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Free, login-free worked solutions to every question in NEET (UG) 2025 — Code 47. Check your answers below, then predict your rank or practise with full mock tests.
The current passing through the battery in the given circuit, is:
Solution: 0.5 A The dotted part of the network is a balanced Wheatstone bridge, so its equivalent resistance is $R' = \dfrac{4 \times 8}{12} = \dfrac{8}{3}\ \Omega$. This is in series with the rest of the circuit: $R_{\text{eq}} = \dfrac{8}{3} + \dfrac{1}{3} + 1.5 + 5.5 = 10\ \Omega$. ∴ Current through the battery, $i = \dfrac{V}{R_{\text{eq}}} = \dfrac{5}{10} = 0.5\ \text{A}$
The electric field in a plane electromagnetic wave is given by $$E_z = 60 \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{V/m}.$$ Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
Solution: $B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$ In an electromagnetic wave $\vec{E}$ and $\vec{B}$ oscillate in phase, perpendicular to each other and to the direction of travel, with $B_0 = \dfrac{E_0}{c}$. The wave travels along the x-axis and $\vec{E}$ is along z, so $\vec{B}$ is along y. $B_0 = \dfrac{60}{3 \times 10^{8}} = 2 \times 10^{-7}\ \text{T}$ ∴ $B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$
A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
Solution: $f$ (unchanged) Pipe open at both ends: $f = \dfrac{v}{2L}$. Dipped to half its length, it becomes a closed pipe of length $\dfrac{L}{2}$: $f' = \dfrac{v}{4(L/2)} = \dfrac{v}{2L}$. ∴ $f' = f$
An electron (mass $9 \times 10^{-31}$ kg and charge $1.6 \times 10^{-19}$ C) moving with speed $c/100$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4}$ T perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^{8}\ \text{ms}^{-1}$)
Solution: $\vec{E}$ perpendicular to $\vec{B}$, magnitude $27 \times 10^{2}\ \text{V m}^{-1}$ For no deflection the net Lorentz force is zero: $q(\vec{E} + \vec{v} \times \vec{B}) = 0 \Rightarrow \vec{E} = -\vec{v} \times \vec{B}$. So $\vec{E}$ is perpendicular to both $\vec{v}$ and $\vec{B}$. ∴ $E = vB = \dfrac{3 \times 10^{8}}{100} \times 9 \times 10^{-4} = 27 \times 10^{2}\ \text{V m}^{-1}$
In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power ($p$) and magnification ($m$) for each lens will be, respectively
Solution: $4p$ and $m^4$ For thin lenses in contact the powers add: $P = p + p + p + p = 4p$. The total magnification is the product of the separate magnifications: $m_{\text{total}} = m \times m \times m \times m = m^4$.
A 2 amp current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be
Solution: 1 : 4 Magnetic moment of a current loop, $m = IA = I\pi r^2$. With the same current, $m \propto r^2$. ∴ $\dfrac{m_1}{m_2} = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$
A constant voltage of 50 V is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $CD$ of the circuit is:
Solution: 2.0 A $R_{AB} = (1\ \Omega \parallel 3\ \Omega) + (2\ \Omega \parallel 4\ \Omega) = \dfrac{3}{4} + \dfrac{4}{3} = \dfrac{25}{12}\ \Omega$ Total current, $I = \dfrac{50}{25/12} = 24\ \text{A}$. It divides as 18 A through 1 Ω and 6 A through 3 Ω, then as 16 A through 2 Ω and 8 A through 4 Ω. ∴ By the junction rule at C, $I_{CD} = 18 - 16 = 2\ \text{A}$ (from C to D).
Two gases $A$ and $B$ are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas $A$ and $B$ are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio $\dfrac{r_A}{r_B}$ is equal to
Solution: $\dfrac{3}{4}$ First law of thermodynamics: $\Delta Q = \Delta U + P\Delta V$. Equal $\Delta Q$ and equal $\Delta U$ give equal work $P\Delta V$, and at equal pressure equal $\Delta V = \pi r^2 d$. $\pi r_A^2 \times 16 = \pi r_B^2 \times 9$ ∴ $\dfrac{r_A}{r_B} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$
A container has two chambers of volumes $V_1 = 2$ litres and $V_2 = 3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1$ atm and $p_2 = 2$ atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of
Solution: 1.6 atm No heat enters and no work is done, so the total internal energy is unchanged; for an ideal gas it is proportional to $pV$. $p(V_1 + V_2) = p_1V_1 + p_2V_2$ $p \times 5 = 1 \times 2 + 2 \times 3 = 8$ ∴ $p = 1.6\ \text{atm}$
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
Solution: 88 Earth days Kepler's law of periods: $T^2 \propto R^3$. $\dfrac{T_{\text{Mars}}}{T_{\text{Mercury}}} = \left(\dfrac{R_{\text{Mars}}}{R_{\text{Mercury}}}\right)^{3/2} = 4^{3/2} = 8$ ∴ $T_{\text{Mercury}} = \dfrac{687}{8} \approx 86$ days, nearest to 88 Earth days (the radius ratio is only about 4).
To an ac power supply of 220 V at 50 Hz, a resistor of 20 $\Omega$, a capacitor of reactance 25 $\Omega$ and an inductor of reactance 45 $\Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively
Solution: 7.8 A and 45° Impedance, $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\ \Omega$ Current, $I = \dfrac{V}{Z} = \dfrac{220}{20\sqrt{2}} \approx 7.8\ \text{A}$ $\tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{20}{20} = 1$ ∴ $\phi = 45^{\circ}$
A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
Solution: $\dfrac{R}{16}$ Each piece has resistance $\dfrac{R}{8}$. Four pieces in parallel: $\dfrac{R/8}{4} = \dfrac{R}{32}$. ∴ Two such sets in series: $R_{\text{eq}} = 2 \times \dfrac{R}{32} = \dfrac{R}{16}$
The output ($Y$) of the given logic implementation is similar to the output of an/a _____ gate.
Solution: NOR The first gate is a NOR and the second a NAND: $Y_1 = \overline{A + B}$, $Y_2 = \overline{A \cdot B}$. Both feed an AND gate: $Y = \overline{A + B} \cdot \overline{A \cdot B}$. By De Morgan's theorem, $Y = \overline{A + B + A \cdot B} = \overline{A + B(1 + A)} = \overline{A + B}$. ∴ The output is that of a NOR gate.
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
Solution: $\dfrac{3F}{8}$ Initially $F = \dfrac{kq^2}{r^2}$. Touching A shares its charge equally: A and the third sphere each get $\dfrac{q}{2}$. Touching B shares $\dfrac{q}{2} + q$ equally: B gets $\dfrac{3q}{4}$. ∴ $F' = \dfrac{k \cdot \dfrac{q}{2} \cdot \dfrac{3q}{4}}{r^2} = \dfrac{3}{8} \cdot \dfrac{kq^2}{r^2} = \dfrac{3F}{8}$
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at $x = 0.1$ cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is $M = 5$ cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
Solution: 4.98 cm Least count $= 1\ \text{MSD} - 1\ \text{VSD} = 0.1 - 0.09 = 0.01\ \text{cm}$ Observed reading $= 5 + 8 \times 0.01 = 5.08\ \text{cm}$ Zero error $= +0.1\ \text{cm}$, which is subtracted. ∴ Diameter $= 5.08 - 0.1 = 4.98\ \text{cm}$
In some appropriate units, time ($t$) and position ($x$) relation of a moving particle is given by $t = x^{2} + x$. The acceleration of the particle is
Solution: $-\dfrac{2}{(2x + 1)^{3}}$ $\dfrac{dt}{dx} = 2x + 1 \Rightarrow v = \dfrac{dx}{dt} = \dfrac{1}{2x + 1}$ Acceleration, $a = v\dfrac{dv}{dx} = \dfrac{1}{2x + 1} \times \dfrac{-2}{(2x + 1)^2}$ ∴ $a = -\dfrac{2}{(2x + 1)^3}$
Which of the following options represent the variation of photoelectric current with property of light shown on the $x$-axis? A. Photoelectric current against intensity of light — a straight line through the origin B. Photoelectric current against intensity of light — a horizontal line C. Photoelectric current against frequency of light — a straight line through the origin D. Photoelectric current against frequency of light — a straight line meeting the axis at a positive frequency
Solution: Graph A only: current against intensity is a straight line through the origin Photoelectric current is directly proportional to the intensity of light, so its graph against intensity is a straight line through the origin. (B) The current does change with intensity, so a horizontal line is wrong. (C) Below the threshold frequency there is no current, and above it the saturation current does not depend on frequency. (D) A straight line cutting the axis at the threshold frequency is the graph of stopping potential against frequency, not of current.
A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the $n^{\text{th}}$ orbit and the particle's speed $\nu$ in the orbit depend on $n$ as
Solution: $r \propto n^{2/3}$; $\nu \propto n^{1/3}$ Constant central force: $\dfrac{m\nu^2}{r} = F \Rightarrow \nu^2 \propto r$. Bohr's quantisation condition: $m\nu r = \dfrac{nh}{2\pi} \Rightarrow \nu r \propto n$. With $\nu \propto r^{1/2}$: $r^{3/2} \propto n$, so $r \propto n^{2/3}$. ∴ $\nu \propto r^{1/2} \propto n^{1/3}$
A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:
Solution: $\left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$ At P the tension is zero, so the weight component along the string gives the centripetal force: $mg\sin\theta = \dfrac{mv^2}{l}$. Conservation of mechanical energy from the lowest point to P (height $l + l\sin\theta$): $v_0^2 = v^2 + 2gl(1 + \sin\theta)$. Putting $gl = \dfrac{v^2}{\sin\theta}$: $v_0^2 = v^2\left(3 + \dfrac{2}{\sin\theta}\right)$ ∴ $\dfrac{v}{v_0} = \left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{1/2}$
A full wave rectifier circuit with diodes ($D_1$) and ($D_2$) is shown in the figure. If input supply voltage $V_{\text{in}} = 220\sin(100\pi t)$ volt, then at $t = 15$ msec
Solution: $D_1$ reverse biased, $D_2$ forward biased $\omega = 100\pi \Rightarrow T = \dfrac{2\pi}{\omega} = 0.02\ \text{s} = 20\ \text{ms}$ $t = 15\ \text{ms} = \dfrac{3T}{4}$, which lies in the negative half cycle ($V_{\text{in}} = 220\sin\tfrac{3\pi}{2} = -220\ \text{V}$). ∴ In the negative half cycle $D_1$ is reverse biased and $D_2$ is forward biased.
