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Free, login-free worked solutions to every question in NEET (UG) 2023 — Code H1. Check your answers below, then predict your rank or practise with full mock tests.
The magnetic energy stored in an inductor of inductance 4 $\mu$H carrying a current of 2 A is
Solution: $8\ \mu\text{J}$ Energy stored in an inductor, $U = \tfrac{1}{2}Li^2$. ∴ $U = \tfrac{1}{2} \times (4 \times 10^{-6}) \times (2)^2\ \text{J} = 8 \times 10^{-6}\ \text{J} = 8\ \mu\text{J}$
A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?
Solution: Capacitor The capacitor charges as the rectified voltage rises and discharges slowly through the load as it falls, so it smooths out the ac ripple; it acts as a filter.
An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5}$ N C$^{-1}$. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.
Solution: $2\ \text{mC}$ Torque on a dipole, $\tau = pE\sin\theta$. $4 = p \times 2 \times 10^{5} \times \sin 30^\circ \Rightarrow p = 4 \times 10^{-5}\ \text{C m}$ ∴ $q = \dfrac{p}{l} = \dfrac{4 \times 10^{-5}}{2 \times 10^{-2}}\ \text{C} = 2 \times 10^{-3}\ \text{C} = 2\ \text{mC}$
The equivalent capacitance of the system shown in the following circuit is A 3 $\mu$F capacitor from terminal $A$ in series with a parallel pair of 3 $\mu$F capacitors, whose far side is terminal $B$.
Solution: $2\ \mu\text{F}$ The two $3\ \mu\text{F}$ capacitors in parallel give $C_p = 3 + 3 = 6\ \mu\text{F}$. This is in series with $3\ \mu\text{F}$, so $C_{\text{eq}} = \dfrac{3 \times 6}{3 + 6}\ \mu\text{F} = 2\ \mu\text{F}$.
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is
Solution: $4\lambda$ The shortest wavelength of a series is for the transition from $n = \infty$: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{\infty}\right) = \dfrac{R}{n_f^2}$. For the Balmer series ($n_f = 2$), $\lambda = \dfrac{4}{R}$. For the Brackett series ($n_f = 4$), $\lambda' = \dfrac{16}{R}$. ∴ $\lambda' = 4\lambda$
Resistance of a carbon resistor determined from colour codes is $(22000 \pm 5\%)\ \Omega$. The colour of third band must be
Solution: Orange $22000\ \Omega = 22 \times 10^{3}\ \Omega$: the first two bands give the digits 2 and 2, and the third band is the multiplier $10^{3}$. The multiplier $10^{3}$ (digit 3) is orange.
Two bodies of mass $m$ and $9m$ are placed at a distance $R$. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ($G$ = gravitational constant)
Solution: $-\dfrac{16Gm}{R}$ The field is zero at distance $x$ from $m$ where $\dfrac{Gm}{x^2} = \dfrac{G(9m)}{(R - x)^2}$, i.e. $R - x = 3x$, so $x = \dfrac{R}{4}$. Potential there, $V = -\dfrac{Gm}{R/4} - \dfrac{G(9m)}{3R/4} = -\dfrac{4Gm}{R} - \dfrac{12Gm}{R}$ ∴ $V = -\dfrac{16Gm}{R}$
If the galvanometer $G$ does not show any deflection in the circuit shown, the value of $R$ is given by The circuit is a 10 V cell in series with a 400 $\Omega$ resistor forming the left loop; $R$ is the middle branch between the two loops; and a 2 V cell in series with the galvanometer $G$ forms the right loop.
Solution: $100\ \Omega$ No deflection means no current in the branch with the 2 V cell, so the potential difference across $R$ is 2 V. The same current flows through $400\ \Omega$ and $R$: $I = \dfrac{10 - 2}{400}\ \text{A} = 0.02\ \text{A}$ ∴ $R = \dfrac{2}{0.02}\ \Omega = 100\ \Omega$
For Young's double slit experiment, two statements are given below: Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant. Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases. In the light of the above statements, choose the correct answer from the options given below:
Solution: Statement I is true but Statement II is false Angular separation of fringes, $\theta = \dfrac{\lambda}{d}$. Statement I is true: $\theta$ does not depend on the distance $D$ of the screen. Statement II is false: $\theta \propto \lambda$, so a larger wavelength increases the angular separation.
The venturi-meter works on
Solution: Bernoulli's principle At the narrow throat the fluid speeds up and its pressure drops; by Bernoulli's principle this pressure difference gives the rate of flow.
Let a wire be suspended from the ceiling (rigid support) and stretched by a weight $W$ attached at its free end. The longitudinal stress at any point of cross-sectional area $A$ of the wire is
Solution: $\dfrac{W}{A}$ Longitudinal stress = restoring force per unit area; neglecting the wire's own weight, the restoring force at every cross-section equals $W$. ∴ Stress $= \dfrac{W}{A}$
The angular acceleration of a body, moving along the circumference of a circle, is
Solution: Along the axis of rotation Angular velocity and angular acceleration are axial vectors; they point along the axis of rotation.
The temperature of a gas is $-50^{\circ}$C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?
Solution: $3295^\circ\text{C}$ $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$, so $v_{\text{rms}} \propto \sqrt{T}$. Increased by 3 times means the new speed is $v + 3v = 4v$, so $T_2 = 4^2\,T_1 = 16\,T_1$. $T_1 = 273 - 50 = 223\ \text{K} \Rightarrow T_2 = 16 \times 223\ \text{K} = 3568\ \text{K}$ ∴ $T_2 = 3568 - 273 = 3295^\circ\text{C}$
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
Solution: Cs only Photoelectrons are emitted only if the photon energy is at least equal to the work function. 2.20 eV is more than the work function of Cs (2.14 eV) but less than that of K (2.30 eV) and Na (2.75 eV).
An ac source is connected to a capacitor $C$. Due to decrease in its operating frequency
Solution: Displacement current decreases Capacitive reactance, $X_C = \dfrac{1}{\omega C}$, increases as $\omega$ decreases, so the current $I = \dfrac{V}{X_C}$ decreases. The displacement current between the plates equals the conduction current in the wires, so it also decreases.
Light travels a distance $x$ in time $t_1$ in air and $10x$ in time $t_2$ in another denser medium. What is the critical angle for this medium?
Solution: $\sin^{-1}\left(\dfrac{10\,t_1}{t_2}\right)$ At the critical angle, $\sin i_c = \dfrac{1}{\mu} = \dfrac{v_{\text{medium}}}{v_{\text{air}}}$. $v_{\text{air}} = \dfrac{x}{t_1}$ and $v_{\text{medium}} = \dfrac{10x}{t_2}$ ∴ $\sin i_c = \dfrac{10x/t_2}{x/t_1} = \dfrac{10\,t_1}{t_2}$, so $i_c = \sin^{-1}\left(\dfrac{10\,t_1}{t_2}\right)$
The net magnetic flux through any closed surface is
Solution: Zero Gauss's law for magnetism, $\oint \vec{B} \cdot d\vec{S} = 0$: magnetic monopoles do not exist, so every field line entering a closed surface also leaves it.
A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?
Solution: $0.27\ \text{A}$ For an ideal transformer, input power = output power = 60 W. ∴ $I_p = \dfrac{60}{220}\ \text{A} \approx 0.27\ \text{A}$
A metal wire has mass $(0.4 \pm 0.002)$ g, radius $(0.3 \pm 0.001)$ mm and length $(5 \pm 0.02)$ cm. The maximum possible percentage error in the measurement of density will nearly be
Solution: 1.6% Density, $\rho = \dfrac{M}{\pi r^2 l}$, so $\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta M}{M} + 2\dfrac{\Delta r}{r} + \dfrac{\Delta l}{l}$. $\dfrac{\Delta\rho}{\rho} = \dfrac{0.002}{0.4} + \dfrac{2 \times 0.001}{0.3} + \dfrac{0.02}{5} = 0.005 + 0.0067 + 0.004 = 0.0157$ ∴ Maximum percentage error $\approx 1.6\%$
Given below are two statements: Statement I: Photovoltaic devices can convert optical radiation into electricity. Statement II: Zener diode is designed to operate under reverse bias in breakdown region. 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: a photovoltaic device, such as a solar cell, converts optical radiation into electricity. Statement II is correct: a Zener diode is designed to work in reverse bias in the breakdown region, which is why it is used as a voltage regulator.
A bullet is fired from a gun at the speed of 280 m s$^{-1}$ in the direction $30^{\circ}$ above the horizontal. The maximum height attained by the bullet is ($g = 9.8$ m s$^{-2}$, $\sin 30^{\circ} = 0.5$)
Solution: $1000\ \text{m}$ Maximum height of a projectile, $H = \dfrac{u^2\sin^2\theta}{2g}$. ∴ $H = \dfrac{(280)^2 \times (0.5)^2}{2 \times 9.8}\ \text{m} = 1000\ \text{m}$
A Carnot engine has an efficiency of 50% when its source is at a temperature 327$^{\circ}$C. The temperature of the sink is
Solution: $27^\circ\text{C}$ Efficiency of a Carnot engine, $\eta = 1 - \dfrac{T_2}{T_1}$, with $T_1 = 327 + 273 = 600\ \text{K}$. $0.5 = 1 - \dfrac{T_2}{600} \Rightarrow T_2 = 300\ \text{K}$ ∴ Sink temperature $= 300 - 273 = 27^\circ\text{C}$
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03 N m$^{-1}$)
Solution: $3.01 \times 10^{-4}\ \text{J}$ A soap bubble has two surfaces, so energy $= S \times 2 \times 4\pi r^2$. ∴ $E = 0.03 \times 2 \times 4\pi \times (2 \times 10^{-2})^2\ \text{J} \approx 3.01 \times 10^{-4}\ \text{J}$
The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to
Solution: $\dfrac{1}{V}$ The shortest wavelength is emitted when the whole kinetic energy $eV$ of an electron goes into one photon: $eV = \dfrac{hc}{\lambda_{\min}}$. ∴ $\lambda_{\min} = \dfrac{hc}{eV} \propto \dfrac{1}{V}$
In a series $LCR$ circuit, the inductance $L$ is 10 mH, capacitance $C$ is 1 $\mu$F and resistance $R$ is 100 $\Omega$. The frequency at which resonance occurs is
Solution: $1.59\ \text{kHz}$ Resonant frequency, $f = \dfrac{1}{2\pi\sqrt{LC}}$. ∴ $f = \dfrac{1}{2\pi\sqrt{10 \times 10^{-3} \times 1 \times 10^{-6}}}\ \text{Hz} = \dfrac{10^{4}}{2\pi}\ \text{Hz} \approx 1.59\ \text{kHz}$
The potential energy of a long spring when stretched by 2 cm is $U$. If the spring is stretched by 8 cm, potential energy stored in it will be
Solution: $16U$ Potential energy of a stretched spring, $U = \tfrac{1}{2}kx^2$, so $U \propto x^2$. ∴ $\dfrac{U'}{U} = \left(\dfrac{8}{2}\right)^2 = 16$, i.e. $U' = 16U$
The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\dfrac{1}{16}\right)^{\text{th}}$ of its initial value?
