Free, login-free worked solutions to every question in NEET (UG) 2022 — Solved paper. Check your answers below, then predict your rank or practise with full mock tests.
If a soap bubble expands, the pressure inside the bubble
Solution: The excess pressure inside a soap bubble of radius $r$ is given as $$p = \frac{4T}{r}$$ where $T$ is the surface tension. $$\Rightarrow\ p \propto \frac{1}{r} \qquad \ldots(i)$$ Therefore, when the soap bubble expands its radius increases, hence from Eq. (i), the pressure inside it will decrease.
The graph which shows the variation of the de-Broglie wavelength ($\lambda$) of a particle and its associated momentum ($p$) is
Solution: The de-Broglie wavelength ($\lambda$) of a moving particle is given as $$\lambda = \frac{h}{p}$$ $$\therefore\ \lambda \propto \frac{1}{p}$$ Hence the graph showing the variation of de-Broglie wavelength ($\lambda$) of a moving particle with its momentum ($p$) is a rectangular hyperbola.
A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is
Solution: Given, mass of body, $m = 60$ g $= 6 \times 10^{-2}$ kg Gravitational force, $F = 3.0$ N Magnitude of gravitational field intensity $$I = \frac{F}{m} = \frac{3}{6 \times 10^{-2}} = 50\ \text{N/kg}$$
Given below are two statements Statement I : Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element ($Idl$) of a current carrying conductor only. Statement II : Biot-Savart's law is analogous to Coulomb's inverse square law of charge $q$, with the former being related to the field produced by a scalar source, $Idl$ while the latter being produced by a vector source, $q$. In light of above statements choose the most appropriate answer from the options given below
Solution: According to Biot-Savart's law, the magnetic field due to a small current element $Id\mathbf{l}$, at a distance $r$ from it is given as $$dB = \frac{\mu_0\left(Id\mathbf{l} \times \mathbf{r}\right)}{4\pi r^{3}}$$ Thus, the current element $Id\mathbf{l}$ is a VECTOR quantity, not a scalar one. Hence statement I is correct and statement II is incorrect.
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second
Solution: Initial speed of a freely falling body, $u = 0$. Distance travelled by a freely falling body in the $n$th second $$s_n = u + \frac{1}{2}g(2n - 1) = 0 + \frac{1}{2}g(2n - 1)$$ $$s_n = \frac{1}{2}g(2n - 1)$$ Distance travelled in the first second ($n = 1$): $s_1 = \dfrac{1}{2}g(2 \times 1 - 1) = \dfrac{g}{2}$ In the 2nd second ($n = 2$): $s_2 = \dfrac{1}{2}g(2 \times 2 - 1) = \dfrac{3g}{2}$ In the 3rd second ($n = 3$): $s_3 = \dfrac{1}{2}g(2 \times 3 - 1) = \dfrac{5g}{2}$ In the 4th second ($n = 4$): $s_4 = \dfrac{1}{2}g(2 \times 4 - 1) = \dfrac{7g}{2}$ $$\therefore\ s_1 : s_2 : s_3 : s_4 = \frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} : \frac{7g}{2} = 1 : 3 : 5 : 7$$
In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be
Solution: In half wave rectification, input frequency $f_{\text{in}} = 60$ Hz. Since the frequency of the output voltage obtained from a half wave rectifier is the same as the frequency of the input AC voltage — only one half of each cycle appears at the output, so one pulse is produced per input cycle. Hence $f_{\text{out}} = 60$ Hz.
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball ($v$) as a function of time ($t$) is The graph carries four curves: $A$ is a straight line rising from the origin, $B$ rises steeply and then levels off at a constant value, $C$ starts high and falls steadily, and $D$ rises and then curves back down.
Solution: When a spherical ball is dropped in a long column of a highly viscous liquid, then its velocity first increases and after some time it moves with constant velocity, known as the terminal velocity ($v$). Hence the correct variation of the speed of the ball in the liquid column with time is the curve that rises and then flattens to a constant value, which is curve $B$.
