium, and $Oscillatoria$ and $Nostoc$ are cyanobacteria, all of which are prokaryotes.

Q99.

The two functional groups characteristic of sugars are

  • A. Hydroxyl and methyl
  • B. Carbonyl and methyl
  • C. Carbonyl and hydroxyl ✓
  • D. Carbonyl and phosphate

Solution: Sugar is a common term used to denote carbohydrate. Carbohydrates are polyhydroxy aldehydes, ketones or their derivatives, which means they have carbonyl and hydroxyl groups.

Q100.

Which of the following is not a product of light reaction of photosynthesis?

  • A. ATP
  • B. NADH ✓
  • C. Oxygen
  • D. NADPH

Solution: ATP, NADPH and oxygen are products of the light reaction. NADH is a product of the respiration process, not of the light reaction.

Q101.

Stomatal movement is not affected by

  • A. Temperature
  • B. Light
  • C. CO$_2$ concentration
  • D. O$_2$ concentration ✓

Solution: Light, temperature and concentration of CO$_2$ affect opening and closing of stomata, while they are not affected by O$_2$ concentration.

Q102.

The Golgi complex participates in

  • A. Fatty acid breakdown
  • B. Formation of secretory vesicles ✓
  • C. Activation of amino acid
  • D. Respiration in bacteria

Solution: Golgi complex, after processing, releases secretory vesicles from its trans-face. It is the site of packaging and dispatch of materials to intracellular targets or outside the cell.

Q103.

Which of the following is true for nucleolus?

  • A. Larger nucleoli are present in dividing cells
  • B. It is a membrane-bound structure
  • C. It is a site for active ribosomal RNA synthesis ✓
  • D. It takes part in spindle formation

Solution: Nucleolus is a non-membranous structure and is a site of r-RNA synthesis. Larger and more numerous nucleoli are present in cells actively carrying out protein synthesis, not in dividing cells.

Q104.

Stomata in grass leaf are

  • A. Dumb-bell shaped ✓
  • B. Kidney shaped
  • C. Barrel shaped
  • D. Rectangular

Solution: Grass, being a monocot, has dumb-bell shaped stomata in its leaves. In dicots the guard cells of the stomata are bean or kidney shaped.

Q105.

The stage during which separation of the paired homologous chromosomes begins is

  • A. Pachytene
  • B. Diplotene ✓
  • C. Zygotene
  • D. Diakinesis

Solution: At diplotene the synaptonemal complex disintegrates and the recombined homologous chromosomes of the bivalents start to separate, remaining attached only at the chiasmata. Terminalisation begins at the diplotene stage, i.e. chiasmata start to shift towards the ends.

Q106.

Which of the following is commonly used as a vector for introducing a DNA fragment in human lymphocytes?

  • A. Retrovirus ✓
  • B. Ti plasmid
  • C. pBR 322
  • D. $\lambda$ phage

Solution: Retrovirus is commonly used as a vector for introducing a DNA fragment in human lymphocytes. In gene therapy, lymphocytes from the blood of the patient are grown in culture outside the body and a functional gene is introduced into these lymphocytes using a retroviral vector.

Q107.

Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called

  • A. Bio-infringement
  • B. Biopiracy ✓
  • C. Bioexploitation
  • D. Biodegradation

Solution: Biopiracy is the term used to refer to the use of bioresources by multinational companies and other organisations without proper authorisation from the countries and people concerned, and without compensatory payment.

Q108.

In India, the organisation responsible for assessing the safety of introducing genetically modified organisms for public use is

  • A. Indian Council of Medical Research (ICMR)
  • B. Council for Scientific and Industrial Research (CSIR)
  • C. Genetic Engineering Appraisal Committee (GEAC) ✓
  • D. Research Committee on Genetic Manipulation (RCGM)

Solution: The Indian Government has set up an organisation, GEAC (Genetic Engineering Appraisal Committee), which makes decisions regarding the validity of GM research and the safety of introducing GM organisms for public services.

Q109.

The correct order of steps in Polymerase Chain Reaction (PCR) is

  • A. Extension, Denaturation, Annealing
  • B. Annealing, Extension, Denaturation
  • C. Denaturation, Annealing, Extension ✓
  • D. Denaturation, Extension, Annealing

Solution: PCR is used for making multiple copies of a gene (or DNA) of interest in vitro. Each cycle has three steps: denaturation, primer annealing and extension of primer.

Q110.

Select the correct match

  • A. Ribozyme – Nucleic acid ✓
  • B. F$_2$ $\times$ Recessive parent – Dihybrid cross
  • C. G. Mendel – Transformation
  • D. T.H. Morgan – Transduction

Solution: Ribozyme is a catalytic RNA, which is a nucleic acid, so that match is correct. A cross of F$_2$ with the recessive parent is a test cross, not a dihybrid cross. Transformation was demonstrated by Griffith, and transduction by Zinder and Lederberg.

Q111.

A 'new' variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to

  • A. Co-667
  • B. Sharbati Sonora
  • C. Basmati ✓
  • D. Lerma Rojo

Solution: There are an estimated 27 documented varieties of Basmati rice grown in India. In 1997, an American company got a patent rights on Basmati rice through the US Patent and Trademark Office, which is a case of biopiracy.

Q112.

Select the correct match

  • A. Alec Jeffreys – $Streptococcus$ $pneumoniae$
  • B. Alfred Hershey and Martha Chase – TMV
  • C. Francois Jacob and Jacques Monod – Lac operon ✓
  • D. Matthew Meselson and F. Stahl – $Pisum$ $sativum$

Solution: Francois Jacob and Jacques Monod proposed the model of gene regulation known as the operon model, i.e. the lac operon. Alec Jeffreys developed the DNA fingerprinting technique. Matthew Meselson and F. Stahl demonstrated semi-conservative DNA replication in $E.$ $coli$. Alfred Hershey and Martha Chase proved DNA, and not protein, is the genetic material, working with bacteriophage.

