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Free, login-free worked solutions to every question in NEET (UG) 2022 — Solved paper. Check your answers below, then predict your rank or practise with full mock tests.
If a soap bubble expands, the pressure inside the bubble
Solution: decreases Excess pressure inside a soap bubble of radius $r$ is $p = \dfrac{4T}{r}$, where $T$ is the surface tension. ∴ $p \propto \dfrac{1}{r}$, so as the bubble expands the pressure inside it decreases.
The graph which shows the variation of the de-Broglie wavelength ($\lambda$) of a particle and its associated momentum ($p$) is
Solution: a rectangular hyperbola de Broglie wavelength $\lambda = \dfrac{h}{p}$, so $\lambda p = h$, a constant. ∴ the graph of $\lambda$ against $p$ is a rectangular hyperbola, falling steeply at first and then flattening.
A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is
Solution: 50 N/kg Gravitational field intensity is the force per unit mass, $I = \dfrac{F}{m}$. $I = \dfrac{3.0\ \text{N}}{60 \times 10^{-3}\ \text{kg}} = 50\ \text{N/kg}$
Given below are two statements Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element ($Idl$) of a current carrying conductor only. Statement II : Biot-Savart's law is analogous to Coulomb's inverse square law of charge $q$, with the former being related to the field produced by a scalar source, $Idl$ while the latter being produced by a vector source, $q$. In light of above statements choose the most appropriate answer from the options given below
Solution: Statement I is correct and Statement II is incorrect Statement I is correct: Biot–Savart law gives the field of a small current element, $d\mathbf{B} = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\mathbf{l} \times \mathbf{r}}{r^{3}}$. Statement II is incorrect: it swaps the sources; $I\,d\mathbf{l}$ is a vector source, while the charge $q$ in Coulomb's law is a scalar source.
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second
Solution: 1 : 3 : 5 : 7 Distance covered in the $n$th second $s_n = u + \dfrac{g}{2}(2n - 1)$, with $u = 0$ for a freely falling body. $s_1 : s_2 : s_3 : s_4 = \dfrac{g}{2} : \dfrac{3g}{2} : \dfrac{5g}{2} : \dfrac{7g}{2}$ ∴ the distances are in the ratio of odd numbers, 1 : 3 : 5 : 7 (Galileo's law of odd numbers).
In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be
Solution: 60 Hz A half-wave rectifier passes only one half of each input cycle, giving one output pulse per cycle. ∴ output frequency = input frequency = 60 Hz.
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball ($v$) as a function of time ($t$) is The graph carries four curves: $A$ is a straight line rising from the origin, $B$ rises steeply and then levels off at a constant value, $C$ starts high and falls steadily, and $D$ rises and then curves back down.
Solution: curve $B$ As the ball speeds up, the viscous drag $6\pi\eta r v$ grows until it balances the weight minus the upthrust. After that the net force is zero and the ball moves with a constant terminal velocity. ∴ the speed rises and then levels off at a constant value, which is curve $B$.
The angle between the electric lines of force and the equipotential surface is
Solution: 90° No work is done in moving a charge along an equipotential surface: $W = q\,\Delta V = 0$. So $\mathbf{E} \cdot d\mathbf{s} = 0$ for every small displacement $d\mathbf{s}$ along the surface. ∴ electric lines of force are perpendicular to the equipotential surface, at 90°.
As the temperature increases, the electrical resistance
Solution: increases for conductors but decreases for semiconductors The temperature coefficient of resistivity is positive for conductors and negative for semiconductors. In a conductor, electrons collide more often at higher temperature, so the relaxation time falls and resistance increases. In a semiconductor, more electrons are freed into the conduction band, so the carrier density rises and resistance decreases.
In the given nuclear reaction, the element $X$ is $$^{22}_{11}\text{Na} \longrightarrow X + e^{+} + \nu$$
Solution: neon-22, ${}^{22}_{10}\text{Ne}$ In a nuclear reaction both mass number and charge (atomic number) are conserved. Charge: $11 = Z + 1 \Rightarrow Z = 10$; mass number: $A = 22$, since the positron and the neutrino add nothing to it. ∴ $X = {}^{22}_{10}\text{Ne}$; the change is a $\beta^{+}$ decay.
Let $T_1$ and $T_2$ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr's model of an atom, the ratio of $T_1 : T_2$ is:
Solution: 9 : 4 Bohr's model: $E_n = -\dfrac{13.6}{n^{2}}\ \text{eV}$. First excited state $n = 2$; second excited state $n = 3$. $\dfrac{T_1}{T_2} = \dfrac{1/2^{2}}{1/3^{2}} = \dfrac{9}{4}$, so $T_1 : T_2 = 9 : 4$
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is
Solution: $\sqrt{2} : 1$ Radius of gyration $K = \sqrt{\dfrac{I}{M}}$, so $\dfrac{K_1}{K_2} = \sqrt{\dfrac{I_1}{I_2}}$. For a thin disc $I_1 = \dfrac{MR^{2}}{2}$ about the central normal axis and $I_2 = \dfrac{MR^{2}}{4}$ about a diameter. $\dfrac{K_1}{K_2} = \sqrt{2}$, so $K_1 : K_2 = \sqrt{2} : 1$
When light propagates through a material medium of relative permittivity $\epsilon_r$ and relative permeability $\mu_r$, the velocity of light, $v$ is given by ($c$-velocity of light in vacuum)
Solution: $v = \tfrac{c}{\sqrt{\epsilon_r \mu_r}}$ Speed of light in a medium $v = \dfrac{1}{\sqrt{\mu\epsilon}} = \dfrac{1}{\sqrt{\mu_r\mu_0\,\epsilon_r\epsilon_0}}$. Since $c = \dfrac{1}{\sqrt{\mu_0\epsilon_0}}$, $v = \dfrac{c}{\sqrt{\epsilon_r\mu_r}}$.
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is :
Solution: $36 \times 10^{7}$ J Energy = power × time, $E = Pt$, with $P = 100\ \text{kW} = 10^{5}\ \text{W}$ and $t = 1\ \text{h} = 3600\ \text{s}$. $E = 10^{5} \times 3600 = 3.6 \times 10^{8}\ \text{J} = 36 \times 10^{7}\ \text{J}$
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms$^{-1}$. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is ($g = 10$ ms$^{-2}$)
Solution: 34500 W At constant speed the motor's pull balances the weight plus friction: $F = mg + f$. $F = 2000 \times 10 + 3000 = 23000\ \text{N}$ Power $P = Fv = 23000 \times 1.5 = 34500\ \text{W}$
Two resistors of resistance, 100 $\Omega$ and 200 $\Omega$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 $\Omega$ to that in 200 $\Omega$ in a given time is
Solution: 2 : 1 In parallel both resistors have the same potential difference $V$, and the heat produced in time $t$ is $H = \dfrac{V^{2}t}{R}$. $\dfrac{H_1}{H_2} = \dfrac{R_2}{R_1} = \dfrac{200}{100}$, so $H_1 : H_2 = 2 : 1$
The peak voltage of the AC source is equal to
Solution: $\sqrt{2}$ times the rms value Peak and rms voltage are related by $V_0 = \sqrt{2}\,V_{\text{rms}}$.
A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$, the speed of the third (lighter) fragment is
Solution: $2\sqrt{2}\,v$ The shell was at rest, so the total momentum of the fragments is zero (conservation of linear momentum). Masses are $\dfrac{2m}{5}$, $\dfrac{2m}{5}$ and $\dfrac{m}{5}$; the two equal fragments have perpendicular momenta of $\dfrac{2mv}{5}$ each. The third fragment's momentum balances their resultant: $\dfrac{m}{5}v_3 = \sqrt{2} \times \dfrac{2mv}{5}$ ∴ $v_3 = 2\sqrt{2}\,v$
If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is
Solution: $1 : \sqrt{2}$ Speed of a transverse wave on a string $v = \sqrt{\dfrac{T}{\mu}}$, where $\mu$ is the mass per unit length, so $v \propto \sqrt{T}$. $\dfrac{v_1}{v_2} = \sqrt{\dfrac{T}{2T}} = \dfrac{1}{\sqrt{2}}$, so $v_1 : v_2 = 1 : \sqrt{2}$
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. On the $p$-$V$ diagram, curve 4 is horizontal, curve 1 is vertical, and curves 2 and 3 fall away from the initial point with curve 3 having the smaller slope. The curve which represents the adiabatic process among 1, 2, 3 and 4 is
Solution: curve 2 Curve 4 (horizontal, constant $p$) is isobaric and curve 1 (vertical, constant $V$) is isochoric. From the same point an adiabatic curve is steeper than an isothermal one: slope $-\gamma\dfrac{p}{V}$ against $-\dfrac{p}{V}$, with $\gamma > 1$. ∴ the steeper curve 2 is adiabatic and curve 3 is isothermal.
The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s$^{2}$ is
Solution: $4\pi\ \text{rad s}^{-2}$ $\omega_0 = 1200\ \text{rpm} = \dfrac{1200 \times 2\pi}{60} = 40\pi\ \text{rad s}^{-1}$ and $\omega = 3120\ \text{rpm} = 104\pi\ \text{rad s}^{-1}$ For uniform angular acceleration, $\omega = \omega_0 + \alpha t$. $\alpha = \dfrac{104\pi - 40\pi}{16} = 4\pi\ \text{rad s}^{-2}$
Plane angle and solid angle have
Solution: units but no dimensions Plane angle = arc/radius (unit radian) and solid angle = area/radius² (unit steradian). Each is a ratio of like quantities, so it is dimensionless even though it has a unit.
In the given circuits (A), (B) and (C) the potential drop across the two $p$-$n$ junctions are equal in
Solution: both circuits (A) and (C) In circuits (A) and (C) both p-n junction diodes are in the same biasing condition, so they offer equal resistance. The two diodes are in series, so equal potential drops across each junction.
In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is
Solution: 12 Fringe width $\beta = \dfrac{\lambda D}{d}$, so the number of fringes in a fixed length is proportional to $\dfrac{1}{\lambda}$. $n_1\lambda_1 = n_2\lambda_2 \Rightarrow n_2 = \dfrac{8 \times 600}{400} = 12$
A square loop of side 1 m and resistance 1 $\Omega$ placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is
Solution: 0.5 Wb Magnetic flux $\phi = BA\cos\theta$, where $\theta$ is the angle between the field and the normal to the loop. The plane of the loop is perpendicular to the field, so $\theta = 0^{\circ}$; area $A = 1\ \text{m}^{2}$. $\phi = 0.5 \times 1 \times \cos 0^{\circ} = 0.5\ \text{Wb}$
Two hollow conducting spheres of radii $R_1$ and $R_2$ ($R_1 > R_2$) have equal charges. The potential would be
Solution: more on the smaller sphere Potential of a charged conducting sphere $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$, so for equal charges $V \propto \dfrac{1}{R}$. ∴ the smaller sphere (radius $R_2$) is at the higher potential.