A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is filled in it) has small area $A$. It is filled with a gas of density $\rho$ and takes a spherical shape of radius $R$. When the gas is allowed to flow freely out of it, its radius $r$ changes from $R$ to 0 (zero) in time $T$. If the speed $v(r)$ of gas coming out of the balloon depends on $r$ as $r^{a}$ and $T \propto S^{\alpha} A^{\beta} \rho^{\gamma} R^{\delta}$ then
Solution: $a = -\tfrac{1}{2}$, $\alpha = -\tfrac{1}{2}$, $\beta = -1$, $\gamma = \tfrac{1}{2}$, $\delta = \tfrac{7}{2}$ Excess pressure inside the balloon $\propto \dfrac{S}{r}$; by Bernoulli's principle $\tfrac{1}{2}\rho v^2 \propto \dfrac{S}{r}$, so $v \propto \sqrt{\dfrac{S}{\rho r}}$ and $a = -\tfrac{1}{2}$. Rate of volume loss equals outflow: $-4\pi r^2\dfrac{dr}{dt} = Av \propto A\sqrt{\dfrac{S}{\rho r}}$ $dt \propto -\dfrac{1}{A}\sqrt{\dfrac{\rho}{S}}\, r^{5/2}\, dr \Rightarrow T \propto \dfrac{1}{A}\sqrt{\dfrac{\rho}{S}}\, R^{7/2}$ ∴ $\alpha = -\tfrac{1}{2}$, $\beta = -1$, $\gamma = \tfrac{1}{2}$, $\delta = \tfrac{7}{2}$
A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is
Solution: 125 Magnification of a compound microscope in normal adjustment, $m = \dfrac{L}{f_o} \times \dfrac{D}{f_e}$. ∴ $m = \dfrac{40}{2} \times \dfrac{25}{4} = 125$
Two identical point masses $P$ and $Q$, suspended from two separate massless springs of spring constants $k_1$ and $k_2$, respectively, oscillate vertically. If their maximum speeds are the same, the ratio $\left(A_Q/A_P\right)$ of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ is
Solution: $\sqrt{\dfrac{k_1}{k_2}}$ Maximum speed in SHM, $v_{\max} = A\omega$, with $\omega = \sqrt{\dfrac{k}{m}}$. Equal maximum speeds: $A_P\sqrt{\dfrac{k_1}{m}} = A_Q\sqrt{\dfrac{k_2}{m}}$ ∴ $\dfrac{A_Q}{A_P} = \sqrt{\dfrac{k_1}{k_2}}$
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
Solution: Non-zero everywhere, maximum at the imaginary cylindrical surface joining the edges of the plates With $\dfrac{d\sigma}{dt}$ constant, the displacement current $i_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = A\dfrac{d\sigma}{dt}$ is constant and spread evenly over the plate area. By the Ampere-Maxwell law, $B \propto r$ inside the plate region and $B \propto \dfrac{1}{r}$ outside it, as for a straight current-carrying wire. ∴ $B$ rises inside, falls outside, and is greatest at $r = R$, the cylinder joining the edges of the plates.
An electric dipole with dipole moment $5 \times 10^{-6}$ Cm is aligned with the direction of a uniform electric field of magnitude $4 \times 10^{5}$ N/C. The dipole is then rotated through an angle of $60^{\circ}$ with respect to the electric field. The change in the potential energy of the dipole is:
Solution: 1.0 J Potential energy of a dipole in a uniform field, $U = -pE\cos\theta$. $\Delta U = pE(\cos 0^{\circ} - \cos 60^{\circ}) = 5 \times 10^{-6} \times 4 \times 10^{5} \times \tfrac{1}{2}$ ∴ $\Delta U = 1.0\ \text{J}$
There are two inclined surfaces of equal length ($L$) and same angle of inclination $45^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction ($\mu_k$) between the object and the rough surface is close to
Solution: 0.75 Starting from rest over the same length, $L = \tfrac{1}{2}at^2$, so $t \propto \dfrac{1}{\sqrt{a}}$. $a_{\text{smooth}} = g\sin\theta$, $a_{\text{rough}} = g(\sin\theta - \mu_k\cos\theta)$ $\dfrac{t_{\text{rough}}}{t_{\text{smooth}}} = 2 \Rightarrow \dfrac{\sin\theta}{\sin\theta - \mu_k\cos\theta} = 4$ At $45^{\circ}$, $\sin\theta = \cos\theta$, so $\dfrac{1}{1 - \mu_k} = 4$. ∴ $\mu_k = 0.75$
De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052$ nm)
Solution: 0.67 nm Radius of the $n$th orbit, $r_n = n^2 a_0 = 4 \times 0.052 = 0.208\ \text{nm}$ Bohr's condition in de Broglie form: $2\pi r_n = n\lambda$. ∴ $\lambda = \dfrac{2\pi \times 0.208}{2} \approx 0.65\ \text{nm}$, closest to 0.67 nm
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
Solution: 108 days No external torque, so angular momentum is conserved: $I_1\omega_1 = I_2\omega_2$, with $I = \tfrac{2}{5}MR^2$. $\tfrac{2}{5}MR^2 \cdot \dfrac{2\pi}{T} = \tfrac{2}{5}M(2R)^2 \cdot \dfrac{2\pi}{T'}$ ∴ $T' = 4T = 4 \times 27 = 108$ days
A physical quantity $P$ is related to four observations $a$, $b$, $c$ and $d$ as follows: $$P = a^{3}b^{2}\,/\,c\sqrt{d}$$ The percentage errors of measurement in $a$, $b$, $c$ and $d$ are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity $P$ is
Solution: 13% Relative errors add, each multiplied by its power: $\dfrac{\Delta P}{P} = 3\dfrac{\Delta a}{a} + 2\dfrac{\Delta b}{b} + \dfrac{\Delta c}{c} + \tfrac{1}{2}\dfrac{\Delta d}{d}$ ∴ Percentage error $= 3(1) + 2(3) + 2 + \tfrac{1}{2}(4) = 13\%$
The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\dfrac{3}{8}d$ and $\dfrac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25\,K_2$, the value of $K_1$ is:
Solution: $K_1 = \tfrac{8}{3} \approx 2.66$ The slabs and the air gap $t_3 = d - \tfrac{3d}{8} - \tfrac{d}{2} = \tfrac{d}{8}$ are in series: $C = \dfrac{\varepsilon_0 A}{\tfrac{t_1}{K_1} + \tfrac{t_2}{K_2} + t_3}$. $C = 2C_0 = \dfrac{2\varepsilon_0 A}{d}$, so $\dfrac{3d}{8K_1} + \dfrac{d}{2} \cdot \dfrac{1.25}{K_1} + \dfrac{d}{8} = \dfrac{d}{2}$ $\dfrac{3}{8K_1} + \dfrac{5}{8K_1} = \dfrac{3}{8} \Rightarrow \dfrac{1}{K_1} = \dfrac{3}{8}$ ∴ $K_1 = \dfrac{8}{3} \approx 2.66$
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8$ m/s$^{2}$)
Solution: 21 N s Speed just before impact, $v_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 40} = 28\ \text{m s}^{-1}$ (downward) Speed just after impact, $v_2 = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 10} = 14\ \text{m s}^{-1}$ (upward) ∴ Impulse = change in momentum $= 0.5 \times [14 - (-28)] = 21\ \text{N s}$
Two cities $X$ and $Y$ are connected by a regular bus service with a bus leaving in either direction every $T$ min. A girl is driving scooty with a speed of 60 km/h in the direction $X$ to $Y$ notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed (assumed constant) of the buses.
Solution: 15 min, 120 km/h Buses going the same way are a distance $vT$ apart. Buses in her direction (relative speed $v - 60$): $(v - 60) \times 30 = vT$ Buses in the opposite direction (relative speed $v + 60$): $(v + 60) \times 10 = vT$ Equating: $30v - 1800 = 10v + 600 \Rightarrow v = 120\ \text{km/h}$ ∴ $T = \dfrac{(120 + 60) \times 10}{120} = 15\ \text{min}$
An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27$^{\circ}$C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given, $R = \dfrac{100}{12}\ \text{J mol}^{-1}\text{K}^{-1}$, and molecular mass of O$_2$ = 32, 1 atm pressure $= 1.01 \times 10^{5}$ N/m$^{2}$]
Solution: 0.116 kg Absolute pressure = gauge pressure + atmospheric pressure = 11 + 1 = 12 atm. Moles left, $n = \dfrac{pV}{RT} = \dfrac{12 \times 1.01 \times 10^{5} \times 30 \times 10^{-3}}{\tfrac{100}{12} \times 300} \approx 14.54\ \text{mol}$ Moles withdrawn $= 18.20 - 14.54 = 3.66\ \text{mol}$ ∴ Mass withdrawn $= 3.66 \times 32 \approx 117\ \text{g}$, nearly 0.116 kg
$AB$ is a part of an electrical circuit (see figure). The potential difference "$V_A - V_B$", at the instant when current $i = 2$ A and is increasing at a rate of 1 amp/second is:
Solution: 10 volt Going from A to B through the inductor, the cell and the resistor: $V_A - L\dfrac{di}{dt} - 5 - iR = V_B$ $V_A - V_B = 1 \times 1 + 5 + 2 \times 2$ ∴ $V_A - V_B = 10\ \text{V}$
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
Solution: $\omega(t)$ increasing with $t$; $A(t)$ decreasing with $t$ Time period $T = 2\pi\sqrt{\dfrac{m}{k}}$; as sand leaks, $m$ falls, so $T$ falls and $\omega = \dfrac{2\pi}{T}$ rises. The leaking sand carries away part of the kinetic energy, so the energy of oscillation $\tfrac{1}{2}kA^2$ falls and the amplitude decreases.
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron)
Solution: $\dfrac{he}{2\pi m}$ Magnetic force gives the centripetal force: $evB = \dfrac{mv^2}{r} \Rightarrow v = \dfrac{eBr}{m}$ Flux condition: $B\pi r^2 = \dfrac{nh}{e}$ Magnetic moment, $\mu = IA = \dfrac{ev}{2\pi r} \cdot \pi r^2 = \dfrac{evr}{2} = \dfrac{e^2Br^2}{2m}$ Substituting $Br^2 = \dfrac{nh}{\pi e}$: $\mu = \dfrac{neh}{2\pi m}$ ∴ For the lowest state ($n = 1$), $\mu = \dfrac{he}{2\pi m}$
A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :
Solution: 27 N Gravitational force varies as $\dfrac{1}{(R + h)^2}$, so $\dfrac{W_h}{W} = \dfrac{R^2}{(R + h)^2}$. With $h = \dfrac{R}{3}$: $\dfrac{W_h}{W} = \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}$ ∴ $W_h = \dfrac{9}{16} \times 48 = 27\ \text{N}$
Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0$ ($\theta_0 \ll 1$) with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is: (take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \dfrac{dy}{dx}$, $g$ is the acceleration due to gravity)
Solution: $\dfrac{d^{2}y}{dx^{2}} = \dfrac{\rho g}{S}y$ With $\dfrac{dy}{dx}$ small, the curvature of the surface is $\dfrac{d^2y/dx^2}{\left[1 + (dy/dx)^2\right]^{3/2}} \approx \dfrac{d^2y}{dx^2}$. The excess pressure across the curved surface, $S \times \text{curvature}$, balances the hydrostatic pressure $\rho g y$. $\rho g y = S\dfrac{d^2y}{dx^2}$ ∴ $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}y$
The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at $22.5^{\circ}$ from the polarization axis of one of the polaroids, is ($I_0$ is the intensity of polarised light after passing through the first polaroid):
Solution: $\dfrac{I_0}{8}$ Malus' law: $I = I_0\cos^2\theta$. After the middle sheet, $I_1 = I_0\cos^2 22.5^{\circ}$. The last sheet is at $67.5^{\circ}$ to the middle one: $I_2 = I_0\cos^2 22.5^{\circ}\cos^2 67.5^{\circ} = I_0\cos^2 22.5^{\circ}\sin^2 22.5^{\circ}$ ∴ $I_2 = \dfrac{I_0}{4}\sin^2 45^{\circ} = \dfrac{I_0}{8}$
A photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\lambda_{\text{photon}}/\lambda_{\text{electron}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)
Solution: $c\sqrt{\dfrac{2m}{E}}$ Photon: $E = \dfrac{hc}{\lambda} \Rightarrow \lambda_{\text{photon}} = \dfrac{hc}{E}$ Electron: $E = \dfrac{p^2}{2m}$ and $\lambda = \dfrac{h}{p} \Rightarrow \lambda_{\text{electron}} = \dfrac{h}{\sqrt{2mE}}$ ∴ $\dfrac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \dfrac{hc}{E} \cdot \dfrac{\sqrt{2mE}}{h} = c\sqrt{\dfrac{2m}{E}}$
An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then
Solution: Reflected light is completely polarised; angle of reflection close to 60° Brewster's law: $\tan i_B = \mu = 1.73 = \sqrt{3} \Rightarrow i_B = 60^{\circ}$, and at this angle the reflected light is completely polarised. (A) The transmitted light is only partially polarised (its angle of refraction is $90^{\circ} - 60^{\circ} = 30^{\circ}$). (C) The reflected light is completely, not partially, polarised, and the angle of reflection is $60^{\circ}$, not $30^{\circ}$. (D) Only the reflected light is completely polarised; the transmitted light is partially polarised.