Q113.

Which of the following has proved helpful in preserving pollen as fossils?

  • A. Pollenkitt
  • B. Cellulosic intine
  • C. Sporopollenin ✓
  • D. Oil content

Solution: Sporopollenin cannot be degraded by enzymes, strong acids or alkali, therefore it is helpful in preserving pollen as fossils. Pollenkitt helps in insect pollination. The inner sporoderm layer of the pollen grain, known as the intine, is made up of cellulose and pectin. Oil content has no role in pollen preservation.

Q114.

The experimental proof for semiconservative replication of DNA was first shown in a

  • A. Fungus
  • B. Bacterium ✓
  • C. Virus
  • D. Plant

Solution: Semi-conservative DNA replication was first shown in the bacterium $Escherichia$ $coli$ by Matthew Meselson and Franklin Stahl.

Q115.

Which of the following pairs is wrongly matched?

  • A. Starch synthesis in pea : Multiple alleles ✓
  • B. ABO blood grouping : Co-dominance
  • C. T.H. Morgan : Linkage
  • D. XO type sex determination : Grasshopper

Solution: Starch synthesis in pea is controlled by a pleiotropic gene, not by multiple alleles. The other options (2), (3) and (4) are correctly matched.

Q116.

Offsets are produced by

  • A. Meiotic divisions
  • B. Mitotic divisions ✓
  • C. Parthenogenesis
  • D. Parthenocarpy

Solution: Offset is a vegetative part of a plant, formed by mitosis. Meiotic divisions do not occur in somatic cells. Parthenogenesis is the formation of an embryo from an ovum or egg without fertilisation. Parthenocarpy is fruit formed without fertilisation, generally seedless.

Q117.

Select the correct statement

  • A. Franklin Stahl coined the term "linkage"
  • B. Punnett square was developed by a British scientist ✓
  • C. Transduction was discovered by S. Altman
  • D. Spliceosomes take part in translation

Solution: The Punnett square was developed by the British geneticist Reginald C. Punnett, so statement (2) is correct. The term 'linkage' was coined by T.H. Morgan. Transduction was discovered by Zinder and Lederberg. Spliceosomes take part in splicing of hnRNA, which is a step of post-transcriptional processing, not translation.

Q118.

Which of the following flowers only once in its life-time?

  • A. Bamboo species ✓
  • B. Jackfruit
  • C. Papaya
  • D. Mango

Solution: Bamboo species are monocarpic, i.e. they flower generally only once in their life-time, after 50-100 years. Jackfruit, papaya and mango are polycarpic, i.e. they produce flowers and fruits many times in their life-time.

Q119.

Niche is

  • A. all the biological factors in the organism's environment
  • B. the physical space where an organism lives
  • C. the functional role played by the organism where it lives ✓
  • D. the range of temperature that the organism needs to live

Solution: Ecological niche was termed by J. Grinnell. It refers to the functional role played by the organism where it lives.

Q120.

In stratosphere, which of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen?

  • A. Carbon
  • B. Cl ✓
  • C. Oxygen
  • D. Fe

Solution: UV rays act on CFCs, releasing Cl atoms. Chlorine reacts with ozone in a sequential manner, converting it into molecular oxygen, and the Cl atom is not consumed, so it acts as a catalyst. Carbon, oxygen and Fe are not related to ozone layer depletion.

Q121.

What type of ecological pyramid would be obtained with the following data? Secondary consumer : 120 g Primary consumer : 60 g Primary producer : 10 g

  • A. Inverted pyramid of biomass ✓
  • B. Pyramid of energy
  • C. Upright pyramid of biomass
  • D. Upright pyramid of numbers

Solution: The given data depicts an inverted pyramid of biomass, usually found in an aquatic ecosystem. The pyramid of energy is always upright. An upright pyramid of biomass or of numbers is not possible here, as the data shows the primary producer is less than the primary consumer, which in turn is less than the secondary consumers.

Q122.

Which of the following is a secondary pollutant?

  • A. CO
  • B. CO$_2$
  • C. O$_3$ ✓
  • D. SO$_2$

Solution: O$_3$ (ozone) is a secondary pollutant, as it is formed by the reaction of primary pollutants in the atmosphere. CO, CO$_2$ and SO$_2$ are all primary pollutants, released directly from a source.

Q123.

World Ozone Day is celebrated on

  • A. 5$^{\text{th}}$ June
  • B. 21$^{\text{st}}$ April
  • C. 22$^{\text{nd}}$ April
  • D. 16$^{\text{th}}$ September ✓

Solution: World Ozone Day is celebrated on 16$^{\text{th}}$ September. 5$^{\text{th}}$ June is World Environment Day and 22$^{\text{nd}}$ April is Earth Day.

Q124.

Natality refers to

  • A. Death rate
  • B. Birth rate ✓
  • C. Number of individuals entering a habitat
  • D. Number of individuals leaving the habitat

Solution: Natality refers to birth rate. Death rate is mortality, the number of individuals entering a habitat is immigration, and the number of individuals leaving the habitat is emigration.

Q125.

Match the items given in Column I with those in Column II and select the correct option given below: Column I: (a) Herbarium (b) Key (c) Museum (d) Catalogue Column II: (i) It is a place having a collection of preserved plants and animals (ii) A list that enumerates methodically all the species found in an area with brief description aiding identification (iii) Is a place where dried and pressed plant specimens mounted on sheets are kept (iv) A booklet containing a list of characters and their alternates which are helpful in identification of various taxa

  • A. (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
  • B. (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
  • C. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) ✓
  • D. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

Solution: Herbarium – a place where dried and pressed plant specimens mounted on sheets are kept. Key – a booklet containing a list of characters and their alternates, helpful in identification of various taxa. Museum – a place having a collection of preserved plant and animal specimens. Catalogue – a list that enumerates methodically all the species found in an area with brief description aiding identification.

Q126.

Which one is wrongly matched?