Solution: 80 minutes After $n$ half-lives the activity is $\dfrac{A_0}{2^n}$; $\dfrac{1}{16} = \dfrac{1}{2^4}$ gives $n = 4$. ∴ $t = 4 \times 20\ \text{min} = 80\ \text{min}$
If $\displaystyle\oint_s \vec{E} \cdot \overrightarrow{dS} = 0$ over a surface, then
Solution: The number of flux lines entering the surface equals the number leaving it Zero net flux, $\oint \vec{E} \cdot d\vec{S} = 0$, means as many field lines leave the surface as enter it. (A) Zero flux only means the net charge enclosed is zero; charges may lie outside the surface. (B) The field inside need not be uniform; only the net flux through the surface is zero. (D) The field can vary over the surface; only its surface integral is zero.
The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is
Solution: 2 : 1 Fundamental frequency of an open pipe, $f_o = \dfrac{v}{2l}$; of a closed pipe, $f_c = \dfrac{v}{4l}$. ∴ $f_o : f_c = 2 : 1$
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are
Solution: Random errors Errors that come from unpredictable fluctuations in experimental conditions, such as temperature or voltage supply, have no systematic cause and are called random errors.
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10}$ Hz and amplitude 48 V m$^{-1}$. Then the amplitude of oscillating magnetic field is (Speed of light in free space $= 3 \times 10^{8}$ m s$^{-1}$)
Solution: $1.6 \times 10^{-7}\ \text{T}$ In an electromagnetic wave, $\dfrac{E_0}{B_0} = c$. ∴ $B_0 = \dfrac{48}{3 \times 10^{8}}\ \text{T} = 1.6 \times 10^{-7}\ \text{T}$
A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is
Solution: Along north-east Force acts along the change in momentum, $\Delta\vec{p} = m(\vec{v}_f - \vec{v}_i)$. Final velocity (east) minus initial velocity (south) is east plus north, two vectors of equal size. ∴ The force acts along north-east.
The magnitude and direction of the current in the following circuit is The single loop $ABCD$ carries, along the top from $A$ to $B$: a 2 $\Omega$ resistor, then a 10 V cell, then a 5 V cell connected in opposition, then a 1 $\Omega$ resistor. The bottom branch from $D$ to $C$ carries a 7 $\Omega$ resistor. The point between the two cells is marked $E$.
Solution: 0.5 A from $A$ to $B$ through $E$ The two cells oppose each other, so the net emf is $10 - 5 = 5\ \text{V}$; the total resistance is $2 + 1 + 7 = 10\ \Omega$. $I = \dfrac{5}{10}\ \text{A} = 0.5\ \text{A}$ The current flows in the direction set by the larger (10 V) cell, from $A$ to $B$ through $E$.
A vehicle travels half the distance with speed $v$ and the remaining distance with speed $2v$. Its average speed is
Solution: $\dfrac{4v}{3}$ Average speed = total distance / total time. With each half of length $s$: $t = \dfrac{s}{v} + \dfrac{s}{2v} = \dfrac{3s}{2v}$ ∴ $v_{\text{av}} = \dfrac{2s}{3s/2v} = \dfrac{4v}{3}$
The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{-11}$ m. What is the radius of third allowed orbit of hydrogen atom?
Solution: 4.77 Å Radius of the $n$th Bohr orbit, $r_n \propto n^2$. ∴ $r_3 = 3^2 \times 5.3 \times 10^{-11}\ \text{m} = 4.77 \times 10^{-10}\ \text{m} = 4.77$ Å
10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is
Solution: 100 In series, $R_s = 10R$, so $I = \dfrac{E}{10R}$. In parallel, $R_p = \dfrac{R}{10}$, so $I' = \dfrac{10E}{R}$. ∴ $n = \dfrac{I'}{I} = 100$
The $x$-$t$ graph of a particle performing simple harmonic motion is shown in the figure. It is a sine curve starting from the origin, reaching a maximum displacement of 1 m, crossing zero at $t = 4$ s and completing one full cycle at $t = 8$ s. The acceleration of the particle at $t = 2$ s is
Solution: $-\dfrac{\pi^{2}}{16}\ \text{m s}^{-2}$ From the graph, $A = 1\ \text{m}$ and $T = 8\ \text{s}$, so $\omega = \dfrac{2\pi}{8} = \dfrac{\pi}{4}\ \text{rad s}^{-1}$ and $x = \sin\dfrac{\pi t}{4}$. In SHM, $a = -\omega^2 x$. At $t = 2\ \text{s}$, $x = \sin\dfrac{\pi}{2} = 1\ \text{m}$. ∴ $a = -\dfrac{\pi^2}{16} \times 1 = -\dfrac{\pi^2}{16}\ \text{m s}^{-2}$
Two thin lenses are of same focal lengths ($f$), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be
Solution: Infinite For thin lenses in contact, $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{1}{f} - \dfrac{1}{f} = 0$. ∴ $F$ is infinite; the combination behaves like a plane glass plate.
A satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\dfrac{3\pi}{Gd}$ represents
Solution: $T^{2}$ For a satellite just above the surface, gravity provides the centripetal force: $\dfrac{GMm}{R^2} = mR\left(\dfrac{2\pi}{T}\right)^2 \Rightarrow T^2 = \dfrac{4\pi^2R^3}{GM}$. Mass of the earth, $M = \tfrac{4}{3}\pi R^3 d$. ∴ $T^2 = \dfrac{4\pi^2R^3}{G \cdot \tfrac{4}{3}\pi R^3 d} = \dfrac{3\pi}{Gd}$
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 ($g = 10$ m s$^{-2}$).
Solution: $1.5\ \text{m s}^{-2}$ The body stays at rest on the floor as long as static friction can supply the force $ma$: $ma_{\max} = \mu_s mg$. ∴ $a_{\max} = \mu_s g = 0.15 \times 10\ \text{m s}^{-2} = 1.5\ \text{m s}^{-2}$
An electric dipole is placed as shown in the figure: a charge $-q$ and a charge $+q$ lie on a line, each 3 cm from the midpoint $O$, and the point $P$ lies on the same line at 5 cm from $O$, on the $+q$ side. The electric potential (in $10^{2}$ V) at point $P$ due to the dipole is ($\varepsilon_0$ = permittivity of free space and $\dfrac{1}{4\pi\varepsilon_0} = K$)
Solution: $\left(\dfrac{3}{8}\right)qK$ Potential due to a point charge, $V = \dfrac{Kq}{r}$; $P$ is 2 cm from $+q$ and 8 cm from $-q$. $V = \dfrac{Kq}{2 \times 10^{-2}} - \dfrac{Kq}{8 \times 10^{-2}} = Kq \times 10^{2}\left(\dfrac{1}{2} - \dfrac{1}{8}\right)$ ∴ $V = \dfrac{3}{8}qK \times 10^{2}\ \text{V}$, i.e. $\left(\dfrac{3}{8}\right)qK$ in units of $10^{2}$ V
The net impedance of circuit (as shown in figure) will be An inductor of $\dfrac{50}{\pi}$ mH, a capacitor of $\dfrac{10^{3}}{\pi}$ $\mu$F and a 10 $\Omega$ resistor are connected in series across a 220 V, 50 Hz source.
Solution: $5\sqrt{5}\ \Omega$ Impedance of a series LCR circuit, $Z = \sqrt{R^2 + (X_C - X_L)^2}$. $X_L = 2\pi fL = 2\pi \times 50 \times \dfrac{50}{\pi} \times 10^{-3}\ \Omega = 5\ \Omega$ $X_C = \dfrac{1}{2\pi fC} = \dfrac{1}{2\pi \times 50 \times \dfrac{10^{3}}{\pi} \times 10^{-6}}\ \Omega = 10\ \Omega$ ∴ $Z = \sqrt{10^2 + (10 - 5)^2}\ \Omega = \sqrt{125}\ \Omega = 5\sqrt{5}\ \Omega$
For the following logic circuit, the truth table is Input $A$ passes through a NOT gate and input $B$ passes through a NOT gate; the two inverted signals then feed a NAND gate whose output is $Y$.
Solution: $Y$ = 0, 1, 1, 1 for $AB$ = 00, 01, 10, 11 (an OR gate) The NOT gates give $\overline{A}$ and $\overline{B}$; the NAND gate then gives $Y = \overline{\overline{A} \cdot \overline{B}}$. By De Morgan's theorem, $Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$, an OR gate. ∴ $Y = 0$ only when $A = B = 0$; otherwise $Y = 1$.
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)? A biconvex lens of refractive index $n_1 = 1.5$ with $R_1 = R_2 = 20$ cm is embedded in a medium of refractive index $n_2 = 1.6$, which forms the two outer layers.
Solution: $-100\ \text{cm}$ Thin lenses in contact: $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \dfrac{1}{f_3}$, each from $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$. Outer layers (plano-concave, $\mu = 1.6$): $\dfrac{1}{f_1} = \dfrac{1}{f_3} = (1.6 - 1)\left(-\dfrac{1}{20}\right) = -\dfrac{0.6}{20}\ \text{cm}^{-1}$ Middle biconvex lens ($\mu = 1.5$): $\dfrac{1}{f_2} = (1.5 - 1)\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = \dfrac{0.5}{10}\ \text{cm}^{-1}$ $\dfrac{1}{F} = -\dfrac{0.6}{20} + \dfrac{0.5}{10} - \dfrac{0.6}{20} = -\dfrac{0.1}{10} = -\dfrac{1}{100}\ \text{cm}^{-1}$ ∴ $F = -100\ \text{cm}$
A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes $\dfrac{u}{3}$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is
Solution: 27 cm Uniform retardation: $v^2 = u^2 - 2as$. Over the first 24 cm, $\dfrac{u^2}{9} = u^2 - 2a(24) \Rightarrow a = \dfrac{u^2}{54}$ Over the whole length $L$ the bullet comes to rest: $0 = u^2 - 2aL \Rightarrow L = \dfrac{u^2}{2a}$ ∴ $L = \dfrac{54}{2}\ \text{cm} = 27\ \text{cm}$
The resistance of platinum wire at 0$^{\circ}$C is 2 $\Omega$ and 6.8 $\Omega$ at 80$^{\circ}$C. The temperature coefficient of resistance of the wire is
Solution: $3 \times 10^{-2}\ ^\circ\text{C}^{-1}$ Variation of resistance with temperature, $R_t = R_0(1 + \alpha t)$. $6.8 = 2(1 + 80\alpha) \Rightarrow 80\alpha = 3.4 - 1 = 2.4$ ∴ $\alpha = \dfrac{2.4}{80} = 3 \times 10^{-2}\ ^\circ\text{C}^{-1}$
A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s$^{-1}$. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take $g = 10$ m s$^{-2}$)
Solution: 64 m Taking upward as positive, displacement $s = ut - \tfrac{1}{2}gt^2$. $s = 4 \times 4 - \tfrac{1}{2} \times 10 \times 4^2 = 16 - 80 = -64\ \text{m}$ ∴ The water surface is 64 m below the point of projection, so the bridge is 64 m high.
