Free, login-free worked solutions to every question in NEET (UG) 2023 — Code H1. Check your answers below, then predict your rank or practise with full mock tests.
The magnetic energy stored in an inductor of inductance 4 $\mu$H carrying a current of 2 A is
Solution: Magnetic energy stored in an inductor $$U = \frac{1}{2}Li^{2}$$ $$= \frac{1}{2} \times 4 \times 10^{-6} \times (2)^{2}$$ $$= 8 \times 10^{-6}\ \text{J}$$ $$U = 8\ \mu\text{J}$$
A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?
Solution: The capacitor removes the ac ripple from the rectified output. A capacitor placed across the load charges on the rising part of the rectified wave and discharges slowly through the load, smoothing the pulsating dc — which is why it is called a filter.
An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5}$ N C$^{-1}$. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.
Solution: $E = 2 \times 10^{5}$ N/C, $l = 2$ cm, $\tau = 4$ N m $$\vec{\tau} = \vec{p} \times \vec{E}$$ $$4 = pE\sin\theta$$ $$4 = p \times 2 \times 10^{5} \times \sin 30^{\circ}$$ $$p = 4 \times 10^{-5}\ \text{C m}$$ $$q = \frac{p}{l} = \frac{4 \times 10^{-5}}{0.02} = 2\ \text{mC}$$
The equivalent capacitance of the system shown in the following circuit is A 3 $\mu$F capacitor from terminal $A$ in series with a parallel pair of 3 $\mu$F capacitors, whose far side is terminal $B$.
Solution: For parallel grouping, $$C_1 = 3 + 3 = 6\ \mu\text{F}$$ For series grouping, $$C_{\text{eq}} = \frac{C_1C_2}{C_1 + C_2} = \frac{3 \times 6}{3 + 6} = \frac{18}{9}$$ $$C_{\text{eq}} = 2\ \mu\text{F}$$
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is
Solution: $$\frac{1}{\lambda} = R\left[\frac{1}{n_2^{2}} - \frac{1}{n_1^{2}}\right]$$ For Balmer, $n_2 = 2$, $n_1 = \infty$: $$\frac{1}{\lambda} = R\left[\frac{1}{4} - \frac{1}{\infty}\right] \quad \Rightarrow \quad \lambda = \frac{4}{R} \qquad \ldots(1)$$ For Bracket, $n_2 = 4$, $n_1 = \infty$: $$\frac{1}{\lambda'} = R\left[\frac{1}{16} - \frac{1}{\infty}\right] \quad \Rightarrow \quad \lambda' = \frac{16}{R} \qquad \ldots(2)$$ Dividing (2) by (1), $$\lambda' = 4\lambda$$
Resistance of a carbon resistor determined from colour codes is $(22000 \pm 5\%)\ \Omega$. The colour of third band must be
Solution: Resistance $= (22 \times 10^{3})\ \Omega \pm 5\%$ The third band corresponds to the decimal multiplier. Decimal multiplier $= 10^{3}$ $$\Rightarrow\ \text{Colour} \rightarrow \text{Orange}$$
Two bodies of mass $m$ and $9m$ are placed at a distance $R$. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ($G$ = gravitational constant)
Solution: Let the gravitational field at point $Q$, at distance $x$ from $m$, be zero. So $$\frac{Gm}{x^{2}} = \frac{G(9m)}{(R - x)^{2}}$$ $$\frac{(R - x)^{2}}{x^{2}} = 9 \quad \Rightarrow \quad x = \frac{R}{4}$$ $$V_P = \frac{-Gm}{x} - \frac{G(9m)}{R - x}$$ $$V_P = \frac{-Gm}{\dfrac{R}{4}} - \frac{G(9m)}{\dfrac{3R}{4}}$$ $$= \frac{-4Gm}{R} - \frac{12Gm}{R}$$ $$= \frac{-16Gm}{R}$$
If the galvanometer $G$ does not show any deflection in the circuit shown, the value of $R$ is given by The circuit is a 10 V cell in series with a 400 $\Omega$ resistor forming the left loop; $R$ is the middle branch between the two loops; and a 2 V cell in series with the galvanometer $G$ forms the right loop.
Solution: Since the galvanometer does not show any deflection, $$\Rightarrow\ i_g = 0$$ So no current flows in the 2 V branch, and the potential difference across $R$ equals 2 V. The whole current from the 10 V cell flows through the 400 $\Omega$ resistor and then through $R$: $$\frac{10 - 2}{400} = \frac{2}{R}$$ $$\Rightarrow\ R = \frac{2 \times 400}{8} = 100\ \Omega$$
For Young's double slit experiment, two statements are given below: Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant. Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases. In the light of the above statements, choose the correct answer from the options given below:
Solution: For YDSE, angular fringe width is given by $\alpha = \dfrac{\lambda}{d}$. It does not depend on the distance of the screen from the slit, so statement I is correct. Angular fringe width $\propto \lambda$. If $\lambda$ increases, the angular separation of fringes increases, not decreases. So statement I is true and statement II is false.
The venturi-meter works on
Solution: Venturi-meter works on Bernoulli's principle: the constriction speeds the fluid up, and the accompanying pressure drop is read off to give the flow rate.
Let a wire be suspended from the ceiling (rigid support) and stretched by a weight $W$ attached at its free end. The longitudinal stress at any point of cross-sectional area $A$ of the wire is
Solution: $$\text{Longitudinal stress} = \frac{\text{Internal restoring force}}{\text{Area}} = \frac{F_{\text{ext}}}{\text{Area}}$$ The internal restoring force at any cross-section equals the weight $W$ hanging below it, so $$\text{Stress} = \frac{W}{A}$$
The angular acceleration of a body, moving along the circumference of a circle, is
Solution: Angular acceleration of a body moving along the circumference of a circle is along the axis of rotation. Angular quantities such as $\vec{\omega}$ and $\vec{\alpha}$ are axial vectors, directed along the axis about which the rotation takes place.
The temperature of a gas is $-50^{\circ}$C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?
Solution: $$v_{\text{rms}} = \sqrt{\frac{3RT}{m}} \quad \Rightarrow \quad v_{\text{rms}} \propto \sqrt{T}$$ $$T_1 = 273 - 50 = 223\ \text{K}$$ $v_{\text{rms}}$ is increased by 3 times, so the final rms speed $= v + 3v = 4v$. $$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$$ $$\frac{v}{4v} = \sqrt{\frac{223}{T_2}} \quad \Rightarrow \quad \frac{1}{16} = \frac{223}{T_2}$$ $$T_2 = 3568\ \text{K}$$ $$T_2 = 3568 - 273 = 3295^{\circ}\text{C}$$
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
Solution: Photoemission occurs only when the incident photon energy exceeds the work function. Energy of incident radiation = 2.20 eV Work function of Cs $\rightarrow$ 2.14 eV Work function of K $\rightarrow$ 2.30 eV Work function of Na $\rightarrow$ 2.75 eV Since the work functions of potassium and sodium are more than the energy of the incident radiation, photoelectrons may be emitted from caesium only.
An ac source is connected to a capacitor $C$. Due to decrease in its operating frequency
Solution: $$X_C = \frac{1}{\omega C}$$ Since $\omega$ is decreasing, $X_C$ will increase, hence the current will decrease. Conduction current = displacement current, therefore the displacement current will decrease.
Light travels a distance $x$ in time $t_1$ in air and $10x$ in time $t_2$ in another denser medium. What is the critical angle for this medium?
Solution: At the critical angle, $\mu_2\sin i_c = \mu_1$, so $$\sin i_c = \frac{\mu_1}{\mu_2}$$ Using $\mu = \dfrac{c}{V}$, $$\sin i_c = \frac{\mu_1}{\mu_2} = \frac{V_2}{V_1}$$ The speed in air is $V_1 = \dfrac{x}{t_1}$ and in the denser medium $V_2 = \dfrac{10x}{t_2}$: $$\sin i_c = \frac{10x\,t_1}{t_2\,x} = \frac{10\,t_1}{t_2}$$ $$i_c = \sin^{-1}\left(\frac{10\,t_1}{t_2}\right)$$
The net magnetic flux through any closed surface is
Solution: $$\oint \vec{B} \cdot \overrightarrow{ds} = \text{zero}$$ Magnetic monopoles do not exist, so every field line that enters a closed surface must also leave it. Hence the net magnetic flux through any closed surface is zero.
A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?
Solution: For an ideal transformer, $$P_{\text{input}} = P_{\text{output}}$$ $$(VI)_{\text{in}} = 60$$ $$220 \times I = 60$$ $$I = 0.27\ \text{A}$$
A metal wire has mass $(0.4 \pm 0.002)$ g, radius $(0.3 \pm 0.001)$ mm and length $(5 \pm 0.02)$ cm. The maximum possible percentage error in the measurement of density will nearly be
Solution: We know, $$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{M}{\pi r^{2}\ell}$$ Using the concept of errors, $$\frac{\Delta\rho}{\rho} = \frac{\Delta M}{M} + \frac{2\Delta r}{r} + \frac{\Delta\ell}{\ell}$$ $$= \left(\frac{0.002}{0.4} + \frac{2 \times 0.001}{0.3} + \frac{0.02}{5}\right)$$ $$\frac{\Delta\rho}{\rho} = 0.0156$$ $$\frac{\Delta\rho}{\rho}\% = 1.56\% \approx 1.6\%$$
Given below are two statements: Statement I: Photovoltaic devices can convert optical radiation into electricity. Statement II: Zener diode is designed to operate under reverse bias in breakdown region. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both statements are correct. I: Photovoltaic devices convert optical radiation into electricity — that is what a solar cell does. II: Zener diode is designed to operate under reverse bias in the breakdown region, e.g. a Zener diode as a voltage regulator.
A bullet is fired from a gun at the speed of 280 m s$^{-1}$ in the direction $30^{\circ}$ above the horizontal. The maximum height attained by the bullet is ($g = 9.8$ m s$^{-2}$, $\sin 30^{\circ} = 0.5$)
Solution: $$H = \frac{u^{2}\sin^{2}\theta}{2g}$$ $$H = \frac{(280)^{2}\left(\sin^{2}30^{\circ}\right)}{2 \times 9.8}$$ $$= \frac{280 \times 280 \times 0.5 \times 0.5}{2 \times 9.8}$$ $$H = 1000\ \text{m}$$
A Carnot engine has an efficiency of 50% when its source is at a temperature 327$^{\circ}$C. The temperature of the sink is
Solution: Efficiency $\eta = \dfrac{50}{100} = \dfrac{1}{2}$ Efficiency of a Carnot engine, $$\eta = 1 - \frac{T_2}{T_1}$$ with $T_1 = 327 + 273 = 600$ K: $$\eta = 1 - \frac{T_2}{600}$$ $$\frac{1}{2} = 1 - \frac{T_2}{600}$$ $$\frac{T_2}{600} = \frac{1}{2} \quad \Rightarrow \quad T_2 = 300\ \text{K}$$ $$T_2 = 300 - 273 = 27^{\circ}\text{C}$$
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03 N m$^{-1}$)
Solution: A soap bubble has two surfaces, so the energy required is $$W = \left[S \times \Delta A\right] \times 2$$ $$\Rightarrow\ \text{Energy required} = \left[0.03 \times 4 \times \pi \times 4 \times 10^{-4}\right] \times 2$$ $$= 3.015 \times 10^{-4}\ \text{J}$$
The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to
Solution: The shortest wavelength corresponds to the whole kinetic energy of the electron being converted into one photon: $$eV = \frac{hc}{\lambda_{\min}}$$ $$\lambda_{\min} = \frac{hc}{eV}$$ $$\lambda_{\min} \propto \frac{1}{V}$$
In a series $LCR$ circuit, the inductance $L$ is 10 mH, capacitance $C$ is 1 $\mu$F and resistance $R$ is 100 $\Omega$. The frequency at which resonance occurs is
Solution: For resonance frequency, $$f = \frac{1}{2\pi\sqrt{LC}}$$ $$\Rightarrow\ f = \frac{1}{2 \times \pi \times \sqrt{10 \times 10^{-3} \times 1 \times 10^{-6}}} = \frac{10^{4}}{2\pi}$$ $$= 1.591 \times 10^{3}$$ $$= 1.591\ \text{kHz}$$
The potential energy of a long spring when stretched by 2 cm is $U$. If the spring is stretched by 8 cm, potential energy stored in it will be
Solution: Potential energy stored in a spring, $U = \dfrac{1}{2}Kx^{2}$. $$U = \frac{1}{2}K(2)^{2} = \frac{1}{2}(K)(4) = 2K \qquad \ldots(i)$$ $$U' = \frac{1}{2}K(8)^{2} = \frac{1}{2}K \times 64 = 32K \qquad \ldots(ii)$$ On dividing (i) by (ii), $$\frac{U}{U'} = \frac{2K}{32K} = \frac{1}{16}$$ $$U' = 16U$$
The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\dfrac{1}{16}\right)^{\text{th}}$ of its initial value?