A Copper wire of length 10 m and radius $\left(\dfrac{10^{-2}}{\sqrt{\pi}}\right)$ m has electrical resistance of 10 $\Omega$. The current density in the wire for an electric field strength of 10 (V/m) is
Solution: $10^{5}$ A/m² Area of cross-section $A = \pi r^{2} = \pi \times \dfrac{10^{-4}}{\pi} = 10^{-4}\ \text{m}^{2}$ Current density $J = \dfrac{I}{A} = \dfrac{V}{RA} = \dfrac{El}{RA}$ $J = \dfrac{10 \times 10}{10 \times 10^{-4}} = 10^{5}\ \text{A/m}^{2}$
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is
Solution: $12.56 \times 10^{-2}$ T Field inside a long solenoid $B = \mu_0 n I$, independent of its radius. $n = 100$ turns per mm $= 10^{5}$ turns per metre $B = 4\pi \times 10^{-7} \times 10^{5} \times 1 = 4\pi \times 10^{-2}\ \text{T} = 12.56 \times 10^{-2}\ \text{T}$
Two objects of mass 10 kg and 20 kg, respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is
Solution: $\tfrac{20}{3}$ m With the 10 kg mass at the origin, $X_{\text{CM}} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$. $X_{\text{CM}} = \dfrac{10 \times 0 + 20 \times 10}{10 + 20} = \dfrac{20}{3}\ \text{m}$
A light ray falls on a glass surface of refractive index $\sqrt{3}$ at an angle 60$^{\circ}$. The angle between the refracted and reflected rays would be
Solution: 90° $\tan 60^{\circ} = \sqrt{3} = \mu$, so the light falls at the polarising (Brewster) angle. Snell's law: $\sin r = \dfrac{\sin 60^{\circ}}{\sqrt{3}} = \dfrac{1}{2}$, so $r = 30^{\circ}$. Angle between the reflected and refracted rays $= 180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}$
The dimensions $[\text{MLT}^{-2}\text{A}^{-2}]$ belong to the
Solution: magnetic permeability Force per unit length between two long parallel currents $\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$, so $\mu_0 = \dfrac{2\pi F d}{I_1 I_2\, l}$. $[\mu_0] = \dfrac{[\text{MLT}^{-2}][\text{L}]}{[\text{A}^{2}][\text{L}]} = [\text{MLT}^{-2}\text{A}^{-2}]$ (A) Self inductance is $[\text{ML}^{2}\text{T}^{-2}\text{A}^{-2}]$. (C) Electric permittivity is $[\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2}]$. (D) Magnetic flux is $[\text{ML}^{2}\text{T}^{-2}\text{A}^{-1}]$.
The displacement-time graph of two moving particles make angles of 30$^{\circ}$ and 45$^{\circ}$ with the $X$-axis as shown in the figure. The ratio of their respective velocity is
Solution: $1 : \sqrt{3}$ The slope of a displacement–time graph gives the velocity, $v = \tan\theta$. $\dfrac{v_1}{v_2} = \dfrac{\tan 30^{\circ}}{\tan 45^{\circ}} = \dfrac{1/\sqrt{3}}{1}$, so $v_1 : v_2 = 1 : \sqrt{3}$
A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is
Solution: +5 D Lens maker's formula: $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$, with $R_1 = +20\ \text{cm}$ and $R_2 = -20\ \text{cm}$. $\dfrac{1}{f} = 0.5 \times \dfrac{2}{20} = \dfrac{1}{20}\ \text{cm}^{-1}$, so $f = 20\ \text{cm} = 0.2\ \text{m}$ Power $P = \dfrac{1}{f\ (\text{in m})} = \dfrac{1}{0.2} = +5\ \text{D}$
Match List-I with List-II List-I (Electromagnetic waves): (A) AM radio waves (B) Microwaves (C) Infrared radiations (D) X-rays List-II (Wavelength): (i) $10^{-10}$ m (ii) $10^{2}$ m (iii) $10^{-2}$ m (iv) $10^{-4}$ m Choose the correct answer from the options given below
Solution: A-ii, B-iii, C-iv, D-i AM radio waves (530–1710 kHz) have wavelengths of a few hundred metres, about $10^{2}$ m: A-ii. Microwaves lie between 1 mm and 0.1 m, so about $10^{-2}$ m: B-iii. Infrared lies between 700 nm and 1 mm, so about $10^{-4}$ m: C-iv. X-rays lie between $10^{-3}$ nm and 1 nm, so about $10^{-10}$ m: D-i.
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei, respectively is:
Solution: 5 : 4 Nuclear radius $R = R_0A^{1/3}$, so $\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}$. $\dfrac{R_1}{R_2} = \left(\dfrac{125}{64}\right)^{1/3} = \dfrac{5}{4}$, so $R_1 : R_2 = 5 : 4$
Two transparent media $A$ and $B$ are separated by a plane boundary. The speed of light in those media are $1.5 \times 10^{8}$ m/s and $2.0 \times 10^{8}$ m/s, respectively. The critical angle for a ray of light for these two media is
Solution: $\sin^{-1}(0.750)$ Total internal reflection occurs for light going from the denser medium A (where light is slower) towards the rarer medium B. Refractive index of A with respect to B $\mu = \dfrac{v_B}{v_A}$, and $\sin i_c = \dfrac{1}{\mu}$. $\sin i_c = \dfrac{v_A}{v_B} = \dfrac{1.5 \times 10^{8}}{2.0 \times 10^{8}} = 0.750$, so $i_c = \sin^{-1}(0.750)$
A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad s$^{-1}$. If the vertical component of earth's magnetic field at that place is $2 \times 10^{-5}$ T and electrical resistance of the coil is 12.56 $\Omega$, then the maximum induced current in the coil will be
Solution: 1 A A coil rotating in a field has maximum emf $e_{\max} = NBA\omega$; the vertical component $B_V$ is perpendicular to the horizontal axis, so it gives the changing flux. $e_{\max} = 1000 \times 2 \times 10^{-5} \times \pi (10)^{2} \times 2 = 4\pi\ \text{V} = 12.56\ \text{V}$ $I_{\max} = \dfrac{e_{\max}}{R} = \dfrac{12.56}{12.56} = 1\ \text{A}$
Match List-I with List-II. List-I: (A) Gravitational constant ($G$) (B) Gravitational potential energy (C) Gravitational potential (D) Gravitational intensity List-II: (i) $[\text{L}^{2}\text{T}^{-2}]$ (ii) $[\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]$ (iii) $[\text{LT}^{-2}]$ (iv) $[\text{ML}^{2}\text{T}^{-2}]$ Choose the correct answer from the options given below
Solution: A-ii, B-iv, C-i, D-iii $G = \dfrac{Fr^{2}}{m_1m_2}$ gives $[\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]$: A-ii. Gravitational potential energy is energy, $[\text{ML}^{2}\text{T}^{-2}]$: B-iv. Gravitational potential is energy per unit mass, $[\text{L}^{2}\text{T}^{-2}]$: C-i. Gravitational intensity is force per unit mass, $[\text{LT}^{-2}]$: D-iii.
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R) A coil spring of copper has more tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: Assertion (A) is true but Reason (R) is false Assertion is true: when a coil spring is stretched, its wire is twisted rather than lengthened, so the wire changes shape without changing volume; this is decided by the shear modulus. Reason is false: a copper spring does not have more tensile strength than a steel spring of the same size; steel has the higher Young's modulus and yield strength.
The area of a rectangular field (in m$^{2}$) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is
Solution: $14 \times 10^{2}$ m² Area $= 55.3\ \text{m} \times 25\ \text{m} = 1382.5\ \text{m}^{2}$ In a product the result keeps as many significant figures as the factor with the fewest; 25 m has two. ∴ area $= 1.4 \times 10^{3}\ \text{m}^{2} = 14 \times 10^{2}\ \text{m}^{2}$
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is
Solution: $5.6\ \text{m}^{3}$ With no intermolecular forces the water behaves as an ideal gas, and at STP 1 mol of it occupies 22.4 L. Moles of water $n = \dfrac{4.5\ \text{kg}}{0.018\ \text{kg mol}^{-1}} = 250\ \text{mol}$ $V = 250 \times 22.4 \times 10^{-3}\ \text{m}^{3} = 5.6\ \text{m}^{3}$
A wheatstone bridge is used to determine the value of unknown resistance $X$ by adjusting the variable resistance $Y$ as shown in the figure. For the most precise measurement of $X$, the resistances $P$ and $Q$
Solution: approximately equal and small A Wheatstone bridge is most sensitive when the resistances in its arms are of the same order, so $P$ and $Q$ are taken nearly equal. Keeping $P$ and $Q$ small keeps the currents large enough for the galvanometer to show a small imbalance, which reduces the error in $X$.
A series $L$-$C$-$R$ circuit with inductance 10 H, capacitance 10 $\mu$F, resistance 50 $\Omega$ is connected to an AC source of voltage, $V = 200\sin(100\,t)$ volt. If the resonant frequency of the LCR circuit is $\nu_0$ and the frequency of the AC source is $\nu$, then
Solution: $\nu_0 = \nu = \tfrac{50}{\pi}$ Hz Source: $\omega = 100\ \text{rad s}^{-1}$, so $\nu = \dfrac{\omega}{2\pi} = \dfrac{100}{2\pi} = \dfrac{50}{\pi}\ \text{Hz}$ Resonant frequency $\nu_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{10 \times 10^{-5}}} = \dfrac{1}{2\pi \times 10^{-2}} = \dfrac{50}{\pi}\ \text{Hz}$ ∴ $\nu_0 = \nu = \dfrac{50}{\pi}\ \text{Hz}$; the circuit is at resonance.
A capacitor of capacitance $C = 900$ pF is charged fully by 100 V battery $B$ as shown in figure (a). Then, it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C = 900$ pF as shown in figure (b). The electrostatic energy stored by the system (b) is
Solution: $2.25 \times 10^{-6}$ J Charge is shared when the capacitors are joined; common potential $V = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2} = \dfrac{C \times 100 + C \times 0}{2C} = 50\ \text{V}$ Energy stored $U = 2 \times \dfrac{1}{2}CV^{2} = CV^{2}$ $U = 900 \times 10^{-12} \times 50^{2} = 2.25 \times 10^{-6}\ \text{J}$
Two point charges $-q$ and $+q$ are placed at a distance of $L$, as shown in the figure. The magnitude of electric field intensity at a distance $R\ (R \gg L)$ on the axis of the dipole varies as
Solution: $\tfrac{1}{R^{3}}$ Charges $-q$ and $+q$ a distance $L$ apart form an electric dipole of moment $p = qL$. On the axis, for $R \gg L$, $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{R^{3}}$, so $E \propto \dfrac{1}{R^{3}}$.
A ball is projected with a velocity, 10 ms$^{-1}$, at an angle of 60$^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be
Solution: Initial speed of ball, $u = 10$ ms$^{-1}$, and the angle of projection is 60$^{\circ}$. At the highest point of the trajectory of the ball, its vertical component of velocity becomes zero and only the horizontal component of velocity exists, which is equal to the horizontal component of velocity at the time of projection. i.e. speed of ball at highest point $$= u\cos\theta = 10\cos 60^{\circ} = 10 \times \frac{1}{2} = 5\ \text{ms}^{-1}$$
From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is :
Solution: increases linearly with $r$ inside the wire and falls as $\tfrac{1}{r}$ outside Ampere's circuital law: $B \times 2\pi r = \mu_0 I_{\text{enclosed}}$. Inside ($r < a$, where $a$ is the radius of the wire) the enclosed current is $I\dfrac{r^{2}}{a^{2}}$, so $B = \dfrac{\mu_0 I r}{2\pi a^{2}}$ and $B \propto r$. Outside ($r > a$) all of $I$ is enclosed, so $B = \dfrac{\mu_0 I}{2\pi r}$ and $B \propto \dfrac{1}{r}$.
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is
Solution: 11 $T = 2\pi\sqrt{\dfrac{l}{g}}$, so $\dfrac{T_1}{T_2} = \sqrt{\dfrac{121}{100}} = \dfrac{11}{10}$ $10\,T_1 = 11\,T_2$: while the longer pendulum makes 10 vibrations, the shorter one makes 11. ∴ they are next in phase at the mean position after 11 vibrations of the shorter pendulum.