The angle between the electric lines of force and the equipotential surface is
Solution: Work done to move a charge ($q$) particle between any two points on an equipotential surface is zero, i.e. $$W = q(V_2 - V_1)$$ $$\Rightarrow\ \mathbf{F} \cdot d\mathbf{s} = q(V - V) = 0 \qquad [\because V_1 = V_2 = V]$$ $$\Rightarrow\ q(\mathbf{E} \cdot d\mathbf{s}) = 0$$ $$\Rightarrow\ \mathbf{E} \cdot d\mathbf{s} = 0$$ Therefore $\mathbf{E} \perp d\mathbf{s}$, i.e. the electric field is perpendicular to the displacement vector $d\mathbf{s}$. Since $d\mathbf{s}$ lies in the plane of the equipotential surface, the angle between the electric lines of force and the equipotential surface is 90$^{\circ}$.
As the temperature increases, the electrical resistance
Solution: The coefficient of thermal resistance $\alpha$ is positive for a conductor and negative for a semiconductor. Thus, on increasing temperature, the resistance of a conductor increases and the resistance of a semiconductor decreases.
In the given nuclear reaction, the element $X$ is $$^{22}_{11}\text{Na} \longrightarrow X + e^{+} + \nu$$
Solution: The given nuclear reaction is represented as $$^{22}_{11}\text{Na} \longrightarrow {}^{M}_{Z}X + e^{+} + \nu$$ Since atomic number and mass number are the same on both sides of the reaction, for atomic number $$11 = Z + 1 \ \Rightarrow\ Z = 10$$ For mass number, $$22 = M \ \Rightarrow\ M = 22$$ $$\therefore\ {}^{M}_{Z}X = {}^{22}_{10}\text{Ne}$$
Let $T_1$ and $T_2$ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr's model of an atom, the ratio of $T_1 : T_2$ is:
Solution: For the first excited state of the hydrogen atom, $n_1 = 2$. For the second excited state of the hydrogen atom, $n_2 = 3$. According to Bohr's model of the H-atom, the energy of an electron revolving in the $n$th orbit is $$E = \frac{-13.6}{n^{2}}\ \text{eV}$$ For the first excited state, $$T_1 = E_1 = \frac{-13.6}{n_1^{2}} = \frac{-13.6}{2^{2}} \qquad [\because n_1 = 2]$$ For the second excited state, $$T_2 = E_2 = \frac{-13.6}{n_2^{2}} = \frac{-13.6}{3^{2}}\ \text{eV} \qquad [\because n_2 = 3]$$ $$\therefore\ \frac{T_1}{T_2} = \frac{-13.6/2^{2}}{-13.6/3^{2}} = \frac{3^{2}}{2^{2}} = \frac{9}{4}$$ $$\Rightarrow\ T_1 : T_2 = 9 : 4$$
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is
Solution: Radius of gyration of a body is given as $$K = \sqrt{\frac{I}{m}}$$ where $I$ is the moment of inertia and $m$ the mass of the body. $$\therefore\ \frac{K_1}{K_2} = \sqrt{\frac{I_1}{I_2}} \qquad \ldots(i)$$ $I_1$ = moment of inertia of a thin uniform disc about an axis passing through its centre and normal to its plane $= \dfrac{MR^{2}}{2}$ $I_2$ = moment of inertia of the disc about its diameter $= \dfrac{MR^{2}}{4}$ From Eq. (i), $$\frac{K_1}{K_2} = \sqrt{\frac{MR^{2}/2}{MR^{2}/4}} = \sqrt{2}$$ $$\therefore\ K_1 : K_2 = \sqrt{2} : 1$$
When light propagates through a material medium of relative permittivity $\epsilon_r$ and relative permeability $\mu_r$, the velocity of light, $v$ is given by ($c$-velocity of light in vacuum)
Solution: When light (an EM wave) propagates through a material medium of relative permittivity $\epsilon_r$ and relative permeability $\mu_r$, the velocity of light (EM wave) is given by $$v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$$ where $c$ is the speed of light in vacuum.