Solution: $$A = \frac{A_0}{2^{n}}$$ $$\frac{A}{A_0} = \frac{1}{2^{n}}$$ $$\frac{1}{16} = \frac{1}{2^{n}} \quad \Rightarrow \quad \frac{1}{2^{4}} = \frac{1}{2^{n}} \quad \Rightarrow \quad n = 4$$ $$n = \frac{t}{T_{1/2}}, \qquad t = 4 \times T_{1/2} = 4 \times 20$$ $$= 80\ \text{minutes}$$
If $\displaystyle\oint_s \vec{E} \cdot \overrightarrow{dS} = 0$ over a surface, then
Solution: $$\phi_{\text{net}} = \oint_s \vec{E} \cdot \overrightarrow{dS} = 0$$ $$\Rightarrow\ \text{Net flux through the surface is zero.}$$ $$\Rightarrow\ \text{Therefore, the number of flux lines entering the surface must be equal to the number of flux lines leaving it.}$$ Zero net flux means zero net enclosed charge; it does not require the field to be uniform or constant, nor that all charges lie inside.
The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is
Solution: $$f_0 = f_{\text{open pipe}} = \frac{v}{2l}$$ $$f_c = f_{\text{closed pipe}} = \frac{v}{4l}$$ $$\frac{f_0}{f_c} = \frac{v}{2l} \times \frac{4l}{v}$$ $$f_0 : f_c = 2 : 1$$
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are
Solution: The errors which cannot be associated with any systematic or constant cause are called random errors. These errors can arise due to unpredictable fluctuations in experimental conditions, e.g. random change in pressure, temperature, voltage supply etc.
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10}$ Hz and amplitude 48 V m$^{-1}$. Then the amplitude of oscillating magnetic field is (Speed of light in free space $= 3 \times 10^{8}$ m s$^{-1}$)
Solution: From the properties of an electromagnetic wave we know that $$C = \frac{E_0}{B_0}$$ where $E_0$ is the amplitude of the oscillating electric field and $B_0$ the amplitude of the oscillating magnetic field. $$\Rightarrow\ B_0 = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{-7}\ \text{T}$$
A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is
Solution: The force is along the change in momentum, $\Delta\vec{p} = \vec{p}_f - \vec{p}_i$. The initial velocity is southward and the final velocity is eastward with the same speed, so $$\Delta\vec{v} = \vec{v}_{\text{east}} - \vec{v}_{\text{south}} = \vec{v}_{\text{east}} + \vec{v}_{\text{north}}$$ The resultant of equal eastward and northward vectors points along the north-east direction, so the force acts along north-east.
The magnitude and direction of the current in the following circuit is The single loop $ABCD$ carries, along the top from $A$ to $B$: a 2 $\Omega$ resistor, then a 10 V cell, then a 5 V cell connected in opposition, then a 1 $\Omega$ resistor. The bottom branch from $D$ to $C$ carries a 7 $\Omega$ resistor. The point between the two cells is marked $E$.
Solution: Using Kirchhoff's law, the two cells oppose each other so the net emf is $10 - 5 = 5$ V, and the total resistance is $2 + 1 + 7 = 10\ \Omega$: $$i = \frac{10 - 5}{10} = \frac{5}{10}$$ $$i = 0.5\ \text{A}$$ in the clockwise direction, that is from $A$ to $B$ through $E$.
A vehicle travels half the distance with speed $v$ and the remaining distance with speed $2v$. Its average speed is
Solution: For equal distances covered at two speeds, the average speed is the harmonic mean: $$v_{\text{avg}} = \frac{2v_1v_2}{v_1 + v_2}$$ $$= \frac{2 \times v \times 2v}{v + 2v}$$ $$= \frac{4v}{3}$$
The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{-11}$ m. What is the radius of third allowed orbit of hydrogen atom?
Solution: $$r_n \propto \frac{n^{2}}{Z}$$ $$\frac{r_1}{r_2} = \left(\frac{1}{3}\right)^{2}$$ $$r_2 = 9r_1 = 5.3 \times 10^{-11} \times 9$$ $$= 47.7 \times 10^{-11}\ \text{m}$$ $$= 4.77\ \text{Å}$$
10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is
Solution: In series combination, $$R_{\text{eq}} = 10R \quad \Rightarrow \quad i = \frac{E}{10R}$$ In parallel combination, $$R_{\text{eq}} = \frac{R}{10} \quad \Rightarrow \quad i' = \frac{E}{\dfrac{R}{10}} = \frac{10E}{R}$$ $$i' = 10 \times 10\,i = 100\,i$$ $$n = 100$$
The $x$-$t$ graph of a particle performing simple harmonic motion is shown in the figure. It is a sine curve starting from the origin, reaching a maximum displacement of 1 m, crossing zero at $t = 4$ s and completing one full cycle at $t = 8$ s. The acceleration of the particle at $t = 2$ s is
Solution: Position of the particle as a function of time, $x = A\sin\omega t$. From the figure, $A = 1$ and the time period is 8 s, so $$\omega = \frac{2\pi}{8} = \frac{\pi}{4}$$ $$x = \sin\frac{\pi}{4}t$$ $$v = \frac{dx}{dt} = \frac{\pi}{4}\cos\frac{\pi}{4}t$$ $$a = \frac{dv}{dt} = -\frac{\pi^{2}}{16}\sin\frac{\pi}{4}t$$ At $t = 2$ s, $\sin\dfrac{\pi}{2} = 1$: $$a = -\frac{\pi^{2}}{16}\ \text{m/s}^{2}$$
Two thin lenses are of same focal lengths ($f$), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be
Solution: Convex lens $f_1 > 0$, concave lens $f_2 < 0$, with equal magnitudes. $$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{f} - \frac{1}{f} = 0$$ $$f_{\text{eq}} = \infty$$ The combination behaves like a plane glass plate.
A satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\dfrac{3\pi}{Gd}$ represents
Solution: For a satellite orbiting just above the surface, the gravitational force supplies the centripetal force: $$\frac{GMm}{R^{2}} = mR\omega^{2} = mR\left(\frac{2\pi}{T}\right)^{2}$$ $$\Rightarrow\ T^{2} = \frac{4\pi^{2}R^{3}}{GM}$$ Writing the mass of the earth as $M = \dfrac{4}{3}\pi R^{3}d$: $$T^{2} = \frac{4\pi^{2}R^{3}}{G \cdot \dfrac{4}{3}\pi R^{3}d} = \frac{3\pi}{Gd}$$ So $\dfrac{3\pi}{Gd}$ represents $T^{2}$.
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 ($g = 10$ m s$^{-2}$).
Solution: Working in the frame of the car, the body stays stationary when the pseudo force is balanced by static friction: $$ma_{\max} = \mu_s mg$$ $$a_{\max} = \mu_s g = 0.15 \times 10 = 1.5\ \text{m/s}^{2}$$
An electric dipole is placed as shown in the figure: a charge $-q$ and a charge $+q$ lie on a line, each 3 cm from the midpoint $O$, and the point $P$ lies on the same line at 5 cm from $O$, on the $+q$ side. The electric potential (in $10^{2}$ V) at point $P$ due to the dipole is ($\varepsilon_0$ = permittivity of free space and $\dfrac{1}{4\pi\varepsilon_0} = K$)
Solution: Electrostatic potential due to a point charge is given by $\dfrac{Kq}{r}$. The distance from $+q$ to $P$ is $5 - 3 = 2$ cm and from $-q$ to $P$ is $5 + 3 = 8$ cm: $$V_{\text{net at point }P} = \frac{Kq}{2 \times 10^{-2}} - \frac{Kq}{8 \times 10^{-2}}$$ $$= \frac{Kq \times 10^{2}}{2}\left(1 - \frac{1}{4}\right)$$ $$= \left(\frac{3}{8}Kq\right) \times 10^{2}\ \text{V} = \frac{3}{8}qK$$
The net impedance of circuit (as shown in figure) will be An inductor of $\dfrac{50}{\pi}$ mH, a capacitor of $\dfrac{10^{3}}{\pi}$ $\mu$F and a 10 $\Omega$ resistor are connected in series across a 220 V, 50 Hz source.
Solution: $$L = \frac{50}{\pi}\ \text{mH}$$ $$X_L = 2\pi \times 50 \times \frac{50}{\pi} \times 10^{-3} = 5\ \Omega$$ $$C = \frac{10^{3}}{\pi} \times 10^{-6}\ \text{F}$$ $$X_C = \frac{1 \times \pi}{2\pi \times 50 \times 10^{3} \times 10^{-6}} = \frac{10^{3}}{100} = 10\ \Omega$$ $$Z = \sqrt{\left(X_C - X_L\right)^{2} + R^{2}}$$ $$Z = \sqrt{(10 - 5)^{2} + 10^{2}} = \sqrt{125} = 5\sqrt{5}\ \Omega$$
For the following logic circuit, the truth table is Input $A$ passes through a NOT gate and input $B$ passes through a NOT gate; the two inverted signals then feed a NAND gate whose output is $Y$.
Solution: The two NOT gates give $\overline{A}$ and $\overline{B}$, and the NAND gate then gives $$Y = \overline{\overline{A} \cdot \overline{B}} = A + B$$ by De Morgan's law. It is an OR gate. So the truth table is $A$=0, $B$=0, $Y$=0 $A$=0, $B$=1, $Y$=1 $A$=1, $B$=0, $Y$=1 $A$=1, $B$=1, $Y$=1
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)? A biconvex lens of refractive index $n_1 = 1.5$ with $R_1 = R_2 = 20$ cm is embedded in a medium of refractive index $n_2 = 1.6$, which forms the two outer layers.
Solution: Effective focal length $\Rightarrow f_{\text{eff}}$ $$\frac{1}{f_{\text{eff}}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$$ Also, $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ $$\frac{1}{f_1} = (1.6 - 1)\left(\frac{1}{\infty} - \frac{1}{20}\right) = \frac{-0.6}{20}$$ $$\frac{1}{f_2} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = \frac{0.5}{10}$$ $$\frac{1}{f_3} = (1.6 - 1)\left(\frac{1}{-20} - \frac{1}{\infty}\right) = \frac{-0.6}{20}$$ $$\frac{1}{f_{\text{eff}}} = \frac{-0.6}{20} + \frac{0.5}{10} - \frac{0.6}{20}$$ $$\frac{1}{f_{\text{eff}}} = \frac{-0.6}{10} + \frac{0.5}{10} = \frac{-0.1}{10} = \frac{-1}{100}$$ $$\therefore\ f_{\text{eff}} = -100\ \text{cm}$$
A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes $\dfrac{u}{3}$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is
Solution: The retardation is uniform, so $v^{2} = u^{2} - 2as$. For the first 24 cm, $$\left(\frac{u}{3}\right)^{2} = u^{2} - 2a(24)$$ $$\frac{u^{2}}{9} = u^{2} - 48a \quad \Rightarrow \quad 48a = \frac{8u^{2}}{9} \quad \Rightarrow \quad a = \frac{u^{2}}{54}$$ For the whole length $L$, the bullet comes to rest: $$0 = u^{2} - 2aL \quad \Rightarrow \quad L = \frac{u^{2}}{2a} = \frac{u^{2}}{2 \times \dfrac{u^{2}}{54}} = 27\ \text{cm}$$
The resistance of platinum wire at 0$^{\circ}$C is 2 $\Omega$ and 6.8 $\Omega$ at 80$^{\circ}$C. The temperature coefficient of resistance of the wire is
Solution: Using $R = R_0(1 + \alpha\Delta T)$, where $\alpha$ is the thermal coefficient of resistance, $$6.8 = 2\{1 + \alpha(80 - 0)\}$$ $$\frac{6.8}{2} - 1 = \alpha \times 80$$ $$\alpha = \frac{3.4 - 1}{80} = \frac{2.4}{80} = 0.03$$ $$\therefore\ \alpha = 3 \times 10^{-2}\ ^{\circ}\text{C}^{-1}$$
A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s$^{-1}$. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take $g = 10$ m s$^{-2}$)
Solution: Taking upward as positive with $u = 4$ m/s and $t = 4$ s, $$s = ut - \frac{1}{2}gt^{2}$$ $$= 4 \times 4 - \frac{1}{2} \times 10 \times 16$$ $$= 16 - 80 = -64\ \text{m}$$ The negative sign means the ball ends 64 m below the point of projection, so the height of the bridge above the water surface is 64 m.
A very long conducting wire is bent in a semi-circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by The wire runs in from the left, carries current $i$ to $A$, turns through a semicircle of radius $R$ centred on $P$ down to $B$, and then runs back out to the left.