  • A. Uniflagellate gametes – $Polysiphonia$ ✓
  • B. Biflagellate zoospores – Brown algae
  • C. Unicellular organism – $Chlorella$
  • D. Gemma cups – $Marchantia$

Solution: $Polysiphonia$ is a genus of red algae, where asexual spores and gametes are non-motile or non-flagellated. So the pairing with uniflagellate gametes is wrong; the other options (2), (3) and (4) are correctly matched.

Q127.

After karyogamy followed by meiosis, spores are produced exogenously in

  • A. $Neurospora$
  • B. $Alternaria$
  • C. $Saccharomyces$
  • D. $Agaricus$ ✓

Solution: In $Agaricus$, a genus of basidiomycetes, basidiospores or meiospores are produced exogenously on the basidium. $Neurospora$, a genus of ascomycetes, produces ascospores as meiospores but endogenously inside the ascus. $Alternaria$, a genus of deuteromycetes, does not produce sexual spores. $Saccharomyces$, a unicellular ascomycete, produces ascospores endogenously.

Q128.

Winged pollen grains are present in

  • A. Mustard
  • B. $Cycas$
  • C. $Pinus$ ✓
  • D. Mango

Solution: Winged pollen grains are present in $Pinus$. The two air sacs (wings) make the pollen buoyant, which aids in wind pollination.

Q129.

Pneumatophores occur in

  • A. Halophytes ✓
  • B. Free-floating hydrophytes
  • C. Submerged hydrophytes
  • D. Carnivorous plants

Solution: Halophytes like mangroves have pneumatophores. These are apogeotropic (negatively geotropic) roots having lenticels called pneumathodes, which help the plant to take up O$_2$ from the air.

Q130.

Plants having little or no secondary growth are

  • A. Grasses ✓
  • B. Deciduous angiosperms
  • C. Cycads
  • D. Conifers

Solution: Grasses are monocots and monocots usually do not have secondary growth. Palm-like monocots have anomalous secondary growth.

Q131.

Casparian strips occur in

  • A. Epidermis
  • B. Pericycle
  • C. Endodermis ✓
  • D. Cortex

Solution: Endodermis has casparian strips on the radial and inner tangential walls. These strips are suberin rich, and force water and solutes to pass through the cell membrane rather than between cells.

Q132.

Secondary xylem and phloem in dicot stem are produced by

  • A. Apical meristems
  • B. Vascular cambium ✓
  • C. Axillary meristems
  • D. Phellogen

Solution: Vascular cambium is partially secondary in origin. It forms secondary xylem towards its inside and secondary phloem towards its outside. 4 – 10 times more secondary xylem is produced than secondary phloem.

Q133.

Select the wrong statement :

  • A. Cell wall is present in members of Fungi and Plantae
  • B. Mushrooms belong to Basidiomycetes
  • C. Mitochondria are the powerhouse of the cell in all kingdoms except Monera
  • D. Pseudopodia are locomotory and feeding structures in Sporozoans ✓

Solution: Pseudopodia are locomotory and feeding structures in sarcodines (amoeboid protozoans), not in sporozoans. So statement (4) is wrong; the other three statements are correct.

Q134.

Which of the following statements is correct?

  • A. Ovules are not enclosed by ovary wall in gymnosperms ✓
  • B. $Selaginella$ is heterosporous, while $Salvinia$ is homosporous
  • C. Stems are usually unbranched in both $Cycas$ and $Cedrus$
  • D. Horsetails are gymnosperms

Solution: Gymnosperms have naked ovules, i.e. the ovules are not enclosed by an ovary wall. They are called phanerogams without a womb or ovary. Both $Selaginella$ and $Salvinia$ are heterosporous. $Cycas$ is unbranched but $Cedrus$ is branched. Horsetails ($Equisetum$) are pteridophytes, not gymnosperms.

Q135.

Sweet potato is a modified

  • A. Stem
  • B. Adventitious root ✓
  • C. Rhizome
  • D. Tap root

Solution: Sweet potato is a modified adventitious root for storage of food. Rhizomes are underground modified stems, and a tap root is the primary root directly elongated from the radicle.

Q136.

The correct order of N-compounds in its decreasing order of oxidation states is

  • A. HNO$_3$, NO, N$_2$, NH$_4$Cl ✓
  • B. HNO$_3$, NO, NH$_4$Cl, N$_2$
  • C. NH$_4$Cl, N$_2$, NO, HNO$_3$
  • D. HNO$_3$, NH$_4$Cl, NO, N$_2$

Solution: Oxidation state of nitrogen in each compound: HNO$_3$ : $+5$ NO : $+2$ N$_2$ : $0$ NH$_4$Cl : $-3$ Hence the decreasing order is HNO$_3$, NO, N$_2$, NH$_4$Cl.

Q137.

The correct order of atomic radii in group 13 elements is

  • A. B < Al < In < Ga < Tl
  • B. B < Al < Ga < In < Tl
  • C. B < Ga < Al < In < Tl ✓
  • D. B < Ga < Al < Tl < In

Solution: The atomic radii in pm are B $= 85$, Ga $= 135$, Al $= 143$, In $= 167$, Tl $= 170$. Gallium has a smaller radius than aluminium because of the poor shielding of the intervening 3d electrons. Hence the correct order is B < Ga < Al < In < Tl.

Q138.

Considering Ellingham diagram, which of the following metals can be used to reduce alumina?

  • A. Fe
  • B. Zn
  • C. Cu
  • D. Mg ✓

Solution: A metal whose oxide has a more negative free energy of formation lies lower in the Ellingham diagram and can reduce the oxide above it. The metal which is more reactive than 'Al' can reduce alumina, i.e. 'Mg' is the correct option.

Q139.

Which one of the following elements is unable to form MF$_6^{3-}$ ion?

  • A. Ga
  • B. Al
  • C. In
  • D. B ✓

Solution: 'B' has no vacant d-orbitals in its valence shell, so it cannot extend its covalency beyond 4. Hence 'B' cannot form an ion like BF$_6^{3-}$.