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is :
Solution: Given, power radiated by transmitter, $$P = 100\ \text{kW} = 10^{5}\ \text{W}$$ $$\therefore\ \text{Time}, t = 1\ \text{h} = 60 \times 60\ \text{s} = 3.6 \times 10^{3}\ \text{s}$$ $$\therefore\ \text{Energy radiated by the transmitter}, E = P \times t$$ $$= 10^{5} \times 3.6 \times 10^{3}$$ $$= 3.6 \times 10^{8}\ \text{J} = 36 \times 10^{7}\ \text{J}$$
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms$^{-1}$. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is ($g = 10$ ms$^{-2}$)
Solution: Given, mass (lift + passengers), $m = 2000$ kg $$\therefore\ \text{Weight (lift + passengers)}, w = mg = 2000\,g\ \text{N}$$ $$v = 1.5\ \text{m/s}$$ Frictional force, $f = 3000$ N Since the lift is moving with constant velocity, the net applied force on the lift = friction force + weight: $$F = 3000 + 2000\,g = 3000 + 2000 \times 10 = 23000\ \text{N}$$ $$\therefore\ \text{Minimum power delivered}, P = F \times v = 23000 \times 1.5 = 34500\ \text{W}$$
Two resistors of resistance, 100 $\Omega$ and 200 $\Omega$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 $\Omega$ to that in 200 $\Omega$ in a given time is
Solution: Let these two resistors be connected to a battery of voltage $V$ volts. In a parallel combination, the potential difference across each resistor is the same. Hence, the potential difference across $R_1$ and $R_2$ is $$V_1 = V_2 = V\ \text{volts}$$ If $H_1$ and $H_2$ are the thermal energies produced through resistances $R_1$ and $R_2$ in time $t$, then $$H_1 = \frac{V_1^{2}}{R_1}t = \frac{V^{2}t}{100}$$ $$H_2 = \frac{V_2^{2}t}{R_2} = \frac{V^{2}t}{200}$$ Hence, $$\frac{H_1}{H_2} = \frac{V^{2}t/100}{V^{2}t/200} = \frac{200}{100} = \frac{2}{1}$$ $$\Rightarrow\ H_1 : H_2 = 2 : 1$$
The peak voltage of the AC source is equal to
Solution: The peak voltage ($V_0$) and the rms voltage ($V_{\text{rms}}$) are related as $$V_0 = \sqrt{2}\,V_{\text{rms}}$$ Thus, the peak voltage of the AC source is equal to $\sqrt{2}$ times the rms value of the AC source.
A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$, the speed of the third (lighter) fragment is
Solution: Let the total mass of the shell be $m$. The ratio of the masses of the fragments is 2 : 2 : 1: $$m_1 + m_2 + m_3 = m$$ $$2k + 2k + k = m \ \Rightarrow\ k = 0.2\,m$$ $$\therefore\ \text{Mass of lighter fragment}, m_3 = k = 0.2\,m$$ $$m_1 = m_2 = 2 \times 0.2\,m = 0.4\,m$$ The two equal fragments fly off along mutually perpendicular directions, so their resultant momentum is $$p_3 = \sqrt{p_1^{2} + p_2^{2}}$$ $$\Rightarrow\ 0.2\,m v_3 = \sqrt{(0.4\,mv)^{2} + (0.4\,mv)^{2}}$$ $$\Rightarrow\ v_3 = 2\sqrt{2}\,v$$
If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is
Solution: The velocity of a transverse wave in a string is $v \propto \sqrt{T}$, where $T$ is the tension in the string. $$\therefore\ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{T_1}{2T_1}} \qquad [\because T_2 = 2T_1]$$ $$= \frac{1}{\sqrt{2}}$$ $$v_1 : v_2 = 1 : \sqrt{2}$$
An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. On the $p$-$V$ diagram, curve 4 is horizontal, curve 1 is vertical, and curves 2 and 3 fall away from the initial point with curve 3 having the smaller slope. The curve which represents the adiabatic process among 1, 2, 3 and 4 is
Solution: When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between it and the surroundings, the process is known as an adiabatic process. Curve 4 is an isobaric process, because pressure is constant. Curve 1 is an isochoric process, because volume is constant. Of curves 2 and 3, curve 3 has a smaller slope than curve 2. Since an adiabatic curve is steeper than an isothermal one, curve 3 represents the isothermal process and curve 2 represents the adiabatic process.