Solution: Splitting the conductor into the upper straight wire (1), the semicircle (2) and the lower straight wire (3): $$B_P \text{ due to wire 1} = \frac{\mu_0 i}{4\pi R}\ \otimes$$ $$B_P \text{ due to wire 3} = \frac{\mu_0 i}{4\pi R}\ \otimes$$ $$B_P \text{ due to wire 2} = \frac{\mu_0 i}{4R}\ \odot$$ $$B_{\text{net}} = -\frac{\mu_0 i}{2\pi R} + \frac{\mu_0 i}{4R} = \frac{\mu_0 i}{4R}\left[-\frac{2}{\pi} + 1\right] = \frac{\mu_0 i}{4R}\left[1 - \frac{2}{\pi}\right]$$ Pointed away from the page.
A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})$ T. The magnitude of the magnetic force acting on the wire is
Solution: Magnetic force acting on a current carrying wire is $$\vec{F} = I\vec{\ell} \times \vec{B}$$ $$= IL\hat{i} \times \left(2\hat{i} + 3\hat{j} - 4\hat{k}\right) = 3IL\hat{k} + 4IL\hat{j}$$ Magnitude of force $$\left|\vec{F}\right| = \sqrt{(3IL)^{2} + (4IL)^{2}}$$ $$= 5\,IL$$
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : A reaction can have zero activation energy. Reason R : The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy. In the light of the above statements, choose the correct answer from the options given below :
Solution: A few reactions can have zero activation energy, for example radical combination reactions — so the Assertion is true. Activation energy is defined as the minimum amount of extra energy absorbed by reactants to achieve threshold energy — so the Reason is also true. But the definition in R does not explain why a reaction may have zero activation energy, so R is not the correct explanation of A.
The correct order of energies of molecular orbitals of N$_2$ molecule, is
Solution: For molecules like B$_2$, C$_2$, N$_2$ etc., where $s$-$p$ mixing is significant, the $\pi 2p$ orbitals lie below the $\sigma 2p_z$. The increasing order of energies of the various molecular orbitals is $$\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma^{*}2p_z$$
The given compound C$_6$H$_5$–CH=CH–CH(X)–CH$_2$–CH$_3$ is an example of ______.
Solution: The halogen X is attached to an $sp^{3}$ hybridised carbon that is directly next to a carbon-carbon double bond. A halide in which the halogen atom is bonded to an $sp^{3}$ hybridised carbon atom adjacent to a carbon-carbon double bond is called an allylic halide. In a vinylic halide the halogen sits on the doubly-bonded carbon itself; in a benzylic halide it sits on the carbon attached to the ring; in an aryl halide it is on the ring carbon.
Homoleptic complex from the following complexes is
Solution: Complexes in which a metal is bound to only one kind of donor group are called homoleptic complexes. Potassium trioxalatoaluminate (III) is K$_3$[Al(ox)$_3$] — the aluminium is bound only to oxalate ligands, so it is a homoleptic complex. Each of the other three carries two or more different ligands in the coordination sphere, making them heteroleptic.
The right option for the mass of CO$_2$ produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40) $$\text{CaCO}_3 \xrightarrow{\ 1200\ \text{K}\ } \text{CaO} + \text{CO}_2$$
Solution: $$\text{CaCO}_3 \xrightarrow{\ 1200\ \text{K}\ } \text{CaO} + \text{CO}_2$$ From 100 g CaCO$_3$ $\rightarrow$ 44 g CO$_2$ produced. As the limestone is 20% pure, the mass of pure CaCO$_3$ is $$20 \times \frac{20}{100} = 4\ \text{g}$$ So 4 g CaCO$_3$ gives $$\frac{44}{100} \times 4 = 1.76\ \text{g CO}_2$$
The relation between $n_m$, ($n_m$ = the number of permissible values of magnetic quantum number ($m$)) for a given value of azimuthal quantum number ($l$), is
Solution: The magnetic quantum number $m$ runs from $-l$ to $+l$ including zero, so the number of permissible values is $$n_m = 2l + 1$$ Rearranging for $l$, $$l = \frac{n_m - 1}{2}$$
Select the correct statements from the following A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is $9.10939 \times 10^{-31}$ kg. C. All the isotopes of a given element show same chemical properties. D. Protons and electrons are collectively known as nucleons. E. Dalton's atomic theory, regarded the atom as an ultimate particles of matter Choose the correct answer from the options given below
Solution: Atoms consist of three fundamental particles: electrons, protons and neutrons — so statement A is wrong. The mass of the electron is $9.10939 \times 10^{-31}$ kg — B is correct. All the isotopes of a given element show the same chemical properties — C is correct. Protons and NEUTRONS present in the nucleus are collectively called nucleons, not protons and electrons — so D is wrong. Dalton's atomic theory regarded the atom as the ultimate particle of matter — E is correct. So the correct statements are B, C, E only.
Which one is an example of heterogenous catalysis?
Solution: Combination of N$_2$ and H$_2$ to form NH$_3$ in the presence of finely divided Fe is an example of heterogeneous catalysis — the catalyst is a solid while the reactants are gases. $$\text{N}_2(g) + 3\text{H}_2(g) \xrightarrow{\ \text{Fe}(s)\ } 2\text{NH}_3(g)$$ All the others are examples of homogeneous catalysis, where catalyst and reactants are in the same phase: $$\text{C}_{12}\text{H}_{22}\text{O}_{11}(aq) + \text{H}_2\text{O}(l) \xrightarrow{\ \text{H}_2\text{SO}_4(l)\ } \text{Glucose}(aq) + \text{Fructose}(aq)$$ $$2\text{SO}_2(g) + \text{O}_2(g) \xrightarrow{\ \text{NO}(g)\ } 2\text{SO}_3(g)$$
Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is :
Solution: This reaction is called soda lime decarboxylation: $$\text{CH}_3\text{COO}^{-}\text{Na}^{+} \xrightarrow[\text{CaO}(s)]{\text{NaOH}} \text{CH}_4(g) + \text{Na}_2\text{CO}_3(s)$$ The organic compound obtained is methane. Molar mass of CH$_4$ = 16 g/mol Weight of 2 moles of CH$_4$ $= 16 \times 2 = 32$ g
The number of $\sigma$ bonds, $\pi$ bonds and lone pair of electrons in pyridine, respectively are:
Solution: Pyridine is C$_5$H$_5$N, a six-membered aromatic ring with five carbons and one nitrogen. $\sigma$ bonds: six ring $\sigma$ bonds plus five C–H $\sigma$ bonds $= 11$ $\pi$ bonds: the aromatic ring carries three $\pi$ bonds $= 3$ Lone pairs: the nitrogen's lone pair lies in an $sp^{2}$ orbital in the plane of the ring $= 1$ So the answer is 11, 3, 1.
Which amongst the following molecules on polymerization produces neoprene?
Solution: Neoprene is formed by free radical polymerisation of chloroprene. $$n\,\text{CH}_2\text{=C(Cl)–CH=CH}_2 \xrightarrow{\ \text{Polymerisation}\ } \left[\text{CH}_2\text{–C(Cl)=CH–CH}_2\right]_n$$ The monomer 2-chloro-1,3-butadiene is called chloroprene, and the polymer is neoprene.
The stability of Cu$^{2+}$ is more than Cu$^{+}$ salts in aqueous solution due to
Solution: The stability of Cu$^{2+}$(aq) is more than Cu$^{+}$(aq) due to the much more negative $\Delta_{\text{hyd}}H^{\circ}$ of Cu$^{2+}$(aq) than Cu$^{+}$(aq), which more than compensates for the second ionisation enthalpy of Cu. $\Delta_{\text{hyd}}H^{\circ}$ of Cu$^{2+}$(aq) $= -2121$ kJ mol$^{-1}$ $\Delta_i H_1^{\circ}$ of Cu $= +745$ kJ mol$^{-1}$ $\Delta_i H_2^{\circ}$ of Cu $= +1960$ kJ mol$^{-1}$
Complete the following reaction Cyclohexanone [A] reacts with HCN to give the cyanohydrin [B], which on treatment with conc. H$_2$SO$_4$ and heat gives [C]. [C] is ________
Solution: HCN adds across the carbonyl of cyclohexanone to give the cyanohydrin, which carries an OH and a CN on the same ring carbon. Concentrated H$_2$SO$_4$ on heating protonates the OH, and loss of water together with a $\beta$-hydrogen gives the $\alpha,\beta$-unsaturated nitrile — the CN now sits on a ring carbon that bears a double bond. The nitrile is then hydrolysed under the aqueous acidic conditions (H$^{+}$/H$_2$O) to the carboxylic acid. So [C] is cyclohex-1-ene-1-carboxylic acid.
Which of the following reactions will NOT give primary amine as the product?
Solution: Reduction of an isocyanide gives a secondary amine, because the nitrogen already carries the methyl group: $$\text{CH}_3\text{NC} \xrightarrow[\text{(ii) H}_3\text{O}^{+}]{\text{(i) LiAlH}_4} \text{CH}_3\text{NHCH}_3 \quad \text{(Secondary amine)}$$ The others all give primary amines: $$\text{CH}_3\text{CONH}_2 \xrightarrow[\text{(ii) H}_3\text{O}^{+}]{\text{(i) LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2$$ $$\text{CH}_3\text{CONH}_2 \xrightarrow{\ \text{Br}_2/\text{KOH}\ } \text{CH}_3\text{NH}_2 \quad \text{(Hoffmann bromamide degradation)}$$ $$\text{CH}_3\text{CN} \xrightarrow[\text{(ii) H}_3\text{O}^{+}]{\text{(i) LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2$$
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Helium is used to dilute oxygen in diving apparatus. Reason R : Helium has high solubility in O$_2$. In the light of the above statements, choose the correct answer from the options given below
Solution: Helium is used as a diluent for oxygen in modern diving apparatus because of its very LOW solubility in blood — which is the opposite of what the Assertion's reason would suggest. Gases diffuse easily with each other, so helium does mix with oxygen; on that reading both statements stand as printed but the Reason does not explain the Assertion.
Identify the product in the following reaction: Benzenediazonium chloride, C$_6$H$_5$N$_2^{+}$Cl$^{-}$, is treated with (i) Cu$_2$Br$_2$/HBr, then (ii) Mg/dry ether, then (iii) H$_2$O.
Solution: Step (i) is the Sandmeyer reaction: the diazonium group is replaced by bromine to give bromobenzene. Step (ii) with Mg in dry ether converts the aryl halide into the Grignard reagent, phenylmagnesium bromide. Step (iii) with water protonates the carbanion-like aryl group, so the Grignard reagent is hydrolysed to benzene. The product is therefore benzene.
Consider the following reaction and identify the product (P). 3-Methylbutan-2-ol, CH$_3$–CH(CH$_3$)–CH(OH)–CH$_3$, is treated with HBr to give Product (P).
Solution: HBr protonates the hydroxyl group, and loss of water gives a secondary carbocation: $$\text{CH}_3\text{–CH(CH}_3\text{)–}\overset{\oplus}{\text{C}}\text{H–CH}_3 \quad (2^{\circ}\ \text{carbocation})$$ A hydride shift from the adjacent tertiary carbon converts this into the more stable tertiary carbocation: $$\text{CH}_3\text{–}\overset{\oplus}{\text{C}}\text{(CH}_3\text{)–CH}_2\text{–CH}_3$$ Bromide then attacks the tertiary carbocation, giving 2-bromo-2-methylbutane.
Identify product (A) in the following reaction: A compound carrying an acetyl group (–COCH$_3$) on a cyclohexane ring and a second acetyl group on an attached benzene ring is treated with Zn–Hg / conc. HCl to give (A) + 2H$_2$O.
Solution: This reaction is the Clemmensen reduction: Zn–Hg with concentrated HCl reduces a carbonyl group $\text{C=O}$ all the way to $\text{CH}_2$, releasing water. Each acetyl group –CO–CH$_3$ therefore becomes –CH$_2$–CH$_3$, an ethyl group. Since there are two carbonyls, 2H$_2$O are released, which matches the equation given.
Taking stability as the factor, which one of the following represents correct relationship?
Solution: As we move down the group, due to the poor shielding effect of the intervening $d$ and $f$ orbitals, the increased effective nuclear charge holds the $ns$ electrons tightly and therefore restricts their participation in bonding — the inert pair effect. So the relative stability of the $+1$ oxidation state increases for heavier elements. $$E^{\circ}\ \text{for In}^{3+}|\text{In}^{+} = -0.16\ \text{V}$$ $$E^{\circ}\ \text{for Tl}^{3+}|\text{Tl}^{+} = +1.6\ \text{V}$$ Hence TlI is more stable than TlI$_3$.
The conductivity of centimolar solution of KCl at 25$^{\circ}$C is 0.0210 ohm$^{-1}$ cm$^{-1}$ and the resistance of the cell containing the solution at 25$^{\circ}$C is 60 ohm. The value of cell constant is
Solution: Conductivity = conductance $\times$ cell constant $$k = G\,G^{*} = \frac{1}{R}G^{*}$$ $$G^{*} = k \times R = 0.0210 \times 60 = 1.26\ \text{cm}^{-1}$$
Which one of the following statements is correct?