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take $g = 10$ m/s$^{2}$)
Solution: $100\sqrt{3}$ N In equilibrium the floor's normal force is $Mg$, and friction $f$ balances the normal force $N_w$ of the smooth wall. The rod makes $30^{\circ}$ with the floor; torques about its foot: $Mg\dfrac{L}{2}\cos 30^{\circ} = N_w L\sin 30^{\circ}$ $f = N_w = \dfrac{Mg}{2}\cot 30^{\circ} = \dfrac{200}{2} \times \sqrt{3}$ ∴ $f = 100\sqrt{3}\ \text{N}$
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1/T_2$ is
Solution: $\dfrac{5}{3}$ Thermal resistances in series: $R_1 = R_3 = \dfrac{l}{2KA}$, $R_2 = \dfrac{l}{KA}$, total $\dfrac{2l}{KA}$. The same heat current flows through every rod: $H = \dfrac{3T - T}{2l/KA} = \dfrac{TKA}{l}$ $T_1 = 3T - HR_1 = 3T - \dfrac{T}{2} = \dfrac{5T}{2}$ $T_2 = T + HR_3 = T + \dfrac{T}{2} = \dfrac{3T}{2}$ ∴ $\dfrac{T_1}{T_2} = \dfrac{5}{3}$
The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car $A$ stops after 1000 m and car $B$ stops after 1500 m. If $F_A$ and $F_B$ are the forces applied by the breaks on cars $A$ and $B$ respectively, then the ratio of $\dfrac{F_A}{F_B}$ is
Solution: $\dfrac{2}{3}$ Work-energy theorem: the braking force removes all the kinetic energy, so $FS = K$. $\dfrac{F_A}{F_B} = \dfrac{K_A}{K_B} \times \dfrac{S_B}{S_A} = \dfrac{100}{225} \times \dfrac{1500}{1000}$ ∴ $\dfrac{F_A}{F_B} = \dfrac{2}{3}$
A sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is:
Solution: $\dfrac{7}{57}$ Whole sphere about the Y-axis: $I = \tfrac{2}{5}M(2R)^2 = \tfrac{8}{5}MR^2$ Same density and one-eighth the volume, so the small sphere has mass $M' = \dfrac{M}{8}$. Its centre is $R$ from the Y-axis; by the parallel axis theorem, $I' = \tfrac{2}{5} \cdot \dfrac{M}{8}R^2 + \dfrac{M}{8}R^2 = \dfrac{7}{40}MR^2$ ∴ $\dfrac{I'}{I - I'} = \dfrac{7/40}{8/5 - 7/40} = \dfrac{7}{64 - 7} = \dfrac{7}{57}$
If the molar conductivity ($\Lambda_m$) of a 0.050 mol L$^{-1}$ solution of a monobasic weak acid is 90 S cm$^{2}$ mol$^{-1}$, its extent (degree) of dissociation will be [Assume $\Lambda^{\circ}_{+} = 349.6$ S cm$^{2}$ mol$^{-1}$ and $\Lambda^{\circ}_{-} = 50.4$ S cm$^{2}$ mol$^{-1}$.]
Solution: 0.225 Degree of dissociation, $\alpha = \dfrac{\Lambda_m}{\Lambda^{\circ}_m}$, where $\Lambda^{\circ}_m = \Lambda^{\circ}_{+} + \Lambda^{\circ}_{-}$ (Kohlrausch's law). $\Lambda^{\circ}_m = 349.6 + 50.4 = 400\ \text{S cm}^{2}\ \text{mol}^{-1}$ ∴ $\alpha = \dfrac{90}{400} = 0.225$
Given below are two statements : Statement I : A hypothetical diatomic molecule with bond order zero is quite stable. Statement II : As bond order increases, the bond length increases. 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 false Statement I is false: a zero bond order means an unstable molecule; only a positive bond order means a stable one. Statement II is false: as bond order increases, bond length decreases.
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n = 2 \rightarrow n = 3$ and $n = 4 \rightarrow n = 6$ transitions, respectively, is
Solution: $\dfrac{1}{4}$ $\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ For $2 \rightarrow 3$: $\dfrac{1}{\lambda_1} = R_H\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = \dfrac{5R_H}{36}$ For $4 \rightarrow 6$: $\dfrac{1}{\lambda_2} = R_H\left(\dfrac{1}{16} - \dfrac{1}{36}\right) = \dfrac{5R_H}{144}$ ∴ $\dfrac{\lambda_1}{\lambda_2} = \dfrac{5/144}{5/36} = \dfrac{1}{4}$
The correct order of the wavelength of light absorbed by the following complexes is, A. $[\text{Co}(\text{NH}_3)_6]^{3+}$ B. $[\text{Co}(\text{CN})_6]^{3-}$ C. $[\text{Cu}(\text{H}_2\text{O})_4]^{2+}$ D. $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ Choose the correct answer from the options given below:
Solution: B < A < D < C A stronger field ligand gives a larger crystal field splitting, so light of shorter wavelength is absorbed. Absorption maxima: [Co(CN)₆]³⁻ 310 nm, [Co(NH₃)₆]³⁺ 475 nm, [Ti(H₂O)₆]³⁺ 498 nm, [Cu(H₂O)₄]²⁺ 600 nm. ∴ Increasing wavelength: B < A < D < C
If the rate constant of a reaction is 0.03 s$^{-1}$, how much time does it take for 7.2 mol L$^{-1}$ concentration of the reactant to get reduced to 0.9 mol L$^{-1}$? (Given: $\log 2 = 0.301$)
Solution: 69.3 s The unit of $k$ (s⁻¹) shows a first order reaction: $t = \dfrac{2.303}{k}\log\dfrac{[R]_0}{[R]}$ $t = \dfrac{2.303}{0.03}\log\dfrac{7.2}{0.9} = \dfrac{2.303}{0.03} \times 3 \times 0.301$ ∴ $t \approx 69.3\ \text{s}$ (three half-lives of 23.1 s each)
Match List I with List II List-I (Mixture): A. CHCl$_3$ + C$_6$H$_5$NH$_2$ B. Crude oil in petroleum industry C. Glycerol from spent-lye D. Aniline - water List-II (Method of separation): (I) Distillation under reduced pressure (II) Steam distillation (III) Fractional distillation (IV) Simple distillation Choose the correct answer from the options given below:
Solution: A-IV, B-III, C-I, D-II A: chloroform (b.p. 334 K) and aniline (b.p. 457 K) boil far apart, so simple distillation (IV). B: crude oil fractions have close boiling points, so fractional distillation (III). C: glycerol decomposes at its boiling point, so it is recovered from spent-lye by distillation under reduced pressure (I). D: aniline is steam volatile and immiscible with water, so steam distillation (II).
The major product of the following reaction is 3-benzoylpropanenitrile (C$_6$H$_5$COCH$_2$CH$_2$CN) treated with (i) CH$_3$MgBr (excess), then (ii) H$_3$O$^{+}$
Solution: 5-hydroxy-5-phenylhexan-2-one, C₆H₅C(CH₃)(OH)CH₂CH₂COCH₃ Excess CH₃MgBr adds to both the ketone C=O and the C≡N group. The ketone gives an alkoxide, which on hydrolysis becomes a tertiary alcohol, C₆H₅C(CH₃)(OH)–. The nitrile gives an imine salt, –C(CH₃)=NMgBr, which H₃O⁺ hydrolyses to a methyl ketone, –COCH₃. ∴ Product: C₆H₅C(CH₃)(OH)CH₂CH₂COCH₃
Which one of the following compounds can exist as cis-trans isomers?
Solution: 1,2-Dimethylcyclohexane 1,2-Dimethylcyclohexane has one methyl on each of two ring carbons, and the ring prevents rotation, so the methyls can be on the same side (cis) or opposite sides (trans). (B) Pent-1-ene has two H atoms on the terminal doubly bonded carbon, so no cis-trans isomers. (C) 2-Methylhex-2-ene has two methyl groups on C-2, so no cis-trans isomers. (D) 1,1-Dimethylcyclopropane has both methyls on the same carbon, so no cis-trans isomers.
Among the following, choose the ones with equal number of atoms. A. 212 g of Na$_2$CO$_3$(s) [molar mass = 106 g] B. 248 g of Na$_2$O(s) [molar mass = 62 g] C. 240 g of NaOH(s) [molar mass = 40 g] D. 12 g of H$_2$(g) [molar mass = 2 g] E. 220 g of CO$_2$(g) [molar mass = 44 g] Choose the correct answer from the options given below :
Solution: A, B and D only Number of atoms = moles × atoms per formula unit × $N_A$. A: $\dfrac{212}{106} \times 6 = 12$; B: $\dfrac{248}{62} \times 3 = 12$; D: $\dfrac{12}{2} \times 2 = 12$ (each $\times N_A$) C: $\dfrac{240}{40} \times 3 = 18$; E: $\dfrac{220}{44} \times 3 = 15$ (each $\times N_A$) ∴ A, B and D contain equal numbers of atoms.