Q140.

Which of the following statements is not true for halogens?

  • A. All form monobasic oxyacids
  • B. All are oxidizing agents
  • C. Chlorine has the highest electron-gain enthalpy
  • D. All but fluorine show positive oxidation states ✓

Solution: Due to high electronegativity and small size, F forms only one oxoacid, HOF, known as fluoric (I) acid. The oxidation number of F in HOF is $+1$, so it is not true that all halogens but fluorine show positive oxidation states.

Q141.

In the structure of ClF$_3$, the number of lone pair of electrons on central atom 'Cl' is

  • A. One
  • B. Two ✓
  • C. Three
  • D. Four

Solution: In ClF$_3$ the central chlorine atom has 7 valence electrons; three are used in bonding to the three fluorine atoms. The remaining four electrons form two lone pairs, giving a trigonal bipyramidal electron geometry with the two lone pairs equatorial, and a T-shaped molecule. So the number of lone pairs on central Cl is 2.

Q142.

The difference between amylose and amylopectin is

  • A. Amylopectin have 1 $\to$ 4 $\alpha$-linkage and 1 $\to$ 6 $\alpha$-linkage ✓
  • B. Amylose have 1 $\to$ 4 $\alpha$-linkage and 1 $\to$ 6 $\beta$-linkage
  • C. Amylose is made up of glucose and galactose
  • D. Amylopectin have 1 $\to$ 4 $\alpha$-linkage and 1 $\to$ 6 $\beta$-linkage

Solution: Amylose and amylopectin are polymers of $\alpha$-D-glucose, so a $\beta$-link is not possible. Amylose is linear with 1 $\to$ 4 $\alpha$-linkage, whereas amylopectin is branched and has both 1 $\to$ 4 and 1 $\to$ 6 $\alpha$-linkages.

Q143.

Regarding cross-linked or network polymers, which of the following statements is incorrect?

  • A. They contain covalent bonds between various linear polymer chains.
  • B. They are formed from bi- and tri-functional monomers.
  • C. They contain strong covalent bonds in their polymer chains. ✓
  • D. Examples are bakelite and melamine.

Solution: Cross-linked or network polymers are formed from bi-functional and tri-functional monomers and contain strong covalent bonds BETWEEN various linear polymer chains, e.g. bakelite, melamine. Option (3) describes bonds within the chains, which is not what defines cross-linking, so it is the incorrect statement.

Q144.

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H$_2$SO$_4$. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

  • A. 1.4
  • B. 3.0
  • C. 4.4
  • D. 2.8 ✓

Solution: HCOOH $\xrightarrow{\text{Conc.H}_2\text{SO}_4}$ CO(g) $+$ H$_2$O(l) 2.3 g of formic acid is $\dfrac{1}{20}$ mol, giving $\dfrac{1}{20}$ mol CO. (COOH)$_2$ $\xrightarrow{\text{Conc.H}_2\text{SO}_4}$ CO(g) $+$ CO$_2$(g) $+$ H$_2$O(l) 4.5 g of oxalic acid is $\dfrac{1}{20}$ mol, giving $\dfrac{1}{20}$ mol CO and $\dfrac{1}{20}$ mol CO$_2$. The gaseous mixture formed is CO and CO$_2$; when it is passed through KOH, only CO$_2$ is absorbed, so the remaining gas is CO. $$\text{Weight of CO} = \frac{2}{20}\times 28 = 2.8\ \text{g}$$

Q145.

Which of the following oxides is most acidic in nature?

  • A. MgO
  • B. BeO ✓
  • C. CaO
  • D. BaO

Solution: Down the group the basic character increases: BeO < MgO < CaO < BaO So the most acidic should be BeO. In fact BeO is an amphoteric oxide, while the other given oxides are basic.

Q146.

Nitration of aniline in strong acidic medium also gives m-nitroaniline because

  • A. Inspite of substituents nitro group always goes to only m-position.
  • B. In electrophilic substitution reactions amino group is meta directive.
  • C. In acidic (strong) medium aniline is present as anilinium ion. ✓
  • D. In absence of substituents nitro group always goes to m-position.

Solution: In strong acidic medium aniline is protonated and exists as the anilinium ion. The –NH$_3^{+}$ group is m-directing, hence besides para (51%) and ortho (2%), the meta product (47%) is also formed in significant yield.

Q147.

The compound A on treatment with Na gives B, and with PCl$_5$ gives C. B and C react together to give diethyl ether. A, B and C are in the order

  • A. C$_2$H$_5$OH, C$_2$H$_6$, C$_2$H$_5$Cl
  • B. C$_2$H$_5$OH, C$_2$H$_5$Cl, C$_2$H$_5$ONa
  • C. C$_2$H$_5$OH, C$_2$H$_5$ONa, C$_2$H$_5$Cl ✓
  • D. C$_2$H$_5$Cl, C$_2$H$_6$, C$_2$H$_5$OH

Solution: C$_2$H$_5$OH (A) $\xrightarrow{\text{Na}}$ C$_2$H$_5$O$^{-}$Na$^{+}$ (B) C$_2$H$_5$OH (A) $\xrightarrow{\text{PCl}_5}$ C$_2$H$_5$Cl (C) B and C then react by Williamson's synthesis: C$_2$H$_5$O$^{-}$Na$^{+}$ $+$ C$_2$H$_5$Cl $\xrightarrow{S_N2}$ C$_2$H$_5$OC$_2$H$_5$ So A, B and C are C$_2$H$_5$OH, C$_2$H$_5$ONa and C$_2$H$_5$Cl.

Q148.