Solution: The daily requirement of Mg and Ca in the human body is estimated to be 200 – 300 mg, that is 0.2 – 0.3 g, so that statement is correct. Bone in the human body is not an inert and unchanging substance — it is continuously being solubilised and redeposited. It is Ca that plays the important role in neuromuscular function, interneuronal transmission, cell membrane integrity and blood coagulation, not Mg. All enzymes that utilise ATP in phosphate transfer require Mg as the co-factor, not Ca.
Match List-I with List-II. List-I: A. Coke B. Diamond C. Fullerene D. Graphite List-II: I. Carbon atoms are $sp^{3}$ hybridised II. Used as a dry lubricant III. Used as a reducing agent IV. Cage like molecules Choose the correct answer from the options given below :
Solution: Coke is largely used as a reducing agent in metallurgy. In diamond, each carbon atom undergoes $sp^{3}$ hybridisation and is linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. Buckminsterfullerene contains six membered and five membered rings and hence is a cage like molecule. Graphite is very soft and slippery, hence it is used as a dry lubricant in machines running at high temperature. So A-III, B-I, C-IV, D-II.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic. Reason R : The deep blue solution is due to the formation of amide. In the light of the above statements, choose the correct answer from the options given below :
Solution: On dissolving an alkali metal (sodium) in liquid ammonia, a deep blue solution is developed due to the ammoniated electron, which absorbs energy in the visible region of light and imparts the blue colour. Due to the unpaired electron, the solution is paramagnetic. $$\text{M} + (x + y)\text{NH}_3 \rightarrow \left[\text{M(NH}_3)_x\right]^{+} + \left[e(\text{NH}_3)_y\right]^{-}$$ So the assertion statement is correct but the reason is incorrect — the colour is not due to amide formation.
Which of the following statements are NOT correct? A. Hydrogen is used to reduce heavy metal oxides to metals. B. Heavy water is used to study reaction mechanism. C. Hydrogen is used to make saturated fats from oils. D. The H–H bond dissociation enthalpy is lowest as compared to a single bond between two atoms of any elements. E. Hydrogen reduces oxides of metals that are more active than iron. Choose the most appropriate answer from the options given below:
Solution: Statements A, B and C are correct. (D) The H–H bond dissociation energy is the MAXIMUM as compared to a single bond between two atoms of any element, not the lowest — so D is not correct. (E) Hydrogen reduces oxides of metals that are LESS active than iron, not more active — so E is not correct.
The element expected to form largest ion to achieve the nearest noble gas configuration is
Solution: To reach the neon configuration, N gains three electrons to give N$^{3-}$, O gains two to give O$^{2-}$, F gains one to give F$^{-}$ and Na loses one to give Na$^{+}$. For isoelectronic species, as the negative charge on the anion increases the ionic size increases, because the same nuclear charge holds more electrons. So N forms the N$^{3-}$ anion with the largest ionic size.
In Lassaigne's extract of an organic compound, both nitrogen and sulphur are present, which gives blood red colour with Fe$^{3+}$ due to the formation of
Solution: In case nitrogen and sulphur are both present in an organic compound, sodium thiocyanate is formed. It gives a blood red colour and no Prussian blue, since there are no free cyanide ions. $$\text{Na} + \text{C} + \text{N} + \text{S} \longrightarrow \text{NaSCN}$$ $$\text{Fe}^{3+} + \text{SCN}^{-} \longrightarrow [\text{Fe(SCN)}]^{2+} \quad \text{(Blood red)}$$
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include : A. dipole - dipole forces B. dipole - induced dipole forces C. hydrogen bonding D. covalent bonding E. dispersion forces Choose the most appropriate answer from the options given below :
Solution: Intermolecular forces are the forces of attraction and repulsion between interacting molecules. This term does not include covalent bonds, as a covalent bond holds the atoms of a molecule together — it is an intramolecular force. Hence dipole-dipole forces, dipole-induced dipole forces, hydrogen bonding and dispersion forces are intermolecular forces.
Amongst the following the total number of species NOT having eight electrons around central atom in its outermost shell, is NH$_3$, AlCl$_3$, BeCl$_2$, CCl$_4$, PCl$_5$ :
Solution: Counting the electrons around the central atom in each species: NH$_3$ — 8 e$^{-}$ in nitrogen (three bond pairs and one lone pair) AlCl$_3$ — 6 e$^{-}$ in Al (electron deficient) BeCl$_2$ — 4 e$^{-}$ in Be (electron deficient) CCl$_4$ — 8 e$^{-}$ in C PCl$_5$ — 10 e$^{-}$ in P (expanded octet) So AlCl$_3$, BeCl$_2$ and PCl$_5$ do not obey the octet rule: 3 species.
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A$_x$B$_y$, then the value of $x + y$ is in option
Solution: Let the number of atoms of element B be $N$. In a cubic close packed structure there are $2N$ tetrahedral voids, and A occupies one third of them: $$\text{number of atoms of element A} = \frac{1}{3} \times 2N = \frac{2N}{3}$$ $$\therefore\ \text{The formula of the compound is } \text{A}_{\frac{2N}{3}}\text{B}_N = \text{A}_2\text{B}_3$$ So $x = 2$, $y = 3$ and $x + y = 5$.
Which amongst the following options are correct graphical representation of Boyle's law?
Solution: According to Boyle's law, $$PV = nRT$$ $$P = nRT\left(\frac{1}{V}\right)$$ $P$ versus $\left(\dfrac{1}{V}\right)$ gives a straight line graph through the origin with slope $nRT$. Since the slope rises with temperature, the steepest line is the highest temperature, giving a fan of straight lines labelled $T_3 > T_2 > T_1$.
Some tranquilizers are listed below. Which one from the following belongs to barbiturates?
Solution: Veronal is the derivative of barbituric acid and is considered as a barbiturate. Meprobamate, valium and chlordiazepoxide are other tranquilizers, but they are not barbiturates.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : In equation $\Delta_r G = -nFE_{\text{cell}}$, value of $\Delta_r G$ depends on $n$. Reason R : $E_{\text{cell}}$ is an intensive property and $\Delta_r G$ is an extensive property. In the light of the above statements, choose the correct answer from the options given below
Solution: The value of $\Delta_r G$ depends on the $n$ value as per the equation $\Delta_r G = -nFE_{\text{cell}}$, where $E$ is the emf of the cell and $nF$ is the amount of charge passed. So the assertion statement is correct. $E_{\text{cell}}$ is an intensive property while $\Delta_r G$ is an extensive thermodynamic property, so the reason is correct too — but it does not explain the assertion.
Given below are two statements : Statement I : A unit formed by the attachment of a base to 1$'$ position of sugar is known as nucleoside. Statement II : When nucleoside is linked to phosphorous acid at 5$'$-position of sugar moiety, we get nucleotide. In the light of the above statements, choose the correct answer from the options given below :
Solution: A unit formed by the attachment of a base to the 1$'$ position of sugar is known as a nucleoside. In nucleosides, the sugar carbons are numbered as 1$'$, 2$'$, 3$'$ etc. in order to distinguish these from the bases. When a nucleoside is linked to PHOSPHORIC acid at the 5$'$-position of the sugar moiety, we get a nucleotide — not phosphorous acid as Statement II says. So Statement I is true and Statement II is false.
For a certain reaction, the rate $= k[A]^{2}[B]$, when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would
Solution: Rate $(r) = k[A]^{2}[B]$ When the concentration of A is tripled, $[A'] = [3A]$: $$r' = k[A']^{2}[B] = k[3A]^{2}[B] = 9k[A]^{2}[B]$$ $$r' = 9r$$
Amongst the given options which of the following molecules/ion acts as a Lewis acid?
Solution: Lewis acids are the ones which accept a lone pair of electrons, due to the presence of a vacant orbital in the outermost shell. Boron in BF$_3$ has only six electrons around it and an empty $p$ orbital, so BF$_3$ is a Lewis acid. H$_2$O, OH$^{-}$ and NH$_3$ all carry lone pairs that they can donate, so each of them is a Lewis base.
Which of the following statements are INCORRECT? A. All the transition metals except scandium form MO oxides which are ionic. B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc$_2$O$_3$ to Mn$_2$O$_7$. C. Basic character increases from V$_2$O$_3$ to V$_2$O$_4$ to V$_2$O$_5$. D. V$_2$O$_4$ dissolves in acids to give VO$_4^{3-}$ salts. E. CrO is basic but Cr$_2$O$_3$ is amphoteric. Choose the correct answer from the options given below:
Solution: Statements A, B and E are correct as printed: All transition metals except Sc form MO oxides which are ionic. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc$_2$O$_3$ to Mn$_2$O$_7$. CrO is basic but Cr$_2$O$_3$ is amphoteric. Statement C is incorrect because the ACIDIC character increases from V$_2$O$_3$ to V$_2$O$_4$ to V$_2$O$_5$, not the basic character. Statement D is incorrect because V$_2$O$_4$ dissolves in acids to give VO$^{2+}$, not VO$_4^{3-}$. So the incorrect statements are C and D only.
What fraction of one edge centred octahedral void lies in one unit cell of fcc?
Solution: The total number of octahedral voids in FCC is four. Octahedral voids in FCC = Edge centres + Body centre An edge is shared between four unit cells, so the contribution of an edge centre is $\dfrac{1}{4}$. $$\therefore\ \text{Fraction of one edge centred octahedral void in one unit cell of FCC} = \frac{1}{4}$$
Match List-I with List-II : List-I (Oxoacids of Sulphur): A. Peroxodisulphuric acid B. Sulphuric acid C. Pyrosulphuric acid D. Sulphurous acid List-II (Bonds): I. Two S–OH, Four S=O, One S–O–S II. Two S–OH, One S=O III. Two S–OH, Four S=O, One S–O–O–S IV. Two S–OH, Two S=O Choose the correct answer from the options given below.
Solution: Peroxodisulphuric acid, H$_2$S$_2$O$_8$ — two S–OH, four S=O and the peroxide linkage S–O–O–S. Sulphuric acid, H$_2$SO$_4$ — two S–OH and two S=O. Pyrosulphuric acid, H$_2$S$_2$O$_7$ — two S–OH, four S=O and one S–O–S bridge. Sulphurous acid, H$_2$SO$_3$ — two S–OH and one S=O. So A–III, B–IV, C–I, D–II.
Identify the major product obtained in the following reaction: 2-Acetylbenzaldehyde, a benzene ring bearing a –COCH$_3$ group and a –CHO group on adjacent carbons, is treated with 2[Ag(NH$_3$)$_2$]$^{+}$ and 3$^{-}$OH on heating to give the major product.
Solution: Ammoniacal silver nitrate solution is Tollens' reagent. Tollens' reagent can be used to distinguish an aldehyde from a ketone, as an aldehyde on warming with Tollens' reagent produces a silver mirror due to formation of silver metal in alkaline medium. The aldehyde is oxidised to the corresponding carboxylate anion. $$\text{2-acetylbenzaldehyde} + 2[\text{Ag(NH}_3)_2]^{+} + 3\text{OH}^{-} \xrightarrow{\ \Delta\ } \text{2-acetylbenzoate} + 2\text{Ag} + 2\text{H}_2\text{O} + 4\text{NH}_3$$ The ketone group is untouched, so the major product carries –COCH$_3$ and –COO$^{-}$.
Consider the following compounds/species: i. Naphthalene ii. Cyclopentadienyl anion iii. Cyclobutadiene iv. Cyclopropenyl anion v. Cyclopropenyl cation vi. Cyclooctatetraene vii. Anthracene The number of compounds/species which obey Huckel's rule is ______.
Solution: Criteria for Huckel's rule: (i) Planarity (ii) Complete delocalisation of $\pi$ electrons (iii) Presence of $(4n + 2)\pi$ electrons in the ring where $n$ is an integer ($n$ = 0, 1, 2, …) The compounds which follow Huckel's rule are naphthalene (10 $\pi$ e$^{-}$), the cyclopentadienyl anion (6 $\pi$ e$^{-}$), the cyclopropenyl cation (2 $\pi$ e$^{-}$) and anthracene (14 $\pi$ e$^{-}$). Cyclobutadiene (4 $\pi$ e$^{-}$) and the cyclopropenyl anion (4 $\pi$ e$^{-}$) are antiaromatic, and cyclooctatetraene is non-planar (tub shaped). So the number is 4.
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
Solution: Enthalpy is defined as $H = U + PV$, so for a reaction at constant temperature involving ideal gases, $$\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_g RT$$ where $\Delta n_g$ is the change in the number of moles of gas.