The given graph is a representation of kinetics of a reaction, drawn at constant temperature $T$ as a horizontal line against the $x$ axis. The $y$ and $x$ axes for zero and first order reactions, respectively are
Solution: zero order: rate against concentration; first order: $t_{1/2}$ against concentration A horizontal line means the plotted quantity does not change with the quantity on the $x$ axis. Zero order: rate $= k[\text{R}]^{0} = k$, the same at every concentration. First order: $t_{1/2} = \dfrac{0.693}{k}$, the same at every initial concentration. (A), (D) For a zero order reaction concentration falls linearly with time, so that plot is not horizontal. (C) The rate of a first order reaction is not constant (rate $= k[\text{R}]$), so a plot of its rate is not a horizontal line.
Which compound amongst the following is not an aromatic compound?
Solution: cycloheptatriene (non-aromatic) Cycloheptatriene has an sp³ CH₂ carbon in the ring, so the ring is not fully conjugated (nor planar) and it is non-aromatic, even with 6 π electrons. (A) The cycloheptatrienyl (tropylium) cation is planar and fully conjugated with 6 π electrons ($4n + 2$, $n = 1$): aromatic. (B) The cyclopropenyl cation is planar and fully conjugated with 2 π electrons ($n = 0$): aromatic. (D) The cyclopentadienyl anion is planar and fully conjugated with 6 π electrons ($n = 1$): aromatic.
$$R\text{Mg}X + \text{CO}_2 \xrightarrow{\ \text{Dry ether}\ } Y \xrightarrow{\ \text{H}_3\text{O}^{+}\ } R\text{COOH}$$ What is $Y$ in the above reaction ?
Solution: RCOO⁻Mg⁺X In a Grignard reagent the alkyl group carries a partial negative charge and adds to the carbon of CO₂ (nucleophilic addition). This gives the halomagnesium salt of the acid, RCOO⁻Mg⁺X, which is Y. Hydrolysis with H₃O⁺ then gives the carboxylic acid RCOOH.
Which of the following statement is not correct about diborane?
Solution: both boron atoms are sp²-hybridised (the incorrect statement) Each boron in diborane is sp³ hybridised, forming two terminal B–H bonds and two bridge bonds. (A) Correct: the four terminal B–H bonds are ordinary 2-centre-2-electron bonds. (B) Correct: the four terminal H atoms and the two B atoms lie in one plane, with the two bridging H atoms above and below it. (D) Correct: the two B–H–B bridges are 3-centre-2-electron (banana) bonds.
Gadolinium has a low value of third ionisation enthalpy because of
Solution: high exchange enthalpy Gd is [Xe]4f⁷5d¹6s²; removing the third electron (the 5d electron) leaves Gd³⁺ as [Xe]4f⁷. The half-filled 4f⁷ subshell is extra stable because of its high exchange enthalpy, so the third electron is removed easily.
Which one is not correct mathematical equation for Dalton's law of partial pressure? Here $p$ = total pressure of gaseous mixture
Solution: $p_i = x_i\,p_i^{\circ}$ (the incorrect statement) Dalton's law relates partial pressure to the total pressure of the mixture; $p_i = x_i\,p_i^{\circ}$ uses the pressure of the pure component and is the form of Raoult's law. (A) Correct: each gas exerts $n_i\dfrac{RT}{V}$, and these add up to $p$. (B) Correct: partial pressure $p_i = x_i\,p$. (D) Correct: total pressure $p = p_1 + p_2 + p_3$.
Given below are two statements. Statement I The boiling points of the following hydrides of group 16 elements increases in the order H$_2$O < H$_2$S < H$_2$Se < H$_2$Te Statement II The boiling points of these hydride increases with increase in molar mass. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: Both Statement I and Statement II are incorrect Statement I is incorrect: the order is H₂S < H₂Se < H₂Te < H₂O, since hydrogen bonding gives water the highest boiling point. Statement II is incorrect: H₂O has the lowest molar mass but the highest boiling point, so boiling point does not simply rise with molar mass.
Given below are two statements. Statement I The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group. Statement II $o$-nitrophenol, $m$-nitrophenol and $p$-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: the electron-withdrawing –NO₂ group stabilises the phenoxide ion, so nitrophenols are stronger acids than phenol. Statement II is incorrect: acid strength depends on the position of –NO₂: p-nitrophenol > o-nitrophenol > m-nitrophenol > phenol. At the ortho and para positions –NO₂ withdraws electrons by the stronger –R effect; at the meta position only by the –I effect. o-Nitrophenol is a little weaker than the para isomer because of intramolecular hydrogen bonding.
The Kjeldahl's method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?
Solution: aniline, C₆H₅NH₂ Kjeldahl's method converts the nitrogen of a compound into ammonium sulphate by heating with concentrated H₂SO₄; the –NH₂ nitrogen of aniline is converted this way. (A) Nitrogen present in a ring, as in pyridine, is not converted into ammonium sulphate. (C) Nitrogen of the azo group (–N=N–) is not converted into ammonium sulphate. (D) Nitrogen of the nitro group is not converted into ammonium sulphate.
Match List-I with List-II. List-I (Drug class): (A) Antacids (B) Antihistamines (C) Analgesics (D) Antimicrobials List-II (Drug molecule): (i) Salvarsan (ii) Morphine (iii) Cimetidine (iv) Seldane Choose the correct answer from the options given below
Solution: A-iii, B-iv, C-ii, D-i Cimetidine stops histamine acting on receptors in the stomach wall, so less acid is released: antacid, A-iii. Seldane (terfenadine) competes with histamine for its receptor sites: antihistamine, B-iv. Morphine is a narcotic analgesic that relieves severe pain: C-ii. Salvarsan, an arsenic compound used to treat syphilis, is an antimicrobial: D-i.
The IUPAC name of an element with atomic number 119 is
Solution: ununennium (Uue) IUPAC names such elements from the numerical roots of their digits plus -ium: 1, 1, 9 → un + un + enn + ium = ununennium (Uue).
Amongst the following which one will have maximum 'lone pair - lone pair' electron repulsions?
Solution: XeF₂ Lone pairs on the central atom: XeF₂ has 3, ClF₃ has 2, and SF₄ and IF₅ have 1 each. The more lone pairs on the central atom, the greater the lone pair–lone pair repulsion, so XeF₂ has the most.
Which of the following sequence of reactions is suitable to synthesise chlorobenzene?
Solution: benzene, Cl₂, anhydrous FeCl₃ Chlorobenzene is made by electrophilic substitution: benzene with Cl₂ in the presence of anhydrous FeCl₃, a Lewis acid that generates the electrophile Cl⁺. C₆H₆ + Cl₂ → C₆H₅Cl + HCl (anhydrous FeCl₃) (A) Phenol does not form a diazonium salt with NaNO₂ and HCl; only an aromatic amine such as aniline does. (B) Benzene does not react with HCl to give chlorobenzene. (C) Heating aniline with HCl only gives the salt anilinium chloride, C₆H₅NH₃⁺Cl⁻.
Which of the following $p$-$V$ curve represents maximum work done? Each option shows an isothermal curve on $p$-$V$ axes: (a) a curve starting at high pressure and running far out to large volume, sweeping out a large area; (b) a nearly vertical line at small volume; (c) a curve that falls steeply and stops at a small volume; (d) a closed elliptical loop.
Solution: curve (a) The work done in a process equals the area under its $p$–$V$ curve. Curve (a) runs from high pressure out to a large volume, so it sweeps out the largest area and gives the most work.
Given below are half-cell reactions $$\text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}, \qquad E^{\circ}_{\text{Mn}^{2+}/\text{MnO}_4^{-}} = -1.510\ \text{V}$$ $$\frac{1}{2}\text{O}_2 + 2\text{H}^{+} + 2e^{-} \longrightarrow \text{H}_2\text{O}, \qquad E^{\circ}_{\text{O}_2/\text{H}_2\text{O}} = +1.223\ \text{V}$$ Will the permanganate ion, MnO$_4^{-}$ liberate O$_2$ from water in the presence of an acid?
Solution: yes, because $E^{\circ}_{\text{cell}} = +0.287$ V The reduction potential of MnO₄⁻/Mn²⁺ is +1.510 V; the −1.510 V given is for the reverse (oxidation) change. MnO₄⁻ is reduced at the cathode and water is oxidised to O₂ at the anode. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.510 - 1.223 = +0.287\ \text{V}$ ∴ $E^{\circ}_{\text{cell}}$ is positive ($\Delta G^{\circ} < 0$), so MnO₄⁻ can liberate O₂ from acidified water.
The IUPAC name of the complex [Ag(H$_2$O)$_2$][Ag(CN)$_2$] is
Solution: diaquasilver(I) dicyanidoargentate(I) The cation is [Ag(H₂O)₂]⁺ and the anion is [Ag(CN)₂]⁻; silver is +1 in both. The cation is named first with the metal's ordinary name (silver); the anion uses the Latin name with -ate (argentate). Ligands come first in each part, aqua for H₂O and cyanido for CN⁻: diaquasilver(I) dicyanidoargentate(I).
Identify the incorrect statement from the following.
Solution: the oxidation number of K in KO₂ is +4 (the incorrect statement) KO₂ is potassium superoxide, K⁺O₂⁻, so the oxidation number of K is +1; alkali metals show only the +1 state. (B) Correct: ionisation enthalpy decreases down the group as atomic size increases. (C) Correct: lithium is the strongest reducing agent, with the most negative E° (−3.04 V), owing to the large hydration enthalpy of Li⁺. (D) Correct: alkali metals react with water to form their hydroxides and hydrogen.
What mass of 95% pure CaCO$_3$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? $$\text{CaCO}_3(s) + 2\text{HCl}(aq) \longrightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$$ [Calculate upto second place of decimal point]
Solution: 1.32 g 1 mol CaCO₃ neutralises 2 mol HCl; moles of HCl $= 0.5 \times \dfrac{50}{1000} = 0.025$, so CaCO₃ needed $= 0.0125$ mol. Mass of pure CaCO₃ $= 0.0125 \times 100 = 1.25\ \text{g}$ Mass of 95% pure sample $= \dfrac{1.25 \times 100}{95} = 1.32\ \text{g}$
Given below are two statements. Statement I Primary aliphatic amines react with HNO$_2$ to give unstable diazonium salts. Statement II Primary aromatic amines react with HNO$_2$ to form diazonium salts which are stable even above 300 K. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: primary aliphatic amines give alkyldiazonium salts that are unstable and decompose at once, giving off N₂ and forming alcohols. Statement II is incorrect: arenediazonium salts are stable only in the cold (273–278 K) and decompose at higher temperatures, so not above 300 K.
Which statement regarding polymers is not correct?
Solution: thermosetting polymers are reusable (the incorrect statement) Thermosetting polymers become heavily cross-linked on heating and cannot be softened again, so they are not reusable. (A) Correct: fibres have strong intermolecular forces such as hydrogen bonds, which give them high tensile strength. (B) Correct: thermoplastics soften on heating and harden on cooling, again and again. (D) Correct: elastomers have chains held by the weakest intermolecular forces, which lets them stretch.
The incorrect statement regarding chirality is
Solution: enantiomers are superimposable mirror images (the incorrect statement) Enantiomers are non-superimposable mirror images of each other; non-superimposability is what makes a molecule chiral. (A) Correct: $\text{S}_{\text{N}}2$ is a one-step backside attack, so the product shows inversion of configuration. (C) Correct: in a racemic mixture the rotations of the two enantiomers cancel, so the net rotation is zero. (D) Correct: $\text{S}_{\text{N}}1$ goes through a planar carbocation that is attacked from both faces, giving racemisation (a 1 : 1 mixture).
Given below are two statements. Statement I In the coagulation of a negative sol, the flocculating power of the three given ions is in the order Al$^{3+}$ > Ba$^{2+}$ > Na$^{+}$ Statement II In the coagulation of a positive sol, the flocculating power of the three given salts is in the order NaCl > Na$_2$SO$_4$ > Na$_3$PO$_4$ In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Hardy–Schulze rule: the greater the charge on the ion of opposite sign to the sol, the greater its coagulating power. Statement I is correct: a negative sol is coagulated by cations, so Al³⁺ > Ba²⁺ > Na⁺. Statement II is incorrect: a positive sol is coagulated by anions, so PO₄³⁻ > SO₄²⁻ > Cl⁻, i.e. Na₃PO₄ > Na₂SO₄ > NaCl.