Among the given compounds I-III, the correct order of bond dissociation energy of C–H bond marked with * is : I. the ring C–H of benzene II. the terminal alkyne C–H of phenylacetylene, C$_6$H$_5$C$\equiv$C–H III. a CH$_2$ hydrogen of cyclopropene's saturated carbon
Solution: II > I > III The more s character in the carbon's hybrid orbital, the shorter and stronger the C–H bond. II: sp carbon (50% s); I: sp² carbon (33% s); III: sp³ carbon (25% s). ∴ Bond dissociation energy: II > I > III
The standard heat of formation, in kcal/mol of Ba$^{2+}$ is : [Given : standard heat of formation of SO$_4^{2-}$ ion (aq) = $-216$ kcal/mol, standard heat of crystallisation of BaSO$_4$(s) = $-4.5$ kcal/mol, standard heat of formation of BaSO$_4$(s) = $-349$ kcal/mol]
Solution: −128.5 kcal/mol Formation of BaSO₄(s) = formation of the aqueous ions + crystallisation, Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Hess's law: ΔfH°(BaSO₄) = ΔfH°(Ba²⁺) + ΔfH°(SO₄²⁻) + ΔH°(crystallisation) $-349 = x - 216 - 4.5$ ∴ Standard heat of formation of Ba²⁺(aq), $x = -128.5\ \text{kcal mol}^{-1}$
Consider the following compounds : KO$_2$, H$_2$O$_2$ and H$_2$SO$_4$ The oxidation state of the underlined elements (K in KO$_2$, O in H$_2$O$_2$, S in H$_2$SO$_4$) in them are, respectively,
Solution: +1, −1 and +6 K in KO₂: an alkali metal is always +1 (O₂⁻ is the superoxide ion). O in H₂O₂: a peroxide, H–O–O–H, so each O is −1. S in H₂SO₄: $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$
Which one of the following reactions does NOT give benzene as the product?
Solution: Benzenediazonium chloride warmed with water: it gives phenol, not benzene Benzenediazonium chloride warmed with water gives phenol: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl. (B) Sodium benzoate heated with soda lime loses CO₂ (decarboxylation) and gives benzene. (C) n-Hexane over molybdenum oxide at 773 K and 10–20 atm is cyclised and dehydrogenated (aromatised) to benzene. (D) Ethyne passed through a red hot iron tube at 873 K undergoes cyclic polymerisation to benzene.
Which of the following are paramagnetic? A. [NiCl$_4$]$^{2-}$ B. Ni(CO)$_4$ C. [Ni(CN)$_4$]$^{2-}$ D. [Ni(H$_2$O)$_6$]$^{2+}$ E. Ni(PPh$_3$)$_4$ Choose the correct answer from the options given below :
Solution: A and D only A: [NiCl₄]²⁻ has Ni²⁺ (3d⁸) with weak field Cl⁻, tetrahedral (sp³), two unpaired electrons: paramagnetic. D: [Ni(H₂O)₆]²⁺ has Ni²⁺ (3d⁸), octahedral (sp³d²), two unpaired electrons: paramagnetic. B and E: Ni(0) in Ni(CO)₄ and Ni(PPh₃)₄ is 3d¹⁰ after pairing (sp³), no unpaired electrons: diamagnetic. C: [Ni(CN)₄]²⁻ is square planar (dsp²) with strong field CN⁻, all electrons paired: diamagnetic.
Which one of the following compounds does not decolourize bromine water?
Solution: Cyclohexane Cyclohexane is a saturated hydrocarbon with no activated ring, so it does not react with bromine water and the colour stays. (A) Aniline is brominated at once to 2,4,6-tribromoaniline, a white precipitate, and the colour is discharged. (C) Phenol gives a white precipitate of 2,4,6-tribromophenol, and the colour is discharged. (D) Styrene adds Br₂ across its C=C bond, and the colour is discharged.
Match List-I with List-II. List-I: A. Haber process B. Wacker oxidation C. Wilkinson catalyst D. Ziegler catalyst List-II: I. Fe catalyst II. PdCl$_2$ III. [(PPh$_3$)$_3$RhCl] IV. TiCl$_4$ with Al(CH$_3$)$_3$ Choose the correct answer from the options given below :
Solution: A-I, B-II, C-III, D-IV A: the Haber process for ammonia uses an iron catalyst (I). B: Wacker oxidation of ethene to ethanal is catalysed by PdCl₂ (II). C: Wilkinson catalyst is [(PPh₃)₃RhCl] (III), used to hydrogenate alkenes. D: TiCl₄ with Al(CH₃)₃ is the Ziegler catalyst (IV), used to make polythene.
Match List-I with List-II. List-I (Name of Vitamin): A. Vitamin B$_{12}$ B. Vitamin D C. Vitamin B$_2$ D. Vitamin B$_6$ List-II (Deficiency disease): I. Cheilosis II. Convulsions III. Rickets IV. Pernicious anaemia Choose the correct answer from the options given below:
Solution: A-IV, B-III, C-I, D-II A: vitamin B₁₂ deficiency causes pernicious anaemia (IV). B: vitamin D deficiency causes rickets (III). C: vitamin B₂ (riboflavin) deficiency causes cheilosis, cracking at the corners of the mouth and lips (I). D: vitamin B₆ (pyridoxine) deficiency causes convulsions (II).
Given below are two statements : Statement I : Ferromagnetism is considered as an extreme form of paramagnetism. Statement II : The number of unpaired electrons in a Cr$^{2+}$ ion (Z = 24) is the same as that of a Nd$^{3+}$ ion (Z = 60). 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: ferromagnetism is an extreme form of paramagnetism. Statement II is false: Cr²⁺ is 3d⁴ with 4 unpaired electrons, while Nd³⁺ is 4f³ with 3 unpaired electrons.
If the half-life ($t_{1/2}$) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to :
Solution: 10 minutes First order: $t = \dfrac{2.303}{k}\log\dfrac{[R]_0}{[R]}$, with $k = \dfrac{0.693}{t_{1/2}}$. For 99.9% completion, $\dfrac{[R]_0}{[R]} = 1000$, so $t = \dfrac{2.303 \times 3}{0.693}\,t_{1/2} \approx 10\,t_{1/2}$. ∴ $t \approx 10 \times 1 = 10\ \text{minutes}$
The correct order of decreasing basic strength of the given amines is:
Solution: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine Lower pKb means a stronger base; pKb values are N-ethylethanamine 3.00, ethanamine 3.29, N-methylaniline 9.30, benzenamine 9.38. Aryl amines are weaker because the lone pair on N is delocalised into the benzene ring; alkyl groups raise basicity by their +I effect. ∴ N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
Match List I with List II List-I (Ion): A. Co$^{2+}$ B. Mg$^{2+}$ C. Pb$^{2+}$ D. Al$^{3+}$ List-II (Group Number in Cation Analysis): I. Group-I II. Group-III III. Group-IV IV. Group-VI Choose the correct answer from the options given below :
Solution: A-III, B-IV, C-I, D-II A: Co²⁺ is precipitated as CoS by H₂S in ammoniacal medium, Group IV (III). B: Mg²⁺ remains to the last, Group VI (IV). C: Pb²⁺ is precipitated as PbCl₂ by dilute HCl, Group I (I). D: Al³⁺ is precipitated as Al(OH)₃ by NH₄OH in the presence of NH₄Cl, Group III (II).
Phosphoric acid ionizes in three steps with their ionization constant values $K_{a_1}$, $K_{a_2}$ and $K_{a_3}$, respectively, while K is the overall ionization constant. Which of the following statements are true? A. $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$ B. H$_3$PO$_4$ is a stronger acid than H$_2$PO$_4^{-}$ and HPO$_4^{2-}$ C. $K_{a_1} > K_{a_2} > K_{a_3}$ D. $K_{a_1} = \dfrac{K_{a_3} + K_{a_2}}{2}$ Choose the correct answer from the options given below :
Solution: Statements A, B and C only The overall constant is the product of the stepwise constants, $K = K_{a_1}K_{a_2}K_{a_3}$, so $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$ (A). Each step removes H⁺ from a more negatively charged species, which is harder, so $K_{a_1} > K_{a_2} > K_{a_3}$ and H₃PO₄ is the strongest acid of the three (B, C). (D) $K_{a_1}$ is not the average of $K_{a_2}$ and $K_{a_3}$; it is far larger than both.
Which of the following statements are true? A. Unlike Ga that has a very high melting point, Cs has a very low melting point. B. On Pauling scale, the electronegativity values of N and Cl are not the same. C. Ar, K$^{+}$, Cl$^{-}$, Ca$^{2+}$, and S$^{2-}$ are all isoelectronic species. D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na. E. The atomic radius of Cs is greater than that of Li and Rb. Choose the correct answer from the options given below :
Solution: Statements C and E only C and E are true: Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ all have 18 electrons, and atomic radius increases down group 1, so Cs is larger than Rb and Li. (A) Ga also has a very low melting point (303 K), close to that of Cs (302 K). (B) N and Cl have the same electronegativity on the Pauling scale, 3.0. (D) The order is Si > Mg > Al > Na: Mg is higher than Al because its 3s electron penetrates nearer the nucleus than the 3p electron of Al.
Given below are two statements : Statement I : Like nitrogen that can form ammonia, arsenic can form arsine. Statement II : Antimony cannot form antimony pentoxide. 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: all group 15 elements form hydrides of the type EH₃, so nitrogen gives NH₃ and arsenic gives arsine, AsH₃. Statement II is incorrect: group 15 elements form oxides E₂O₃ and E₂O₅, and antimony forms antimony pentoxide, Sb₂O₅.
Which of the following aqueous solution will exhibit highest boiling point?
Solution: 0.01 M Na₂SO₄ Elevation of boiling point, $\Delta T_b = iK_bm$, so the solution with the largest $i \times m$ boils highest. Glucose: $1 \times 0.015 = 0.015$; urea: $1 \times 0.01 = 0.01$ KNO₃: $2 \times 0.01 = 0.02$; Na₂SO₄: $3 \times 0.01 = 0.03$ ∴ 0.01 M Na₂SO₄ has the highest boiling point.
Given below are two statements : Statement-I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273 – 278 K. It decomposes easily in the dry state. Statement-II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI. 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: aniline with nitrous acid (NaNO₂ + HCl) at 273–278 K gives benzenediazonium chloride, which decomposes easily in the dry state. Statement II is correct: iodine is not easily introduced into the benzene ring directly, so iodobenzene is made by treating the diazonium salt with KI: C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + KCl + N₂.
Identify the suitable reagent for the following conversion. Methyl benzoate, C$_6$H$_5$COOCH$_3$, is converted to benzaldehyde, C$_6$H$_5$CHO.
Solution: (i) AlH(iBu)₂ (DIBAL-H), (ii) H₂O DIBAL-H, AlH(iBu)₂, reduces an ester only as far as the aldehyde: C₆H₅COOCH₃ → C₆H₅CHO. (A) H₂/Pd-BaSO₄ (Rosenmund reduction) reduces acid chlorides, not esters. (B) LiAlH₄ reduces the ester all the way to the primary alcohol, C₆H₅CH₂OH. (D) NaBH₄ does not reduce esters.
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The primary alkyl iodide shown undergoes S$_N$2 reaction faster than the corresponding primary alkyl chloride of the same carbon skeleton. Reason (R) : Iodine is a better leaving group because of its large size. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both A and R are true and R is the correct explanation of A Assertion is true: in S$_{\text{N}}$2 the C–X bond breaks as the nucleophile attacks, so for the same alkyl group the order of reactivity is R–I > R–Br > R–Cl. Reason is true and explains it: the large iodide ion carries its charge well, so I⁻ is the best leaving group among the halides.
The correct order of decreasing acidity of the following aliphatic acids is
Solution: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH Electron donating alkyl groups (+I effect) destabilise the carboxylate ion and lower the acidity. HCOOH has no alkyl group, and the number of methyl groups then rises: CH₃– < (CH₃)₂CH– < (CH₃)₃C–. ∴ Acidity: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH
Which one of the following reactions does NOT belong to "Lassaigne's test"?