Given below are two statements : Statement I : The nutrient deficient water bodies lead to eutrophication Statement II : Eutrophication leads to decrease in the level of oxygen in the water bodies. In the light of the above statements, choose the correct answer from the options given below:
Solution: Eutrophication is caused by nutrient ENRICHED water bodies, not nutrient deficient ones — the excess nitrate and phosphate feed heavy algal growth. So Statement I is incorrect. That algal bloom then dies and decomposes, and the decomposers consume dissolved oxygen, so eutrophication does lead to a decrease in the level of oxygen in the water body. Statement II is true.
Pumice stone is an example of
Solution: Pumice stone is a solid sol. Dispersed phase : Gas Dispersed medium : Solid A gas dispersed in a solid is classed as a solid sol (also called solid foam in some texts), and pumice is the standard example.
Which complex compound is most stable?
Solution: Chelating ligands in general form more stable complexes than their monodentate analogues, because of the chelate effect. Only [CoCl$_2$(en)$_2$]NO$_3$ contains the bidentate chelating ligand ethylenediamine (en). $$\therefore\ \text{The most stable complex is } [\text{CoCl}_2(\text{en})_2]\text{NO}_3$$
The reaction that does NOT take place in a blast furnace between 900 K to 1500 K temperature range during extraction of iron is :
Solution: At 900–1500 K, the higher temperature range in the blast furnace, the reactions which take place are: $$\text{C} + \text{CO}_2 \rightarrow 2\text{CO}$$ $$\text{FeO} + \text{CO} \rightarrow \text{Fe} + \text{CO}_2$$ $$\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3 \quad \text{(Slag formation)}$$ The reduction $\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{FeO} + \text{CO}_2$ takes place at 500–800 K, the lower temperature range, so it is the one that does not occur in this range.
The equilibrium concentrations of the species in the reaction A + B $\rightleftharpoons$ C + D are 2, 3, 10 and 6 mol L$^{-1}$, respectively at 300 K. $\Delta$G$^{\circ}$ for the reaction is (R = 2 cal/mol K)
Solution: At equilibrium the concentrations are A = 2, B = 3, C = 10, D = 6. $$k_{\text{eq}} = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} = \frac{10 \times 6}{2 \times 3} = 10$$ $$\Delta\text{G}^{\circ} = -RT\ln K = -2.303\,RT\log K$$ $$= -2.303 \times 2 \times 300 \times \log 10$$ $$= -1381.8\ \text{cal}$$
Consider the following reaction : Benzyl phenyl ether, C$_6$H$_5$CH$_2$–O–C$_6$H$_5$, is heated with HI to give A + B. Identify products A and B.
Solution: HI first protonates the ether oxygen. The C–O bond that breaks is the one that gives the more stable carbocation: here the benzyl carbocation, which is stabilised by extended conjugation with the ring. The bond to the phenyl carbon cannot break, because an aryl carbocation is far too unstable. Iodide then attacks the benzyl cation, so the products are benzyl iodide and phenol. $$\text{C}_6\text{H}_5\text{CH}_2\text{–O–C}_6\text{H}_5 \xrightarrow{\ \text{HI},\ \Delta\ } \text{C}_6\text{H}_5\text{CH}_2\text{I} + \text{C}_6\text{H}_5\text{OH}$$
Which amongst the following will be most readily dehydrated under acidic conditions?
Solution: Acid-catalysed dehydration goes through a carbocation, so the substrate that gives the most stabilised carbocation and the most stabilised alkene reacts fastest. In structure (4), protonation of one hydroxyl and loss of water gives a carbocation, and elimination gives an alkene. The second hydroxyl group is then lost in the same way, and the final product is a conjugated diene, which is extra stable. The strongly electron-withdrawing nitro group in the other three destabilises the intermediate carbocation and slows the dehydration down.
Identify the final product [D] obtained in the following sequence of reactions. CH$_3$CHO with (i) LiAlH$_4$, (ii) H$_3$O$^{+}$ gives [A]; [A] with H$_2$SO$_4$ and heat gives [B]; [B] with HBr gives [C]; [C] with bromobenzene and Na/dry ether gives [D].
Solution: $$\text{CH}_3\text{CHO} \xrightarrow[\text{(ii) H}_3\text{O}^{+}]{\text{(i) LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH}\ [A] \quad \text{(Reduction)}$$ $$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\Delta]{\text{H}_2\text{SO}_4} \text{CH}_2\text{=CH}_2\ [B] \quad \text{(Dehydration)}$$ $$\text{CH}_2\text{=CH}_2 \xrightarrow{\ \text{HBr}\ } \text{CH}_3\text{CH}_2\text{Br}\ [C] \quad \text{(Bromination)}$$ $$\text{CH}_3\text{CH}_2\text{Br} + \text{C}_6\text{H}_5\text{Br} \xrightarrow{\ \text{Na/dry ether}\ } \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\ [D] \quad \text{(Wurtz-Fittig Reaction)}$$ So [D] is ethylbenzene.
On balancing the given redox reaction, $$a\text{Cr}_2\text{O}_7^{2-} + b\text{SO}_3^{2-}(aq) + c\text{H}^{+}(aq) \rightarrow 2a\text{Cr}^{3+}(aq) + b\text{SO}_4^{2-}(aq) + \frac{c}{2}\text{H}_2\text{O}(l)$$ the coefficients $a$, $b$ and $c$ are found to be, respectively-
Solution: Using the ion electron method: Reduction half reaction: $\text{Cr}_2\text{O}_7^{2-} + 6e^{-} \longrightarrow 2\text{Cr}^{3+}$ Oxidation half reaction: $\text{SO}_3^{2-} \longrightarrow \text{SO}_4^{2-} + 2e^{-}$, taken $\times 3$ Overall reaction: $\text{Cr}_2\text{O}_7^{2-} + 3\text{SO}_3^{2-} \longrightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-}$ To balance 'O' atoms, adding H$_2$O on the RHS: $$\text{Cr}_2\text{O}_7^{2-} + 3\text{SO}_3^{2-} \longrightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + 4\text{H}_2\text{O}$$ To balance 'H' atoms, adding H$^{+}$ on the LHS: $$\text{Cr}_2\text{O}_7^{2-} + 3\text{SO}_3^{2-} + 8\text{H}^{+} \longrightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-} + 4\text{H}_2\text{O}$$ $$\therefore\ a = 1,\quad b = 3,\quad c = 8$$
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Late wood has fewer xylary elements with narrow vessels. Reason R : Cambium is less active in winters. In the light of the above statements, choose the correct answer from the options given below :
Solution: In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels, and this wood is called autumn wood or late wood. So both statements are true, and the reduced cambial activity in winter is precisely the reason the late wood has fewer, narrower elements.
The historic Convention on Biological Diversity, 'The Earth Summit' was held in Rio de Janeiro in the year
Solution: The historic Convention on Biological Diversity, "The Earth Summit", was held in Rio de Janeiro in the year 1992. It called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits.
In the equation GPP $-$ R $=$ NPP GPP is Gross Primary Productivity NPP is Net Primary Productivity R here is ________.
Solution: A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus respiration losses (R) is the net primary productivity. So R = Respiratory loss.
What is the function of tassels in the corn cob?
Solution: Tassels in the corn cob represent the stigma and style, which wave in the wind to trap pollen grains. Maize is wind pollinated, so the long feathery styles hanging out of the cob greatly increase the chance of catching airborne pollen.
What is the role of RNA polymerase III in the process of transcription in Eukaryotes?
Solution: In eukaryotes there are three major types of RNA polymerases. RNA polymerase I transcribes : 5.8S, 18S, 28S rRNAs RNA polymerase II transcribes : hnRNAs (precursor of mRNA) RNA polymerase III transcribes : tRNAs, ScRNA, 5S rRNA and snRNA
Upon exposure to UV radiation, DNA stained with ethidium bromide will show
Solution: In recombinant DNA technology the separated DNA fragments can be visualised only after staining the DNA with a substance known as ethidium bromide, followed by exposure to UV radiation. Bright orange coloured bands of DNA are then seen in an ethidium bromide stained gel exposed to UV light.
The thickness of ozone in a column of air in the atmosphere is measured in terms of :
Solution: The thickness of the ozone in a column of air from the ground to the top of the atmosphere is measured in terms of Dobson units (DU). Noise is measured in decibels, and kilobase is a unit of nucleic acid length.
The phenomenon of pleiotropism refers to
Solution: When a single gene affects multiple phenotypic expressions, the gene is called a pleiotropic gene and the phenomenon is called pleiotropism. Phenylketonuria is a standard example, where mutation in one gene produces mental retardation together with reduced hair and skin pigmentation.
Spraying of which of the following phytohormone on juvenile conifers helps hastening the maturity period, that leads early seed production?
Solution: Spraying juvenile conifers with gibberellins (GAs) hastens the maturity period, thus leading to early seed production.
Given below are two statements : One labelled as Assertion A and the other labelled as Reason R: Assertion A : The first stage of gametophyte in the life cycle of moss is protonema stage. Reason R : Protonema develops directly from spores produced in capsule. In the light of the above statements, choose the most appropriate answer from options given below:
Solution: The predominant stage of the life cycle of a moss is the gametophyte, which consists of two stages. The first stage is the protonema stage, which develops directly from a spore. The capsule of the sporophyte contains spores which give rise to protonema. Thus, the reason correctly explains the assertion.
During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out
Solution: During isolation of the genetic material, purified DNA ultimately precipitates out after the addition of chilled ethanol. Proteins such as histones are removed earlier by treatment with proteases, and RNA is removed by treatment with ribonuclease — so neither is what the ethanol brings down.
The reaction centre in PS II has an absorption maxima at
Solution: In PS-I, the reaction centre chlorophyll $a$ has an absorption peak at 700 nm, while in PS-II the reaction centre has an absorption maxima at 680 nm. That is why the two are written as P700 and P680 respectively.
Family Fabaceae differs from Solanaceae and Liliaceae. With respect to the stamens, pick out the characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae.
Solution: Fabaceae $\rightarrow$ Diadelphous and dithecous anther. Solanaceae $\rightarrow$ Polyandrous, epipetalous and dithecous anther. Liliaceae $\rightarrow$ Polyandrous, epiphyllous and dithecous anther. The diadelphous condition, in which the stamens are united into two bundles, is the feature unique to Fabaceae among these three.
Which micronutrient is required for splitting of water molecule during photosynthesis?
Solution: Manganese plays a major role in the splitting of water to liberate oxygen during photosynthesis. Copper is essential for the overall metabolism in plants. Molybdenum is included in nitrogen metabolism. Magnesium activates several enzymes involved in photosynthesis and respiration.
Among 'The Evil Quartet', which one is considered the most important cause driving extinction of species?
Solution: Habitat loss and fragmentation is the most important cause driving animals and plants to extinction. It heads the list of the four causes known as the Evil Quartet, the others being over-exploitation, alien species invasions and co-extinctions.
Movement and accumulation of ions across a membrane against their concentration gradient can be explained by
Solution: Movement and accumulation of ions across a membrane against their concentration gradient can be explained by active transport. It uses energy to transport molecules from a lower concentration to a higher concentration. Passive transport, osmosis and facilitated diffusion all move substances down a gradient and need no energy input.
In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as
Solution: In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as dedifferentiation. Dedifferentiation is a phenomenon by which living differentiated plant cells, that by now have lost the capacity to divide, can regain the capacity of division under certain conditions.
The process of appearance of recombination nodules occurs at which sub stage of prophase I in meiosis?
Solution: The process of recombination occurs at the pachytene stage of prophase I. This stage is characterised by the appearance of recombination nodules, the sites at which crossing over takes place.
Given below are two statements : Statement I : The forces generated transpiration can lift a xylem-sized column of water over 130 meters height. Statement II : Transpiration cools leaf surfaces sometimes 10 to 15 degrees evaporative cooling. In the light of the above statements, choose the most appropriate answer from the options given below :
Solution: Statement I is correct, as measurements reveal that the forces generated by transpiration can create pressures sufficient to lift a xylem sized column of water up to 130 metres high. Statement II is also correct, as transpiration cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling.
Axile placentation is observed in
Solution: China rose, Tomato, Petunia and Lemon show axile placentation. Dianthus and Primrose show free central placentation. Pea, Lupin and Beans show marginal placentation. Cucumber and mustard show parietal placentation. So the set in which every member shows axile placentation is China rose, Petunia and Lemon.
Which of the following stages of meiosis involves division of centromere?
Solution: Splitting of the centromere occurs during anaphase of mitosis or anaphase II of meiosis. During Metaphase I and II, chromosomes align at the equator. During telophase, chromosomes reach the respective poles.
Frequency of recombination between gene pairs on same chromosome as a measure of the distance between genes to map their position on chromosome, was used for the first time by
Solution: Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and 'mapped' their position on the chromosome. Sutton and Boveri proposed the chromosomal theory of inheritance. Henking discovered the X-chromosome. Thomas Hunt Morgan proved the chromosomal theory of inheritance and proposed the concept of linkage.