Match List-I with List-II. List-I (Hydrides): (A) MgH$_2$ (B) GeH$_4$ (C) B$_2$H$_6$ (D) HF List-II (Nature): (i) Electron precise (ii) Electron deficient (iii) Electron rich (iv) Ionic Choose the correct answer from the options given below
Solution: A-iv, B-i, C-ii, D-iii MgH₂ is a saline (ionic) hydride of an s-block metal: A-iv. GeH₄ (group 14) has exactly enough electrons for its four bonds and no lone pair: electron precise, B-i. B₂H₆ (group 13) has too few electrons for ordinary 2-electron bonds: electron deficient, C-ii. HF (group 17) has lone pairs on fluorine: electron rich, D-iii.
Match List-I with List-II. List-I: (A) Li (B) Na (C) KOH (D) Cs List-II: (i) Absorbent for carbon dioxide (ii) Electrochemical cells (iii) Coolant in fast breeder reactors (iv) Photoelectric cell Choose the correct answer from the options given below
Solution: A-ii, B-iii, C-i, D-iv Lithium is used in electrochemical cells (lithium batteries): A-ii. Liquid sodium is the coolant in fast breeder nuclear reactors: B-iii. KOH, a strong base, absorbs the acidic gas CO₂: C-i. Caesium has the lowest ionisation enthalpy and gives off electrons when light falls on it, so it is used in photoelectric cells: D-iv.
Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells. Reason (R) In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both A and R are correct but R is not the correct explanation of A Assertion is correct: when cations are missing (as in a Schottky defect), an equal charge of anions is also missing, so the solid stays neutral. Reason is correct: in a Frenkel defect a cation moves from its lattice site to an interstitial site, so the solid stays neutral. R does not explain A: missing ions describe the Schottky defect, while R describes the Frenkel defect, in which no ion is lost.
The incorrect statement regarding enzymes is
Solution: enzymes are polysaccharides (the incorrect statement) Almost all enzymes are proteins, made of one or more polypeptide chains of amino acids; they are not polysaccharides. (A) Correct: like chemical catalysts, enzymes lower the activation energy of reactions. (C) Correct: each enzyme is highly specific for its substrate and reaction. (D) Correct: enzymes are biological catalysts (biocatalysts).
Match List-I with List-II. List-I (Products formed): (A) Cyanohydrin (B) Acetal (C) Schiff's base (D) Oxime List-II (Reaction of carbonyl compound with): (i) NH$_2$OH (ii) $R$NH$_2$ (iii) Alcohol (iv) HCN Choose the correct answer from the options given below
Solution: A-iv, B-iii, C-ii, D-i Carbonyl compound + HCN gives a cyanohydrin, with –OH and –CN on the same carbon: A-iv. Aldehyde + two molecules of an alcohol (dry HCl) gives an acetal, with two –OR groups on the same carbon: B-iii. Carbonyl compound + a primary amine RNH₂ gives a Schiff's base, >C=N–R: C-ii. Carbonyl compound + hydroxylamine NH₂OH gives an oxime, >C=N–OH: D-i.
In one molal solution that contains 0.5 mole of a solute, there is
Solution: 500 g of solvent Molality = moles of solute per kilogram of solvent. Mass of solvent $= \dfrac{0.5\ \text{mol}}{1\ \text{mol kg}^{-1}} = 0.5\ \text{kg} = 500\ \text{g}$
Choose the correct statement.
Solution: diamond is sp³-hybridised and graphite is sp²-hybridised In diamond each carbon is bonded to four others (sp³, tetrahedral); in graphite each carbon is bonded to three others in a flat layer (sp²). (A) Both are covalent network solids; graphite is not ionic. (C) Only graphite is a dry lubricant, as its layers slide over one another; diamond is the hardest natural substance. (D) Diamond has a three-dimensional network; only graphite has two-dimensional layers.
Which amongst the following is incorrect statement?
Solution: O₂⁺ ion is diamagnetic (the incorrect statement) O₂⁺ has 15 electrons, with one unpaired electron in a π*2p orbital, so it is paramagnetic. (A) Correct: in C₂ (12 electrons) the last four electrons fill the two degenerate π2p bonding orbitals. (B) Correct: H₂⁺ has two protons and only one electron. (D) Correct: bond order $= \tfrac{1}{2}(N_b - N_a)$ gives 2.5, 2, 1.5 and 1 for O₂⁺, O₂, O₂⁻ and O₂²⁻.
Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) ICl is more reactive than I$_2$. Reason (R) I—Cl bond is weaker than I—I bond. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both A and R are correct and R is the correct explanation of A Assertion is correct: interhalogen compounds such as ICl are more reactive than the halogens (except fluorine). Reason is correct and explains it: the polar I–Cl bond is weaker than the I–I bond, so ICl reacts more readily.
Given below are two statements. Statement I The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole-dipole interactions. Statement II The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: Both Statement I and Statement II are correct Statement I is correct: the polar C=O group gives dipole–dipole attraction (weak molecular association), so aldehydes and ketones boil higher than hydrocarbons of similar mass. Statement II is correct: they cannot form hydrogen bonds with one another, so they boil lower than alcohols of similar mass.
The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given p$K_a$ of CH$_3$COOH = 4.57]
Solution: 5.57 Acetic acid with sodium acetate is an acidic buffer: $\text{pH} = \text{p}K_a + \log\dfrac{[\text{Salt}]}{[\text{Acid}]}$. Equal volumes are mixed, so the concentration ratio stays $\dfrac{0.10}{0.01} = 10$. $\text{pH} = 4.57 + \log 10 = 5.57$
Identify the incorrect statement from the following.
Solution: $d_{x^{2}-y^{2}}$ and $d_{z^{2}}$ have similar shapes (the incorrect statement) $d_{xy}$, $d_{yz}$, $d_{zx}$ and $d_{x^{2}-y^{2}}$ all have four lobes (double dumb-bell); only $d_{z^{2}}$ differs, with two lobes along the $z$-axis and a ring around the middle. (A) Correct: 4d orbitals have the same shapes as the corresponding 3d orbitals. (B) Correct: in a free atom all five 3d orbitals have the same energy (degenerate). (D) Correct: 5d orbitals are larger than the corresponding 4d orbitals, as size increases with $n$.
At 298 K, the standard electrode potentials of Cu$^{2+}$/Cu, Zn$^{2+}$/Zn, Fe$^{2+}$/Fe and Ag$^{+}$/Ag are 0.34 V, $-0.76$ V, $-0.44$ V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction can not occur?
Solution: 2CuSO₄(aq) + 2Ag(s) → 2Cu(s) + Ag₂SO₄(aq), which cannot occur A metal with a lower E° displaces a metal with a higher E° from its salt solution; reactivity order Zn (−0.76 V) > Fe (−0.44 V) > Cu (0.34 V) > Ag (0.80 V). Ag has a higher E° than Cu, so silver cannot displace copper: $E^{\circ}_{\text{cell}} = 0.34 - 0.80 = -0.46\ \text{V}$. (A) Fe displaces Cu: $E^{\circ}_{\text{cell}} = 0.34 - (-0.44) = +0.78\ \text{V}$. (B) Zn displaces Fe: $E^{\circ}_{\text{cell}} = -0.44 - (-0.76) = +0.32\ \text{V}$. (D) Zn displaces Cu: $E^{\circ}_{\text{cell}} = 0.34 - (-0.76) = +1.10\ \text{V}$.
The order of energy absorbed which is responsible for the colour of complexes (A) [Ni(H$_2$O)$_2$(en)$_2$]$^{2+}$ (B) [Ni(H$_2$O)$_4$(en)]$^{2+}$ and (C) [Ni(en)$_3$]$^{2+}$ is
Solution: [Ni(en)₃]²⁺ > [Ni(H₂O)₂(en)₂]²⁺ > [Ni(H₂O)₄(en)]²⁺, i.e. (C) > (A) > (B) en is a stronger field ligand than H₂O, so each H₂O replaced by en increases the crystal field splitting $\Delta_o$. The energy absorbed for the colour equals $\Delta_o$, so more en ligands mean more energy absorbed. ∴ [Ni(en)₃]²⁺ (3 en) > [Ni(H₂O)₂(en)₂]²⁺ (2 en) > [Ni(H₂O)₄(en)]²⁺ (1 en).
A 10.0 L flask contains 64 g of oxygen at 27$^{\circ}$C. (Assume O$_2$ gas is behaving ideally). The pressure inside the flask in bar is (Given, $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$)
Solution: 4.9 bar Moles of O₂ $= \dfrac{64}{32} = 2$; $T = 27 + 273 = 300\ \text{K}$ Ideal gas equation: $p = \dfrac{nRT}{V} = \dfrac{2 \times 0.0831 \times 300}{10.0}$ ∴ $p = 4.986\ \text{bar}$ (given as 4.9 bar)
Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?
Solution: structure (a) Acetone and 2-pentanone with dilute NaOH and heat undergo aldol condensation. Self-condensation of acetone gives 4-methylpent-3-en-2-one; cross aldol between acetone and 2-pentanone gives the mixed condensation products. Structure (a), which carries an ethyl branch on the alkene carbon, cannot arise from any of these enolate combinations, so it is not formed.
The product formed from the following reaction sequence is Cyanobenzene, C$_6$H$_5$CN, treated with (i) LiAlH$_4$, H$_2$O; (ii) NaNO$_2$, HCl; (iii) H$_2$O.
Solution: benzyl alcohol, C₆H₅CH₂OH LiAlH₄ reduces the nitrile to a primary amine: C₆H₅CN → C₆H₅CH₂NH₂ (benzylamine). The –NH₂ of benzylamine is on an sp³ carbon, so with NaNO₂ + HCl it behaves like an aliphatic amine and gives an unstable diazonium salt, C₆H₅CH₂N₂⁺Cl⁻. With water this salt loses N₂ and gives benzyl alcohol, C₆H₅CH₂OH.
Copper crystallises in fcc unit cell with cell edge length of $3.608 \times 10^{-8}$ cm. The density of copper is 8.92 g cm$^{-3}$. Calculate the atomic mass of copper.
Solution: 63.1 u Density of a cubic crystal $d = \dfrac{Z M}{N_A a^{3}}$, with $Z = 4$ for fcc. $M = \dfrac{d N_A a^{3}}{Z} = \dfrac{8.92 \times 6.022 \times 10^{23} \times (3.608 \times 10^{-8})^{3}}{4}$ $M = 63.1\ \text{g mol}^{-1}$, so the atomic mass of copper is 63.1 u.
$$3\text{O}_2(g) \rightleftharpoons 2\text{O}_3(g)$$ For the above reaction at 298 K, $K_C$ is found to be $3.0 \times 10^{-59}$. If the concentration of O$_2$ at equilibrium is 0.040 M then concentration of O$_3$ in M is
Solution: $4.38 \times 10^{-32}$ M $K_C = \dfrac{[\text{O}_3]^{2}}{[\text{O}_2]^{3}}$ $[\text{O}_3]^{2} = 3.0 \times 10^{-59} \times (0.040)^{3} = 1.92 \times 10^{-63}$ ∴ $[\text{O}_3] = 4.38 \times 10^{-32}\ \text{M}$
Given below are two statements. Statement I In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnCl$_2$, known as Lucas reagent. Statement II Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas reagent. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: Lucas reagent (conc. HCl + anhydrous ZnCl₂) tells 1°, 2° and 3° alcohols apart by how fast turbidity appears. Statement II is incorrect: tertiary alcohols react fastest and give turbidity at once; primary alcohols give no turbidity at room temperature.