A very long conducting wire is bent in a semi-circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by The wire runs in from the left, carries current $i$ to $A$, turns through a semicircle of radius $R$ centred on $P$ down to $B$, and then runs back out to the left.
Solution: $\dfrac{\mu_0 i}{4R}\left[1 - \dfrac{2}{\pi}\right]$ pointed away from the page Each long straight part gives $\dfrac{\mu_0 i}{4\pi R}$ at $P$, into the page; together $\dfrac{\mu_0 i}{2\pi R}$ into the page. The semicircle gives $\dfrac{\mu_0 i}{4R}$ out of the page. ∴ $B = \dfrac{\mu_0 i}{4R} - \dfrac{\mu_0 i}{2\pi R} = \dfrac{\mu_0 i}{4R}\left[1 - \dfrac{2}{\pi}\right]$, pointing away from the page since $\dfrac{1}{4} > \dfrac{1}{2\pi}$
A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})$ T. The magnitude of the magnetic force acting on the wire is
Solution: $5\,IL$ Force on a current-carrying wire, $\vec{F} = I\vec{L} \times \vec{B}$. $\vec{F} = IL\,\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k}) = IL(3\hat{k} + 4\hat{j})$ ∴ $|\vec{F}| = IL\sqrt{3^2 + 4^2} = 5\,IL$
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : A reaction can have zero activation energy. Reason R : The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy. In the light of the above statements, choose the correct answer from the options given below :
Solution: Both A and R are true and R is NOT the correct explanation of A A is true: some reactions, such as the combination of free radicals, have zero activation energy. R is true: activation energy is the extra energy reactant molecules must absorb so that their energy reaches the threshold value. R only defines activation energy; it does not explain why it can be zero, so R is not the correct explanation of A.
The correct order of energies of molecular orbitals of N$_2$ molecule, is
Solution: $\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma^{*}2p_z$ In N₂ (as in B₂ and C₂), s–p mixing raises $\sigma 2p_z$ above the two $\pi 2p$ orbitals. Only from O₂ onwards does $\sigma 2p_z$ lie below $\pi 2p_x = \pi 2p_y$.
The given compound C$_6$H$_5$–CH=CH–CH(X)–CH$_2$–CH$_3$ is an example of ______.
Solution: Allylic halide X is on an sp³ carbon next to the C=C double bond, which makes it an allylic halide. In a vinylic halide X is on the doubly bonded carbon itself, in a benzylic halide on the carbon attached to the ring, and in an aryl halide on the ring carbon.
Homoleptic complex from the following complexes is
Solution: Potassium trioxalatoaluminate (III) A homoleptic complex has only one kind of ligand; in K₃[Al(C₂O₄)₃] aluminium is bound only to oxalate. (A) [Co(NH₃)₅(CO₃)]Cl has NH₃ and CO₃²⁻ ligands: heteroleptic. (B) [Cr(NH₃)₃(H₂O)₃]Cl₃ has NH₃ and H₂O ligands: heteroleptic. (D) [Pt(NH₃)₂Cl(NO₂)] has NH₃, Cl⁻ and NO₂⁻ ligands: heteroleptic.
The right option for the mass of CO$_2$ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40) $$\text{CaCO}_3 \xrightarrow{\ 1200\ \text{K}\ } \text{CaO} + \text{CO}_2$$
Solution: 1.76 g CaCO₃ → CaO + CO₂: 100 g of CaCO₃ gives 44 g of CO₂. Mass of pure CaCO₃ $= 20 \times \dfrac{20}{100} = 4\ \text{g}$ ∴ Mass of CO₂ $= \dfrac{44}{100} \times 4\ \text{g} = 1.76\ \text{g}$
The relation between $n_m$, ($n_m$ = the number of permissible values of magnetic quantum number ($m$)) for a given value of azimuthal quantum number ($l$), is
Solution: $l = \dfrac{n_m - 1}{2}$ For a given $l$, $m$ takes the values $-l, \ldots, 0, \ldots, +l$, so $n_m = 2l + 1$. ∴ $l = \dfrac{n_m - 1}{2}$
Select the correct statements from the following A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is $9.10939 \times 10^{-31}$ kg. C. All the isotopes of a given element show same chemical properties. D. Protons and electrons are collectively known as nucleons. E. Dalton's atomic theory, regarded the atom as an ultimate particles of matter Choose the correct answer from the options given below
Solution: B, C and E only B, C and E are correct: the electron's mass is $9.10939 \times 10^{-31}$ kg, isotopes have the same chemical properties, and Dalton regarded the atom as the ultimate particle of matter. Statement A is wrong: atoms are made of three fundamental particles, electrons, protons and neutrons (only ordinary hydrogen, protium, has no neutron). Statement D is wrong: nucleons are the protons and neutrons in the nucleus, not protons and electrons.
Which one is an example of heterogenous catalysis?
Solution: Combination of dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron N₂(g) + 3H₂(g) → 2NH₃(g) over solid iron: the catalyst is in a different phase from the gaseous reactants, so the catalysis is heterogeneous. (A) NO and O₃ are both gases: homogeneous catalysis. (C) Oxides of nitrogen, SO₂ and O₂ are all gases: homogeneous catalysis. (D) H⁺ ions and sugar are in the same aqueous solution: homogeneous catalysis.
Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is :
Solution: 32 g Decarboxylation with soda lime: CH₃COONa + NaOH → CH₄ + Na₂CO₃ (in the presence of CaO). ∴ Mass of 2 mol of CH₄ $= 2 \times 16\ \text{g} = 32\ \text{g}$
The number of $\sigma$ bonds, $\pi$ bonds and lone pair of electrons in pyridine, respectively are:
Solution: 11, 3, 1 Pyridine, C₅H₅N, has six ring σ bonds and five C–H σ bonds: 11 σ bonds. The aromatic ring has three π bonds, and nitrogen has one lone pair (in an sp² orbital in the plane of the ring).
Which amongst the following molecules on polymerization produces neoprene?
Solution: H₂C=C(Cl)–CH=CH₂ (chloroprene) Neoprene is formed by free radical polymerisation of chloroprene, CH₂=C(Cl)–CH=CH₂ (2-chloro-1,3-butadiene).
The stability of Cu$^{2+}$ is more than Cu$^{+}$ salts in aqueous solution due to
Solution: Hydration energy The much more negative hydration enthalpy of Cu²⁺(aq) than Cu⁺(aq) more than compensates for the second ionisation enthalpy of copper.
Complete the following reaction Cyclohexanone [A] reacts with HCN to give the cyanohydrin [B], which on treatment with conc. H$_2$SO$_4$ and heat gives [C]. [C] is ________
Solution: Cyclohex-1-ene-1-carboxylic acid HCN adds to the C=O of cyclohexanone to give the cyanohydrin, with –OH and –CN on the same ring carbon. Hot conc. H₂SO₄ removes water from that carbon and its neighbour, giving a ring C=C next to the –CN group. The –CN group is hydrolysed in the acid to –COOH, so [C] is cyclohex-1-ene-1-carboxylic acid.
Which of the following reactions will NOT give primary amine as the product?
Solution: CH₃NC with (i) LiAlH₄, (ii) H₃O⁺ Reduction of methyl isocyanide gives a secondary amine, CH₃NHCH₃, because the nitrogen already carries a methyl group. (B) LiAlH₄ reduces CH₃CONH₂ to ethanamine, CH₃CH₂NH₂: a primary amine. (C) Hoffmann bromamide degradation of CH₃CONH₂ gives methanamine, CH₃NH₂: a primary amine. (D) LiAlH₄ reduces CH₃CN to ethanamine, CH₃CH₂NH₂: a primary amine.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Helium is used to dilute oxygen in diving apparatus. Reason R : Helium has high solubility in O$_2$. In the light of the above statements, choose the correct answer from the options given below
Solution: Both A and R are true and R is NOT the correct explanation of A A is true: helium is used to dilute oxygen in diving apparatus because of its very low solubility in blood. R is true: like any gas, helium mixes with oxygen in all proportions. Mixing with oxygen is not why helium is used; the reason is its low solubility in blood, so R is not the correct explanation of A.
Identify the product in the following reaction: Benzenediazonium chloride, C$_6$H$_5$N$_2^{+}$Cl$^{-}$, is treated with (i) Cu$_2$Br$_2$/HBr, then (ii) Mg/dry ether, then (iii) H$_2$O.
Solution: Benzene (i) Sandmeyer reaction: Cu₂Br₂/HBr replaces –N₂⁺Cl⁻ by –Br, giving bromobenzene. (ii) Mg in dry ether converts bromobenzene into phenylmagnesium bromide, C₆H₅MgBr. (iii) Water hydrolyses the Grignard reagent: C₆H₅MgBr + H₂O → C₆H₆ + Mg(OH)Br, so the product is benzene.
Consider the following reaction and identify the product (P). 3-Methylbutan-2-ol, CH$_3$–CH(CH$_3$)–CH(OH)–CH$_3$, is treated with HBr to give Product (P).
Solution: 2-Bromo-2-methylbutane HBr protonates the –OH; loss of water gives a secondary carbocation at C-2. A hydride shift from the neighbouring tertiary carbon gives the more stable tertiary carbocation, (CH₃)₂C⁺–CH₂CH₃. Br⁻ attacks this carbocation, giving 2-bromo-2-methylbutane.
Identify product (A) in the following reaction: A compound carrying an acetyl group (–COCH$_3$) on a cyclohexane ring and a second acetyl group on an attached benzene ring is treated with Zn–Hg / conc. HCl to give (A) + 2H$_2$O.
Solution: Both acetyl groups reduced to ethyl, –CH₂CH₃ Zn–Hg with conc. HCl (Clemmensen reduction) reduces each C=O group to CH₂. Each acetyl group, –COCH₃, becomes an ethyl group, –CH₂CH₃; the two C=O groups release 2H₂O, as the equation shows.
Taking stability as the factor, which one of the following represents correct relationship?
Solution: TlI > TlI₃ Down group 13 the inert pair effect makes the +1 state more stable than +3 for thallium, so TlI is more stable than TlI₃. (A) For aluminium the +3 state is the stable one, so AlCl₃ is more stable than AlCl. (C) For thallium +1 is more stable than +3, so TlCl is more stable than TlCl₃. (D) Indium shows both +1 and +3 states; the inert pair effect makes +1 clearly the more stable state only for thallium.
The conductivity of centimolar solution of KCl at 25$^{\circ}$C is 0.0210 ohm$^{-1}$ cm$^{-1}$ and the resistance of the cell containing the solution at 25$^{\circ}$C is 60 ohm. The value of cell constant is
Solution: $1.26\ \text{cm}^{-1}$ Conductivity, $\kappa = \dfrac{G^*}{R}$, so cell constant $G^* = \kappa R$. ∴ $G^* = 0.0210 \times 60\ \text{cm}^{-1} = 1.26\ \text{cm}^{-1}$
Which one of the following statements is correct?
Solution: The daily requirement of Mg and Ca in the human body is estimated to be 0.2–0.3 g The daily requirement of Mg and Ca in the human body is estimated at 200–300 mg, i.e. 0.2–0.3 g. (A) Bone is not inert; it is continuously dissolved and redeposited, about 400 mg a day in man. (B) It is calcium, not magnesium, that plays roles in neuromuscular function and interneuronal transmission. (D) Enzymes that use ATP in phosphate transfer need magnesium, not calcium, as the cofactor.