The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad/s$^{2}$ is
Solution: Given, initial angular velocity of the fly wheel, $$\omega_0 = 2\pi\ \text{rad} \times 1200\ \text{rpm} = 2\pi\ \text{rad} \times \frac{1200}{60}\ \text{rps} = 40\pi\ \text{rad/s}$$ Similarly, final angular velocity of the fly wheel, $$\omega = 2\pi\ \text{rad} \times 3120\ \text{rpm} = 2\pi\ \text{rad} \times \frac{3120}{60}\ \text{rps} = 104\pi\ \text{rad/s}$$ Time, $t = 16$ s. If $\alpha$ is the angular acceleration, then using the equation of rotational motion, $$\omega = \omega_0 + \alpha t$$ $$\Rightarrow\ 104\pi = 40\pi + \alpha \times 16$$ $$\Rightarrow\ \alpha = \frac{104\pi - 40\pi}{16} = \frac{64\pi}{16} = 4\pi\ \text{rad/s}^{2}$$
Plane angle and solid angle have
Solution: Plane angle is measured in radian whereas solid angle is measured in steradian. Both of these are ratios of two like quantities — an arc to a radius, and an area to a radius squared — so they carry no dimensions. Therefore, plane angle and solid angle have units but no dimensions.
In the given circuits (A), (B) and (C) the potential drop across the two $p$-$n$ junctions are equal in
Solution: In the circuit diagrams (A) and (C), both the $p$-$n$ junction diodes are in the same biasing conditions. So, both diodes offer equal resistances. Since both diodes are in series, therefore equal potential will drop across the junction.
In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is
Solution: In Young's double slit experiment, the wavelength of light used is inversely proportional to the number of fringes observed in a fixed segment of the screen, i.e. $$\lambda_1 n_1 = \lambda_2 n_2$$ $$n_2 = \frac{n_1\lambda_1}{\lambda_2}$$ Here, $\lambda_1 = 600$ nm $= 6 \times 10^{-7}$ m, $\lambda_2 = 400$ nm $= 4 \times 10^{-7}$ m and $n_1 = 8$. $$\therefore\ n_2 = \frac{8 \times 6 \times 10^{-7}}{4 \times 10^{-7}} = 12$$
A square loop of side 1 m and resistance 1 $\Omega$ placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is
Solution: Side of the square loop, $a = 1$ m. $$\therefore\ \text{Area of square loop}, A = a^{2} = 1^{2} = 1\ \text{m}^{2}$$ Resistance of square loop, $R = 1\ \Omega$; magnetic field, $B = 0.5$ T. Since the plane of the loop is perpendicular to the direction of the magnetic field, the angle made by the normal to the loop with the direction of the magnetic field will be $0^{\circ}$, i.e. $\theta = 0$. $$\therefore\ \phi = BA\cos\theta = 0.5 \times 1 \times \cos 0^{\circ} = 0.5\ \text{Wb}$$
Two hollow conducting spheres of radii $R_1$ and $R_2$ ($R_1 > R_2$) have equal charges. The potential would be
Solution: Potential on a hollow conducting sphere of radius $R$ due to charge $Q$ is $$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R}$$ $$\Rightarrow\ V \propto \frac{1}{R} \qquad \text{[for the same charge } Q\text{]}$$ $$\Rightarrow\ \frac{V_1}{V_2} = \frac{R_2}{R_1}$$ Since $R_1 > R_2$, $\dfrac{R_1}{R_2} > 1 \Rightarrow \dfrac{R_2}{R_1} < 1$, so $\dfrac{V_1}{V_2} < 1$, i.e. $V_1 < V_2$. Hence, the potential will be more on the smaller sphere.