Solution: 2CuO + C → 2Cu + CO₂ (detection of carbon) Heating with copper(II) oxide oxidises carbon to CO₂; this detects carbon and is not part of Lassaigne's test, in which the compound is fused with sodium metal. (B) Na + C + N → NaCN converts nitrogen to sodium cyanide. (C) 2Na + S → Na₂S converts sulphur to sodium sulphide. (D) Na + X → NaX converts a halogen to the sodium halide.
How many products (including stereoisomers) are expected from monochlorination of the following compound? (CH$_3$)$_2$CH–CH$_2$–CH$_3$
Solution: 6 2-Methylbutane has four kinds of hydrogen, so there are four structural monochloro products. 1-Chloro-2-methylbutane, ClCH₂CH(CH₃)CH₂CH₃, has a chiral carbon: 2 stereoisomers. 2-Chloro-3-methylbutane, (CH₃)₂CHCHClCH₃, has a chiral carbon: 2 stereoisomers. 2-Chloro-2-methylbutane and 1-chloro-3-methylbutane have no chiral carbon: 1 each. ∴ Total $= 2 + 2 + 1 + 1 = 6$
Sugar 'X' A. is found in honey B. is a keto sugar C. exists in $\alpha$ and $\beta$ – anomeric forms. D. Is laevorotatory. 'X' is :
Solution: D-Fructose D-Fructose is a ketohexose found in honey and fruits. Its five-membered (furanose) ring exists as two anomers, α-D-(–)-fructofuranose and β-D-(–)-fructofuranose. It is laevorotatory, hence written D-(–)-fructose. Sucrose and maltose are disaccharides, and D-glucose is an aldose that is dextrorotatory, so none of them fits.
Dalton's Atomic theory could not explain which of the following?
Solution: Law of gaseous volumes Dalton's theory explained the laws of conservation of mass, constant proportions and multiple proportions, but could not explain the law of gaseous volumes.
Higher yield of NO in $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$ can be obtained at [$\Delta$H of the reaction = +180.7 kJ mol$^{-1}$] A. Higher temperature B. Lower temperature C. Higher concentration of N$_2$ D. Higher concentration of O$_2$ Choose the correct answer from the options given below :
Solution: Higher temperature and higher concentrations of N₂ and O₂ (A, C, D only) Le Chatelier's principle: the reaction is endothermic (ΔH = +180.7 kJ mol⁻¹), so a higher temperature shifts the equilibrium forward (A). Adding more of either reactant, N₂ or O₂, also shifts the equilibrium forward (C, D). (B) A lower temperature favours the backward, exothermic reaction and lowers the yield of NO.
Match List-I with List-II List-I: A. XeO$_3$ B. XeF$_2$ C. XeOF$_4$ D. XeF$_6$ List-II: (I) $sp^{3}d$; linear (II) $sp^{3}$; pyramidal (III) $sp^{3}d^{3}$; distorted octahedral (IV) $sp^{3}d^{2}$; square pyramidal Choose the correct answer from the options given below :
Solution: A-II, B-I, C-IV, D-III A: XeO₃ has 3 bond pairs and 1 lone pair: sp³, pyramidal (II). B: XeF₂ has 2 bond pairs and 3 lone pairs: sp³d, linear (I). C: XeOF₄ has 5 bond pairs and 1 lone pair: sp³d², square pyramidal (IV). D: XeF₆ has 6 bond pairs and 1 lone pair: sp³d³, distorted octahedral (III).
Match List-I with List-II List-I (Example): A. Humidity B. Alloys C. Amalgams D. Smoke List-II (Type of Solution): I. Solid in solid II. Liquid in gas III. Solid in gas IV. Liquid in solid Choose the correct answer from the options given below:
Solution: A-II, B-I, C-IV, D-III A: humidity is water in air, a liquid in gas (II). B: an alloy is a solid in solid (I). C: an amalgam is mercury dissolved in a metal, a liquid in solid (IV). D: smoke is solid particles in air, a solid in gas (III).
Energy and radius of first Bohr orbit of He$^{+}$ and Li$^{2+}$ are [Given R$_H$ = $2.18 \times 10^{-18}$ J, $a_0$ = 52.9 pm]
Solution: Li²⁺: $E_1 = -19.62 \times 10^{-18}$ J, $r_1 = 17.6$ pm; He⁺: $E_1 = -8.72 \times 10^{-18}$ J, $r_1 = 26.4$ pm For hydrogen-like species, $E_n = -R_H\dfrac{Z^2}{n^2}$ and $r_n = \dfrac{a_0 n^2}{Z}$. He⁺ ($Z = 2$): $E_1 = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18}\ \text{J}$; $r_1 = \dfrac{52.9}{2} = 26.45\ \text{pm}$ Li²⁺ ($Z = 3$): $E_1 = -2.18 \times 10^{-18} \times 9 = -19.62 \times 10^{-18}\ \text{J}$; $r_1 = \dfrac{52.9}{3} = 17.63\ \text{pm}$
Which among the following electronic configurations belong to main group elements? A. [Ne]3s$^{1}$ B. [Ar]3d$^{3}$4s$^{2}$ C. [Kr]4d$^{10}$5s$^{2}$5p$^{5}$ D. [Ar]3d$^{10}$4s$^{1}$ E. [Rn]5f$^{0}$6d$^{2}$7s$^{2}$ Choose the correct answer from the option given below :
Solution: A and C only Main group elements are the s- and p-block elements: [Ne]3s¹ is Na (s-block) and [Kr]4d¹⁰5s²5p⁵ is I (p-block). (B) [Ar]3d³4s² is V, a d-block element. (D) [Ar]3d¹⁰4s¹ is Cu, a d-block element. (E) [Rn]5f⁰6d²7s² is Th, an f-block element (actinoid).
C(s) + 2H$_2$(g) $\rightarrow$ CH$_4$(g); $\Delta$H = $-74.8$ kJ mol$^{-1}$. Which of the following diagrams gives an accurate representation of the above reaction? [R $\rightarrow$ reactants; P $\rightarrow$ products]
Solution: Energy profile with the products below the reactants, the gap marked 74.8 ΔH = −74.8 kJ mol⁻¹, so the reaction is exothermic and the products lie below the reactants. The 74.8 kJ mol⁻¹ is the drop from the reactant level to the product level, after the rise over the activation barrier.
Predict the major product 'P' in the following sequence of reactions- 1-Methylcyclopentene, treated with (i) HBr, benzoyl peroxide; (ii) KCN; (iii) Na(Hg)/C$_2$H$_5$OH, gives P (Major).
Solution: (2-Methylcyclopentyl)methanamine: CH₃ on one ring carbon and CH₂NH₂ on the adjacent one HBr with benzoyl peroxide adds against Markovnikov's rule (peroxide effect), so Br goes to the ring carbon next to the one carrying CH₃: 1-bromo-2-methylcyclopentane. KCN replaces Br by CN (nucleophilic substitution): 2-methylcyclopentanecarbonitrile. Na(Hg)/C₂H₅OH reduces –CN to –CH₂NH₂. ∴ P is (2-methylcyclopentyl)methanamine.
Identify the correct orders against the property mentioned A. H$_2$O > NH$_3$ > CHCl$_3$ – dipole moment B. XeF$_4$ > XeO$_3$ > XeF$_2$ – number of lone pairs on central atom C. O–H > C–H > N–O – bond length D. N$_2$ > O$_2$ > H$_2$ – bond enthalpy Choose the correct answer from the options given below:
Solution: A, D only A is correct: dipole moments are H₂O 1.85 D, NH₃ 1.47 D, CHCl₃ 1.04 D. D is correct: bond enthalpies are N₂ 946 kJ mol⁻¹, O₂ 498 kJ mol⁻¹, H₂ 435.8 kJ mol⁻¹. (B) Lone pairs on Xe are XeF₂ 3, XeF₄ 2, XeO₃ 1, so the given order is wrong. (C) The bond length order is N–O > C–H > O–H, the reverse of the given order.
Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C$_4$H$_8$O is :
Solution: 10 Five-membered ring: tetrahydrofuran (oxolane), 1. Four-membered rings: 2-methyloxetane (chiral, 2) and 3-methyloxetane (1). Three-membered rings: 2-ethyloxirane (chiral, 2), 2,2-dimethyloxirane (1), cis-2,3-dimethyloxirane (meso, 1) and trans-2,3-dimethyloxirane (chiral, 2). ∴ Total $= 1 + 3 + 6 = 10$
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given : R = 0.0831 L atm mol$^{-1}$ K$^{-1}$] K$_P$ for the reaction at 1000 K is
Solution: 0.033 At equilibrium, $K_c = \dfrac{k_f}{k_b} = \dfrac{1}{2500}$. $K_p = K_c(RT)^{\Delta n_g}$, with $\Delta n_g = 2 - 1 = 1$. ∴ $K_p = \dfrac{1}{2500} \times 0.0831 \times 1000 \approx 0.033$
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
Solution: The solution shows negative deviation By Raoult's law an ideal solution would have $p = \tfrac{5}{15} \times 63 + \tfrac{10}{15} \times 78 = 21 + 52 = 73$ torr; the observed 70 torr is lower, so the deviation is negative. (A) Negative deviation comes from stronger X–Y attractions, so the volume of the solution is slightly less, not greater, than the sum. (B) Positive deviation would need a vapour pressure above 73 torr. (D) An ideal solution would show exactly 73 torr.
Which of the following is the unit of productivity of an Ecosystem?
Solution: (kcal m⁻²) yr⁻¹ Productivity is the rate of biomass production, so it is expressed per unit area per unit time: g m⁻² yr⁻¹ or (kcal m⁻²) yr⁻¹.
The first menstruation is called :
Solution: Menarche The first menstruation, which begins at puberty, is called menarche.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : All vertebrates are chordates but all chordates are not vertebrate. Reason (R) : The members of subphylum vertebrata possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults. 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 Assertion is true: every vertebrate has a notochord at some stage and is therefore a chordate, but protochordates are chordates that are not vertebrates. Reason is true and explains it: in vertebrates the embryonic notochord is replaced by a cartilaginous or bony vertebral column in the adult.
Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?
Solution: 9 : 3 : 3 : 1 RRYY × rryy gives F₁ RrYy (round yellow), which forms four kinds of gametes, RY, Ry, rY and ry, in equal numbers because the genes assort independently. ∴ F₂: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green
Given below are two statements : Statement I : The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA. Statement II : Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel. 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: separated DNA bands are cut out of the agarose gel and extracted (elution), and these purified fragments are joined to cloning vectors to make recombinant DNA. Statement II is correct: negatively charged DNA moves towards the anode, and smaller fragments move farther, so larger ones stay near the wells.
What is the main function of the spindle fibers during mitosis?
Solution: To separate the chromosomes Spindle fibres attach to the kinetochores of the chromosomes and pull the chromatids apart to opposite poles.
How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?
Solution: 1 meiosis and 3 mitosis The megaspore mother cell divides once by meiosis to form four megaspores, of which one is functional. The nucleus of the functional megaspore divides by mitosis three times in succession, giving the 8-nucleate, 7-celled embryo sac.
Identify the statement that is NOT correct.
Solution: Antigen binding site is located at C-terminal region of antibody molecules (the incorrect statement) The antigen binding sites are formed by the variable regions at the N-terminal ends of the heavy and light chains, not at the C-terminal end. (A) The constant regions of the heavy and light chains do lie towards the C-terminal end, so this statement is correct. (B) Each antibody has two light and two heavy chains (H₂L₂), so this statement is correct. (C) The chains are held together by disulphide bonds, so this statement is correct.