Match List-I with List-II. List-I: A. Coke B. Diamond C. Fullerene D. Graphite List-II: I. Carbon atoms are $sp^{3}$ hybridised II. Used as a dry lubricant III. Used as a reducing agent IV. Cage like molecules Choose the correct answer from the options given below :
Solution: A-III, B-I, C-IV, D-II A-III: coke is used as a reducing agent in metallurgy. B-I: in diamond every carbon is sp³ hybridised and bonded tetrahedrally to four others. C-IV: fullerenes are cage-like molecules made of five- and six-membered rings. D-II: graphite is soft and slippery, so it is used as a dry lubricant.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic. Reason R : The deep blue solution is due to the formation of amide. In the light of the above statements, choose the correct answer from the options given below :
Solution: A is true but R is false A is true: the blue colour comes from ammoniated electrons, which absorb visible light, and their unpaired spins make the solution paramagnetic. R is false: the colour is not due to amide; amide forms only when the solution stands and slowly liberates hydrogen.
Which of the following statements are NOT correct? A. Hydrogen is used to reduce heavy metal oxides to metals. B. Heavy water is used to study reaction mechanism. C. Hydrogen is used to make saturated fats from oils. D. The H–H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any elements. E. Hydrogen reduces oxides of metals that are more active than iron. Choose the most appropriate answer from the options given below:
Solution: D, E only (the incorrect statements) Statement D is incorrect: the H–H bond dissociation enthalpy is the highest for a single bond between two atoms of any element. Statement E is incorrect: hydrogen reduces oxides of metals less active than iron, not more active. Statements A, B and C are correct: hydrogen reduces heavy metal oxides, heavy water is used to study reaction mechanisms, and hydrogen converts oils into saturated fats.
The element expected to form largest ion to achieve the nearest noble gas configuration is
Solution: N To reach the neon configuration N forms N³⁻, O forms O²⁻, F forms F⁻ and Na forms Na⁺; all have 10 electrons. In an isoelectronic series the ion with the fewest protons is the largest, so N³⁻ (7 protons) is the largest ion.
In Lassaigne's extract of an organic compound, both nitrogen and sulphur are present, which gives blood red colour with Fe$^{3+}$ due to the formation of
Solution: [Fe(SCN)]²⁺ With both N and S present, sodium fusion gives NaSCN; Fe³⁺ + SCN⁻ → [Fe(SCN)]²⁺, which is blood red.
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include : A. dipole - dipole forces B. dipole - induced dipole forces C. hydrogen bonding D. covalent bonding E. dispersion forces Choose the most appropriate answer from the options given below :
Solution: A, B, C, E are correct Dipole–dipole forces, dipole–induced dipole forces, hydrogen bonding and dispersion forces all act between molecules. Covalent bonding (D) holds atoms together within a molecule, so it is not an intermolecular force.
Amongst the following the total number of species NOT having eight electrons around central atom in its outermost shell, is NH$_3$, AlCl$_3$, BeCl$_2$, CCl$_4$, PCl$_5$ :
Solution: 3 Electrons around the central atom: NH₃ 8, CCl₄ 8, AlCl₃ 6, BeCl₂ 4, PCl₅ 10. ∴ AlCl₃, BeCl₂ and PCl₅ do not have eight: 3 species.
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A$_x$B$_y$, then the value of $x + y$ is in option
Solution: 5 In ccp, $N$ atoms of B give $2N$ tetrahedral voids; A fills one-third of them, i.e. $\dfrac{2N}{3}$. A : B $= \dfrac{2N}{3} : N = 2 : 3$, so the formula is A₂B₃. ∴ $x + y = 2 + 3 = 5$
Which amongst the following options are correct graphical representation of Boyle's law?
Solution: Straight lines of $P$ against $1/V$ through the origin, labelled $T_3 > T_2 > T_1$ Boyle's law: at constant temperature $PV = nRT$ is constant, so $P = nRT \cdot \dfrac{1}{V}$. $P$ against $\dfrac{1}{V}$ is a straight line through the origin with slope $nRT$; a higher temperature gives a steeper line, hence $T_3 > T_2 > T_1$. (A) $P$ against $\dfrac{1}{V}$ is a straight line, not a hyperbola. (B) $P$ against $T$ at constant volume is Gay-Lussac's law, not Boyle's law.
Some tranquilizers are listed below. Which one from the following belongs to barbiturates?
Solution: Veronal Veronal is a derivative of barbituric acid, so it is a barbiturate; valium, chlordiazepoxide and meprobamate are tranquilizers but not barbiturates.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : In equation $\Delta_r G = -nFE_{\text{cell}}$, value of $\Delta_r G$ depends on $n$. Reason R : $E_{\text{cell}}$ is an intensive property and $\Delta_r G$ is an extensive property. In the light of the above statements, choose the correct answer from the options given below
Solution: Both A and R are true and R is NOT the correct explanation of A A is true: in $\Delta_r G = -nFE_{\text{cell}}$, $nF$ is the charge passed for the reaction as written, so $\Delta_r G$ changes with $n$. R is true: $E_{\text{cell}}$ is an intensive property and $\Delta_r G$ is an extensive property. R is a general statement about the two quantities; the dependence on $n$ comes from the charge passed in the cell reaction, so R is not the correct explanation of A.
Given below are two statements : Statement I : A unit formed by the attachment of a base to 1$'$ position of sugar is known as nucleoside. Statement II : When nucleoside is linked to phosphorous acid at 5$'$-position of sugar moiety, we get nucleotide. In the light of the above statements, choose the correct answer from the options given below :
Solution: Statement I is true but Statement II is false Statement I is true: a base attached at the 1′ position of the sugar forms a nucleoside. Statement II is false: a nucleotide forms when the nucleoside is linked to phosphoric acid (not phosphorous acid) at the 5′ position of the sugar.
For a certain reaction, the rate $= k[A]^{2}[B]$, when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would
Solution: Increase by a factor of nine Rate $= k[A]^2[B]$, so with [B] fixed, rate $\propto [A]^2$. ∴ Tripling [A] multiplies the rate by $3^2 = 9$.
Amongst the given options which of the following molecules/ion acts as a Lewis acid?
Solution: BF₃ A Lewis acid accepts an electron pair; boron in BF₃ has only six valence electrons and a vacant p orbital. (B) OH⁻ has lone pairs to donate: a Lewis base. (C) NH₃ has a lone pair on nitrogen: a Lewis base. (D) H₂O has two lone pairs on oxygen: a Lewis base.
Which of the following statements are INCORRECT? A. All the transition metals except scandium form MO oxides which are ionic. B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc$_2$O$_3$ to Mn$_2$O$_7$. C. Basic character increases from V$_2$O$_3$ to V$_2$O$_4$ to V$_2$O$_5$. D. V$_2$O$_4$ dissolves in acids to give VO$_4^{3-}$ salts. E. CrO is basic but Cr$_2$O$_3$ is amphoteric. Choose the correct answer from the options given below:
Solution: C and D only (the incorrect statements) Statement C is incorrect: acidic character, not basic, increases from V₂O₃ to V₂O₄ to V₂O₅. Statement D is incorrect: V₂O₄ dissolves in acids to give VO²⁺ salts, not VO₄³⁻. Statements A, B and E are correct: all transition metals except Sc form ionic MO oxides, the group oxidation state is reached from Sc₂O₃ to Mn₂O₇, and CrO is basic while Cr₂O₃ is amphoteric.
What fraction of one edge centred octahedral void lies in one unit cell of fcc?
Solution: $\dfrac{1}{4}$ In fcc the octahedral voids are at the body centre and the 12 edge centres. An edge is shared by four unit cells, so one edge-centred void contributes $\dfrac{1}{4}$ to a unit cell.
Match List-I with List-II : List-I (Oxoacids of Sulphur): A. Peroxodisulphuric acid B. Sulphuric acid C. Pyrosulphuric acid D. Sulphurous acid List-II (Bonds): I. Two S–OH, Four S=O, One S–O–S II. Two S–OH, One S=O III. Two S–OH, Four S=O, One S–O–O–S IV. Two S–OH, Two S=O Choose the correct answer from the options given below.
Solution: A–III, B–IV, C–I, D–II A–III: peroxodisulphuric acid, H₂S₂O₈, has two S–OH, four S=O and a peroxide link, S–O–O–S. B–IV: sulphuric acid, H₂SO₄, has two S–OH and two S=O. C–I: pyrosulphuric acid, H₂S₂O₇, has two S–OH, four S=O and one S–O–S bridge. D–II: sulphurous acid, H₂SO₃, has two S–OH and one S=O.
Identify the major product obtained in the following reaction: 2-Acetylbenzaldehyde, a benzene ring bearing a –COCH$_3$ group and a –CHO group on adjacent carbons, is treated with 2[Ag(NH$_3$)$_2$]$^{+}$ and 3$^{-}$OH on heating to give the major product.
Solution: The ring bearing –COCH₃ and –COO⁻ Tollens' reagent, [Ag(NH₃)₂]⁺ in alkali, oxidises the –CHO group to –COO⁻ and deposits silver. It does not attack ketones, so the –COCH₃ group is unchanged.
Consider the following compounds/species: i. Naphthalene ii. Cyclopentadienyl anion iii. Cyclobutadiene iv. Cyclopropenyl anion v. Cyclopropenyl cation vi. Cyclooctatetraene vii. Anthracene The number of compounds/species which obey Huckel's rule is ______.
Solution: 4 Hückel's rule: a planar, fully conjugated ring with $(4n + 2)\ \pi$ electrons is aromatic. Naphthalene (10 π electrons), the cyclopentadienyl anion (6), the cyclopropenyl cation (2) and anthracene (14) obey the rule. Cyclobutadiene and the cyclopropenyl anion (4 π electrons each) do not, nor does cyclooctatetraene (8 π electrons, tub-shaped, not planar). ∴ 4 species obey Hückel's rule.
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
Solution: $\Delta H = \Delta U + \Delta n_g RT$ Enthalpy, $H = U + pV$, so $\Delta H = \Delta U + \Delta(pV)$. For ideal gases at constant temperature, $\Delta(pV) = \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas. ∴ $\Delta H = \Delta U + \Delta n_g RT$
Given below are two statements : Statement I : The nutrient deficient water bodies lead to eutrophication Statement II : Eutrophication leads to decrease in the level of oxygen in the water bodies. In the light of the above statements, choose the correct answer from the options given below:
Solution: Statement I is incorrect but Statement II is true Statement I is incorrect: eutrophication is caused by nutrient enrichment (nitrates and phosphates), not nutrient deficiency. Statement II is true: the algal bloom dies and decomposes, and the decomposition uses up the dissolved oxygen.
Pumice stone is an example of
Solution: Solid sol Pumice stone is a gas (dispersed phase) in a solid (dispersion medium); such a colloid is a solid sol.
Which complex compound is most stable?