A Copper wire of length 10 m and radius $\left(\dfrac{10^{-2}}{\sqrt{\pi}}\right)$ m has electrical resistance of 10 $\Omega$. The current density in the wire for an electric field strength of 10 (V/m) is
Solution: Given, radius $r = \dfrac{10^{-2}}{\sqrt{\pi}}$ m, $R = 10\ \Omega$, $E = 10$ V/m and $l = 10$ m. $$\therefore\ \text{Area of cross-section of copper wire}, A = \pi r^{2} = \pi \times \left(\frac{10^{-2}}{\sqrt{\pi}}\right)^{2} = 10^{-4}\ \text{m}^{2}$$ $$\therefore\ \text{Current density}, J = \frac{i}{A} = \frac{V/R}{A} \qquad \left[\because i = \frac{V}{R}\right]$$ $$= \frac{V}{RA} = \frac{El}{RA} \qquad [\because V = El]$$ $$= \frac{10 \times 10}{10 \times 10^{-4}} = 10^{5}\ \text{A/m}^{2}$$
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is
Solution: Given, radius of solenoid $r = 1$ mm. Number of turns per mm $= \dfrac{100}{1\ \text{mm}}$ $$\therefore\ \text{Number of turns per metre} = \frac{100}{1 \times 10^{-3}\ \text{m}}$$ i.e. $n = 10^{5}$ turns/m, and current flowing through the solenoid $I = 1$ A. $$\therefore\ \text{Magnetic field strength at the centre of the solenoid}$$ $$B = \mu_0 n I = 4\pi \times 10^{-7} \times 10^{5} \times 1 = 4\pi \times 10^{-2}\ \text{T} = 12.56 \times 10^{-2}\ \text{T}$$
Two objects of mass 10 kg and 20 kg, respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is
Solution: Place the 10 kg mass at the origin $(0, 0)$ and the 20 kg mass at $(10, 0)$. Let $X_{\text{CM}}$ be the position of the centre of mass from the 10 kg mass. $$\therefore\ X_{\text{CM}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$ $$= \frac{10 \times 0 + 20 \times 10}{10 + 20}$$ $$= \frac{200}{30} = \frac{20}{3}\ \text{m}$$
A light ray falls on a glass surface of refractive index $\sqrt{3}$ at an angle 60$^{\circ}$. The angle between the refracted and reflected rays would be
Solution: Given, refractive index of glass surface $\mu = \sqrt{3}$ and angle of incidence $i = 60^{\circ}$. Since the above situation follows Brewster's law, because $$\mu = \tan i = \tan 60^{\circ} = \sqrt{3}$$ Hence, angle $i$ is equal to the polarising angle $i_p$, i.e. $i_p = i = 60^{\circ}$. By Snell's law, $$\mu = \frac{\sin i_p}{\sin r} \ \Rightarrow\ \tan i_p = \frac{\sin i_p}{\sin r} \qquad [\text{since } \mu = \tan i_p]$$ $$\Rightarrow\ \frac{\sin i_p}{\cos i_p} = \frac{\sin i_p}{\sin r} \ \Rightarrow\ \cos i_p = \sin r$$ $$\Rightarrow\ \cos i_p = \cos(90^{\circ} - r) \ \Rightarrow\ i_p = 90^{\circ} - r \ \Rightarrow\ r + i_p = 90^{\circ}$$ Hence, the angle between the refracted and reflected rays $$= 180^{\circ} - (r + i_p) = 180^{\circ} - 90^{\circ} = 90^{\circ}$$
The dimensions $[\text{MLT}^{-2}\text{A}^{-2}]$ belong to the
Solution: The magnetic field due to a current carrying solenoid is $B = \mu_0 nI$, so $$\mu_0 = \frac{B}{nI}$$ $$\Rightarrow\ [\mu_0] = \frac{[B]}{[n][I]} = \frac{[F/IL]}{[n][I]} \qquad \left[\because F = IBL \Rightarrow B = \frac{F}{IL}\right]$$ $$= \frac{[F]}{[n][L][I^{2}]} = \frac{[\text{MLT}^{-2}]}{[\text{L}^{-1}][\text{L}][\text{A}^{2}]} = [\text{MLT}^{-2}\text{A}^{-2}]$$ For comparison, the dimensional formula of self inductance is $[\text{M}^{1}\text{L}^{2}\text{T}^{-2}\text{A}^{-2}]$, of electric permittivity $[\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2}]$ and of magnetic flux $[\text{ML}^{2}\text{T}^{-2}\text{A}^{-1}]$.