Consider the following : A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis. B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females. C. The first polar body is associated with the formation of the primary oocyte. D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding. Choose the correct answer from the options given below:
Solution: A and B are true A is true: meiosis begins in primary oocytes before birth, but in spermatocytes only after puberty. B is true: a secondary spermatocyte soon completes the second meiotic division, while the secondary oocyte completes it only after a sperm enters. (C) The first polar body forms along with the secondary oocyte, when the primary oocyte completes meiosis I. (D) The LH surge causes ovulation; menstruation follows when the corpus luteum degenerates and progesterone falls.
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus. Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells. In light of the above statements, choose the most appropriate answer from the options given below:
Solution: A is true but R is false Assertion is true: tapetal cells have dense cytoplasm and generally more than one nucleus. Reason is false: the tapetum nourishes the developing pollen grains, not the microspore mother cells.
The blue and white selectable markers have been developed which differentiate recombinant colonies from non-recombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate. Given below are two statements about this method: Statement I : The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies. Statement II : The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies. 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: an insert in the β-galactosidase gene inactivates it (insertional inactivation), so recombinant colonies give no colour; blue colonies are non-recombinants. Statement II is correct: colonies without blue colour carry the insert and are identified as recombinants.
In bryophytes, the gemmae help in which one of the following?
Solution: Asexual reproduction Gemmae are green, multicellular, asexual buds formed in gemma cups; they detach and grow into new plants.
Match List I with List II: List-I: A. Adenosine B. Adenylic acid C. Adenine D. Alanine List-II: I. Nitrogen base II. Nucleotide III. Nucleoside IV. Amino acid Choose the option with all correct matches.
Solution: A-III, B-II, C-I, D-IV A: adenosine is a nucleoside, adenine joined to a sugar (III). B: adenylic acid is a nucleotide, a nucleoside with a phosphate esterified to the sugar (II). C: adenine is a nitrogen base, a purine (I). D: alanine is an amino acid with a methyl group as its R group (IV).
With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in F$_3$ generation.
Solution: 1/4 In F₁ a carrier female and a normal male have an affected son, so the disorder is X-linked recessive. The consanguineous mating is between a carrier female, $X^{c}X$, and an affected male, $X^{c}Y$. Children: $X^{c}X^{c}$ (affected female), $X^{c}Y$ (affected male), $X^{c}X$ (carrier female), $XY$ (normal male). ∴ Probability of a child with no disease who is a carrier $= \dfrac{1}{4}$
Consider the following statements regarding function of adrenal medullary hormones : (A) It causes pupilary constriction. (B) It is a hyperglycemic hormone. (C) It causes piloerection. (D) It increases strength of heart contraction. Choose the correct answer from the options given below :
Solution: B, C and D only Adrenaline and noradrenaline raise blood glucose by breaking down glycogen (B), cause piloerection (C) and increase the strength of heart contraction (D). (A) They cause dilation of the pupils, not constriction.
Which of the following is an example of a zygomorphic flower?
Solution: Pea A zygomorphic flower can be divided into two similar halves in only one vertical plane; pea has such flowers. Chilli, petunia and datura have actinomorphic flowers, which can be divided into two similar halves by any radial plane through the centre.
Who proposed that the genetic code for amino acids should be made up of three nucleotides?
Solution: George Gamow George Gamow, a physicist, argued that the code must use three nucleotides, since four bases taken three at a time give $4^3 = 64$ combinations, enough for 20 amino acids.
Given below are two statements: Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers. Statement II: Ecosystems are exempted from 2$^{\text{nd}}$ law of thermodynamics. 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: energy flows one way, from the sun to producers and then to consumers. Statement II is incorrect: ecosystems obey the second law of thermodynamics and need a constant supply of energy to counter the tendency towards increasing disorder.
Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.
Solution: Analogy, convergent Sweet potato is a modified root and potato a modified stem, yet both store food: different structures, same function. Such analogous organs are the result of convergent evolution.
All living members of the class Cyclostomata are :
Solution: Ectoparasite All living cyclostomes are ectoparasites on some fishes, holding on with their circular, jawless sucking mouth.
Histones are enriched with -
Solution: Lysine & Arginine Histones are positively charged, basic proteins rich in lysine and arginine, whose positive side chains bind the negatively charged DNA.
Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?
Solution: $\dfrac{dN}{dt} = rN\left(\dfrac{K - N}{K}\right)$ Verhulst-Pearl logistic growth: $\dfrac{dN}{dt} = rN\left(\dfrac{K - N}{K}\right)$, where $N$ is population density, $r$ the intrinsic rate of natural increase and $K$ the carrying capacity. Growth slows as $N$ approaches $K$ and stops when $N = K$.
Given below are two statements : one is labelled as Assertion (A), and the other is labelled as Reason (R). Assertion (A) : The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell. Reason (R) : Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus. 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 Assertion is true: the Golgi apparatus principally packages materials made by the ER and delivers them to intracellular targets or outside the cell. Reason is true: ER vesicles fuse with the cis (forming) face, and the modified materials leave from the trans (maturing) face. The Reason only describes the route through the Golgi; it does not explain why packaging and delivery is its primary function.
Which of the following statements about RuBisCO is true?
Solution: It catalyses the carboxylation of RuBP RuBisCO (RuBP carboxylase-oxygenase) catalyses the carboxylation of RuBP, the first step of the Calvin cycle. (B) It is not active only in the dark: the Calvin cycle goes on in the light, using ATP and NADPH from the light reaction. (C) RuBisCO has a much greater affinity for CO₂ than for O₂; their relative concentrations decide which binds. (D) Photolysis of water is linked to photosystem II, not to RuBisCO.
Match List-I with List-II. List-I: A. Progesterone B. Relaxin C. Melanocyte stimulating hormone D. Catecholamines List-II: I. Pars intermedia II. Ovary III. Adrenal Medulla IV. Corpus luteum Choose the correct answer from the options given below :
Solution: A-IV, B-II, C-I, D-III A: progesterone is secreted mainly by the corpus luteum (IV). B: relaxin is secreted by the ovary in the later phase of pregnancy (II). C: melanocyte stimulating hormone comes from the pars intermedia of the pituitary (I). D: catecholamines (adrenaline and noradrenaline) come from the adrenal medulla (III).
The protein portion of an enzyme is called:
Solution: Apoenzyme When a non-protein cofactor is needed for activity, the protein portion of the enzyme is called the apoenzyme.
Which of the following enzyme(s) are NOT essential for gene cloning? A. Restriction enzymes B. DNA ligase C. DNA mutase D. DNA recombinase E. DNA polymerase Choose the correct answer from the options given below:
Solution: C and D only: DNA mutase and DNA recombinase DNA mutase and DNA recombinase play no part in the steps of gene cloning. (A) Restriction enzymes cut the DNA at specific sites, so they are essential. (B) DNA ligase joins the cut gene to the vector, so it is essential. (E) DNA polymerase copies the DNA, for example in PCR, so it is essential.
Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?
Solution: Innate Immunity Innate immunity is the non-specific defence present at birth; acquired immunity, both humoral and cell-mediated, is pathogen-specific and develops later.
Which factor is important for termination of transcription?
Solution: $\rho$ (rho) In prokaryotes, RNA polymerase associates transiently with the $\sigma$ factor to start transcription and with the $\rho$ factor to terminate it.
Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?
Solution: Anti-diuretic hormone (ADH) ADH (vasopressin) is synthesised in the hypothalamus and carried along axons to the posterior pituitary, which stores and releases it; ACTH, LH and FSH are made in the anterior pituitary itself.
Which of the following microbes is NOT involved in the preparation of household products? A. Aspergillus niger B. Lactobacillus C. Trichoderma polysporum D. Saccharomyces cerevisiae E. Propionibacterium sharmanii Choose the correct answer from the options given below:
Solution: A and C only: Aspergillus niger and Trichoderma polysporum Aspergillus niger is used in industry to produce citric acid and Trichoderma polysporum to produce cyclosporin A, so neither makes household products. (B) Lactobacillus turns milk into curd. (D) Saccharomyces cerevisiae (baker's yeast) ferments the dough for bread. (E) Propionibacterium sharmanii makes the large holes in Swiss cheese.
Given below are two statements : Statement I : Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it. Statement II : Fig wasp and fig tree exhibit mutual relationship as fig wasp completes its life cycle in fig fruit and fig fruit gets pollinated by fig wasp. In the light of the above statements, choose the most appropriate answer from the options given below :
Solution: Both statement I and statement II are incorrect Statement I is incorrect: the fig is a plant structure; the wasp only lays its eggs in it and pollinates it. Statement II is incorrect: the wasp pollinates the fig inflorescence (its flowers), not the fruit, while using the developing fruit as its egg-laying site.
Role of the water vascular system in Echinoderms is : A. Respiration and Locomotion B. Excretion and Locomotion C. Capture and transport of food D. Digestion and Respiration E. Digestion and Excretion Choose the correct answer from the options given below :
Solution: A and C only The water vascular system of echinoderms helps in locomotion, capture and transport of food, and respiration (A, C). (B) Echinoderms have no excretory system, and the water vascular system does not excrete. (D) Digestion takes place in the digestive tract, not in the water vascular system. (E) The water vascular system does neither digestion nor excretion.
After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s) / tissue(s) like A. thymus B. bone marrow C. spleen D. lymph nodes E. Peyer's patches Choose the correct answer from the options given below
Solution: Spleen, lymph nodes and Peyer's patches (C, D, E only) After maturing in the primary lymphoid organs, lymphocytes move to secondary lymphoid organs such as the spleen, lymph nodes, tonsils, Peyer's patches and appendix, where they interact with antigens. (A) The thymus is a primary lymphoid organ, where T-lymphocytes mature. (B) The bone marrow is a primary lymphoid organ, where all blood cells, including lymphocytes, are produced.
Match List I with List II: List I: A. The Evil Quartet B. Ex situ conservation C. Lantana camara D. Dodo List II: I. Cryopreservation II. Alien species invasion III. Causes of biodiversity losses IV. Extinction Choose the option with all correct matches.
Solution: A-III, B-I, C-II, D-IV A: the 'Evil Quartet' is the name for the four major causes of biodiversity loss (III). B: cryopreservation of gametes at very low temperature is an ex situ method (I). C: Lantana camara is an invasive alien weed in India (II). D: the dodo of Mauritius is a recent extinction (IV).
Read the following statements on plant growth and development. (A) Parthenocarpy can be induced by auxins. (B) Plant growth regulators can be involved in promotion as well as inhibition of growth. (C) Dedifferentiation is a pre-requisite for re-differentiation. (D) Abscisic acid is a plant growth promoter. (E) Apical dominance promotes the growth of lateral buds. Choose the option with all correct statements.
Solution: A, B, C only A, B and C are correct: auxins induce parthenocarpy (e.g. in tomatoes), plant growth regulators may promote or inhibit growth, and redifferentiation follows dedifferentiation. (D) Abscisic acid is a growth inhibitor, not a promoter. (E) In apical dominance the growing apical bud suppresses the growth of the lateral buds.