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is

  • A. CH $\equiv$ CH
  • B. CH$_2$ $=$ CH$_2$
  • C. CH$_4$ ✓
  • D. CH$_3$ – CH$_3$

Solution: CH$_4$ (A) $\xrightarrow{\text{Br}_2/h\nu}$ CH$_3$Br CH$_3$Br $\xrightarrow[\text{Wurtz reaction}]{\text{Na/dry ether}}$ CH$_3$–CH$_3$ Ethane is a gaseous hydrocarbon containing less than four carbon atoms, so (A) is CH$_4$. Alkenes and alkynes react with bromine by addition, not substitution.

Q149.

The compound C$_7$H$_8$ undergoes the following reactions: C$_7$H$_8$ $\xrightarrow{3\text{Cl}_2/\Delta}$ A $\xrightarrow{\text{Br}_2/\text{Fe}}$ B $\xrightarrow{\text{Zn/HCl}}$ C The product 'C' is

  • A. m-bromotoluene ✓
  • B. o-bromotoluene
  • C. p-bromotoluene
  • D. 3-bromo-2,4,6-trichlorotoluene

Solution: C$_7$H$_8$ is toluene. With 3Cl$_2$ under heat the side chain is chlorinated to give benzotrichloride, C$_6$H$_5$CCl$_3$ (A). The –CCl$_3$ group is meta directing, so bromination with Br$_2$/Fe gives m-bromobenzotrichloride (B). Zn/HCl reduces the –CCl$_3$ group back to –CH$_3$, giving m-bromotoluene (C).

Q150.

Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?

  • A. N$_2$O$_5$ ✓
  • B. NO$_2$
  • C. NO
  • D. N$_2$O

Solution: NO, NO$_2$ and N$_2$O are all introduced into the atmosphere by natural processes as well as human activity. N$_2$O$_5$ is not a common atmospheric pollutant from either source.

Q151.

Which of the following molecules represents the order of hybridisation sp$^{2}$, sp$^{2}$, sp, sp from left to right atoms?

  • A. HC $\equiv$ C – C $\equiv$ CH
  • B. CH$_2$ = CH – C $\equiv$ CH ✓
  • C. CH$_3$ – CH = CH – CH$_3$
  • D. CH$_2$ = CH – CH = CH$_2$

Solution: The number of orbitals required in hybridisation equals the number of $\sigma$-bonds around each carbon atom. For CH$_2$ $=$ CH – C $\equiv$ CH the carbons are, from left to right, sp$^{2}$, sp$^{2}$, sp, sp.

Q152.

Which of the following carbocations is expected to be most stable?

  • A. Structure (1)
  • B. Structure (2)
  • C. Structure (3)
  • D. Structure (4) ✓

Solution: The –NO$_2$ group exhibits a –I effect, and it decreases with increase in distance. In option (4) the positive charge is present on the carbon atom at the maximum distance from the nitro group, so the –I effect reaching it is minimum and the stability is maximum.

Q153.

In the reaction Phenol $+$ CHCl$_3$ $+$ NaOH $\longrightarrow$ sodium salicylaldehyde (the ortho-hydroxy benzaldehyde salt) The electrophile involved is

  • A. Dichloromethyl cation ($\overset{\oplus}{\text{C}}$HCl$_2$)
  • B. Formyl cation ($\overset{\oplus}{\text{C}}$HO)
  • C. Dichlorocarbene (: CCl$_2$) ✓
  • D. Dichloromethyl anion ($\overset{\ominus}{\text{C}}$HCl$_2$)

Solution: This is the Reimer-Tiemann reaction. The electrophile formed is :CCl$_2$ (dichlorocarbene), according to the following reaction: CHCl$_3$ $+$ OH$^{-}$ $\rightleftharpoons$ $^{\ominus}$CCl$_3$ $+$ H$_2$O $^{\ominus}$CCl$_3$ $\longrightarrow$ :CCl$_2$ $+$ Cl$^{-}$

Q154.

Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their

  • A. Formation of intramolecular H-bonding
  • B. Formation of carboxylate ion
  • C. Formation of intermolecular H-bonding ✓
  • D. More extensive association of carboxylic acid via van der Waals force of attraction

Solution: Due to formation of intermolecular H-bonding in carboxylic acids, association occurs and dimers are formed. Hence the boiling point increases and becomes more than the boiling point of aldehydes, ketones and alcohols of comparable molecular mass.

Q155.

Compound A, C$_8$H$_{10}$O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectively

  • A. CH$_3$–C$_6$H$_4$–CH$_2$–OH and I$_2$
  • B. C$_6$H$_5$–CH$_2$–CH$_2$–OH and I$_2$
  • C. (CH$_3$)$_2$C$_6$H$_3$–OH and I$_2$
  • D. C$_6$H$_5$–CH(OH)–CH$_3$ and I$_2$ ✓

Solution: 2NaOH $+$ I$_2$ $\longrightarrow$ NaOI $+$ NaI $+$ H$_2$O, so Y is I$_2$. Option (4) is a secondary alcohol which on oxidation gives phenyl methyl ketone (acetophenone). This on reaction with I$_2$ and NaOH forms iodoform, CHI$_3$, a yellow precipitate with a characteristic smell, along with sodium benzoate.

Q156.

Identify the major products P, Q and R in the following sequence of reactions: Benzene $+$ CH$_3$CH$_2$CH$_2$Cl $\xrightarrow{\text{Anhydrous AlCl}_3}$ P P $\xrightarrow[\text{(ii) H}_3\text{O}^{+}/\Delta]{\text{(i) O}_2}$ Q $+$ R

  • A. P = propylbenzene, Q = benzaldehyde, R = ethanol
  • B. P = propylbenzene, Q = benzaldehyde, R = benzoic acid
  • C. P = isopropylbenzene (cumene), Q = phenol, R = acetone ✓
  • D. P = isopropylbenzene (cumene), Q = phenol, R = propan-2-ol

Solution: In the Friedel-Crafts alkylation the incipient n-propyl carbocation undergoes a 1,2-hydride shift to the more stable isopropyl carbocation. So the product P is isopropylbenzene (cumene). Cumene on oxidation with O$_2$ gives cumene hydroperoxide, and acid hydrolysis with H$_3$O$^{+}$ brings about the hydroperoxide rearrangement to give phenol (Q) and acetone (R).