In the neutral or faintly alkaline medium, KMnO$_4$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from
Solution: +7 to +4 In neutral or faintly alkaline solution MnO₄⁻ is reduced to MnO₂: 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻ Mn changes from +7 in MnO₄⁻ to +4 in MnO₂, while iodine goes from −1 to +5.
Match List-I with List-II. List-I (Ores): (A) Haematite (B) Magnetite (C) Calamine (D) Kaolinite List-II (Composition): (i) Fe$_3$O$_4$ (ii) ZnCO$_3$ (iii) Fe$_2$O$_3$ (iv) [Al$_2$(OH)$_4$Si$_2$O$_5$] Choose the correct answer from the options given below
Solution: A-iii, B-i, C-ii, D-iv Haematite, an ore of iron, is Fe₂O₃: A-iii. Magnetite, the magnetic ore of iron, is Fe₃O₄: B-i. Calamine, an ore of zinc, is ZnCO₃: C-ii. Kaolinite, a form of clay (aluminium ore), is [Al₂(OH)₄Si₂O₅]: D-iv.
The correct IUPAC name of the compound drawn as a six-carbon chain carrying Cl on one carbon, OH on the next-but-one, a methyl branch between them, and Br at the far end is
Solution: 1-bromo-5-chloro-4-methylhexan-3-ol –OH is the principal functional group, so the chain is numbered from the end that gives it the lower locant: C-3 from the Br end, against C-4 from the other end. This puts Br on C-1, CH₃ on C-4 and Cl on C-5. Prefixes are cited in alphabetical order (bromo, chloro, methyl): 1-bromo-5-chloro-4-methylhexan-3-ol. (A) and (C) give –OH the higher locant 4; (B) does not cite the prefixes alphabetically.
If radius of second Bohr orbit of the He$^{+}$ ion is 105.8 pm, what is the radius of third Bohr orbit of Li$^{2+}$ ion?
Solution: 158.7 pm Bohr radius $r_n = 52.9\,\dfrac{n^{2}}{Z}\ \text{pm}$, so $r \propto \dfrac{n^{2}}{Z}$. For Li²⁺ ($n = 3$, $Z = 3$) and He⁺ ($n = 2$, $Z = 2$): $\dfrac{r_3}{r_2} = \dfrac{3^{2}/3}{2^{2}/2} = \dfrac{3}{2}$ $r_3 = \dfrac{3}{2} \times 105.8\ \text{pm} = 158.7\ \text{pm}$
Compound $X$ on reaction with O$_3$ followed by Zn/H$_2$O gives formaldehyde and 2-methyl propanal as products. The compound $X$ is
Solution: 3-methylbut-1-ene Ozonolysis (O₃, then Zn/H₂O) breaks the C=C bond and turns each alkene carbon into a C=O group. Joining HCHO and (CH₃)₂CHCHO through their carbonyl carbons gives (CH₃)₂CH–CH=CH₂, 3-methylbut-1-ene. (A) 2-Methylbut-1-ene gives HCHO and butan-2-one. (B) 2-Methylbut-2-ene gives acetone and acetaldehyde. (C) Pent-2-ene gives acetaldehyde and propanal.
For a first order reaction $A \rightarrow$ products, initial concentration of $A$ is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min$^{-1}$ is
Solution: $0.9212\ \text{min}^{-1}$ For a first order reaction $k = \dfrac{2.303}{t}\log\dfrac{[\text{A}]_0}{[\text{A}]}$. $k = \dfrac{2.303}{5}\log\dfrac{0.1}{0.001} = \dfrac{2.303 \times 2}{5}$ ∴ $k = 0.9212\ \text{min}^{-1}$
The pollution due to oxides of sulphur gets enhanced due to the presence of (I) particulate matter (II) ozone (III) hydrocarbons (IV) hydrogen peroxide Choose the most appropriate answer from the options given below
Solution: I, II and IV only Particulate matter in polluted air catalyses the oxidation of SO₂ to SO₃: 2SO₂ + O₂ → 2SO₃. Ozone and hydrogen peroxide also promote this oxidation: SO₂ + O₃ → SO₃ + O₂ and SO₂ + H₂O₂ → H₂SO₄. Hydrocarbons are linked with nitrogen oxides in photochemical smog, not with the oxidation of SO₂.
Which of the following is not a method of $ex$ $situ$ conservation?
Solution: national parks National parks protect species in their natural habitat, so they are in situ, not ex situ, conservation. (B) Micropropagation raises plants by tissue culture outside their habitat: ex situ. (C) Cryopreservation stores gametes, seeds or tissues in liquid nitrogen at −196 °C: ex situ. (D) In vitro fertilisation is done in the laboratory: ex situ.
Given below are two statements Statement I The primary CO$_2$ acceptor in C$_4$ plants is phosphoenolpyruvate and is found in the mesophyll cells. Statement II Mesophyll cells of C$_4$ plants lack RuBisCO enzyme. In the light of the above statements, choose the correct answer from the options given below.
Solution: Both Statement I and Statement II are correct Statement I is correct: in C₄ plants the primary CO₂ acceptor is the 3-carbon phosphoenol pyruvate (PEP) in the mesophyll cells, and PEP carboxylase fixes CO₂. Statement II is correct: mesophyll cells of C₄ plants lack RuBisCO; it is in the bundle sheath cells, where the Calvin cycle runs.
XO type of sex-determination can be found in
Solution: grasshoppers Grasshoppers show XO sex determination: females are XX and males have only one X chromosome (XO).
In old trees, the greater part of secondary xylem is dark brown and resistant to insect attack due to I. secretion of secondary metabolites and their deposition in the lumen of vessels. II. deposition of organic compounds like tannins and resins in the central layers of stem. III. deposition of suberin and aromatic substances in the outer layer of stem. IV. deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem. V. presence of parenchyma cells, functionally active xylem elements and essential oils. Choose the correct answer from the options given below.
Solution: I and II The dark heartwood of old trees forms when secondary metabolites such as tannins, resins, oils, gums and aromatic substances are deposited in the central (innermost) layers of the stem, including the lumen of vessels. These deposits make heartwood hard, durable and resistant to microorganisms and insects. III describes cork, the suberised outer layer; IV and V refer to the peripheral wood (sapwood), which is light in colour and still conducts water.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) Polymerase chain reaction is used in DNA amplification. Reason (R) The ampicillin resistant gene is used as a selectable marker to check transformation. In the light of the above statements, choose the correct answer from the options given below.
Solution: Both A and R are correct, but R is not the correct explanation of A Assertion is correct: the polymerase chain reaction makes many copies of a DNA fragment or gene in vitro (DNA amplification). Reason is correct: when cells are spread on ampicillin plates only transformants survive, so the ampicillin resistance gene is a selectable marker. R does not explain A: amplification of DNA and selection of transformants are separate steps.
Which one of the following statement is not true regarding gel electrophoresis technique?
Solution: a chromogenic substrate gives blue DNA bands on the gel (the incorrect statement) A chromogenic substrate giving a blue colour is used in blue-white screening of recombinant colonies, not in gel electrophoresis. (A) Correct: separated DNA fragments are stained with ethidium bromide. (C) Correct: stained DNA is seen as bright orange bands under UV light. (D) Correct: cutting out the bands and extracting the DNA from the gel piece is called elution.
The gaseous plant growth regulator is used in plants to
Solution: promote root growth and root hair formation The gaseous plant growth regulator is ethylene; it promotes root growth and root hair formation, increasing the absorption surface. (B) Overcoming apical dominance is done by cytokinins. (C) Killing dicotyledonous weeds is done by auxins such as 2,4-D. (D) Speeding up malting in the brewing industry is done by gibberellins (GA₃).
What amount of energy is released from glucose during lactic acid fermentation?
Solution: less than 7% In fermentation glucose is only partly broken down, so less than 7 per cent of its energy is released, and not all of that is trapped in ATP.
Read the following statements about the vascular bundles. I. In roots, xylem and phloem in a vascular bundle are arranged in an alternate manner along the different radii. II. Conjoint closed vascular bundles do not possess cambium. III. In open vascular bundles, cambium is present in between xylem and phloem. IV. The vascular bundles of dicotyledonous stem possess endarch protoxylem. V. In monocotyledonous root, usually there are more than six xylem bundles present Choose the correct answer from the options given below
Solution: I, II, III, IV and V All five statements are correct. I: in roots xylem and phloem lie on alternate radii (radial vascular bundles). II and III: closed bundles have no cambium; open bundles have cambium between xylem and phloem. IV: dicot stem bundles have endarch protoxylem (protoxylem towards the centre). V: monocot roots usually have more than six xylem bundles (polyarch).
Identify the correct set of statements. I. The leaflets are modified into pointed hard thorns in $Citrus$ and $Bougainvillea$. II. Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin. III. Stem is flattened and fleshy in $Opuntia$ and modified to perform the function of leaves. IV. $Rhizophora$ shows vertically upward growing roots that help to get oxygen for respiration. V. Subaerially growing stems in grasses and strawberry help in vegetative propagation. Choose the correct answer from the options given below.
Solution: II, III, IV and V II: tendrils of cucumber and pumpkin develop from axillary buds and are slender and spirally coiled. III: Opuntia has a flattened, fleshy green stem that does the work of leaves. IV: Rhizophora has pneumatophores, roots that grow vertically upwards to get oxygen for respiration. V: subaerial stems of grasses and strawberry spread and form new plants (vegetative propagation). I is wrong: in Citrus and Bougainvillea the thorns are modified axillary buds, not leaflets.
Which one of the following plants does not show plasticity?
Solution: maize Plasticity is the ability to form different structures in response to environment or phase of life, as in heterophylly; maize does not show it. (A) Coriander: juvenile leaves differ in shape from mature leaves. (B) Buttercup: leaves formed in air differ in shape from those formed in water. (D) Cotton: juvenile leaves differ in shape from mature leaves.
Given below are two statements, Statement I Cleistogamous flowers are invariably autogamous. Statement II Cleistogamy is disadvantageous as there is no chance for cross-pollination. In the light of the above statements, choose the correct answer from the options given below.
Solution: Both Statement I and Statement II are correct Statement I is correct: cleistogamous flowers never open and their anthers and stigma lie close together, so only their own pollen reaches the stigma. Statement II is correct: with no chance of cross-pollination there is little genetic variation, a disadvantage, though seed-set is assured even without pollinators.
'Girdling Experiment' was performed by plant physiologists to identify the plant tissue through which
Solution: food is transported In girdling, a ring of bark down to the phloem is removed from the trunk. After a few weeks the bark just above the ring swells, because food moving down in the phloem cannot pass the ring. ∴ phloem is the tissue that transports food, here towards the roots.
Hydrocolloid carrageen is obtained from
Solution: Rhodophyceae only Carrageen is a hydrocolloid from red algae (Rhodophyceae); brown algae (Phaeophyceae) give algin instead.
Given below are two statements, Statement I Mendel studied seven pairs of contrasting traits in pea plants and proposed the law of inheritance. Statement II Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height. In the light of the above statements, choose the correct answer from the options given below
Solution: Both Statement I and Statement II are correct Statement I is correct: Mendel studied seven pairs of contrasting traits in garden pea (1856–1863) and proposed the laws of inheritance. Statement II is correct: the seven characters were seed shape, seed colour, flower colour, pod shape, pod colour, flower position and stem height.
Production of cucumber has increased manifold in recent years. Application of which of the following phytohormones has resulted in this increased yield as the hormone is known to produce female flowers in the plants?
Solution: ethylene Ethylene promotes female flowers in cucumber, which increases the yield.