Match List-I with List-II. List-I: A. Pteridophyte B. Bryophyte C. Angiosperm D. Gymnosperm List-II: (I) Salvia (II) Ginkgo (III) Polytrichum (IV) Salvinia Choose the option with all correct matches.
Solution: A-IV, B-III, C-I, D-II A: Salvinia is a water fern, a pteridophyte (IV). B: Polytrichum is a moss, a bryophyte (III). C: Salvia is a flowering plant, an angiosperm (I). D: Ginkgo is a gymnosperm (II).
Why can't insulin be given orally to diabetic patients?
Solution: It will be digested in the gastro-intestinal (GI) tract Insulin is a protein, so taken by mouth it would be digested in the gut before it could act.
Which one of the following is the characteristic feature of gymnosperms?
Solution: Seeds are naked In gymnosperms the ovules are not enclosed by an ovary wall, so the seeds formed after fertilisation remain uncovered (naked).
Frogs respire in water by skin and buccal cavity and on land by skin, buccal cavity and lungs. Choose the correct answer from the following :
Solution: The statement is false for water but true for land In water, frogs respire only through the skin (cutaneous respiration), so the statement is false for water. On land, the buccal cavity, skin and lungs all act as respiratory organs, so it is true for land.
Silencing of specific mRNA is possible via RNAi because of
Solution: Complementary dsRNA In RNA interference a complementary dsRNA molecule binds to a specific mRNA and prevents its translation (silencing).
Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?
Solution: They are fraternal twins A boy and a girl cannot come from one zygote, so they must be fraternal (dizygotic) twins from two separately fertilised eggs. (A) Fraternal twins share on average about 50% of their genes, like ordinary siblings, not 75%. (B) Monozygotic twins come from one zygote and are always of the same sex. (D) Fraternal twins arise naturally when two eggs are released and fertilised; IVF is not needed.
Match List I with List II : List I: A. Scutellum B. Non-albuminous seed C. Epiblast D. Perisperm List II: I. Persistent nucellus II. Cotyledon of Monocot seed III. Groundnut IV. Rudimentary cotyledon Choose the option with all correct matches.
Solution: A-II, B-III, C-IV, D-I A: the scutellum is the single cotyledon of a monocot (grass) embryo (II). B: groundnut seeds have no residual endosperm, so they are non-albuminous (III). C: the epiblast is a rudimentary cotyledon in some grass embryos (IV). D: perisperm is the residual, persistent nucellus, as in black pepper and beet (I).
In frog, the Renal portal system is a special venous connection that acts to link :
Solution: Kidney and lower part of body In frogs the renal portal system links the kidney with the lower parts of the body, while the hepatic portal system links the liver and the intestine.
Match List-I with List-II. List-I: A. Heart B. Kidney C. Gastro-intestinal tract D. Adrenal Cortex List-II: (I) Erythropoietin (II) Aldosterone (III) Atrial natriuretic factor (IV) Secretin Choose the correct answer from the options given below :
Solution: A-III, B-I, C-IV, D-II A: the atrial wall of the heart secretes atrial natriuretic factor (III). B: the juxtaglomerular cells of the kidney secrete erythropoietin (I). C: the gastro-intestinal tract secretes secretin (IV). D: the adrenal cortex secretes aldosterone (II).
Cardiac activities of the heart are regulated by: A. Nodal tissue B. A special neural centre in the medulla oblongata C. Adrenal medullary hormones D. Adrenal cortical hormones Choose the correct answer from the options given below :
Solution: A, B and C only The heart is myogenic: its normal activity is auto-regulated by the nodal tissue (A). A special neural centre in the medulla oblongata moderates cardiac function through the autonomic nervous system (B). Adrenal medullary hormones increase the cardiac output (C). (D) Adrenal cortical hormones do not directly regulate cardiac activity; among adrenal hormones, the medullary ones raise heart rate and output.
Streptokinase produced by bacterium Streptococcus is used for
Solution: Removing clots from blood vessels Streptokinase from Streptococcus, modified by genetic engineering, is used as a 'clot buster' to remove clots from the blood vessels of patients after myocardial infarction.
Who is known as the father of Ecology in India?
Solution: Ramdeo Misra Ramdeo Misra is revered as the Father of Ecology in India; he established teaching and research in ecology at Banaras Hindu University.
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled. Reason (R) : The egg apparatus has 2 polar nuclei. In the light of the above statements, choose the correct answer from the options given below :
Solution: A is true but R is false Assertion is true: the mature embryo sac has 8 nuclei in 7 cells (3 antipodals, 2 synergids, 1 egg cell and 1 central cell with 2 polar nuclei). Reason is false: the egg apparatus is the egg cell and two synergids; the 2 polar nuclei lie in the large central cell.
Neoplastic characteristics of cells refer to : A. A mass of proliferating cell B. Rapid growth of cells C. Invasion and damage to the surrounding tissue D. Those confined to original location Choose the correct answer from the options given below:
Solution: A, B, C only Neoplastic (malignant tumour) cells form a mass of proliferating cells that grow very rapidly and invade and damage the surrounding normal tissue (A, B, C). (D) Staying confined to the original location is a feature of benign tumours, not of neoplastic cells.
Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence. A. Prothallus stage B. Meiosis in spore mother cells C. Fertilisation D. Formation of archegonia and antheridia in gametophyte. E. Transfer of antherozoids to the archegonia in presence of water. Choose the correct answer from the options given below:
Solution: Sequence B, A, D, E, C Spore mother cells in the sporangia undergo meiosis to form spores (B). A spore germinates into the prothallus, the gametophyte (A), which bears antheridia and archegonia (D). Water carries the antherozoids to the archegonia (E), where fertilisation forms the zygote (C). ∴ B, A, D, E, C
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Both wind and water pollinated flowers are not very colourful and do not produce nectar. Reason (R) : The flowers produce enormous amount of pollen grains in wind and water pollinated flowers. 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 Assertion is true: wind and water pollinated flowers need not attract animals, so they are not very colourful and do not produce nectar. Reason is true: they produce enormous numbers of pollen grains, since pollination by wind or water is a chance event. Plenty of pollen does not cause the lack of colour and nectar; both come from not depending on animal pollinators.
Which one of the following enzymes contains 'Haem' as the prosthetic group?
Solution: Catalase In catalase and peroxidase, which break down hydrogen peroxide to water and oxygen, haem is the prosthetic group and part of the active site. (B) RuBisCO, the most abundant protein in the biosphere, has no haem group. (C) Carbonic anhydrase has zinc as its cofactor. (D) Succinate dehydrogenase uses FAD as its prosthetic group.
Match List-I with List-II. List-I: A. Emphysema B. Angina Pectoris C. Glomerulonephritis D. Tetany List-II: I. Rapid spasms in muscle due to low Ca$^{++}$ in body fluid II. Damaged alveolar walls and decreased respiratory surface III. Acute chest pain when not enough oxygen is reaching to heart muscle IV. Inflammation of glomeruli of kidney Choose the correct answer from the options given below :
Solution: A-II, B-III, C-IV, D-I A: in emphysema the alveolar walls are damaged, so the respiratory surface decreases (II). B: angina pectoris is acute chest pain when not enough oxygen reaches the heart muscle (III). C: glomerulonephritis is inflammation of the glomeruli of the kidney (IV). D: tetany is rapid spasms in muscle due to low Ca²⁺ in body fluid (I).
Find the statement that is NOT correct with regard to the structure of monocot stem.
Solution: Hypodermis is parenchymatous (the incorrect statement) In a monocot stem the hypodermis is sclerenchymatous, not parenchymatous. (A) Phloem parenchyma is absent in the monocot stem, so this statement is correct. (C) The vascular bundles are scattered in the ground tissue, so this statement is correct. (D) The vascular bundles are conjoint and closed (no cambium), so this statement is correct.
Which of the following statement is correct about location of the male frog copulatory pad?
Solution: First digit of the fore limb Male frogs have a copulatory pad on the first digit of each fore limb; it is absent in females.
Given below are two statements: Statement I : The primary source of energy in an ecosystem is solar energy. Statement II : The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP). 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: solar energy is the primary source of energy in an ecosystem. Statement II is incorrect: the rate of production of organic matter during photosynthesis is gross primary productivity; NPP is what remains after respiration losses (NPP = GPP − R).
Identify the part of a bio-reactor which is used as a foam braker from the given figure.
Solution: Part C, the foam breaker In the stirred-tank bioreactor shown, the part labelled C is the foam breaker. The other labels are B, the motor; A, the flat bladed impeller; and D, the sterile air supply.
Polymerase chain reaction (PCR) amplifies DNA following the equation.
Solution: $2^{n}$ Each PCR cycle doubles the DNA, so after $n$ cycles one molecule gives $2^n$ copies (3 cycles give $2^3 = 8$).
Match List - I with List - II. List - I: A. Head B. Middle piece C. Acrosome D. Tail List - II: (I) Enzymes (II) Sperm motility (III) Energy (IV) Genetic material Choose the correct answer from the options given below :
Solution: A-IV, B-III, C-I, D-II A: the head contains the elongated haploid nucleus, the genetic material (IV). B: the middle piece has numerous mitochondria that produce energy for movement (III). C: the acrosome is a cap filled with enzymes that help fertilisation (I). D: the tail gives the sperm its motility (II).
Given below are two statements: Statement I: In a floral formula $\oplus$ stands for zygomorphic nature of the flower, and $\underline{\text{G}}$ stands for inferior ovary. Statement II: In a floral formula $\oplus$ stands for actinomorphic nature of the flower and $\underline{\text{G}}$ stands for superior ovary. 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 correct Statement I is incorrect: $\oplus$ stands for an actinomorphic flower (% for zygomorphic), and an inferior ovary is shown by a line above G, $\overline{\text{G}}$. Statement II is correct: $\oplus$ means actinomorphic, and G with a line below it, $\underline{\text{G}}$, means a superior ovary.
From the statements given below choose the correct option : A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S. B. Each ribosome has two sub-units. C. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S. D. The two sub-units of 80S ribosome are 60S and 20S and that of 70S are 50S and 20S. E. The two sub-units of 80S are 60S and 30S and that of 70S are 50S and 30S.
Solution: A, B, C are true Eukaryotic ribosomes are 80S and prokaryotic ones 70S (A); each has two subunits (B), 60S + 40S in the 80S type and 50S + 30S in the 70S type (C). (D) The smaller subunits are 40S and 30S, not 20S. (E) The smaller subunit of the 80S ribosome is 40S, not 30S.
Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization. A. Multicellular heterotrophs with cell wall made of chitin. B. Heterotrophs with tissue/organ/organ system level of body organization. C. Prokaryotes with cell wall made of polysaccharides and amino acids. D. Eukaryotic autotrophs with tissue/organ level of body organization. E. Eukaryotes with cellular body organization. Choose the correct answer from the options given below :
Solution: Sequence C, E, A, D, B C is Monera (prokaryotes; cell wall of polysaccharides and amino acids) and E is Protista (unicellular eukaryotes), both at the cellular level. A is Fungi (multicellular heterotrophs with chitin walls) and D is Plantae (eukaryotic autotrophs with tissues and organs). B is Animalia (heterotrophs with tissue, organ and organ system levels). ∴ Increasing complexity: C, E, A, D, B
The correct sequence of events in the life cycle of bryophytes is A. Fusion of antherozoid with egg. B. Attachment of gametophyte to substratum. C. Reduction division to produce haploid spores. D. Formation of sporophyte. E. Release of antherozoids into water. Choose the correct answer from the options given below :
Solution: Sequence B, E, A, D, C The gametophyte is attached to the substratum by rhizoids (B) and releases antherozoids into water (E). An antherozoid fuses with the egg (A), and the zygote grows into the sporophyte (D). Cells in the sporophyte's capsule undergo reduction division to form haploid spores (C). ∴ B, E, A, D, C
Which are correct: A. Computed tomography and magnetic resonance imaging detect cancers of internal organs. B. Chemotheraputics drugs are used to kill non-cancerous cells. C. $\alpha$-interferon activate the cancer patients' immune system and helps in destroying the tumour. D. Chemotherapeutic drugs are biological response modifiers. E. In the case of leukaemia blood cell counts are decreased. Choose the correct answer from the options given below:
Solution: Statements A and C only A and C are correct: CT and MRI detect cancers of internal organs, and α-interferon activates the patient's immune system and helps destroy the tumour. (B) Chemotherapeutic drugs are used to kill cancerous cells, not non-cancerous ones. (D) Biological response modifiers are substances such as α-interferon, not chemotherapeutic drugs. (E) Leukaemia is detected by increased, not decreased, blood cell counts.
Name the class of enzyme that usually catalyze the following reaction : $$\text{S} - \text{G} + \text{S}^{\#} \rightarrow \text{S} + \text{S}^{\#} - \text{G}$$ Where, G $\rightarrow$ a group other than hydrogen; S $\rightarrow$ a substrate; S$^{\#}$ $\rightarrow$ another substrate
Solution: Transferase Transferases catalyse the transfer of a group G (other than hydrogen) between a pair of substrates: S–G + S# → S + S#–G. (A) Ligases join two compounds, e.g. by C–O, C–S or C–N bonds. (B) Hydrolases break bonds by hydrolysis, e.g. ester, peptide or glycosidic bonds. (C) Lyases remove groups from substrates by means other than hydrolysis, leaving double bonds.
Find the correct statement : (A) In human pregnancy, the major organ systems are formed at the end of 12 weeks. (B) In human pregnancy the major organ systems are formed at the end of 8 weeks. (C) In human pregnancy heart is formed after one month of gestation. (D) In human pregnancy, limbs and digits develop by the end of second month. (E) In human pregnancy the appearance of hair is usually observed in the fifth month. Choose the correct answer from the options given below :
Solution: Statements A, C, D and E only After one month of pregnancy the heart is formed (C), and by the end of the second month the foetus develops limbs and digits (D). By the end of 12 weeks (first trimester) most of the major organ systems are formed (A). The first movements of the foetus and the appearance of hair on the head are usually seen in the fifth month (E). (B) The major organ systems are formed by the end of 12 weeks, not 8 weeks.
Which of the following is an example of non-distilled alcoholic beverage produced by yeast?
Solution: Beer Wine and beer are produced without distillation, whereas whisky, brandy and rum are made by distilling the fermented broth.
Given below are two statements : Statement I : In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable. Statement II : DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism. 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: essential life processes evolved around RNA, which acted both as genetic material and as a catalyst, but being reactive it is unstable. Statement II is correct: DNA evolved from RNA and is more stable; being double stranded with complementary strands, it resists changes by evolving a process of repair.
Given below are two statements : Statement I : Transfer RNAs and ribosomal RNA do not interact with mRNA. Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence. 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: tRNA reads the codons of mRNA through its anticodon, and the ribosome, built of rRNA, binds the mRNA during translation. Statement II is correct: RNAi takes place in all eukaryotic organisms as a method of cellular defence.
Which of the following diagrams is correct with regard to the proximal (P) and distal (D) tubule of the Nephron.
Solution: Diagram set (3) PCT: reabsorbs HCO₃⁻, NaCl and water, and selectively secretes H⁺, NH₃ and K⁺ into the filtrate. DCT: reabsorbs HCO₃⁻ and selectively secretes H⁺, K⁺ and NH₃. This tubular secretion keeps the ionic and acid-base balance of the body fluids.
What is the pattern of inheritance for polygenic trait?
Solution: Non-Mendelian inheritance pattern A polygenic trait is controlled by several genes with additive effects, giving continuous variation instead of Mendel's clear-cut ratios, so it shows a non-Mendelian pattern.
In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called :
Solution: Aleurone layer In cereal seeds the outer covering of the endosperm separates the embryo by a proteinous layer called the aleurone layer.
Match List I with List II: List-I: A. Chlorophyll a B. Chlorophyll b C. Xanthophylls D. Carotenoids List-II: (I) Yellow-green (II) Yellow (III) Blue-green (IV) Yellow to Yellow-orange Choose the option with all correct matches.
Solution: A-III, B-I, C-II, D-IV In a chromatogram of leaf pigments, chlorophyll a is bright or blue-green (III) and chlorophyll b yellow-green (I). Xanthophylls are yellow (II) and carotenoids yellow to yellow-orange (IV).
Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?
Solution: Bacterium In 1983 Eli Lilly introduced DNA sequences for the A and B chains of human insulin into plasmids of E. coli, a bacterium, to produce the insulin chains.
Which of the following are the post-transcriptional events in an eukaryotic cell? A. Transport of pre-mRNA to cytoplasm prior to splicing. B. Removal of introns and joining of exons. C. Addition of methyl group at 5' end of hnRNA. D. Addition of adenine residues at 3' end of hnRNA. E. Base pairing of two complementary RNAs. Choose the correct answer from the options given below :
Solution: B, C, D only The primary transcript (hnRNA) is processed by splicing, which removes introns and joins exons (B). Capping adds methyl guanosine triphosphate at the 5′ end (C), and tailing adds adenine residues at the 3′ end (D). (A) Splicing takes place in the nucleus; only the processed mRNA is transported to the cytoplasm. (E) Base pairing of two complementary RNAs is not a step in processing hnRNA; it is how RNA interference silences mRNA.
Match List-I with List-II. List-I: A. Centromere B. Cilium C. Cristae D. Cell membrane List-II: I. Mitochondrion II. Cell division III. Cell movement IV. Phospholipid Bilayer Choose the correct answer from the options given below :
Solution: A-II, B-III, C-I, D-IV A: the centromere bears the kinetochores to which spindle fibres attach during cell division (II). B: cilia are hair-like outgrowths that move the cell or the fluid around it (III). C: cristae are infoldings of the inner membrane of the mitochondrion (I). D: the cell membrane is a phospholipid bilayer with proteins (IV).
Match List I with List II: List-I: A. Alfred Hershey and Martha Chase B. Euchromatin C. Frederick Griffith D. Heterochromatin List-II: I. Streptococcus pneumoniae II. Densely packed and dark-stained III. Loosely packed and light-stained IV. DNA as genetic material confirmation Choose the correct answer from the options given below:
Solution: A-IV, B-III, C-I, D-II A: Hershey and Chase gave unequivocal proof that DNA is the genetic material (IV). B: euchromatin is loosely packed and stains light (III). C: Griffith worked with strains of Streptococcus pneumoniae (I). D: heterochromatin is densely packed and stains dark (II).
Which chromosome in the human genome has the highest number of genes?
Solution: Chromosome 1 Chromosome 1 has the most genes in the human genome (2968), while the Y chromosome has the fewest (231).
What are the potential drawbacks in adoption of the IVF method? A. High fatality risk to mother B. Expensive instruments and reagents C. Husband/wife necessary for being donors D. Less adoption of orphans E. Not available in India F. Possibility that the early embryo does not survive Choose the correct answer from the options given below:
Solution: Drawbacks B, D and F only IVF needs expensive instruments and reagents (B), may reduce the adoption of orphans (D), and the early embryo may not survive (F). (A) IVF does not carry a high fatality risk to the mother. (C) The gametes need not come from the husband or wife; donor ova or sperm can be used. (E) IVF is available in India.
Which one of the following is an example of ex-situ conservation?
Solution: Zoos and botanical gardens Zoological parks, botanical gardens and wildlife safari parks protect organisms away from their natural habitat (ex situ conservation). Protected areas, national parks and wildlife sanctuaries protect species in their natural habitat (in situ conservation).
A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is
Solution: Mesosome The mesosome, an extension of the plasma membrane into a prokaryotic cell, helps in cell wall formation, DNA replication and respiration.
The plasmid shown carries an origin of replication (Ori), an ampicillin resistance gene (amp), a tetracycline resistance gene (Tet R), a BamH1 site, and an EcoRI site that lies within the $\beta$-galactosidase gene. In the above represented plasmid, an alien piece of DNA is inserted at EcoRI site. Which of the following strategies will be chosen to select the recombinant colonies?
Solution: White colour colonies will be selected An insert at the EcoRI site inactivates the β-galactosidase gene (insertional inactivation), so recombinant colonies give no colour with the chromogenic substrate and appear white. (A) Ampicillin only shows that a cell carries the plasmid; blue colonies on it still have an active β-galactosidase gene and are non-recombinants. (B) The amp and Tet R genes stay intact in both recombinant and non-recombinant plasmids, so these antibiotics cannot tell them apart. (C) Blue colonies are the non-recombinants.
What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?
Solution: Vena cava The vena cava brings deoxygenated blood from the body to the frog's heart; the pulmonary artery also carries deoxygenated blood, but away from the heart to the lungs.
Which of following organisms cannot fix nitrogen? A. Azotobacter B. Oscillatoria C. Anabaena D. Volvox E. Nostoc Choose the correct answer from the options given below:
Solution: Volvox (D only) Volvox is a colonial green alga, a eukaryote, and cannot fix nitrogen. Azotobacter (a free-living bacterium) and the cyanobacteria Oscillatoria, Anabaena and Nostoc can fix atmospheric nitrogen.
While trying to find out the characteristic of a newly found animal, a researcher did the histology of adult animal and observed a cavity with presence of mesodermal tissue towards the body wall but no mesodermal tissue was observed towards the alimentary canal. What could be the possible coelome of that animal?
Solution: Pseudocoelomate A body cavity with mesoderm towards the body wall but none around the alimentary canal is not fully lined by mesoderm: a pseudocoelom. (A) The spongocoel is the central cavity of sponges, not a body cavity lined by mesoderm. (B) Acoelomates have no body cavity at all. (D) In schizocoelomates the coelom forms by splitting of the mesoderm, so it is lined by mesoderm on both sides (a true coelom).
Which one of the following statements refers to Reductionist Biology?
Solution: Physico-chemical approach to study and understand living organisms Studying living organisms by the physico-chemical approach, using the methods of physics and chemistry, is called reductionist biology.
Epiphytes that are growing on a mango branch is an example of which of the following?
Solution: Commensalism An epiphyte on a mango branch gains support and light while the mango tree is neither harmed nor benefited, which is commensalism.
Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?
Solution: Cytokinin Cytokinins promote nutrient mobilisation, which helps delay leaf senescence; they also help overcome apical dominance.
The complex II of mitochondrial electron transport chain is also known as
Solution: Succinate dehydrogenase Complex II is succinate dehydrogenase; complex I is NADH dehydrogenase, complex III cytochrome bc₁ and complex IV cytochrome c oxidase.
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