Q157.

Which of the following compounds can form a zwitterion?

  • A. Aniline
  • B. Acetanilide
  • C. Glycine ✓
  • D. Benzoic acid

Solution: Glycine has both an acidic –COOH group and a basic –NH$_2$ group. Internal proton transfer gives the zwitterion form H$_3$N$^{+}$–CH$_2$–COO$^{-}$. The other three compounds carry only one type of ionisable group and cannot form a zwitterion.

Q158.

For the redox reaction MnO$_4^{-}$ $+$ C$_2$O$_4^{2-}$ $+$ H$^{+}$ $\longrightarrow$ Mn$^{2+}$ $+$ CO$_2$ $+$ H$_2$O The correct coefficients of the reactants for the balanced equation are (MnO$_4^{-}$, C$_2$O$_4^{2-}$, H$^{+}$ respectively)

  • A. 16, 5, 2
  • B. 2, 5, 16 ✓
  • C. 5, 16, 2
  • D. 2, 16, 5

Solution: Mn goes from $+7$ to $+2$, so the n-factor of MnO$_4^{-}$ is 5. Carbon goes from $+3$ to $+4$ for two carbons, so the n-factor of C$_2$O$_4^{2-}$ is 2. The ratio of n-factors of MnO$_4^{-}$ and C$_2$O$_4^{2-}$ is 5 : 2, so the molar ratio in the balanced reaction is 2 : 5. The balanced equation is $$2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$$

Q159.

Which one of the following conditions will favour maximum formation of the product in the reaction, A$_2$(g) $+$ B$_2$(g) $\rightleftharpoons$ X$_2$(g), $\Delta_r$H $= -$X kJ?

  • A. Low temperature and high pressure ✓
  • B. Low temperature and low pressure
  • C. High temperature and low pressure
  • D. High temperature and high pressure

Solution: Two moles of gas give one mole of gas, so on increasing pressure the equilibrium shifts in the direction where the number of gas moles decreases, i.e. the forward direction. The reaction is exothermic, so on decreasing temperature the equilibrium shifts in the exothermic, i.e. forward, direction. So high pressure and low temperature favour maximum formation of product.

Q160.

When initial concentration of the reactant is doubled, the half-life period of a zero order reaction

  • A. Is halved
  • B. Is doubled ✓
  • C. Remains unchanged
  • D. Is tripled

Solution: Half life of a zero order reaction is $$t_{1/2} = \frac{[A_0]}{2K}$$ So $t_{1/2}$ will be doubled on doubling the initial concentration.

Q161.

The correction factor 'a' to the ideal gas equation corresponds to

  • A. Density of the gas molecules
  • B. Volume of the gas molecules
  • C. Forces of attraction between the gas molecules ✓
  • D. Electric field present between the gas molecules

Solution: In the real gas equation, $$\left(P + \frac{an^{2}}{V^{2}}\right)(V - nb) = nRT$$ the van der Waals constant 'a' signifies intermolecular forces of attraction, while 'b' corrects for the volume of the gas molecules.

Q162.

The bond dissociation energies of X$_2$, Y$_2$ and XY are in the ratio of 1 : 0.5 : 1. $\Delta$H for the formation of XY is –200 kJ mol$^{-1}$. The bond dissociation energy of X$_2$ will be

  • A. 200 kJ mol$^{-1}$
  • B. 100 kJ mol$^{-1}$
  • C. 400 kJ mol$^{-1}$
  • D. 800 kJ mol$^{-1}$ ✓

Solution: The reaction for $\Delta_f$H$^{\circ}$(XY) is $$\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \longrightarrow XY(g)$$ Bond energies of X$_2$, Y$_2$ and XY are X, $\dfrac{X}{2}$ and X respectively. $$\Delta H = \left(\frac{X}{2} + \frac{X}{4}\right) - X = -200$$ $$-\frac{X}{2} + \frac{X}{4} = -200$$ $$X = 800\ \text{kJ/mole}$$

Q163.

Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s$^{2}$ 2s$^{2}$ 2p$^{3}$, the simplest formula for this compound is

  • A. Mg$_2$X$_3$
  • B. MgX$_2$
  • C. Mg$_3$X$_2$ ✓
  • D. Mg$_2$X

Solution: Element (X) has the electronic configuration 1s$^{2}$ 2s$^{2}$ 2p$^{3}$, so its valency will be 3. Valency of Mg is 2. Hence the formula of the compound formed by Mg and X will be Mg$_3$X$_2$.

Q164.

Iron exhibits bcc structure at room temperature. Above 900$^{\circ}$C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900$^{\circ}$C (assuming molar mass and atomic radii of iron remains constant with temperature) is

  • A. $\dfrac{\sqrt{3}}{\sqrt{2}}$
  • B. $\dfrac{4\sqrt{3}}{3\sqrt{2}}$
  • C. $\dfrac{1}{2}$
  • D. $\dfrac{3\sqrt{3}}{4\sqrt{2}}$ ✓

Solution: For a BCC lattice, $Z = 2$ and $a = \dfrac{4r}{\sqrt{3}}$. For an FCC lattice, $Z = 4$ and $a = 2\sqrt{2}\,r$. $$\frac{d_{25^{\circ}\text{C}}}{d_{900^{\circ}\text{C}}} = \frac{\left(\frac{ZM}{N_Aa^{3}}\right)_{BCC}}{\left(\frac{ZM}{N_Aa^{3}}\right)_{FCC}} = \frac{2}{4}\left(\frac{2\sqrt{2}\,r}{\frac{4r}{\sqrt{3}}}\right)^{3} = \frac{3\sqrt{3}}{4\sqrt{2}}$$

Q165.

Consider the following species : CN$^{+}$, CN$^{-}$, NO and CN Which one of these will have the highest bond order?