Solution: [CoCl₂(en)₂]NO₃ Chelating ligands form more stable complexes than monodentate ones (chelate effect); only [CoCl₂(en)₂]NO₃ has a chelating ligand, ethane-1,2-diamine. (B) [Co(NH₃)₆]₂(SO₄)₃ has only monodentate NH₃ ligands. (C) [Co(NH₃)₄(H₂O)Br](NO₃)₂ has only monodentate NH₃, H₂O and Br⁻ ligands. (D) [Co(NH₃)₃(NO₃)₃] has only monodentate NH₃ and NO₃⁻ ligands.
The reaction that does NOT take place in a blast furnace between 900 K to 1500 K temperature range during extraction of iron is :
Solution: Fe₂O₃ + CO → 2FeO + CO₂ (it occurs at 500–800 K) Fe₂O₃ + CO → 2FeO + CO₂ takes place in the lower temperature range of the furnace, 500–800 K. (A) C + CO₂ → 2CO takes place at 900–1500 K. (B) CaO + SiO₂ → CaSiO₃, slag formation, also takes place in this range. (D) FeO + CO → Fe + CO₂ takes place at 900–1500 K.
The equilibrium concentrations of the species in the reaction A + B $\rightleftharpoons$ C + D are 2, 3, 10 and 6 mol L$^{-1}$, respectively at 300 K. $\Delta$G$^{\circ}$ for the reaction is (R = 2 cal/mol K)
Solution: $-1381.80\ \text{cal}$ $\Delta G^\circ = -2.303\,RT\log K$ $K = \dfrac{[C][D]}{[A][B]} = \dfrac{10 \times 6}{2 \times 3} = 10$ ∴ $\Delta G^\circ = -2.303 \times 2 \times 300 \times \log 10\ \text{cal} = -1381.8\ \text{cal}$
Consider the following reaction : Benzyl phenyl ether, C$_6$H$_5$CH$_2$–O–C$_6$H$_5$, is heated with HI to give A + B. Identify products A and B.
Solution: A = C₆H₅CH₂I and B = C₆H₅OH HI protonates the ether oxygen; the benzyl–oxygen bond breaks, because the benzyl carbocation is stabilised by the ring, while the aryl–oxygen bond has partial double bond character. I⁻ takes the benzyl group, giving benzyl iodide, C₆H₅CH₂I, and the phenyl group leaves as phenol, C₆H₅OH.
Which amongst the following will be most readily dehydrated under acidic conditions?
Solution: Structure (4), the 1,3-diol with no nitro group Acid-catalysed dehydration goes through a carbocation, so the compound giving the most stable carbocation and alkene reacts fastest. In structure (4), loss of water from one –OH gives a carbocation and then an alkene; the second –OH is lost the same way, giving a conjugated diene, which is extra stable. In the other three, the strongly electron-withdrawing nitro group destabilises the carbocation and slows dehydration.
Identify the final product [D] obtained in the following sequence of reactions. CH$_3$CHO with (i) LiAlH$_4$, (ii) H$_3$O$^{+}$ gives [A]; [A] with H$_2$SO$_4$ and heat gives [B]; [B] with HBr gives [C]; [C] with bromobenzene and Na/dry ether gives [D].
Solution: Ethylbenzene, C₆H₅CH₂CH₃ LiAlH₄ reduces CH₃CHO to ethanol [A]; hot conc. H₂SO₄ dehydrates it to ethene [B]. HBr adds to ethene to give bromoethane [C]. Wurtz–Fittig reaction: CH₃CH₂Br + C₆H₅Br + 2Na → C₆H₅CH₂CH₃ + 2NaBr, so [D] is ethylbenzene.
On balancing the given redox reaction, $$a\text{Cr}_2\text{O}_7^{2-} + b\text{SO}_3^{2-}(aq) + c\text{H}^{+}(aq) \rightarrow 2a\text{Cr}^{3+}(aq) + b\text{SO}_4^{2-}(aq) + \frac{c}{2}\text{H}_2\text{O}(l)$$ the coefficients $a$, $b$ and $c$ are found to be, respectively-
Solution: 1, 3, 8 Reduction: Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺; oxidation: SO₃²⁻ → SO₄²⁻ + 2e⁻, taken three times. Balancing O with H₂O and H with H⁺: Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O ∴ $a = 1$, $b = 3$, $c = 8$; this matches $2a = 2$ Cr³⁺ and $\tfrac{c}{2} = 4$ H₂O.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Late wood has fewer xylary elements with narrow vessels. Reason R : Cambium is less active in winters. In the light of the above statements, choose the correct answer from the options given below :
Solution: Both A and R are true and R is the correct explanation of A A is true: late (autumn) wood has fewer xylary elements with narrow vessels. R is true and explains A: in winter the cambium is less active, so it forms fewer xylary elements with narrow vessels.
The historic Convention on Biological Diversity, 'The Earth Summit' was held in Rio de Janeiro in the year
Solution: 1992 The Earth Summit on biological diversity was held in Rio de Janeiro in 1992; it called on all nations to conserve biodiversity and use its benefits sustainably.
In the equation GPP $-$ R $=$ NPP GPP is Gross Primary Productivity NPP is Net Primary Productivity R here is ________.
Solution: Respiratory loss Plants use a considerable part of GPP in respiration; GPP minus these respiration losses (R) gives NPP.
What is the function of tassels in the corn cob?
Solution: To trap pollen grains The tassels of the corn cob are the stigma and style, which wave in the wind to trap pollen grains (maize is wind pollinated).
What is the role of RNA polymerase III in the process of transcription in Eukaryotes?
Solution: Transcription of tRNA, 5S rRNA and snRNA RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs. (A) The precursor of mRNA (hnRNA) is transcribed by RNA polymerase II. (B) RNA polymerase III transcribes tRNA and 5S rRNA as well, not only snRNAs. (C) 28S, 18S and 5.8S rRNAs are transcribed by RNA polymerase I.
Upon exposure to UV radiation, DNA stained with ethidium bromide will show
Solution: Bright orange colour DNA stained with ethidium bromide shows bright orange bands when the gel is exposed to UV light.
The thickness of ozone in a column of air in the atmosphere is measured in terms of :
Solution: Dobson units The thickness of ozone in a column of air from the ground to the top of the atmosphere is measured in Dobson units (DU).
The phenomenon of pleiotropism refers to
Solution: A single gene affecting multiple phenotypic expression Pleiotropy: a single gene affects multiple phenotypic expressions. In phenylketonuria, for example, a mutation in one gene causes mental retardation and reduced hair and skin pigmentation.
Spraying of which of the following phytohormone on juvenile conifers helps hastening the maturity period, that leads early seed production?
Solution: Gibberellic acid Spraying juvenile conifers with gibberellins (GAs) hastens their maturity period, leading to early seed production.
Given below are two statements : One labelled as Assertion A and the other labelled as Reason R: Assertion A : The first stage of gametophyte in the life cycle of moss is protonema stage. Reason R : Protonema develops directly from spores produced in capsule. In the light of the above statements, choose the most appropriate answer from options given below:
Solution: Both A and R are correct and R is the correct explanation of A A is correct: the moss gametophyte has two stages, and the first is the protonema. R is correct and explains A: the protonema develops directly from a spore released from the capsule, so it is the first gametophytic stage.
During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out
Solution: DNA After RNA is removed with ribonuclease and proteins such as histones with protease, purified DNA precipitates out on adding chilled ethanol.
The reaction centre in PS II has an absorption maxima at
Solution: 680 nm The reaction centre chlorophyll a of PS II absorbs maximally at 680 nm (P680); that of PS I absorbs at 700 nm (P700).
Family Fabaceae differs from Solanaceae and Liliaceae. With respect to the stamens, pick out the characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae.
Solution: Diadelphous and dithecous anthers Fabaceae: stamens ten, diadelphous (united into two bundles), anthers dithecous. Solanaceae has free epipetalous stamens and Liliaceae free epiphyllous stamens; both have dithecous anthers, but neither is diadelphous.
Which micronutrient is required for splitting of water molecule during photosynthesis?
Solution: Manganese Manganese is needed for the splitting of water to liberate oxygen during photosynthesis.
Among 'The Evil Quartet', which one is considered the most important cause driving extinction of species?
Solution: Habitat loss and fragmentation Habitat loss and fragmentation is the most important cause driving animals and plants to extinction; the other causes in the Evil Quartet are over-exploitation, alien species invasions and co-extinctions.
Movement and accumulation of ions across a membrane against their concentration gradient can be explained by
Solution: Active transport Moving ions against their concentration gradient needs energy (ATP), which is active transport; passive transport, osmosis and facilitated diffusion all go down a gradient without using energy.
In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as
Solution: Dedifferentiation Differentiated mesophyll cells, which had lost the capacity to divide, regain it and form callus: this is dedifferentiation.
The process of appearance of recombination nodules occurs at which sub stage of prophase I in meiosis?
Solution: Pachytene Recombination nodules, the sites of crossing over between non-sister chromatids of homologous chromosomes, appear during pachytene.
Given below are two statements : Statement I : The forces generated transpiration can lift a xylem-sized column of water over 130 meters height. Statement II : Transpiration cools leaf surfaces sometimes 10 to 15 degrees evaporative cooling. 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 forces generated by transpiration can lift a xylem-sized column of water over 130 metres high. Statement II is correct: transpiration cools leaf surfaces, sometimes by 10 to 15 degrees, by evaporative cooling.
Axile placentation is observed in
Solution: China rose, Petunia and Lemon China rose, petunia and lemon (like tomato) have axile placentation. Pea, beans and lupin have marginal placentation; Dianthus and primrose free central; mustard and cucumber parietal.
Which of the following stages of meiosis involves division of centromere?
Solution: Anaphase II In anaphase II the centromere of each chromosome splits, so the sister chromatids separate and move to opposite poles. (B) In telophase the chromosomes have already reached the poles. (C) In metaphase I homologous pairs align at the equator; centromeres do not divide in meiosis I. (D) In metaphase II chromosomes align at the equator; the centromere splits only at the next stage, anaphase II.
Frequency of recombination between gene pairs on same chromosome as a measure of the distance between genes to map their position on chromosome, was used for the first time by
Solution: Alfred Sturtevant Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between them and mapped their positions on the chromosome.
Cellulose does not form blue colour with Iodine because
Solution: It does not contain complex helices and hence cannot hold iodine molecules Starch forms helical secondary structures that hold I₂ and give a blue colour; cellulose has no complex helices, so it cannot hold iodine. Cellulose is a polysaccharide, not a disaccharide, and iodine does not break it down.
In gene gun method used to introduce alien DNA into host cells, microparticles of ________ metal are used.
Solution: Tungsten or gold In biolistics (gene gun), cells are bombarded at high velocity with microparticles of gold or tungsten coated with DNA.
Identify the pair of heterosporous pteridophytes among the following :
Solution: Selaginella and Salvinia Selaginella and Salvinia are heterosporous: they produce two kinds of spores, large megaspores and small microspores. Psilotum, Lycopodium and Equisetum are homosporous, producing spores of one kind.