The process of translation of $m$RNA to proteins begins as soon as
Solution: the small subunit of ribosome encounters mRNA An inactive ribosome exists as a large and a small subunit; translation begins when the small subunit encounters an mRNA.
Habitat loss and fragmentation, overexploitation, alien-species invasion and co-extinction are causes for
Solution: biodiversity loss Habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions are the four major causes of biodiversity loss, the 'Evil Quartet'.
Read the following statements and choose the set of correct statements. I. Euchromatin is loosely packed chromatin. II. Heterochromatin is transcriptionally active. III. Histone octamer is wrapped by negatively charged DNA in nucleosome. IV. Histones are rich in lysine and arginine. V. A typical nucleosome contains 400 bp of DNA helix. Choose the correct answer from the options given below.
Solution: I, III and IV I: euchromatin is loosely packed and stains light. III and IV: histones are rich in the basic amino acids lysine and arginine, so the positively charged histone octamer is wrapped by negatively charged DNA. II is wrong: heterochromatin is densely packed and transcriptionally inactive; euchromatin is the active form. V is wrong: a typical nucleosome contains 200 bp of DNA helix.
Which of the following is incorrectly matched?
Solution: Ulothrix – mannitol (the incorrect statement) Ulothrix is a green alga (Chlorophyceae) and stores starch; mannitol is stored by brown algae (Phaeophyceae). (B) Correct: Porphyra, a red alga, stores floridean starch. (C) Correct: Volvox, a green alga, stores starch. (D) Correct: Ectocarpus, a brown alga, has the pigment fucoxanthin.
The device which can remove particulate matter present in the exhaust from a thermal power plant is
Solution: electrostatic precipitator An electrostatic precipitator charges the dust particles with electrons from a corona around high-voltage wires and collects them on grounded plates, removing over 99 per cent of the particulate matter.
DNA polymorphism forms the basis of
Solution: both genetic mapping and DNA fingerprinting Variation in DNA sequence between individuals (DNA polymorphism) is the basis of genetic mapping of the human genome as well as of DNA fingerprinting.
Which one of the following statements cannot be connected to predation?
Solution: both the interacting species are negatively impacted (the incorrect statement) In predation one species (the predator) benefits and the other (the prey) is harmed, + and −; harm to both species is competition. (A) Predation can drive a prey species to extinction, for example when a very efficient predator is introduced. (C) Predators keep prey populations under control, which maintains ecological balance. (D) Predators reduce competition among prey species and so help maintain species diversity.
Which one of the following never occurs during mitotic cell division?
Solution: pairing of homologous chromosomes Pairing of homologous chromosomes (synapsis) happens only in zygotene of prophase I of meiosis, never in mitosis. (A) Centrioles move towards opposite poles in prophase of mitosis. (C) Chromatids coil and condense in prophase of mitosis. (D) Spindle fibres attach to the kinetochores of chromosomes in metaphase of mitosis.
Exoskeleton of arthropods is composed of
Solution: chitin The arthropod exoskeleton is a cuticle of chitin, a complex structural polysaccharide built from N-acetylglucosamine units.
What is the net gain of ATP when each molecule of glucose is converted to two molecules of pyruvic acid?
Solution: 2 Glycolysis uses 2 ATP and produces 4 ATP by substrate-level phosphorylation, so the net gain is 2 ATP per glucose converted to two molecules of pyruvic acid.
Which one of the following plants shows vexillary aestivation and diadelphous stamens?
Solution: Pisum sativum Pea (Pisum sativum, Fabaceae) has vexillary aestivation: the large standard petal overlaps the two wings, which overlap the two keel petals. Its ten stamens are diadelphous, nine fused into one bundle and one free.
Identify the incorrect statement related to pollination.
Solution: moths and butterflies are the most dominant insect pollinators (the incorrect statement) Among insects, bees are the dominant biotic pollinating agents, not moths and butterflies. (A) Correct: among abiotic agents, pollination by wind is more common than by water. (B) Correct: some flowers produce foul odours to attract flies and beetles. (D) Correct: pollination by water is quite rare, limited to about 30 genera, mostly monocots.
Given below are two statements. Statement I Decomposition is a process in which the detritus is degraded into simpler substances by microbes. Statement II Decomposition is faster if the detritus is rich in lignin and chitin. In the light of the above statements, choose the correct answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: decomposers break detritus down into simpler inorganic substances such as CO₂, water and nutrients. Statement II is incorrect: decomposition is slower when detritus is rich in lignin and chitin, and quicker when it is rich in nitrogen and water-soluble substances like sugars.
The appearance of recombination nodules on homologous chromosomes during meiosis characterises
Solution: sites at which crossing over occurs In pachytene, recombination nodules appear on the bivalents; they are the sites at which crossing over occurs between non-sister chromatids of homologous chromosomes.
Which one of the following produces nitrogen-fixing nodules on the roots of $Alnus$ ?
Solution: Frankia Frankia produces nitrogen-fixing nodules on the roots of non-leguminous plants such as Alnus.
Which one of the following is not true regarding the release of energy during ATP synthesis through chemiosmosis? It involves
Solution: breakdown of electron gradient (the incorrect statement) Chemiosmosis depends on a proton gradient across the thylakoid membrane; there is no electron gradient to break down. (B) True: protons move across the membrane to the stroma through the CF₀ channel of ATP synthase, and this drives ATP synthesis. (C) True: NADP⁺ is reduced on the stroma side, which removes protons from the stroma and adds to the gradient. (D) True: breakdown of the proton gradient releases the energy used to make ATP.
The flowers are zygomorphic in I. Mustard II. Gulmohar III. $Cassia$ IV. $Datura$ V. Chilly Choose the correct answer from the options given below.
Solution: gulmohar and Cassia (II and III) A zygomorphic flower can be divided into two similar halves in only one vertical plane, as in gulmohar and Cassia. Mustard, Datura and chilli have actinomorphic flowers, which can be divided into two equal halves by any radial plane through the centre.
Match List-I with List-II. List-I: A. Manganese B. Magnesium C. Boron D. Iron List-II: (i) Activates the enzyme catalase (ii) Required for pollen germination (iii) Activates enzymes of respiration (iv) Functions in splitting of water during photosynthesis Choose the correct answer from the options given below.
Solution: A-iv, B-iii, C-ii, D-i Manganese is needed for the splitting of water that liberates oxygen in photosynthesis: A-iv. Magnesium activates the enzymes of respiration and photosynthesis: B-iii. Boron is required for pollen germination: C-ii. Iron activates the enzyme catalase: D-i.
Addition of more solutes in a given solution will
Solution: lower its water potential Dissolved solute lowers the free energy of water in the solution, so the solute potential $\psi_s$ is always negative. The more solute molecules, the more negative $\psi_s$, so the water potential falls further below that of pure water (zero).
Read the following statements on lipids and find out correct set of statements I. Lecithin found in the plasma membrane is a glycolipid. II. Saturated fatty acids possess one or more C=C bonds. III. Gingelly oil has lower melting point, hence remains as oil in winter. IV. Lipids are generally insoluble in water, but soluble in some organic solvents. V. When fatty acid is esterified with glycerol, monoglycerides are formed. Choose the correct answer from the options given below.
Solution: III, IV and V III: gingelly oil has a low melting point, so it stays liquid in winter. IV: lipids are generally insoluble in water but soluble in some organic solvents. V: fatty acids esterified with glycerol form monoglycerides (and di- and triglycerides). I is wrong: lecithin is a phospholipid, not a glycolipid. II is wrong: saturated fatty acids have no C=C bond; unsaturated ones have one or more.
What is the role of large bundle sheath cells found around the vascular bundles in C$_4$ plants?
Solution: to increase the number of chloroplasts for the Calvin cycle In C₄ plants the Calvin cycle runs in the large bundle sheath cells (Kranz anatomy), which have a large number of chloroplasts and contain RuBisCO. Their thick walls are impervious to gaseous exchange and there are no intercellular spaces, so the CO₂ released inside stays concentrated around RuBisCO. (D) is wrong: this high CO₂ concentration prevents photorespiration rather than providing a site for it.
The entire fleet of buses in Delhi were converted to CNG from diesel. In reference to this, which one of the following statements is false?
Solution: the same diesel engine is used in CNG buses, so conversion is cheap (the incorrect statement) A diesel engine cannot simply run on CNG; the buses needed engines built for CNG, so this statement is false. (B) True: CNG is cheaper than petrol or diesel. (C) True: CNG cannot be siphoned off by thieves or adulterated like petrol or diesel. (D) True: CNG burns most efficiently, and very little of it is left unburnt.
Transposons can be used during which one of the following ?
Solution: gene silencing In RNA interference (RNAi), a complementary dsRNA binds a specific mRNA and prevents its translation, silencing the gene. This dsRNA can come from infection by viruses with RNA genomes or from mobile genetic elements (transposons) that replicate via an RNA intermediate.
Match List-I with List-II. List-I: A. Metacentric chromosome B. Acrocentric chromosome C. Sub-metacentric chromosome D. Telocentric chromosome List-II: (i) Centromere situated close to the end forming one extremely short and one very long arms (ii) Centromere at the terminal end (iii) Centromere in the middle forming two equal arms of chromosomes (iv) Centromere slightly away from the middle forming one shorter arm and one longer arm Choose the correct answer from the options given below.
Solution: A-iii, B-i, C-iv, D-ii A metacentric chromosome has the centromere in the middle, giving two equal arms: A-iii. An acrocentric chromosome has it close to the end, giving one extremely short and one very long arm: B-i. A sub-metacentric chromosome has it slightly away from the middle, giving one shorter and one longer arm: C-iv. A telocentric chromosome has the centromere at the terminal end: D-ii.
Match the plant with the kind of life cycle it exhibits. List-I: A. $Spirogyra$ B. Fern C. $Funaria$ D. $Cycas$ List-II: (i) Dominant diploid sporophyte vascular plant, with highly reduced male or female gametophyte (ii) Dominant haploid free-living gametophyte (iii) Dominant diploid sporophyte alternating with reduced gametophyte called prothallus (iv) Dominant haploid leafy gametophyte alternating with partially dependent multicellular sporophyte Choose the correct answer from the options given below.
Solution: A-ii, B-iii, C-iv, D-i Spirogyra is haplontic, with the free-living haploid gametophyte as the dominant phase: A-ii. Ferns have a dominant sporophyte alternating with a small gametophyte, the prothallus: B-iii. Funaria, a moss, has a dominant haploid leafy gametophyte with a partly dependent sporophyte: C-iv. Cycas, a gymnosperm, has a dominant diploid sporophyte with highly reduced male and female gametophytes: D-i.
Which of the following occurs due to the presence of autosome linked dominant trait?
Solution: myotonic dystrophy Myotonic dystrophy is an autosomal dominant disorder: one copy of the altered gene is enough to cause it. (B) Haemophilia is X-linked recessive. (C), (D) Thalassemia and sickle cell anaemia are autosomal recessive.
The anatomy of spring wood shows some peculiar features. Identify the correct set of statement about springwood. I. It is also called as the early wood. II. In spring season cambium produces xylem elements with narrow vessels. III. It is light in colour. IV. The springwood along with autumnwood shows alternate concentric rings forming annual rings. V. It has lower density. Choose the correct answer from the options given below.
Solution: I, III, IV and V In spring the cambium is very active and forms many xylem elements with wide vessels; this spring wood (early wood) is light in colour and has lower density. With the darker, denser autumn wood (late wood) it forms alternate concentric rings, the annual rings. II is wrong: spring wood has vessels with wider cavities, not narrow ones.
Which part of the fruit, labelled in the given figure makes it a false fruit?
Solution: thalamus (labelled C) The part labelled C is the thalamus, and it makes the fruit a false fruit. A false fruit (pseudocarp or accessory fruit) develops from floral parts other than the ovary, such as the thalamus.