  • A. NO
  • B. CN$^{-}$ ✓
  • C. CN
  • D. CN$^{+}$

Solution: NO has 15 electrons, giving $$\text{BO} = \frac{10 - 5}{2} = 2.5$$ CN$^{-}$ has 14 electrons, with configuration $(\sigma 1s)^{2}$, $(\sigma^{*}1s)^{2}$, $(\sigma 2s)^{2}$, $(\sigma^{*}2s)^{2}$, $(\pi 2p_x)^{2} = (\pi 2p_y)^{2}$, $(\sigma 2p_z)^{2}$, giving $$\text{BO} = \frac{10 - 4}{2} = 3$$ CN has 13 electrons (BO $= 2.5$) and CN$^{+}$ has 12 (BO $= 2$). So CN$^{-}$ has the highest bond order.

Q166.

Which one is a wrong statement?

  • A. Total orbital angular momentum of electron in 's' orbital is equal to zero
  • B. An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
  • C. The value of m for d$_{z^{2}}$ is zero
  • D. The electronic configuration of N atom is 1s$^{2}$ 2s$^{2}$ 2p$_x^{1}$ 2p$_y^{1}$ 2p$_z^{1}$ with the three 2p electrons drawn as two spins up and one spin down ✓

Solution: According to Hund's rule of maximum multiplicity, the three 2p electrons of nitrogen must occupy the three degenerate orbitals singly with PARALLEL spins, i.e. all three arrows up (or all three down). The configuration in option (4), with one of the three 2p electrons of opposite spin, violates Hund's rule, so that statement is wrong.

Q167.

The correct difference between first and second order reactions is that

  • A. The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
  • B. The half-life of a first-order reaction does not depend on [A]$_0$; the half-life of a second-order reaction does depend on [A]$_0$ ✓
  • C. The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
  • D. A first-order reaction can catalyzed; a second-order reaction cannot be catalyzed

Solution: For a first order reaction, $$t_{1/2} = \frac{0.693}{k}$$ which is independent of the initial concentration of reactant. For a second order reaction, $$t_{1/2} = \frac{1}{k[A_0]}$$ which depends on the initial concentration of reactant.

Q168.

In which case is number of molecules of water maximum?

  • A. 18 mL of water ✓
  • B. 0.18 g of water
  • C. 10$^{-3}$ mol of water
  • D. 0.00224 L of water vapours at 1 atm and 273 K

Solution: (1) Mass of water $= 18 \times 1 = 18$ g, so molecules $= \dfrac{18}{18}N_A = N_A$ (2) Molecules $= \dfrac{0.18}{18}N_A = 10^{-2}N_A$ (3) Molecules $= 10^{-3}N_A$ (4) Moles $= \dfrac{0.00224}{22.4} = 10^{-4}$, so molecules $= 10^{-4}N_A$ Hence 18 mL of water has the maximum number of molecules.

Q169.

Among CaH$_2$, BeH$_2$, BaH$_2$, the order of ionic character is

  • A. BeH$_2$ < CaH$_2$ < BaH$_2$ ✓
  • B. CaH$_2$ < BeH$_2$ < BaH$_2$
  • C. BaH$_2$ < BeH$_2$ < CaH$_2$
  • D. BeH$_2$ < BaH$_2$ < CaH$_2$

Solution: For 2$^{\text{nd}}$ group hydrides, on moving down the group the metallic character of the metal increases, so the ionic character of the metal hydride increases. Hence the order is BeH$_2$ < CaH$_2$ < BaH$_2$.

Q170.

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below : BrO$_4^{-}$ $\xrightarrow{1.82\ \text{V}}$ BrO$_3^{-}$ $\xrightarrow{1.5\ \text{V}}$ HBrO $\xrightarrow{1.595\ \text{V}}$ Br$_2$ $\xrightarrow{1.0652\ \text{V}}$ Br$^{-}$ Then the species undergoing disproportionation is

  • A. BrO$_3^{-}$
  • B. BrO$_4^{-}$
  • C. HBrO ✓
  • D. Br$_2$

Solution: HBrO $\longrightarrow$ Br$_2$ is a reduction, $E^{\circ}_{\text{HBrO/Br}_2} = 1.595$ V. HBrO $\longrightarrow$ BrO$_3^{-}$ is an oxidation, $E^{\circ}_{\text{BrO}_3^{-}/\text{HBrO}} = 1.5$ V. For the disproportionation of HBrO, $$E^{\circ}_{cell} = E^{\circ}_{\text{HBrO/Br}_2} - E^{\circ}_{\text{BrO}_3^{-}/\text{HBrO}} = 1.595 - 1.5 = 0.095\ \text{V}$$ Since $E^{\circ}_{cell}$ is positive, HBrO disproportionates.

Q171.

The solubility of BaSO$_4$ in water is $2.42 \times 10^{-3}$ gL$^{-1}$ at 298 K. The value of its solubility product (K$_{sp}$) will be (Given molar mass of BaSO$_4$ = 233 g mol$^{-1}$)

  • A. $1.08 \times 10^{-10}$ mol$^{2}$L$^{-2}$ ✓
  • B. $1.08 \times 10^{-12}$ mol$^{2}$L$^{-2}$
  • C. $1.08 \times 10^{-8}$ mol$^{2}$L$^{-2}$
  • D. $1.08 \times 10^{-14}$ mol$^{2}$L$^{-2}$

Solution: $$s = \frac{2.42 \times 10^{-3}}{233} = 1.04 \times 10^{-5}\ \text{mol L}^{-1}$$ BaSO$_4$(s) $\rightleftharpoons$ Ba$^{2+}$(aq) $+$ SO$_4^{2-}$(aq), each at concentration s. $$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s^{2} = (1.04 \times 10^{-5})^{2}$$ $$K_{sp} = 1.08 \times 10^{-10}\ \text{mol}^{2}\text{L}^{-2}$$

Q172.