Unequivocal proof that DNA is the genetic material was first proposed by
Solution: Alfred Hershey and Martha Chase Hershey and Chase (1952), using bacteriophages labelled with radioactive phosphorus and sulphur, gave the unequivocal proof that DNA is the genetic material. Griffith discovered transformation, Avery, MacLeod and McCarty showed the transforming principle was DNA, and Wilkins and Franklin produced X-ray diffraction data of DNA.
How many ATP and NADPH$_2$ are required for the synthesis of one molecule of Glucose during Calvin cycle?
Solution: 18 ATP and 12 NADPH₂ Each CO₂ fixed in the Calvin cycle needs 3 ATP and 2 NADPH; one glucose needs 6 turns of the cycle. ∴ $6 \times 3 = 18$ ATP and $6 \times 2 = 12$ NADPH₂
Identify the correct statements: A. Detrivores perform fragmentation. B. The humus is further degraded by some microbes during mineralization. C. Water soluble inorganic nutrients go down into the soil and get precipitated by a process called leaching. D. The detritus food chain begins with living organisms. E. Earthworms break down detritus into smaller particles by a process called catabolism. Choose the correct answer from the options given below:
Solution: A, B, C only A, B and C are correct: detritivores fragment detritus, humus is degraded further by microbes during mineralisation, and leaching carries soluble nutrients down to be precipitated as unavailable salts. Statement D is wrong: the detritus food chain begins with dead organic matter (detritus), not living organisms. Statement E is wrong: earthworms break detritus into smaller particles by fragmentation; catabolism is the enzymatic breakdown by bacteria and fungi.
In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are :
Solution: Synergids, Zygote and Primary endosperm nucleus Synergids are cells of the female gametophyte: haploid. The zygote forms by fusion of two haploid gametes: diploid. The primary endosperm nucleus forms by fusion of a male gamete with the two polar nuclei (the diploid secondary nucleus): triploid.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : ATP is used at two steps in glycolysis. Reason R : First ATP is used in converting glucose into glucose-6-phosphate and second ATP is used in conversion of fructose-6-phosphate into fructose-1, 6-diphosphate. In the light of the above statements, choose the correct answer from the options given below :
Solution: Both A and R are true and R is the correct explanation of A A is true: ATP is used at two steps of glycolysis. R is true and explains A: the two ATPs phosphorylate glucose to glucose-6-phosphate and fructose-6-phosphate to fructose-1,6-bisphosphate.
Expressed Sequence Tags (ESTs) refers to
Solution: All genes that are expressed as RNA Expressed Sequence Tags are all the genes that are expressed as RNA; identifying them was one of the two approaches of the Human Genome Project.
Among eukaryotes, replication of DNA takes place in :
Solution: S phase DNA replication takes place in the S (synthesis) phase of interphase.
Given below are two statements : Statement I : Endarch and exarch are the terms often used for describing the position of secondary xylem in the plant body. Statement II : Exarch condition is the most common feature of the root system. In the light of the above statements, choose the correct answer from the options given below:
Solution: Statement I is incorrect but Statement II is true Statement I is incorrect: endarch and exarch describe the position of primary xylem (protoxylem relative to metaxylem), not secondary xylem. Statement II is true: roots have exarch primary xylem, with the protoxylem towards the periphery.
Large, colourful, fragrant flowers with nectar are seen in
Solution: Insect pollinated plants Large, colourful, fragrant flowers rich in nectar attract insects, so they are found in insect pollinated plants.
Which hormone promotes internode/petiole elongation in deep water rice?
Solution: Ethylene Ethylene promotes rapid internode and petiole elongation in deep water rice, keeping the leaves and upper shoot above water.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral meristem. Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at successive node instead of leaves. In the light of the above statements, choose the correct answer from the options given below :
Solution: Both A and R are true and R is the correct explanation of A A is true: a flower is a modified shoot in which the shoot apical meristem changes to a floral meristem. R is true and explains A: the internodes do not elongate, the axis is condensed, and the apex produces floral appendages laterally at successive nodes instead of leaves.
Match List I with List II : List I: A. M Phase B. G$_2$ Phase C. Quiescent stage D. G$_1$ Phase List II: I. Proteins are synthesized II. Inactive phase III. Interval between mitosis and initiation of DNA replication IV. Equational division Choose the correct answer from the options given below :
Solution: A-IV, B-I, C-II, D-III A-IV: M phase (mitosis) is the equational division. B-I: in G₂ the cell synthesises proteins in preparation for mitosis. C-II: the quiescent stage (G₀) is an inactive phase. D-III: G₁ is the interval between mitosis and initiation of DNA replication.
How many different proteins does the ribosome consist of?
Solution: 80 The ribosome is made of structural RNAs and about 80 different proteins.
Which one of the following statements is NOT correct?
Solution: Algal blooms improve water quality and promote fisheries (the incorrect statement) Algal blooms colour the water, deteriorate its quality and cause fish mortality; they do not improve water quality or fisheries. (A) True: water hyacinth grows abundantly in eutrophic water and upsets the ecosystem dynamics of the water body. (B) True: this is biomagnification of toxic substances such as mercury and DDT. (C) True: microbes decomposing sewage use up dissolved oxygen (high BOD), killing aquatic organisms.
Which of the following combinations is required for chemiosmosis?
Solution: Membrane, proton pump, proton gradient, ATP synthase Chemiosmosis needs a membrane, a proton pump, a proton gradient across the membrane and ATP synthase. (A) The gradient is of protons, not electrons, and a membrane is needed. (B) The gradient is of protons, not electrons, and the enzyme is ATP synthase, not NADP synthase. (D) The enzyme is ATP synthase, not NADP synthase.
Match List I with List II : List I: A. Cohesion B. Adhesion C. Surface tension D. Guttation List II: I. More attraction in liquid phase II. Mutual attraction among water molecules III. Water loss in liquid phase IV. Attraction towards polar surfaces Choose the correct answer from the options given below :
Solution: A – II, B – IV, C – I, D – III A – II: cohesion is mutual attraction among water molecules. B – IV: adhesion is attraction of water to polar surfaces. C – I: surface tension arises because water molecules attract each other more in the liquid phase than in the gas phase. D – III: guttation is loss of water in the liquid phase.
Match List I with List II: List I: A. Iron B. Zinc C. Boron D. Molybdenum List II: I. Synthesis of auxin II. Component of nitrate reductase III. Activator of catalase IV. Cell elongation and differentiation Choose the correct answer from the options given below:
Solution: A-III, B-I, C-IV, D-II A-III: iron activates catalase. B-I: zinc is needed for the synthesis of auxin. C-IV: boron is required for cell elongation and differentiation. D-II: molybdenum is a component of nitrogenase and nitrate reductase.
Given below are two statements: Statement I : Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and competitively inferior one will be eliminated eventually. Statement II : In general, carnivores are more adversely affected by competition than herbivores. In the light of the above statements, choose the correct answer from the options given below:
Solution: Statement I is correct, Statement II is false Statement I is correct: this is Gause's competitive exclusion principle. Statement II is false: in general, herbivores and plants are more adversely affected by competition than carnivores.
Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of
Solution: Succinic dehydrogenase Malonate closely resembles the substrate succinate in structure, so it competitively inhibits succinic dehydrogenase. Such competitive inhibitors are often used to control bacterial pathogens.
Identify the correct statements: A. Lenticels are the lens-shaped openings permitting the exchange of gases. B. Bark formed early in the season is called hard bark. C. Bark is a technical term that refers to all tissues exterior to vascular cambium. D. Bark refers to periderm and secondary phloem. E. Phellogen is single-layered in thickness. Choose the correct answer from the options given below:
Solution: A and D only A and D are correct: lenticels are lens-shaped openings for exchange of gases, and bark refers to periderm and secondary phloem. Statement B is wrong: bark formed early in the season is early or soft bark; hard bark forms late in the season. Statement C is wrong: bark is a non-technical term for all tissues outside the vascular cambium. Statement E is wrong: phellogen is a couple of layers thick.
Given below are two statements : One labelled as Assertion A and the other labelled as Reason R : Assertion A : In gymnosperms the pollen grains are released from the microsporangium and carried by air currents. Reason R : Air currents carry the pollen grains to the mouth of the archegonia where the male gametes are discharged and pollen tube is not formed. In the light of the above statements, choose the correct answer from the options given below :
Solution: A is true but R is false A is true: in gymnosperms the pollen grains are released from the microsporangium and carried by air currents. R is false: the pollen grains reach the opening of the ovules, and a pollen tube carrying the male gametes grows towards the archegonia and discharges its contents near their mouth.
Which of the following statements are correct about Klinefelter's Syndrome? A. This disorder was first described by Langdon Down (1866). B. Such an individual has overall masculine development. However, the feminine development is also expressed. C. The affected individual is short statured. D. Physical, psychomotor and mental development is retarded. E. Such individuals are sterile. Choose the correct answer from the options given below:
Solution: B and E only Klinefelter's syndrome (47, XXY): overall masculine development with some feminine development (such as gynaecomastia), and the individuals are sterile, so B and E are correct. Statement A is wrong: Langdon Down (1866) described Down's syndrome, not Klinefelter's syndrome. Statements C and D are wrong: short stature and retarded physical, psychomotor and mental development are features of Down's syndrome.
Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence. A. Insertion of recombinant DNA into the host cell B. Cutting of DNA at specific location by restriction enzyme C. Isolation of desired DNA fragment D. Amplification of gene of interest using PCR Choose the correct answer from the options given below :
Solution: B, C, D, A DNA is first cut at specific sites by restriction enzymes (B), and the desired fragment is isolated (C). The gene of interest is amplified by PCR (D), ligated into a vector, and the recombinant DNA is inserted into the host cell (A).
Match List I with List II : List I: A. Oxidative decarboxylation B. Glycolysis C. Oxidative phosphorylation D. Tricarboxylic acid cycle List II: I. Citrate synthase II. Pyruvate dehydrogenase III. Electron transport system IV. EMP pathway Choose the correct answer from the options given below :
Solution: A – II, B – IV, C – III, D – I A – II: oxidative decarboxylation of pyruvate is catalysed by pyruvate dehydrogenase. B – IV: glycolysis is the EMP (Embden–Meyerhof–Parnas) pathway. C – III: oxidative phosphorylation is driven by the electron transport system. D – I: the TCA cycle starts with citrate synthase joining acetyl CoA and OAA to form citric acid.
Match List I with List II : List I (Interaction): A. Mutualism B. Commensalism C. Amensalism D. Parasitism List II (Species A and B): I. +(A), 0(B) II. $-$(A), 0(B) III. +(A), $-$(B) IV. +(A), +(B) Choose the correct answer from the options given below:
Solution: A-IV, B-I, C-II, D-III A-IV: mutualism (+, +), both species benefit. B-I: commensalism (+, 0), one benefits and the other is unaffected. C-II: amensalism (−, 0), one is harmed and the other is unaffected. D-III: parasitism (+, −), the parasite benefits and the host is harmed.
Broad palm with single palm crease is visible in a person suffering from-
Solution: Down's syndrome Down's syndrome (trisomy of chromosome 21) shows a broad palm with a characteristic palm crease, short stature, a small round head and a furrowed tongue.