In the following palindromic base sequences of DNA, which one can be cut easily by particular restriction enzyme?
Solution: 5′ GAATTC 3′ / 3′ CTTAAG 5′ A restriction enzyme cuts at a palindrome, a sequence that reads the same on both strands when each is read in the same orientation (5′ → 3′). 5′ GAATTC 3′ also reads GAATTC on the other strand from 5′ to 3′, so it is a palindrome; it is the EcoRI site. In the other three sequences the two strands read differently from 5′ to 3′, so they are not palindromes.
While explaining interspecific interaction of population, (+) sign is assigned for beneficial interaction, ($-$) sign is assigned for detrimental interaction and (0) for neutral interaction. Which of the following interactions can be assigned (+) for one species and ($-$) for another species involved in the interaction ?
Solution: predation In predation the predator benefits (+) and the prey is harmed (−). (A) Amensalism is − for one species and 0 for the other. (B) Commensalism is + for one species and 0 for the other. (C) Competition is − for both species.
Which one of the following will accelerate phosphrus cycle?
Solution: weathering of rocks Rock is the natural reservoir of phosphorus, as phosphates; weathering dissolves small amounts into the soil solution, from which plant roots absorb them. ∴ weathering of rocks speeds up the phosphorus cycle; input through rainfall is very small and phosphorus has no significant gaseous exchange.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) Mendel's law of independent assortment does not hold good for the genes that are located closely on the same chromosome. Reason (R) Closely located genes assort independently. In the light of the above statements, choose the correct answer from the options given below.
Solution: Assertion (A) is correct but Reason (R) is not correct Assertion is correct: genes located close together on the same chromosome are linked, so the law of independent assortment does not hold for them. Reason is incorrect: closely located genes are inherited together; only genes far apart on a chromosome recombine often enough to appear to assort independently.
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as,
Solution: sequence annotation Sequencing the whole genome, coding and non-coding, and later assigning functions to its regions is called sequence annotation.
Given below are two statements. Statement I Autoimmune disorder is a condition where body defence mechanism recognises its own cells as foreign bodies. Statement II Rheumatoid arthritis is a condition where body does not attack self cells. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: in an autoimmune disorder the immune system fails to tell self from non-self and attacks the body's own cells. Statement II is incorrect: rheumatoid arthritis is an autoimmune disease, in which the body does attack its own cells (in the joints).
Given below are two statements. Statement I The coagulum is formed of network of threads called thrombins. Statement II Spleen is the graveyard of erythrocytes. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: Statement I is incorrect but Statement II is correct Statement I is incorrect: the coagulum (clot) is a network of threads called fibrins; thrombin is the enzyme that forms fibrin from fibrinogen. Statement II is correct: worn-out RBCs are destroyed in the spleen, the graveyard of erythrocytes.
Identify the asexual reproductive structure associated with $Penicillium$.
Solution: conidia Penicillium reproduces asexually by conidia, non-motile spores produced exogenously.
Given below are two statements. Statement I The release of sperms into the seminiferous tubules is called spermiation. Statement II Spermiogenesis is the process of formation of sperms from spermatogonia. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: spermiation is the release of sperms from the Sertoli cells into the lumen of the seminiferous tubules. Statement II is incorrect: spermiogenesis is the change of spermatids into sperms; forming sperms from spermatogonia is spermatogenesis.
Under normal physiological condition in human being every 100 mL of oxygenated blood can deliver ......... mL of O$_2$ to the tissues.
Solution: 5 mL 100 mL of oxygenated blood carries about 20 mL of O₂, and under normal physiological conditions about 5 mL of it is delivered to the tissues.
$In$-$situ$ conservation refers to
Solution: protect and conserve the whole ecosystem In situ (on site) conservation protects species in their natural habitat, so the whole ecosystem is conserved, as in national parks, wildlife sanctuaries, biosphere reserves and sacred groves.
Natural selection where more individuals acquire specific character value other than the mean character value, leads to
Solution: directional change When more individuals acquire a character value other than the mean, the distribution shifts one way: directional change (stabilisation favours the mean, disruption both ends).
Breeding crops with higher levels of vitamins and minerals or higher proteins and healthier fats is called
Solution: biofortification Breeding crops with higher levels of vitamins and minerals, or higher protein and healthier fats, is biofortification, e.g. Atlas 66, a high-protein wheat.
Which of the following is present between the adjacent bones of the vertebral column?
Solution: cartilage Cartilage, a flexible connective tissue, lies between adjacent bones of the vertebral column and cushions them.
Nitrogenous waste is excreted in the form of pellet or paste by
Solution: Pavo Pavo (peacock) is a bird, and birds are uricotelic: they excrete uric acid as a pellet or paste with very little loss of water.
Which of the following statements with respect to endoplasmic reticulum is incorrect?
Solution: in prokaryotes, only RER are present (the incorrect statement) Prokaryotes have no membrane-bound organelles, so they have no endoplasmic reticulum at all, rough or smooth. (A) Correct: SER has no ribosomes on its surface. (C) Correct: SER is the major site for lipid synthesis. (D) Correct: RER has ribosomes attached to its outer surface.
Which of the following statements are true for spermatogenesis but do not hold true for oogenesis? I. It results in the formation of haploid gametes. II. Differentiation of gamete occurs after the completion of meiosis. III. Meiosis occurs continuously in a mitotically dividing stem cell population. IV. It is controlled by the Luteinising hormone (LH) and Follicle Stimulating Hormone (FSH) secreted by the anterior pituitary. V. It is initiated at puberty. Choose the most appropriate answer from the options given below.
Solution: II, III and V II: spermatids formed after meiosis differentiate into sperms (spermiogenesis); in oogenesis meiosis II is completed only after the sperm enters the secondary oocyte. III: spermatogonia keep dividing by mitosis and entering meiosis throughout reproductive life; oogonia form only in the foetus and none are added after birth. V: spermatogenesis starts at puberty, whereas oogenesis begins during embryonic development. I and IV hold for both: each forms haploid gametes and is controlled by FSH and LH from the anterior pituitary.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) All vertebrates are chordates but all chordates are not vertebrates. Reason (R) Notochord is replaced by vertebral column in the adult vertebrates. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both A and R are correct and R is the correct explanation of A Assertion is correct: all chordates have a notochord at some stage, but only vertebrates have a vertebral column. Reason is correct and explains it: in vertebrates the embryonic notochord is replaced by a cartilaginous or bony vertebral column in the adult. In urochordates the notochord is present only in the larval tail and in cephalochordates it persists throughout life, so they are chordates but not vertebrates.
In which of the following animals, digestive tract has additional chambers like crop and gizzard?
Solution: Pavo, Psittacula, Corvus Crop and gizzard are additional chambers of the digestive tract of birds, and Pavo (peacock), Psittacula (parrot) and Corvus (crow) are all birds. (A) Bufo (toad), Balaenoptera (blue whale) and Bungarus (krait) are not birds. (B) Catla is a fish and Crocodylus a reptile, though Columba is a bird. (D) Chameleon is a reptile.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) Osteoporosis is characterised by decreased bone mass and increased chances of fractures. Reason (R) Common cause of osteoporosis is increased levels of oestrogen. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Assertion (A) is correct but Reason (R) is not correct Assertion is correct: osteoporosis is an age-related disorder with decreased bone mass and increased chances of fractures. Reason is incorrect: a common cause is a decreased, not increased, level of oestrogen.
In gene therapy of Adenosine Deaminase (ADA) deficiency, the patient requires periodic infusion of genetically engineered lymphocytes because
Solution: genetically engineered lymphocytes are not immortal cells In ADA gene therapy, lymphocytes from the patient's blood are grown in culture, given a functional ADA cDNA through a retroviral vector and returned to the patient. These cells are not immortal, so the patient needs periodic infusions of them. A permanent cure would need the ADA gene introduced into cells at early embryonic stages.
Select the incorrect statement with reference to mitosis.
Solution: spindle fibres attach to centromere of chromosomes (the incorrect statement) Spindle fibres attach to the kinetochores, small disc-shaped structures on the surface of the centromere, not to the centromere itself. (B) Correct: at telophase the chromosomes decondense and lose their individuality. (C) Correct: at anaphase the centromeres split and the chromatids separate. (D) Correct: at metaphase all the chromosomes lie at the equator (metaphase plate).
In the taxonomic categories, which hierarchical arrangement in ascending order is correct in case of animals ?
Solution: Kingdom, Phylum, Class, Order, Family, Genus, Species The taxonomic categories run kingdom, phylum, class, order, family, genus, species; only this sequence keeps the ranks in their correct order. (A), (B) and (C) put class or order above phylum.
Given below are two statements. Statement I Restriction endonucleases recognise specific sequence to cut DNA known as palindromic nucleotide sequence. Statement II Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both Statement I and Statement II are correct Statement I is correct: each restriction endonuclease recognises a specific palindromic nucleotide sequence. Statement II is correct: it cuts each strand a little away from the centre of the palindrome, between the same two bases on both strands, leaving sticky ends.
Given below are two statements. Statement I $Mycoplasma$ can pass through less than 1 micron filter size. Statement II $Mycoplasma$ are bacteria with cell wall. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but Statement II is incorrect Statement I is correct: Mycoplasma are the smallest living cells known, so they pass through filters finer than 1 micron (bacteria-proof filters). Statement II is incorrect: Mycoplasma completely lack a cell wall.
Which of the following is not a connective tissue?
Solution: neuroglia Neuroglia are supporting cells of neural tissue that protect and support the neurons; they are not connective tissue. Adipose tissue (loose connective), cartilage (specialised connective) and blood (fluid connective) are all connective tissues.
Lippes loop is a type of contraceptive used as
Solution: non-medicated IUD Lippes loop is a non-medicated IUD; CuT, Cu7 and Multiload 375 are copper-releasing IUDs, and Progestasert and LNG-20 are hormone-releasing IUDs.
At which stage of life, the oogenesis process is initiated?
Solution: embryonic development stage Oogenesis begins during embryonic development, when millions of oogonia form in each foetal ovary; no more are added after birth.
Identify the microorganism which is responsible for the production of an immunosuppressive molecule cyclosporin-A.
Solution: Trichoderma polysporum Cyclosporin A, an immunosuppressive agent used in organ transplant patients, is produced by the fungus Trichoderma polysporum.
Given below are two statements. Statement I Fatty acids and glycerols cannot be absorbed into the blood. Statement II Specialised lymphatic capillaries called lacteals carry chylomicrons into lymphatic vessels and ultimately into the blood. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both Statement I and Statement II are correct Statement I is correct: fatty acids and glycerol are insoluble, so they cannot be absorbed into the blood directly; they first form micelles that enter the intestinal mucosa. Statement II is correct: there they are re-formed into protein-coated fat globules, chylomicrons, which lacteals carry into the lymph and finally into the blood.
In an $E.coli$ strain, $i$ gene gets mutated and its product cannot bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome?
Solution: $z$, $y$, $a$ genes will not be translated The $i$ gene codes for the repressor, which binds the operator and stops RNA polymerase from transcribing the lac operon. Normally the inducer (allolactose, from lactose) binds and inactivates the repressor; the mutant repressor cannot bind the inducer, so it stays on the operator even with lactose present. ∴ $z$, $y$ and $a$ are not transcribed, so they are not translated either.
Tegmina in cockroach, arises from
Solution: mesothorax Tegmina, the opaque, dark, leathery forewings of cockroach, arise from the mesothorax; the hindwings arise from the metathorax.
Which of the following is not the function of conducting part of respiratory system?
Solution: provides surface for diffusion of O₂ and CO₂ (the incorrect statement) Diffusion of O₂ and CO₂ takes place in the exchange (respiratory) part, the alveoli and their ducts, not in the conducting part. The conducting part (external nostrils to terminal bronchioles) carries air to the alveoli, clears it of foreign particles, humidifies it and brings it to body temperature.