Cellulose does not form blue colour with Iodine because
Solution: Cellulose does not contain complex helices and hence cannot hold iodine molecules, so it gives no blue colour. Starch, by contrast, forms helical secondary structures that can trap iodine and give the familiar blue-black colour. The other options are not correct, as cellulose is a polysaccharide, not a disaccharide, and it is not helical.
In gene gun method used to introduce alien DNA into host cells, microparticles of ________ metal are used.
Solution: In the gene gun method, microparticles of tungsten or gold are used. Gold and tungsten are inert in nature, so they do not alter the chemical composition of the cells they are fired into. The cells are bombarded with these DNA-coated microparticles at high velocity, which is why the method is also called biolistics.
Identify the pair of heterosporous pteridophytes among the following :
Solution: Selaginella and Salvinia are heterosporous pteridophytes. They produce two different kinds of spores — microspores and megaspores. Psilotum, Lycopodium and Equisetum are homosporous pteridophytes, producing only one kind of spore.
Unequivocal proof that DNA is the genetic material was first proposed by
Solution: The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase. Avery, Macleoid and McCarty gave the biochemical characterisation of the Transforming Principle. The transformation experiments using Pneumococcus were conducted by Frederick Griffith. Wilkins and Franklin produced X-ray diffraction data of DNA.
How many ATP and NADPH$_2$ are required for the synthesis of one molecule of Glucose during Calvin cycle?
Solution: For every CO$_2$ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH$_2$ are required. To make one molecule of glucose, 6 turns of the cycle are required. Thus, $$6 \times \left[3\ \text{ATP} + 2\ \text{NADPH}_2\right] = 18\ \text{ATP and } 12\ \text{NADPH}_2$$
Identify the correct statements: A. Detrivores perform fragmentation. B. The humus is further degraded by some microbes during mineralization. C. Water soluble inorganic nutrients go down into the soil and get precipitated by a process called leaching. D. The detritus food chain begins with living organisms. E. Earthworms break down detritus into smaller particles by a process called catabolism. Choose the correct answer from the options given below:
Solution: The detritus food chain begins with detritus, that is dead organic matter, not with living organisms — so D is wrong. Earthworms break down detritus into smaller particles by fragmentation. It is the saprotrophic bacteria and fungi that break detritus into simpler inorganic substances by a process called catabolism — so E is wrong. A, B and C are correct as stated.
In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are :
Solution: Synergids are the cells of the gametophyte and hence these are haploid. Zygote is formed by fusion of two gametes and thus it is diploid. Primary endosperm nucleus is formed by the fusion of the diploid secondary nucleus with a male gamete. Therefore, it is triploid. So the haploid, diploid and triploid structures in order are synergids, zygote and primary endosperm nucleus.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : ATP is used at two steps in glycolysis. Reason R : First ATP is used in converting glucose into glucose-6-phosphate and second ATP is used in conversion of fructose-6-phosphate into fructose-1, 6-diphosphate. In the light of the above statements, choose the correct answer from the options given below :
Solution: ATP in glycolysis is used at two steps of conversion, which are Glucose $\rightarrow$ Glucose-6-phosphate Fructose-6-phosphate $\rightarrow$ Fructose-1,6-bisphosphate The reason for the utilisation of ATP is the phosphorylation of the substrates, so the Reason names exactly the two steps the Assertion refers to and explains it correctly.
Expressed Sequence Tags (ESTs) refers to
Solution: All the genes that are expressed as RNA are referred to as Expressed Sequence Tags (ESTs). This was one of the two approaches used in the Human Genome Project, the other being sequence annotation, in which the whole genome is sequenced first and functions assigned afterwards.
Among eukaryotes, replication of DNA takes place in :
Solution: Replication of DNA takes place in the S-phase of the cell cycle in eukaryotes. Most of the cell organelles duplicate in G$_1$ phase.
Given below are two statements : Statement I : Endarch and exarch are the terms often used for describing the position of secondary xylem in the plant body. Statement II : Exarch condition is the most common feature of the root system. In the light of the above statements, choose the correct answer from the options given below:
Solution: Endarch and exarch are the terms often used for describing the position of PRIMARY xylem in the plant body, not secondary xylem — so Statement I is false. Primary xylem is of two types, protoxylem and metaxylem. On the basis of the relative position of protoxylem and metaxylem in the organ, the arrangement of primary xylem can be endarch or exarch. Exarch type of primary xylem is seen in roots, so Statement II is true.
Large, colourful, fragrant flowers with nectar are seen in
Solution: Large, colourful, fragrant flowers with nectar attract biotic pollinators (insects), thus they are seen in insect pollinated plants. Wind pollinated flowers are small and dull with no nectar, and the reward of nectar together with scent and colour is the classic entomophilous set of adaptations.
Which hormone promotes internode/petiole elongation in deep water rice?
Solution: Ethylene promotes rapid internode/petiole elongation in deep water rice plants. This helps the leaves and the upper parts of the shoot to remain above water, so the plant is not drowned as the water rises.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R : Assertion A : A flower is defined as modified shoot wherein the shoot apical meristem changes to floral meristem. Reason R : Internode of the shoot gets condensed to produce different floral appendages laterally at successive node instead of leaves. In the light of the above statements, choose the correct answer from the options given below :
Solution: A flower is a modified shoot wherein the shoot apical meristem changes to floral meristem. Internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral appendages laterally at the successive nodes instead of leaves. Therefore, both A and R are true and R is the correct explanation of A.
Match List I with List II : List I: A. M Phase B. G$_2$ Phase C. Quiescent stage D. G$_1$ Phase List II: I. Proteins are synthesized II. Inactive phase III. Interval between mitosis and initiation of DNA replication IV. Equational division Choose the correct answer from the options given below :
Solution: M phase or mitosis is the phase where the actual cell division occurs. Mitosis is also called equational division. During G$_2$ phase DNA synthesis stops but the cell synthesises RNA, proteins etc. for the next phase. Quiescent stage is the inactive phase which non-dividing cells enter. G$_1$ phase is the interval between mitosis and initiation of DNA replication. So A-IV, B-I, C-II, D-III.
How many different proteins does the ribosome consist of?
Solution: The ribosome consists of structural RNAs and about 80 different proteins.
Which one of the following statements is NOT correct?
Solution: Algal bloom imparts a distinct colour to water bodies. It causes deterioration of the water quality and fish mortality, so it does not improve water quality or promote fisheries — that statement is not correct. The other three describe real consequences: water hyacinth choking eutrophic water bodies, biomagnification of toxins up trophic levels, and the oxygen depletion caused by decomposers in sewage-polluted water.
Which of the following combinations is required for chemiosmosis?
Solution: Chemiosmosis requires a membrane, a proton pump, a proton gradient and ATP synthase. The gradient that matters is a proton gradient, not an electron gradient, and the enzyme that uses it is ATP synthase.
Match List I with List II : List I: A. Cohesion B. Adhesion C. Surface tension D. Guttation List II: I. More attraction in liquid phase II. Mutual attraction among water molecules III. Water loss in liquid phase IV. Attraction towards polar surfaces Choose the correct answer from the options given below :
Solution: Cohesion represents mutual attraction between water molecules. Adhesion represents attraction of water molecules to polar surfaces. Surface tension represents water molecules being attracted to each other in the liquid phase more than to water in the gas phase. Guttation represents loss of water in liquid phase. Thus A – II, B – IV, C – I, D – III.
Match List I with List II: List I: A. Iron B. Zinc C. Boron D. Molybdenum List II: I. Synthesis of auxin II. Component of nitrate reductase III. Activator of catalase IV. Cell elongation and differentiation Choose the correct answer from the options given below:
Solution: Iron activates catalase enzyme. Zinc is needed in the synthesis of auxin. Boron is required for cell elongation and cell differentiation. Molybdenum is a component of nitrogenase and nitrate reductase enzyme. Therefore A-III, B-I, C-IV, D-II.
Given below are two statements: Statement I : Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and competitively inferior one will be eliminated eventually. Statement II : In general, carnivores are more adversely affected by competition than herbivores. In the light of the above statements, choose the correct answer from the options given below:
Solution: Gause's 'Competitive Exclusion Principle' states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually. Thus, statement I is correct. Statement II is incorrect, as in general herbivores and plants appear to be more adversely affected by competition than carnivores.
Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of
Solution: Malonate is a competitive inhibitor of the enzyme succinate dehydrogenase. Inhibition of succinic dehydrogenase by malonate occurs due to the close resemblance of malonate with the substrate succinate in structure. Competitive inhibitors are often used in the control of bacterial pathogens.
Identify the correct statements: A. Lenticels are the lens-shaped openings permitting the exchange of gases. B. Bark formed early in the season is called hard bark. C. Bark is a technical term that refers to all tissues exterior to vascular cambium. D. Bark refers to periderm and secondary phloem. E. Phellogen is single-layered in thickness. Choose the correct answer from the options given below:
Solution: Lenticels are lens shaped openings permitting exchange of gases between the outer atmosphere and internal tissue of the stem — A is correct. Bark that is formed early in the season is called early or soft bark; towards the end of the season late or hard bark is formed — so B is wrong. Bark is a NON-technical term that refers to all tissues exterior to the vascular cambium — so C is wrong. Bark refers to a number of tissue types, viz. periderm and secondary phloem — D is correct. Phellogen is a couple of layers thick, not single-layered — so E is wrong. Therefore, only statements A and D are correct.
Given below are two statements : One labelled as Assertion A and the other labelled as Reason R : Assertion A : In gymnosperms the pollen grains are released from the microsporangium and carried by air currents. Reason R : Air currents carry the pollen grains to the mouth of the archegonia where the male gametes are discharged and pollen tube is not formed. In the light of the above statements, choose the correct answer from the options given below :
Solution: The Assertion is correct: in gymnosperms the pollen grains are released from the microsporangium and they are carried in air currents. The Reason is false. The pollen grains come in contact with the opening of the ovules borne on megasporophylls, and a POLLEN TUBE carrying the male gametes grows towards the archegonia in the ovules and discharges its contents near the mouth of the archegonia.
Which of the following statements are correct about Klinefelter's Syndrome? A. This disorder was first described by Langdon Down (1866). B. Such an individual has overall masculine development. However, the feminine development is also expressed. C. The affected individual is short statured. D. Physical, psychomotor and mental development is retarded. E. Such individuals are sterile. Choose the correct answer from the options given below:
Solution: Klinefelter's syndrome is caused due to the presence of an additional copy of the X-chromosome, resulting in a karyotype of 47, XXY. Such an individual has overall masculine development; however, the feminine development is also expressed. Such individuals are sterile. Thus statements B and E are correct regarding Klinefelter's syndrome. Statements A, C and D are incorrect with respect to Klinefelter's syndrome, as they are associated with Down's syndrome.
Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence. A. Insertion of recombinant DNA into the host cell B. Cutting of DNA at specific location by restriction enzyme C. Isolation of desired DNA fragment D. Amplification of gene of interest using PCR Choose the correct answer from the options given below :
Solution: Recombinant DNA technology involves several steps in a specific sequence: isolation of DNA, fragmentation of DNA by restriction endonucleases, isolation of the desired DNA fragment, ligation of the DNA fragment into a vector, transferring the recombinant DNA into the host, culturing the host cells in a medium at large scale and extraction of the desired product. So of the four steps listed, the order is B, C, D, A.
Match List I with List II : List I: A. Oxidative decarboxylation B. Glycolysis C. Oxidative phosphorylation D. Tricarboxylic acid cycle List II: I. Citrate synthase II. Pyruvate dehydrogenase III. Electron transport system IV. EMP pathway Choose the correct answer from the options given below :
Solution: Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters the mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalyzed by pyruvate dehydrogenase. The scheme of glycolysis was given by Gustav Embden, Otto Meyrhof and J. Parnas, and is often referred to as the EMP pathway. In the electron transport system, the energy of oxidation-reduction is utilised for the production of the proton gradient required for phosphorylation; thus this process is also called oxidative phosphorylation. The TCA cycle starts with the condensation of the acetyl group with oxaloacetic acid (OAA) and water to yield citric acid. The reaction is catalysed by the enzyme citrate synthase. Thus A – II, B – IV, C – III, D – I.
Match List I with List II : List I (Interaction): A. Mutualism B. Commensalism C. Amensalism D. Parasitism List II (Species A and B): I. +(A), 0(B) II. $-$(A), 0(B) III. +(A), $-$(B) IV. +(A), +(B) Choose the correct answer from the options given below:
Solution: (+, +) Mutualism : In this interaction, both the interacting species are benefitted. (+, 0) Commensalism : Only one species is benefitted and the other species remains unharmed. ($-$, 0) Amensalism : Neither species is benefitted. One remains unharmed and the other is harmed. (+, $-$) Parasitism : One species is benefitted and the other is negatively affected. So A-IV, B-I, C-II, D-III.