Given below are two statements: Statement I : A protein is imagined as a line, the left end represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal). Statement II : Adult human haemoglobin, consists of 4 subunits (two subunits of $\alpha$ type and two subunits of $\beta$ type.) In the light of the above statements, choose the correct answer from the options given below:
Solution: Statement I is false but Statement II is true Statement I is false: the first amino acid (left end) is the N-terminal amino acid and the last (right end) is the C-terminal amino acid. Statement II is true: adult human haemoglobin has four subunits, two of α type and two of β type.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Nephrons are of two types: Cortical & Juxta medullary, based on their relative position in cortex and medulla. Reason R: Juxta medullary nephrons have short loop of Henle whereas, cortical nephrons have longer loop of Henle. In the light of the above statements, choose the correct answer from the options given below:
Solution: A is true but R is false A is true: nephrons are cortical or juxtamedullary according to their position in the cortex and medulla. R is false: juxtamedullary nephrons have a very long loop of Henle running deep into the medulla, while cortical nephrons have a short loop.
Match List I with List II. List I: A. Heroin B. Marijuana C. Cocaine D. Morphine List II: I. Effect on cardiovascular system II. Slow down body function III. Painkiller IV. Interfere with transport of dopamine Choose the correct answer from the options given below:
Solution: A-II, B-I, C-IV, D-III A-II: heroin is a depressant and slows down body functions. B-I: marijuana (a cannabinoid) affects the cardiovascular system. C-IV: cocaine interferes with the transport of the neurotransmitter dopamine. D-III: morphine is a very effective sedative and painkiller.
Vital capacity of lung is _________.
Solution: IRV + ERV + TV Vital capacity is the maximum volume of air a person can breathe in after a forced expiration. ∴ VC = ERV + TV + IRV; residual volume is not included because it can never be expelled.
Match List I with List II. List I: A. P-wave B. Q-wave C. QRS complex D. T-wave List II: I. Beginning of systole II. Repolarisation of ventricles III. Depolarisation of atria IV. Depolarisation of ventricles Choose the correct answer from the options given below :
Solution: A-III, B-I, C-IV, D-II A-III: the P-wave is depolarisation of the atria. B-I: the Q-wave marks the beginning of systole. C-IV: the QRS complex is depolarisation of the ventricles. D-II: the T-wave is repolarisation of the ventricles.
Match List I with List II List I (Cells): A. Peptic cells B. Goblet cells C. Oxyntic cells D. Hepatic cells List II (Secretion): I. Mucus II. Bile juice III. Proenzyme pepsinogen IV. HCl and intrinsic factor for absorption of vitamin B$_{12}$ Choose the correct answer from the options given below:
Solution: A-III, B-I, C-IV, D-II A-III: peptic (chief) cells secrete the proenzyme pepsinogen. B-I: goblet cells secrete mucus. C-IV: oxyntic (parietal) cells secrete HCl and the intrinsic factor needed to absorb vitamin B₁₂. D-II: hepatic cells secrete bile.
Match List I with List II. List I: A. Vasectomy B. Coitus interruptus C. Cervical caps D. Saheli List II: I. Oral method II. Barrier method III. Surgical method IV. Natural method Choose the correct answer from the options given below:
Solution: A-III, B-IV, C-II, D-I A-III: vasectomy is a surgical method (sterilisation). B-IV: coitus interruptus is a natural method. C-II: cervical caps are a barrier method. D-I: Saheli is an oral, non-steroidal pill taken once a week.
Which one of the following common sexually transmitted diseases is completely curable when detected early and treated properly?
Solution: Gonorrhoea Except hepatitis-B, genital herpes and HIV infection, STIs are completely curable if detected early and treated properly; gonorrhoea is a bacterial infection and can be cured.
Match List I with List II. List I: A. Ringworm B. Filariasis C. Malaria D. Pneumonia List II: I. Haemophilus influenzae II. Trichophyton III. Wuchereria bancrofti IV. Plasmodium vivax Choose the correct answer from the options given below:
Solution: A-II, B-III, C-IV, D-I A-II: ringworm is caused by fungi such as Trichophyton. B-III: filariasis is caused by Wuchereria bancrofti. C-IV: malaria is caused by Plasmodium, e.g. P. vivax. D-I: pneumonia is caused by Haemophilus influenzae (and Streptococcus pneumoniae).
Which of the following are NOT considered as the part of endomembrane system? A. Mitochondria B. Endoplasmic reticulum C. Chloroplasts D. Golgi complex E. Peroxisomes Choose the most appropriate answer from the options given below:
Solution: A, C and E only The endomembrane system is the ER, Golgi complex, lysosomes and vacuoles, whose functions are coordinated. Mitochondria, chloroplasts and peroxisomes do not work in coordination with these, so they are not part of it.
Which of the following is not a cloning vector?
Solution: Probe A probe is a single-stranded DNA or RNA tagged with a radioactive molecule, used to detect a complementary sequence; pBR322, BAC and YAC are cloning vectors.
Match List I with List II. List I (Interacting species): A. A Leopard and a Lion in a forest/grassland B. A Cuckoo laying egg in a Crow's nest C. Fungi and root of a higher plant in Mycorrhizae D. A cattle egret and a Cattle in a field List II (Name of interaction): I. Competition II. Brood parasitism III. Mutualism IV. Commensalism Choose the correct answer from the options given below.
Solution: A-I, B-II, C-III, D-IV A-I: a leopard and a lion use the same prey: competition. B-II: the cuckoo lays its eggs in the crow's nest and the crow raises the young: brood parasitism. C-III: in mycorrhizae the fungus helps the root absorb nutrients and gets carbohydrates in return: mutualism. D-IV: the egret eats insects flushed out by grazing cattle, which are unaffected: commensalism.
Which one of the following symbols represents mating between relatives in human pedigree analysis?
Solution: A square joined to a circle by a double horizontal line In a pedigree, a single line joining a square (male) and a circle (female) is a mating; a double line is a mating between relatives (consanguineous mating).
Match List I with List II. List I: A. Taenia B. Paramoecium C. Periplaneta D. Pheretima List II: I. Nephridia II. Contractile vacuole III. Flame cells IV. Urecose gland Choose the correct answer from the options given below:
Solution: A-III, B-II, C-IV, D-I A-III: Taenia (a flatworm) excretes through flame cells (protonephridia). B-II: Paramoecium uses a contractile vacuole. C-IV: Periplaneta (cockroach) has urecose glands, which help in excretion. D-I: Pheretima (earthworm) excretes through nephridia.
Once the undigested and unabsorbed substances enter the caecum, their backflow is prevented by
Solution: Ileo-caecal valve The ileo-caecal valve, at the junction of the ileum and the caecum, prevents backflow of the faecal matter.
Match List I with List II. List I: A. Gene 'a' B. Gene 'y' C. Gene 'i' D. Gene 'z' List II: I. $\beta$-galactosidase II. Transacetylase III. Permease IV. Repressor protein Choose the correct answer from the options given below:
Solution: A-II, B-III, C-IV, D-I A-II: in the lac operon gene a codes for transacetylase. B-III: gene y codes for permease. C-IV: gene i codes for the repressor protein. D-I: gene z codes for β-galactosidase.
Which one of the following techniques does not serve the purpose of early diagnosis of a disease for its early treatment?
Solution: Serum and Urine analysis Conventional serum and urine analysis detects a disease only after the pathogen has multiplied and symptoms appear, so it does not allow early diagnosis. PCR, ELISA and recombinant DNA technology can detect a pathogen at very low concentration, before symptoms appear.
Given below are two statements : Statement I : Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat. Statement II : When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. In the light of the above statements, choose the correct answer from the options given below :
Solution: Both Statement I and Statement II are true Statement I is true: low temperature keeps an enzyme temporarily inactive, while high temperature denatures it and destroys its activity. Statement II is true: an inhibitor that closely resembles the substrate competes for the substrate-binding site and is called a competitive inhibitor.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Endometrium is necessary for implantation of blastocyst. Reason R: In the absence of fertilization, the corpus luteum degenerates that causes disintegration of endometrium. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both A and R are true but R is NOT the correct explanation of A A is true: the blastocyst implants in the endometrium of the uterus. R is true: without fertilisation the corpus luteum degenerates, progesterone falls and the endometrium disintegrates (menstruation). R describes what happens when there is no pregnancy, so it does not explain why the endometrium is needed for implantation.
Given below are two statements: one is labelled as Assertion A and other is labelled as Reason R. Assertion A : Amniocentesis for sex determination is one of the strategies of Reproductive and Child Health Care Programme. Reason R : Ban on amniocentesis checks increasing menace of female foeticide. In the light of the above statements, choose the correct answer from the options given below.
Solution: A is false but R is true A is false: amniocentesis for sex determination is not an RCH strategy; the statutory ban on it is. R is true: the ban on amniocentesis for sex determination legally checks the increasing menace of female foeticide.
Which of the following statements is correct?
Solution: Biomagnification refers to increase in concentration of the toxicant at successive trophic levels Biomagnification is the increase in concentration of a toxicant at successive trophic levels. (A) A large amount of nutrients causes algal bloom; it does not restrict it. (B) Algal bloom increases fish mortality. (C) Eutrophication is the natural ageing of a lake by nutrient enrichment of its water.
Which of the following statements are correct regarding female reproductive cycle? A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle. B. First menstrual cycle begins at puberty and is called menopause. C. Lack of menstruation may be indicative of pregnancy. D. Cyclic menstruation extends between menarche and menopause. Choose the most appropriate answer from the options given below.
Solution: A, C and D only A, C and D are correct: non-primate mammals have an oestrus cycle, absence of menstruation may indicate pregnancy, and cyclic menstruation lasts from menarche to menopause. Statement B is wrong: the first menstruation at puberty is menarche; menopause is when menstruation stops, around 50 years of age.
Match List I with List II. List I: A. CCK B. GIP C. ANF D. ADH List II: I. Kidney II. Heart III. Gastric gland IV. Pancreas Choose the correct answer from the options given below :
Solution: A-IV, B-III, C-II, D-I A-IV: CCK acts on the pancreas (and gall bladder), stimulating secretion of pancreatic enzymes (and bile). B-III: GIP inhibits gastric secretion and motility, acting on the gastric glands. C-II: ANF is secreted by the atrial wall of the heart. D-I: ADH acts mainly on the kidney, stimulating reabsorption of water by the distal tubules.
Given below are two statements: Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid. Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome. In the light of the above statements, choose the correct answer from the options given below:
Solution: Statement I is incorrect but Statement II is true Statement I is incorrect: in prokaryotes the DNA is negatively charged and is held with some positively charged proteins in the nucleoid. Statement II is true: in eukaryotes the negatively charged DNA wraps around the positively charged histone octamer to form a nucleosome.
Match List I with List II. List I (Type of Joint): A. Cartilaginous Joint B. Ball and Socket Joint C. Fibrous Joint D. Saddle Joint List II (Found between): I. Between flat skull bones II. Between adjacent vertebrae in vertebral column III. Between carpal and metacarpal of thumb IV. Between Humerus and Pectoral girdle Choose the correct answer from the options given below:
Solution: A-II, B-IV, C-I, D-III A-II: cartilaginous joints lie between adjacent vertebrae. B-IV: a ball and socket joint lies between the humerus and the pectoral girdle. C-I: fibrous joints join the flat skull bones (sutures). D-III: a saddle joint lies between the carpal and metacarpal of the thumb.