Which of the following is a correct match for disease and its symptoms?
Solution: arthritis – inflamed joints Arthritis is inflammation of the joints, so this pair is correctly matched. (A) Tetany is rapid spasms (wild contractions) in muscle due to low, not high, Ca²⁺ in body fluid. (B) Myasthenia gravis is an autoimmune disorder affecting the neuromuscular junction, not a genetic one. (C) Muscular dystrophy is a genetic disorder causing progressive degeneration of skeletal muscle, not an autoimmune one.
Regarding meiosis, which of the statements is incorrect?
Solution: DNA replication occurs in S-phase of meiosis-II (the incorrect statement) DNA replicates only once, in the S phase of the interphase before meiosis I; there is no S phase before meiosis II. (B) Correct: homologous chromosomes pair and recombine in prophase I of meiosis I. (C) Correct: meiosis II ends with four haploid cells. (D) Correct: meiosis has two sequential divisions, meiosis I and meiosis II.
Detritivores breakdown detritus into smaller particles. This process is called
Solution: fragmentation Detritivores such as earthworms break detritus down into smaller particles; this first step of decomposition is fragmentation.
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs?
Solution: $3.3 \times 10^{9}$ bp A human diploid cell has $6.6 \times 10^{9}$ bp, and at 0.34 nm per base pair this DNA is about 2.2 m long. 1.1 m is half of 2.2 m, so the number of base pairs $= \dfrac{6.6 \times 10^{9}}{2} = 3.3 \times 10^{9}\ \text{bp}$
Which of the following functions is not performed by secretions from salivary glands?
Solution: digestion of disaccharides (the incorrect statement) Salivary amylase breaks starch down to the disaccharide maltose; saliva has no enzyme that digests disaccharides. (A) Performed: salivary amylase hydrolyses about 30 per cent of starch, a complex carbohydrate, in the mouth. (B) Performed: the mucus in saliva lubricates the oral cavity and binds food into a bolus. (D) Performed: lysozyme in saliva acts as an antibacterial agent and prevents infections.
A dehydration reaction links two glucose molecules to produce maltose. If the formula for glucose is C$_6$H$_{12}$O$_6$ then what is the formula for maltose?
Solution: C₁₂H₂₂O₁₁ A dehydration reaction removes one water molecule as the two glucose units join by a glycosidic bond. C₆H₁₂O₆ + C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O
If '8' $Drosophila$ in a laboratory population of '80' died during a week, the death rate in the population is ............ individuals per $Drosophila$ per week.
Solution: 0.1 Death rate $= \dfrac{\text{number of deaths}}{\text{population size}}$ in a given time Death rate $= \dfrac{8}{80} = 0.1$ individuals per Drosophila per week
Given below are two statements. Statement I In a scrubber, the exhaust from the thermal plant is passed through the electric wires to charge the dust particles. Statement II Particulate matter (Pm 2.5) cannot be removed by scrubber but can be removed by an electrostatic precipitator. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is incorrect but Statement II is correct Statement I is incorrect: in a scrubber the exhaust passes through a spray of water or lime, which removes gases such as SO₂; charging dust particles with electrode wires happens in an electrostatic precipitator. Statement II is correct: a scrubber does not remove fine particulate matter, while an electrostatic precipitator removes over 99 per cent of it.
Which of the following is a correct statement?
Solution: cyanobacteria are autotrophs classified under kingdom Monera Cyanobacteria (blue-green algae) are photosynthetic autotrophs with chlorophyll a, and as prokaryotes they are placed in kingdom Monera. (A) Bacteria are not exclusively heterotrophic; some are photosynthetic or chemosynthetic autotrophs. (B) Slime moulds are saprophytic protists, classified under kingdom Protista, not Monera. (C) Mycoplasma have DNA and ribosomes but completely lack a cell wall.
Ten $E.coli$ cells with $^{15}$N-$ds$DNA are incubated in medium containing $^{14}$N nucleotide. After 60 minutes, how many $E.coli$ cells will have DNA totally free from $^{15}$N?
Solution: 60 cells E. coli divides every 20 minutes, so 60 minutes gives 3 generations and each cell gives $2^{3} = 8$ cells. Replication is semiconservative: the two original ¹⁵N strands stay in 2 of the 8 DNA molecules (hybrid ¹⁵N–¹⁴N), and the other 6 are fully ¹⁴N. ∴ cells with DNA free of ¹⁵N $= 10 \times 6 = 60$
Select the incorrect statement with respect to acquired immunity.
Solution: acquired immunity is a non-specific defence present at birth (the incorrect statement) Innate immunity is the non-specific defence present at birth; acquired immunity is pathogen-specific and develops after exposure. (A) Correct: a later encounter with the same pathogen gives a highly intensified secondary (anamnestic) response. (B) Correct: the anamnestic response is due to memory of the first encounter. (D) Correct: the first encounter with a pathogen produces a low-intensity primary response.
Statements related to human insulin are given below. Which statement(s) is/are correct about genetically engineered insulin? I. Pro-hormone insulin contains extra stretch of C-peptide. II. A-peptide and B-peptide chains of insulin were produced separately in $E.coli$, extracted and combined by creating disulphide bond between them. III. Insulin used for treating diabetes was extracted from cattles and pigs. IV. Pro-hormone insulin needs to be processed for converting into a mature and functional hormone. V. Some patients develop allergic reactions to the foreign insulin. Choose the most appropriate answer from the options given below.
Solution: only II II describes genetically engineered insulin: chains A and B were produced separately in E. coli, extracted and joined by disulphide bonds. I and IV describe natural insulin, made as a pro-hormone whose extra C-peptide is removed during processing; the engineered route makes the A and B chains directly. III and V describe the older insulin from cattle and pigs, to which some patients were allergic.
Which of the following are not the effects of parathyroid hormone? I. Stimulates the process of bone resorption. II. Decreases Ca$^{2+}$ level in blood. III. Reabsorption of Ca$^{2+}$ by renal tubules. IV. Decreases the absorption of Ca$^{2+}$ from digested food. V. Increases metabolism of carbohydrates. Choose the most appropriate answer from the options given below.
Solution: II, IV and V Parathyroid hormone raises blood Ca²⁺: it stimulates bone resorption (I), reabsorption of Ca²⁺ by the renal tubules (III) and absorption of Ca²⁺ from digested food. ∴ decreasing blood Ca²⁺ (II) and decreasing Ca²⁺ absorption from food (IV) are not its effects. Increasing the metabolism of carbohydrates (V) is an effect of thyroid hormones, not of parathyroid hormone.
Select the incorrect statement regarding synapses.
Solution: transmission across a chemical synapse is always faster than across an electrical one (the incorrect statement) Transmission across an electrical synapse is always faster than across a chemical synapse, where a neurotransmitter must be released and diffuse across the cleft. (A) Correct: at an electrical synapse current flows directly from one neuron into the other. (B) Correct: chemical synapses use neurotransmitters to carry the impulse across the synaptic cleft. (D) Correct: at an electrical synapse the pre- and post-synaptic membranes are in very close proximity.
Match List-I with List-II. List-I: A. Bronchioles B. Goblet cell C. Tendons D. Adipose tissue List-II: (i) Dense regular connective tissue (ii) Loose connective tissue (iii) Glandular tissue (iv) Ciliated epithelium Choose the correct answer from the options given below.
Solution: A-iv, B-iii, C-i, D-ii Bronchioles are lined by ciliated epithelium: A-iv. A goblet cell is a unicellular gland, so it is glandular tissue: B-iii. Tendons, which attach skeletal muscles to bones, are dense regular connective tissue: C-i. Adipose tissue, found mainly beneath the skin, is a loose connective tissue: D-ii.
The recombination frequency between the genes $a$ and $c$ is 5%, $b$ and $c$ is 15%, $b$ and $d$ is 9%, $a$ and $b$ is 20%, $c$ and $d$ is 24% and $a$ and $d$ is 29%. What will be the sequence of these genes on a linear chromosome?
Solution: $a$, $c$, $b$, $d$ Recombination frequency is proportional to the distance between genes, so distances along the chromosome must add up. $a$ to $c$ is 5% and $c$ to $b$ is 15%, giving $a$ to $b$ = 20%, as observed. $b$ to $d$ is 9%, giving $c$ to $d$ = 15% + 9% = 24% and $a$ to $d$ = 20% + 9% = 29%, both as observed. ∴ the gene order is $a$, $c$, $b$, $d$.
If a colourblind female marries a man whose mother was also colourblind, what are the chances of her progeny having colour blindness?
Solution: 100% Colour blindness is X-linked recessive; the colour-blind woman is $X^{c}X^{c}$. A son gets his X chromosome from his mother, so the man whose mother was colour-blind is $X^{c}Y$, i.e. colour-blind. $X^{c}X^{c} \times X^{c}Y$ gives only $X^{c}X^{c}$ daughters and $X^{c}Y$ sons, so all the children (100%) are colour-blind.
Match List-I with List-II. List-I (Biological molecules): A. Glycogen B. Globulin C. Steroids D. Thrombin List-II (Biological functions): (i) Hormone (ii) Biocatalyst (iii) Antibody (iv) Storage product Choose the correct answer from the options given below.
Solution: A-iv, B-iii, C-i, D-ii Glycogen is the storage polysaccharide of animals: A-iv. Antibodies are immunoglobulins (gamma globulins): B-iii. Steroids such as cortisol, testosterone and oestrogen act as hormones: C-i. Thrombin is an enzyme that converts fibrinogen to fibrin in clotting, i.e. a biocatalyst: D-ii.
Which of the following is not a desirable feature of a cloning vector?
Solution: presence of two or more recognition sites, which is not desirable A cloning vector should have a single recognition site for a given restriction enzyme; two or more sites would cut it into several fragments and complicate gene cloning. (A) Desirable: a selectable marker gene helps to pick out the transformants. (B) Desirable: a single restriction site lets the foreign DNA be inserted at one place. (D) Desirable: the vector needs an origin of replication (ori), which also controls its copy number.
Which of the following statements is not true?
Solution: flippers of penguins and dolphins are a pair of homologous organs (the incorrect statement) Flippers of penguins and dolphins do the same job but have different origins, so they are analogous, not homologous. (A) True: sweet potato (a root) and potato (a stem) are analogous storage organs. (B) True: homology indicates common ancestry (divergent evolution). (D) True: analogous structures result from convergent evolution.
Which one of the following statements is correct?
Solution: blood moves freely from atrium to the ventricle during joint diastole In joint diastole all four chambers are relaxed and the tricuspid and bicuspid valves are open, so blood flows freely from the atria into the ventricles. (A) The AV valves are already open during joint diastole; atrial contraction only increases the flow into the ventricles by about 30 per cent. (C) Rising ventricular pressure opens the semilunar valves; they close when ventricular pressure falls. (D) The SA node, not the AV node, generates the action potential that makes the atria contract.
Match List-I with List-II with respect to methods of contraception and their respective actions. List-I: A. Diaphragms B. Contraceptive pills C. Intra Uterine Devices D. Lactational amenorrhea List-II: (i) Inhibit ovulation and implantation (ii) Increase phagocytosis of sperm within uterus (iii) Absence of menstrual cycle and ovulation following parturition (iv) They cover the cervix blocking the entry of sperms Choose the correct answer from the options given below.
Solution: A-iv, B-i, C-ii, D-iii Diaphragms are rubber barriers that cover the cervix during coitus and block the entry of sperms: A-iv. Contraceptive pills inhibit ovulation and implantation: B-i. IUDs increase phagocytosis of sperms within the uterus: C-ii. Lactational amenorrhea is the absence of menstruation and ovulation during intense lactation after parturition: D-iii.
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