Broad palm with single palm crease is visible in a person suffering from-
Solution: Down's syndrome is caused by an additional copy of chromosome number 21. Its symptoms include: a. Broad palm with characteristic palm crease b. Short statured with small round head c. Furrowed tongue and partially open mouth, etc.
Given below are two statements: Statement I : A protein is imagined as a line, the left end represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal). Statement II : Adult human haemoglobin, consists of 4 subunits (two subunits of $\alpha$ type and two subunits of $\beta$ type.) In the light of the above statements, choose the correct answer from the options given below:
Solution: A protein is imagined as a line, the left end represented by the first amino acid and the right end represented by the last amino acid. But the FIRST amino acid is called the N-terminal amino acid and the LAST amino acid is called the C-terminal amino acid — Statement I has them the wrong way round, so it is false. Statement II is true: adult human haemoglobin consists of 4 subunits, two of $\alpha$ type and two of $\beta$ type.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Nephrons are of two types: Cortical & Juxta medullary, based on their relative position in cortex and medulla. Reason R: Juxta medullary nephrons have short loop of Henle whereas, cortical nephrons have longer loop of Henle. In the light of the above statements, choose the correct answer from the options given below:
Solution: The Assertion is true, as there are two types of nephrons, i.e. cortical nephrons and juxtamedullary nephrons, based on their relative position in the cortex and medulla. The Reason is not correct, as the loop of Henle in juxtamedullary nephrons is very long and runs deep into the medulla, while in cortical nephrons it is short. Therefore, Assertion is true but Reason is false.
Match List I with List II. List I: A. Heroin B. Marijuana C. Cocaine D. Morphine List II: I. Effect on cardiovascular system II. Slow down body function III. Painkiller IV. Interfere with transport of dopamine Choose the correct answer from the options given below:
Solution: Heroin is a depressant and slows down body functions. Marijuana affects the cardiovascular system of the body. Cocaine interferes with the transport of the neurotransmitter dopamine. Morphine is a very effective sedative and painkiller. So A-II, B-I, C-IV, D-III.
Vital capacity of lung is _________.
Solution: Vital capacity is the maximum volume of air a person can breathe in after a forced expiration. This includes ERV, TV and IRV: $$\text{VC} = \text{IRV} + \text{ERV} + \text{TV}$$ Residual volume is excluded, because it can never be expelled from the lungs.
Match List I with List II. List I: A. P-wave B. Q-wave C. QRS complex D. T-wave List II: I. Beginning of systole II. Repolarisation of ventricles III. Depolarisation of atria IV. Depolarisation of ventricles Choose the correct answer from the options given below :
Solution: In a standard ECG, the P-wave represents the electrical excitation (or depolarisation) of the atria, which leads to the contraction of both the atria. The Q-wave marks the beginning of systole. The QRS complex represents the depolarisation of the ventricles, which initiates the ventricular contraction. The T-wave represents the return of the ventricles from excited to normal state, i.e. repolarisation. So A-III, B-I, C-IV, D-II.
Match List I with List II List I (Cells): A. Peptic cells B. Goblet cells C. Oxyntic cells D. Hepatic cells List II (Secretion): I. Mucus II. Bile juice III. Proenzyme pepsinogen IV. HCl and intrinsic factor for absorption of vitamin B$_{12}$ Choose the correct answer from the options given below:
Solution: Gastric glands have three major types of cells, namely: (i) Mucus neck cells, which secrete mucus — these are the goblet cells of the list (ii) Peptic or chief cells, which secrete the proenzyme pepsinogen (iii) Parietal or oxyntic cells, which secrete HCl and intrinsic factor for absorption of vitamin B$_{12}$ Hepatic cells secrete bile juice. So A-III, B-I, C-IV, D-II.
Match List I with List II. List I: A. Vasectomy B. Coitus interruptus C. Cervical caps D. Saheli List II: I. Oral method II. Barrier method III. Surgical method IV. Natural method Choose the correct answer from the options given below:
Solution: (i) Vasectomy is a surgical method of contraception. (ii) Coitus interruptus is a natural method of contraception. (iii) Cervical cap is a barrier method of contraception. (iv) Saheli is an oral method of contraception, which is a non-steroidal pill. So A-III, B-IV, C-II, D-I.
Which one of the following common sexually transmitted diseases is completely curable when detected early and treated properly?
Solution: Except for hepatitis-B, genital herpes and HIV infection, other STIs are completely curable if detected early and treated properly. Gonorrhoea is a bacterial disease which can be treated and cured completely; the other diseases mentioned are viral diseases.
Match List I with List II. List I: A. Ringworm B. Filariasis C. Malaria D. Pneumonia List II: I. Haemophilus influenzae II. Trichophyton III. Wuchereria bancrofti IV. Plasmodium vivax Choose the correct answer from the options given below:
Solution: (i) Ringworm is caused by Trichophyton. (ii) Filariasis is caused by Wuchereria bancrofti. (iii) Malaria is caused by Plasmodium species. (iv) Pneumonia is caused by Haemophilus influenzae. So A-II, B-III, C-IV, D-I.
Which of the following are NOT considered as the part of endomembrane system? A. Mitochondria B. Endoplasmic reticulum C. Chloroplasts D. Golgi complex E. Peroxisomes Choose the most appropriate answer from the options given below:
Solution: The endomembrane system includes endoplasmic reticulum (ER), golgi complex, lysosomes and vacuoles. Since the functions of the mitochondria, chloroplast and peroxisomes are not coordinated with the above components, these are not considered as part of the endomembrane system. So A, C and E only.
Which of the following is not a cloning vector?
Solution: A single stranded DNA or RNA tagged with a radioactive molecule is called a probe, and it helps in the detection of a mutated gene. It is not a cloning vector. YAC, BAC and pBR322 are all vectors.
Match List I with List II. List I (Interacting species): A. A Leopard and a Lion in a forest/grassland B. A Cuckoo laying egg in a Crow's nest C. Fungi and root of a higher plant in Mycorrhizae D. A cattle egret and a Cattle in a field List II (Name of interaction): I. Competition II. Brood parasitism III. Mutualism IV. Commensalism Choose the correct answer from the options given below.
Solution: A leopard and a lion in a forest/grassland exemplify competition, where both the species are competing for the same resources. A cuckoo laying an egg in a crow's nest is brood parasitism, where the cuckoo is the parasitic bird that lays its egg in the nest of the crow (host bird). Fungi and root of a higher plant in mycorrhizae exemplify mutualism, where both the species are benefitted. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates. A cattle egret and cattle in a field exemplify commensalism, where one species benefits and the other remains unaffected. The egrets always forage close to where cattle are grazing, because the cattle, as they move, stir up and flush out insects from the vegetation that otherwise might be difficult for the egrets to find and catch. So A-I, B-II, C-III, D-IV.
Which one of the following symbols represents mating between relatives in human pedigree analysis?
Solution: In a human pedigree chart, a mating between two individuals is drawn as a single horizontal line joining them. When the two individuals are relatives, the mating is consanguineous, and this is shown by a DOUBLE horizontal line joining the square to the circle. So the symbol representing mating between relatives is the square joined to the circle by a double line.
Match List I with List II. List I: A. Taenia B. Paramoecium C. Periplaneta D. Pheretima List II: I. Nephridia II. Contractile vacuole III. Flame cells IV. Urecose gland Choose the correct answer from the options given below:
Solution: Protonephridia or flame cells are the excretory structures in platyhelminthes such as Taenia. Nephridia are the tubular excretory structures of earthworms (Pheretima) and other annelids. Single celled organisms like Paramoecium have contractile vacuoles for excretion. Uricose glands are present in the cockroach, Periplaneta. So A-III, B-II, C-IV, D-I.
Once the undigested and unabsorbed substances enter the caecum, their backflow is prevented by
Solution: The undigested food (faeces) enters into the caecum of the large intestine through the ileo-caecal valve, which prevents the backflow of the faecal matter. The gastro-oesophageal sphincter regulates the opening of the oesophagus into the stomach. The pyloric sphincter regulates the opening between the stomach and duodenum. The opening of the common hepato-pancreatic duct is guarded by the sphincter of Oddi.
Match List I with List II. List I: A. Gene 'a' B. Gene 'y' C. Gene 'i' D. Gene 'z' List II: I. $\beta$-galactosidase II. Transacetylase III. Permease IV. Repressor protein Choose the correct answer from the options given below:
Solution: In a lac operon, Gene a codes for the enzyme transacetylase. Gene y codes for the enzyme permease. Gene i codes for the repressor protein. Gene z codes for the enzyme $\beta$-galactosidase. So A-II, B-III, C-IV, D-I.
Which one of the following techniques does not serve the purpose of early diagnosis of a disease for its early treatment?
Solution: Using conventional methods of diagnosis like serum and urine analysis does not help in early diagnosis — these detect a disease only after the symptoms have appeared. Recombinant DNA technology, Polymerase Chain Reaction (PCR) and Enzyme Linked Immuno-Sorbent Assay (ELISA) are some of the techniques that serve the purpose of early diagnosis.
Given below are two statements : Statement I : Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat. Statement II : When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. In the light of the above statements, choose the correct answer from the options given below :
Solution: Low temperature preserves the enzyme in a temporarily inactive state, whereas high temperature destroys enzymatic activity because proteins are denatured by heat — Statement I is true. A competitive inhibitor, due to its close structural similarity with the substrate, competes with the substrate for the substrate-binding site of the enzyme — Statement II is true as well.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Endometrium is necessary for implantation of blastocyst. Reason R: In the absence of fertilization, the corpus luteum degenerates that causes disintegration of endometrium. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both Assertion and Reason are true. Implantation is the embedding of the blastocyst into the endometrium of the uterus, so the endometrium is necessary for it. Corpus luteum secretes a large amount of progesterone which is essential for maintenance of the endometrium of the uterus. In absence of fertilisation, the corpus luteum degenerates, hence the decrease in the level of progesterone hormone will cause disintegration of the endometrium leading to menstruation. But that describes what happens when implantation does NOT occur, so it does not explain why the endometrium is necessary for implantation.
Given below are two statements: one is labelled as Assertion A and other is labelled as Reason R. Assertion A : Amniocentesis for sex determination is one of the strategies of Reproductive and Child Health Care Programme. Reason R : Ban on amniocentesis checks increasing menace of female foeticide. In the light of the above statements, choose the correct answer from the options given below.
Solution: The 'Reproductive and Child Health Care (RCH) programme' deals with creating awareness among people about various reproduction related aspects and providing facilities and support for building up a reproductively healthy society. Amniocentesis is basically used to test for the presence of certain genetic disorders such as Down's syndrome, haemophilia etc., to determine the survivability of the foetus. Amniocentesis is not a sex determination technique in India and is not a strategy of RCH — so the Assertion is false, while the Reason is true.
Which of the following statements is correct?
Solution: Increase in the concentration of the toxicant at successive trophic levels is called biomagnification. A large amount of nutrients in water promotes the growth of algal bloom rather than restricting it, and algal bloom increases fish mortality. Eutrophication refers to the natural ageing of a lake by nutrient enrichment of its water.
Which of the following statements are correct regarding female reproductive cycle? A. In non-primate mammals cyclical changes during reproduction are called oestrus cycle. B. First menstrual cycle begins at puberty and is called menopause. C. Lack of menstruation may be indicative of pregnancy. D. Cyclic menstruation extends between menarche and menopause. Choose the most appropriate answer from the options given below.
Solution: The first menstrual cycle that begins at puberty is called menarche, not menopause — so statement B is wrong. In non-primate mammals, cyclical changes during reproduction are called the oestrus cycle; in primates they are called the menstrual cycle. Lack of menstruation may be indicative of pregnancy. Cyclic menstruation is an indicator of the normal reproductive phase and extends between menarche and menopause. Hence A, C and D only.
Match List I with List II. List I: A. CCK B. GIP C. ANF D. ADH List II: I. Kidney II. Heart III. Gastric gland IV. Pancreas Choose the correct answer from the options given below :
Solution: Cholecystokinin (CCK) acts on both the gall bladder and the pancreas and stimulates the secretion of bile juice and pancreatic enzymes respectively. GIP inhibits gastric secretion and motility, so it acts on the gastric gland. Atrial Natriuretic Factor (ANF) is released from the atrial wall of our heart. Anti-diuretic hormone (ADH) acts mainly on the kidney and stimulates resorption of water and electrolytes by the distal tubules. So A-IV, B-III, C-II, D-I.
Given below are two statements: Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid. Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome. In the light of the above statements, choose the correct answer from the options given below:
Solution: In prokaryotes, the NEGATIVELY charged DNA is held with some POSITIVELY charged proteins in a region termed as nucleoid — Statement I has the charges the wrong way round, so it is incorrect. In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome — Statement II is correct.