Given below are two statements: Statement I: Ligaments are dense irregular tissue. Statement II: Cartilage is dense regular tissue. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both Statement I and Statement II are false Statement I is false: ligaments are dense regular connective tissue. Statement II is false: cartilage is a specialised connective tissue, not dense regular tissue.
In which blood corpuscles, the HIV undergoes replication and produces progeny viruses?
Solution: $T_H$ cells HIV enters helper T-lymphocytes ($T_H$ cells), replicates and produces progeny viruses, which then attack other helper T-cells.
Radial symmetry is NOT found in adults of phylum ______.
Solution: Hemichordata Hemichordates are bilaterally symmetrical, so radial symmetry is not found in them. Coelenterates and ctenophores are radially symmetrical, and adult echinoderms are radially symmetrical (their larvae are bilateral).
Which of the following functions is carried out by cytoskeleton in a cell?
Solution: Motility The cytoskeleton (microtubules, microfilaments and intermediate filaments) gives mechanical support and motility and maintains the shape of the cell.
Given below are two statements: Statement I: Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct. Statement II: The cavity of the cervix is called cervical canal which along with vagina forms birth canal. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both Statement I and Statement II are true Statement I is true: the vas deferens receives a duct from the seminal vesicle and opens into the urethra as the ejaculatory duct. Statement II is true: the cavity of the cervix, the cervical canal, together with the vagina forms the birth canal.
Select the correct group/set of Australian Marsupials exhibiting adaptive radiation.
Solution: Numbat, Spotted cuscus, Flying phalanger Numbat (banded anteater), spotted cuscus and flying phalanger are Australian marsupials that arose by adaptive radiation. (A) Mole and flying squirrel are placental mammals. (B) Lemur, anteater and wolf are placental mammals. (C) Bobcat is a placental mammal.
Given below are two statements: Statement I : Electrostatic precipitator is most widely used in thermal power plant Statement II : Electrostatic precipitator in thermal power plant removes ionising radiations 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: electrostatic precipitators are the most widely used devices in thermal power plants. Statement II is incorrect: they remove particulate matter (over 99 per cent) from the exhaust, not ionising radiations.
Given below are two statements: Statement I: RNA mutates at a faster rate. Statement II: Viruses having RNA genome and shorter life span mutate and evolve faster. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both Statement I and Statement II are true Statement I is true: RNA is unstable and mutates at a faster rate. Statement II is true: so viruses with an RNA genome and a short life span mutate and evolve faster.
Match List I with List II with respect to human eye. List I: A. Fovea B. Iris C. Blind spot D. Sclera List II: I. Visible coloured portion of eye that regulates diameter of pupil. II. External layer of eye formed of dense connective tissue. III. Point of greatest visual acuity or resolution. IV. Point where optic nerve leaves the eyeball and photoreceptor cells are absent. Choose the correct answer from the options given below:
Solution: A-III, B-I, C-IV, D-II A-III: the fovea is the point of greatest visual acuity. B-I: the iris is the visible coloured part that regulates the diameter of the pupil. C-IV: the blind spot is where the optic nerve leaves the eye; it has no photoreceptor cells. D-II: the sclera is the outer layer of dense connective tissue.
Select the correct statements. A. Tetrad formation is seen during Leptotene. B. During Anaphase, the centromeres split and chromatids separate. C. Terminalization takes place during Pachytene. D. Nucleolus, Golgi complex and ER are reformed during Telophase. E. Crossing over takes place between sister chromatids of homologous chromosome. Choose the correct answer from the options given below:
Solution: B and D only B and D are correct: in anaphase the centromeres split and chromatids separate, and in telophase the nucleolus, Golgi complex and ER reform. Statement A is wrong: tetrads (bivalents) form in zygotene, after synapsis; they are not seen in leptotene. Statement C is wrong: terminalisation of chiasmata takes place in diakinesis, not pachytene. Statement E is wrong: crossing over takes place between non-sister chromatids of homologous chromosomes.
Select the correct statements with reference to chordates. A. Presence of a mid-dorsal, solid and double nerve cord. B. Presence of closed circulatory system. C. Presence of paired pharyngeal gill slits. D. Presence of dorsal heart E. Triploblastic pseudocoelomate animals. Choose the correct answer from the options given below:
Solution: B and C only B and C are correct: chordates have a closed circulatory system and paired pharyngeal gill slits. Statement A is wrong: the nerve cord is dorsal, hollow and single. Statement D is wrong: the heart is ventral. Statement E is wrong: chordates are triploblastic and coelomate.
In cockroach, excretion is brought about by- A. Phallic gland B. Urecose gland C. Nephrocytes D. Fat body E. Collaterial glands Choose the correct answer from the options given below :
Solution: B, C and D only Excretion in the cockroach is by Malpighian tubules, helped by the fat body, nephrocytes and urecose glands: B, C and D. The phallic gland (male) and collaterial glands (female, forming the ootheca) are reproductive, not excretory.
Match List I with List II. List I: A. Mast cells B. Inner surface of bronchiole C. Blood D. Tubular parts of nephron List II: I. Ciliated epithelium II. Areolar connective tissue III. Cuboidal epithelium IV. Specialised connective tissue Choose the correct answer from the options give below:
Solution: A-II, B-I, C-IV, D-III A-II: mast cells are found in areolar connective tissue. B-I: the inner surface of bronchioles is lined by ciliated epithelium. C-IV: blood is a specialised connective tissue. D-III: the tubular parts of the nephron are lined by cuboidal epithelium.
Given below are two statements: Statement I : During G$_0$ phase of cell cycle, the cell is metabolically inactive. Statement II : The centrosome undergoes duplication during S phase of interphase. 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: cells in G₀ remain metabolically active but stop proliferating unless called on to do so. Statement II is correct: in animal cells the centriole duplicates in the cytoplasm during the S phase.
Which one of the following is NOT an advantage of inbreeding?
Solution: Decrease in productivity after continuous inbreeding Continued inbreeding usually reduces fertility and even productivity (inbreeding depression); this is a drawback, not an advantage. (A) Eliminating less desirable genes and accumulating superior genes is an advantage of inbreeding. (C) Inbreeding increases homozygosity, so this statement does not describe inbreeding at all. (D) Exposing harmful recessive genes so that selection can eliminate them is an advantage of inbreeding.
Which of the following is characteristic feature of cockroach regarding sexual dimorphism?
Solution: Presence of anal styles Anal styles are present only in male cockroaches, so they mark sexual dimorphism. Sclerites, anal cerci and the dark brown body colour are found in both sexes.
The parts of human brain that helps in regulation of sexual behaviour, expression of excitement, pleasure, rage, fear etc. are:
Solution: Limbic system and hypothalamus The limbic system, along with the hypothalamus, regulates sexual behaviour and the expression of emotional reactions such as excitement, pleasure, rage and fear.
Which of the following statements are correct? A. Basophils are most abundant cells of the total WBCs B. Basophils secrete histamine, serotonin and heparin C. Basophils are involved in inflammatory response D. Basophils have kidney shaped nucleus E. Basophils are agranulocytes Choose the correct answer from the options given below:
Solution: B and C only B and C are correct: basophils secrete histamine, serotonin and heparin, and take part in inflammatory reactions. Statement A is wrong: basophils are the least abundant WBCs (0.5–1%); neutrophils are the most abundant (60–65%). Statement D is wrong: a kidney-shaped nucleus is a feature of monocytes. Statement E is wrong: basophils are granulocytes.
Which one of the following is the sequence on corresponding coding strand, if the sequence on mRNA formed is as follows 5$'$AUCGAUCGAUCGAUCGAUCGAUCG AUCG 3$'$?
Solution: 5′ ATCGATCGATCGATCGATCGATCGATCG 3′ The coding strand has the same sequence and polarity as the mRNA, with thymine in place of uracil. ∴ Coding strand: 5′ ATCGATCGATCGATCGATCGATCGATCG 3′
Match List I with List II. List I: A. Logistic growth B. Exponential growth C. Expanding age pyramid D. Stable age pyramid List II: I. Unlimited resource availability condition II. Limited resource availability condition III. The percent individuals of pre-reproductive age is largest followed by reproductive and post reproductive age groups IV. The percent individuals of pre-reproductives and reproductive age group are same Choose the correct answer from the options given below:
Solution: A-II, B-I, C-III, D-IV A-II: logistic growth occurs when resources are limited. B-I: exponential growth occurs when resources are unlimited. C-III: an expanding (growing) population has the largest share of pre-reproductive individuals, followed by reproductive and post-reproductive. D-IV: a stable population has equal shares of pre-reproductive and reproductive individuals.
Which of the following are NOT under the control of thyroid hormone? A. Maintenance of water and electrolyte balance B. Regulation of basal metabolic rate C. Normal rhythm of sleep-wake cycle D. Development of immune system E. Support the process of RBCs formation Choose the correct answer from the options given below:
Solution: C and D only Thyroid hormones regulate the basal metabolic rate, maintain water and electrolyte balance and support RBC formation (A, B and E). The sleep–wake rhythm is maintained by melatonin from the pineal gland, and the immune system develops under thymosins from the thymus, so C and D are not under thyroid control.
Which of the following statements are correct? A. An excessive loss of body fluid from the body switches off osmoreceptors. B. ADH facilitates water reabsorption to prevent diuresis. C. ANF causes vasodilation. D. ADH causes increase in blood pressure. E. ADH is responsible for decrease in GFR. Choose the correct answer from the options given below:
Solution: B, C and D only B, C and D are correct: ADH increases water reabsorption and prevents diuresis, ADH constricts blood vessels and raises blood pressure, and ANF causes vasodilation. Statement A is wrong: excessive loss of body fluid activates the osmoreceptors, which stimulate the hypothalamus to release ADH. Statement E is wrong: the rise in blood pressure caused by ADH increases glomerular blood flow and hence GFR.
Which of the following statements are correct regarding skeletal muscle? A. Muscle bundles are held together by collagenous connective tissue layer called fascicle. B. Sarcoplasmic reticulum of muscle fibre is a store house of calcium ions. C. Striated appearance of skeletal muscle fibre is due to distribution pattern of actin and myosin proteins. D. M line is considered as functional unit of contraction called sarcomere. Choose the most appropriate answer from the options given below:
Solution: B and C only B and C are correct: the sarcoplasmic reticulum stores Ca²⁺, and the striated appearance comes from the distribution pattern of actin and myosin. Statement A is wrong: the collagenous layer holding the bundles together is the fascia; the bundles themselves are fascicles. Statement D is wrong: the sarcomere, the functional unit of contraction, is the part of a myofibril between two successive Z lines.
The unique mammalian characteristics are:
Solution: Hairs, pinna and mammary glands Hairs, external ears (pinnae) and mammary glands are found only in mammals. (A) Development in mammals is direct, not indirect. (B) Mammals have a dicondylic skull; a monocondylic skull is found in reptiles and birds. (C) A tympanic membrane is also found in amphibians and reptiles.
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