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations : (a) 60 mL $\dfrac{M}{10}$ HCl $+$ 40 mL $\dfrac{M}{10}$ NaOH (b) 55 mL $\dfrac{M}{10}$ HCl $+$ 45 mL $\dfrac{M}{10}$ NaOH (c) 75 mL $\dfrac{M}{5}$ HCl $+$ 25 mL $\dfrac{M}{5}$ NaOH (d) 100 mL $\dfrac{M}{10}$ HCl $+$ 100 mL $\dfrac{M}{10}$ NaOH pH of which one of them will be equal to 1?

  • A. (b)
  • B. (a)
  • C. (c) ✓
  • D. (d)

Solution: For mixture (c): Meq of HCl $= 75 \times \dfrac{1}{5} \times 1 = 15$ Meq of NaOH $= 25 \times \dfrac{1}{5} \times 1 = 5$ Meq of HCl left in the resulting solution $= 10$ Molarity of [H$^{+}$] in the resulting mixture $= \dfrac{10}{100} = \dfrac{1}{10}$ $$\text{pH} = -\log[\text{H}^{+}] = -\log\left[\frac{1}{10}\right] = 1.0$$

Q173.

On which of the following properties does the coagulating power of an ion depend?

  • A. The magnitude of the charge on the ion alone
  • B. Size of the ion alone
  • C. The sign of charge on the ion alone
  • D. Both magnitude and sign of the charge on the ion ✓

Solution: Coagulation of a colloidal solution by an electrolyte depends on the charge present (positive or negative) on the colloidal particles, since the effective ion must carry the opposite sign. The coagulating power of an electrolyte also depends on the magnitude of charge present on the effective ion, as stated by the Hardy-Schulze rule. So both magnitude and sign of the charge matter.

Q174.

Given van der Waals constant for NH$_3$, H$_2$, O$_2$ and CO$_2$ are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?

  • A. NH$_3$ ✓
  • B. H$_2$
  • C. CO$_2$
  • D. O$_2$

Solution: The van der Waals constant 'a' signifies intermolecular forces of attraction. Higher the value of 'a', easier will be the liquefaction of the gas. NH$_3$ has the highest value of 'a' (4.17), so it is most easily liquefied.

Q175.

Iron carbonyl, Fe(CO)$_5$ is

  • A. Tetranuclear
  • B. Mononuclear ✓
  • C. Dinuclear
  • D. Trinuclear

Solution: Based on the number of metal atoms present in a complex, carbonyls are classified into mononuclear, dinuclear, trinuclear and so on. Fe(CO)$_5$ has one Fe atom, so it is mononuclear. Co$_2$(CO)$_8$ is dinuclear and Fe$_3$(CO)$_{12}$ is trinuclear.

Q176.

The type of isomerism shown by the complex [CoCl$_2$(en)$_2$] is

  • A. Geometrical isomerism ✓
  • B. Coordination isomerism
  • C. Linkage isomerism
  • D. Ionization isomerism

Solution: In [CoCl$_2$(en)$_2$] the coordination number of Co is 6 and the compound has octahedral geometry. It exists as a trans-form (optically inactive) and a cis-form (optically active), i.e. it shows geometrical isomerism. As per the given options, the type of isomerism is geometrical isomerism.

Q177.

Which one of the following ions exhibits d-d transition and paramagnetism as well?

  • A. CrO$_4^{2-}$
  • B. Cr$_2$O$_7^{2-}$
  • C. MnO$_4^{2-}$ ✓
  • D. MnO$_4^{-}$

Solution: CrO$_4^{2-}$ gives Cr$^{6+}$ = [Ar], unpaired electrons $n = 0$, diamagnetic. Cr$_2$O$_7^{2-}$ gives Cr$^{6+}$ = [Ar], $n = 0$, diamagnetic. MnO$_4^{2-}$ gives Mn$^{6+}$ = [Ar] 3d$^{1}$, $n = 1$, paramagnetic and capable of a d-d transition. MnO$_4^{-}$ gives Mn$^{7+}$ = [Ar], $n = 0$, diamagnetic.

Q178.

The geometry and magnetic behaviour of the complex [Ni(CO)$_4$] are

  • A. Square planar geometry and diamagnetic
  • B. Tetrahedral geometry and diamagnetic ✓
  • C. Tetrahedral geometry and paramagnetic
  • D. Square planar geometry and paramagnetic

Solution: Ni(28) : [Ar] 3d$^{8}$ 4s$^{2}$ CO is a strong field ligand, so the 4s electrons shift into the 3d orbitals, giving 3d$^{10}$ 4s$^{0}$ with no unpaired electrons. For four 'CO' ligands the hybridisation would be sp$^{3}$, and thus the complex would be diamagnetic and of tetrahedral geometry.

Q179.

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code : Column I: (a) Co$^{3+}$ (b) Cr$^{3+}$ (c) Fe$^{3+}$ (d) Ni$^{2+}$ Column II: (i) $\sqrt{8}$ BM (ii) $\sqrt{35}$ BM (iii) $\sqrt{3}$ BM (iv) $\sqrt{24}$ BM (v) $\sqrt{15}$ BM

  • A. (a)-(iv), (b)-(v), (c)-(ii), (d)-(i) ✓
  • B. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
  • C. (a)-(iii), (b)-(v), (c)-(i), (d)-(ii)
  • D. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Solution: Co$^{3+}$ = [Ar] 3d$^{6}$, unpaired e$^{-}$ $n = 4$, so $\mu = \sqrt{4(4+2)} = \sqrt{24}$ BM. Cr$^{3+}$ = [Ar] 3d$^{3}$, $n = 3$, so $\mu = \sqrt{3(3+2)} = \sqrt{15}$ BM. Fe$^{3+}$ = [Ar] 3d$^{5}$, $n = 5$, so $\mu = \sqrt{5(5+2)} = \sqrt{35}$ BM. Ni$^{2+}$ = [Ar] 3d$^{8}$, $n = 2$, so $\mu = \sqrt{2(2+2)} = \sqrt{8}$ BM.

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