Match List I with List II. List I (Type of Joint): A. Cartilaginous Joint B. Ball and Socket Joint C. Fibrous Joint D. Saddle Joint List II (Found between): I. Between flat skull bones II. Between adjacent vertebrae in vertebral column III. Between carpal and metacarpal of thumb IV. Between Humerus and Pectoral girdle Choose the correct answer from the options given below:
Solution: Cartilaginous joint is present in between the adjacent vertebrae in the vertebral column. Ball and socket joint is present between the humerus and the pectoral girdle. Fibrous joint is present between the flat skull bones. Saddle joint is present between the carpal and metacarpal of the thumb. So A-II, B-IV, C-I, D-III.
Given below are two statements: Statement I: Ligaments are dense irregular tissue. Statement II: Cartilage is dense regular tissue. In the light of the above statements, choose the correct answer from the options given below:
Solution: A ligament is an example of dense REGULAR connective tissue, not dense irregular tissue, so Statement I is incorrect. Cartilage is an example of SPECIALISED connective tissue and not dense regular tissue, therefore Statement II is also incorrect.
In which blood corpuscles, the HIV undergoes replication and produces progeny viruses?
Solution: HIV enters into helper T-lymphocytes (T$_H$), replicates and produces progeny viruses. The progeny viruses released into the blood attack other helper lymphocytes, which is why the helper T-cell count falls progressively in AIDS.
Radial symmetry is NOT found in adults of phylum ______.
Solution: Hemichordates are bilaterally symmetrical animals, so radial symmetry is not found in their adults. Coelenterates are radially symmetrical organisms. Adult echinoderms are radially symmetrical in the adult stage. Ctenophores are radially symmetrical organisms.
Which of the following functions is carried out by cytoskeleton in a cell?
Solution: An elaborate network of filamentous proteinaceous structures consisting of microtubules, microfilaments and intermediate filaments present in the cytoplasm is collectively referred to as the cytoskeleton. It is involved in many functions such as mechanical support, motility, and maintenance of the shape of the cell.
Given below are two statements: Statement I: Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct. Statement II: The cavity of the cervix is called cervical canal which along with vagina forms birth canal. In the light of the above statements, choose the correct answer from the options given below:
Solution: Both statements are correct. Vas deferens receives a duct from the seminal vesicle and opens into the urethra as the ejaculatory duct. The cavity of the cervix is called the cervical canal, which along with the vagina forms the birth canal.
Select the correct group/set of Australian Marsupials exhibiting adaptive radiation.
Solution: Numbat, spotted cuscus and flying phalanger are Australian marsupials exhibiting adaptive radiation. Option (1) is incorrect because mole and flying squirrel are placental mammals. Option (2) is incorrect because lemur and wolf are placental mammals. Option (3) is incorrect because bobcat is a placental mammal.
Given below are two statements: Statement I : Electrostatic precipitator is most widely used in thermal power plant Statement II : Electrostatic precipitator in thermal power plant removes ionising radiations In the light of the above statements, choose the most appropriate answer from the options given below:
Solution: Electrostatic precipitator is most widely used in thermal power plants, so Statement I is correct. It can remove over 99 per cent of the particulate matter present in the exhaust from a thermal power plant. It removes particulate matter, not ionising radiations, so Statement II is incorrect.
Given below are two statements: Statement I: RNA mutates at a faster rate. Statement II: Viruses having RNA genome and shorter life span mutate and evolve faster. In the light of the above statements, choose the correct answer from the options given below:
Solution: RNA, being unstable, mutates at a faster rate. Consequently, viruses having an RNA genome and a shorter life span mutate and evolve faster. Both statements are therefore true.
Match List I with List II with respect to human eye. List I: A. Fovea B. Iris C. Blind spot D. Sclera List II: I. Visible coloured portion of eye that regulates diameter of pupil. II. External layer of eye formed of dense connective tissue. III. Point of greatest visual acuity or resolution. IV. Point where optic nerve leaves the eyeball and photoreceptor cells are absent. Choose the correct answer from the options given below:
Solution: (i) Fovea is the point of greatest visual acuity or resolution. (ii) Iris is the visible coloured portion of the eye that regulates the diameter of the pupil. (iii) Blind spot is the point where the optic nerve leaves the eyeball and photoreceptor cells are absent. (iv) Sclera is the external layer of eye formed of dense connective tissue. So A-III, B-I, C-IV, D-II.
Select the correct statements. A. Tetrad formation is seen during Leptotene. B. During Anaphase, the centromeres split and chromatids separate. C. Terminalization takes place during Pachytene. D. Nucleolus, Golgi complex and ER are reformed during Telophase. E. Crossing over takes place between sister chromatids of homologous chromosome. Choose the correct answer from the options given below:
Solution: Tetrad formation is seen during the zygotene stage, not leptotene — A is wrong. During anaphase, the centromeres split and chromatids separate — B is correct. Terminalisation of chiasmata takes place during diakinesis, not pachytene — C is wrong. Nucleolus, golgi complex and ER are reformed during telophase — D is correct. Crossing over takes place between NON-sister chromatids of homologous chromosomes — E is wrong. So B and D only.
Select the correct statements with reference to chordates. A. Presence of a mid-dorsal, solid and double nerve cord. B. Presence of closed circulatory system. C. Presence of paired pharyngeal gill slits. D. Presence of dorsal heart E. Triploblastic pseudocoelomate animals. Choose the correct answer from the options given below:
Solution: The chordate characters are presence of a closed circulatory system and presence of pharyngeal gill slits, so B and C only are correct. The nerve cord is dorsal, hollow and single, not solid and double. The heart is ventral, not dorsal. They are triploblastic and coelomate, not pseudocoelomate.
In cockroach, excretion is brought about by- A. Phallic gland B. Urecose gland C. Nephrocytes D. Fat body E. Collaterial glands Choose the correct answer from the options given below :
Solution: In cockroach, excretion is brought about by Malpighian tubules, fat body, nephrocytes and urecose glands. Urecose glands are present in male cockroaches of some species. They synthesise uric acid. Nephrocytes are large, colourless, ovoid, binucleate cells attached to the dorsal diaphragm in the body cavity. Fat body accumulates, produces and stores uric acid. Phallic gland is a structure of the male reproductive system of cockroach and it secretes the outer layer of the spermatophore. Collaterial gland is a structure of the female reproductive system and secretes the hard egg-case or ootheca around fertilised eggs — neither is excretory.
Match List I with List II. List I: A. Mast cells B. Inner surface of bronchiole C. Blood D. Tubular parts of nephron List II: I. Ciliated epithelium II. Areolar connective tissue III. Cuboidal epithelium IV. Specialised connective tissue Choose the correct answer from the options give below:
Solution: Areolar connective tissue contains fibroblasts (cells that produce and secrete fibres), macrophages and mast cells. Inner surface of bronchioles is lined by ciliated epithelium. Blood is a specialised connective tissue. Tubular parts of nephron are lined by cuboidal epithelium. So A-II, B-I, C-IV, D-III.
Given below are two statements: Statement I : During G$_0$ phase of cell cycle, the cell is metabolically inactive. Statement II : The centrosome undergoes duplication during S phase of interphase. In the light of the above statements, choose the most appropriate answer from the options given below:
Solution: Cells in the G$_0$ stage remain metabolically ACTIVE but no longer proliferate unless called on to do so depending on the requirement of the organism — so Statement I is incorrect. In animal cells, during the S-phase, DNA replication begins in the nucleus and the centriole duplicates in the cytoplasm — so Statement II is correct.
Which one of the following is NOT an advantage of inbreeding?
Solution: Decreasing the productivity of an inbred population is not an advantage of inbreeding — continued inbreeding usually reduces fertility and productivity, which is inbreeding depression. Options (1) and (4) are genuine advantages of inbreeding. Option (3) is an incorrect statement in itself, because inbreeding INCREASES homozygosity.
Which of the following is characteristic feature of cockroach regarding sexual dimorphism?
Solution: Anal styles are present in male cockroaches and absent in female cockroaches, so their presence is the feature that distinguishes the sexes. Sclerites, anal cerci and dark brown body colour are common features of both male and female cockroaches.
The parts of human brain that helps in regulation of sexual behaviour, expression of excitement, pleasure, rage, fear etc. are:
Solution: The limbic system along with the hypothalamus regulates sexual behaviour, expression of excitement, pleasure, rage, fear etc. Corpora quadrigemina is a part of the midbrain and consists of four round swellings. Corpus callosum is a tract of nerve fibres that connects the right and left cerebral hemispheres. Thalamus is a major coordinating centre in the forebrain for sensory and motor signalling. Midbrain, pons and medulla oblongata together form the brain stem.
Which of the following statements are correct? A. Basophils are most abundant cells of the total WBCs B. Basophils secrete histamine, serotonin and heparin C. Basophils are involved in inflammatory response D. Basophils have kidney shaped nucleus E. Basophils are agranulocytes Choose the correct answer from the options given below:
Solution: Basophils secrete histamine, serotonin, heparin etc. and are involved in the inflammatory response, so B and C are correct. Basophils are granulocytes, not agranulocytes, so E is wrong. Neutrophils are the most abundant cells (60–65%) of the total WBCs, whereas basophils are the least (0.5–1%) abundant of all WBCs, so A is wrong. Monocytes have a kidney-shaped nucleus, not basophils, so D is wrong.
Which one of the following is the sequence on corresponding coding strand, if the sequence on mRNA formed is as follows 5$'$AUCGAUCGAUCGAUCGAUCGAUCG AUCG 3$'$?
Solution: The sequence of the coding strand is the same as the RNA except thymine in the place of uracil, and it keeps the same 5$'$ to 3$'$ polarity. Template strand $\rightarrow$ 3$'$-TAGCTAGCTAGCTAGCTAGCTAGCTAGC-5$'$ Coding strand $\rightarrow$ 5$'$-ATCGATCGATCG ATCGATCGATCGATCG-3$'$ $\downarrow$ Transcription mRNA $\rightarrow$ 5$'$ AUCGAUCGAUCGAUCGAUCGAUCG AUCG 3$'$
Match List I with List II. List I: A. Logistic growth B. Exponential growth C. Expanding age pyramid D. Stable age pyramid List II: I. Unlimited resource availability condition II. Limited resource availability condition III. The percent individuals of pre-reproductive age is largest followed by reproductive and post reproductive age groups IV. The percent individuals of pre-reproductives and reproductive age group are same Choose the correct answer from the options given below:
Solution: Logistic growth occurs when there is a limited resource availability condition. Exponential growth occurs when there is an unlimited resource availability condition. Expanding age pyramid reflects a growing population, where the percent individuals of pre-reproductive age is largest followed by reproductive and post-reproductive age groups. Stable age pyramid shows a stable population, where the percent individuals of pre-reproductive and reproductive age group are same. So A-II, B-I, C-III, D-IV.
Which of the following are NOT under the control of thyroid hormone? A. Maintenance of water and electrolyte balance B. Regulation of basal metabolic rate C. Normal rhythm of sleep-wake cycle D. Development of immune system E. Support the process of RBCs formation Choose the correct answer from the options given below:
Solution: Thyroid hormones play an important role in the regulation of basal metabolic rate, maintenance of water and electrolyte balance and support the process of RBC formation. This hormone is not involved in regulating the normal rhythm of the sleep-wake cycle or the development of the immune system. So C and D only are not under thyroid control.
Which of the following statements are correct? A. An excessive loss of body fluid from the body switches off osmoreceptors. B. ADH facilitates water reabsorption to prevent diuresis. C. ANF causes vasodilation. D. ADH causes increase in blood pressure. E. ADH is responsible for decrease in GFR. Choose the correct answer from the options given below:
Solution: Statements B, C and D are true. ADH facilitates water reabsorption from the DCT of the nephron to prevent diuresis, which causes an increase in blood pressure. ANF, which is secreted by the heart, is a vasodilator. Statements A and E are false: excessive loss of body fluid from the body switches ON the osmoreceptors, and the increase in blood pressure caused by ADH increases the glomerular blood flow and thus the GFR rather than decreasing it.
Which of the following statements are correct regarding skeletal muscle? A. Muscle bundles are held together by collagenous connective tissue layer called fascicle. B. Sarcoplasmic reticulum of muscle fibre is a store house of calcium ions. C. Striated appearance of skeletal muscle fibre is due to distribution pattern of actin and myosin proteins. D. M line is considered as functional unit of contraction called sarcomere. Choose the most appropriate answer from the options given below:
Solution: Statements B and C are the only correct statements, while A and D are incorrect. Muscle bundles are held together by a collagenous connective tissue layer called fascia. The muscle BUNDLES themselves are called fascicles, so A misnames the layer. The portion of the myofibril between two successive 'Z' lines is considered the functional unit of contraction, called a sarcomere — not the M line, so D is wrong.
The unique mammalian characteristics are:
Solution: The presence of hairs, pinna and mammary glands are unique features of mammals. Monocondylic skull is present in reptiles and aves, whereas mammals have a dicondylic skull. Tympanic membrane is present in amphibians also, so it is not considered a unique feature. Indirect development is not seen in mammals.
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