The displacement-time graph of two moving particles make angles of 30$^{\circ}$ and 45$^{\circ}$ with the $X$-axis as shown in the figure. The ratio of their respective velocity is
Solution: The slope of a displacement-time graph for a moving object gives the velocity of the moving object. $$\therefore\ \text{Velocity of first moving particle}, v_1 = \tan\alpha_1 = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$$ $$\text{Velocity of second moving particle}, v_2 = \tan\alpha_2 = \tan 45^{\circ} = 1$$ $$\therefore\ \frac{v_1}{v_2} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$$ $$\therefore\ v_1 : v_2 = 1 : \sqrt{3}$$
A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is
Solution: For a biconvex lens, $R_1 = +20$ cm, $R_2 = -20$ cm and $\mu = 1.5$. By using the lens maker's formula, $$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right) = 0.5\left(\frac{2}{20}\right)$$ $$\Rightarrow\ \frac{1}{f} = 0.5 \times \frac{1}{10} \ \Rightarrow\ f = \frac{10}{0.5} = 20\ \text{cm} = 0.2\ \text{m}$$ $$\therefore\ \text{Power of the lens}, P = \frac{1}{f}\ (\text{in metre}) = \frac{1}{0.2} = +5\ \text{D}$$
Match List-I with List-II List-I (Electromagnetic waves): (A) AM radio waves (B) Microwaves (C) Infrared radiations (D) X-rays List-II (Wavelength): (i) $10^{-10}$ m (ii) $10^{2}$ m (iii) $10^{-2}$ m (iv) $10^{-4}$ m Choose the correct answer from the options given below
Solution: A. Wavelength range of AM radio waves lies between $10^{2}$ m to $10^{3}$ m. B. Wavelength range of microwaves lies between $10^{-3}$ m to $10^{-1}$ m. C. Wavelength range of infrared radiations lies between $10^{-5}$ m to $10^{-3}$ m. D. Wavelength range of X-rays lies between $10^{-12}$ m to $10^{-8}$ m. Therefore, the correct matching pairs are (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i).
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei, respectively is:
Solution: Given, mass number of parent nucleus $A = 189$; mass numbers of daughter nuclei $A_1 = 125$ and $A_2 = 64$. We know that $R = R_0\left(A^{1/3}\right)$, where $R_0$ is a constant. $$\therefore\ \frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{125}{64}\right)^{1/3} = \left[\left(\frac{5}{4}\right)^{3}\right]^{1/3} = \frac{5}{4}$$ $$\therefore\ R_1 : R_2 = 5 : 4$$
Two transparent media $A$ and $B$ are separated by a plane boundary. The speed of light in those media are $1.5 \times 10^{8}$ m/s and $2.0 \times 10^{8}$ m/s, respectively. The critical angle for a ray of light for these two media is
Solution: Given, speed of light in transparent medium $A$, $v_A = 1.5 \times 10^{8}$ m/s, and in transparent medium $B$, $v_B = 2 \times 10^{8}$ m/s. Refractive index ($\mu$) and critical angle ($i_c$) are related as $$\mu = \frac{1}{\sin i_c} \ \Rightarrow\ \sin i_c = \frac{1}{\mu}$$ $$= \frac{1}{(v_B/v_A)} \qquad \left[\because \mu = \frac{v_B}{v_A}\right]$$ $$= \frac{v_A}{v_B} = \frac{1.5 \times 10^{8}}{2 \times 10^{8}} = 0.750$$ $$\Rightarrow\ i_c = \sin^{-1}(0.750)$$
A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad s$^{-1}$. If the vertical component of earth's magnetic field at that place is $2 \times 10^{-5}$ T and electrical resistance of the coil is 12.56 $\Omega$, then the maximum induced current in the coil will be
Solution: Given, number of turns in the big circular coil, $N = 1000$; average radius, $r = 10$ m; vertical component of Earth's magnetic field, $B_V = 2 \times 10^{-5}$ T; resistance of coil, $R = 12.56\ \Omega$; angular velocity, $\omega = 2$ rad/s. Maximum induced emf in the coil $$e_{\max} = NBA\omega = NB_v A\omega$$ $$= 1000 \times 2 \times 10^{-5} \times \pi(10)^{2} \times 2 \qquad [\because A = \pi r^{2}]$$ $$e_{\max} = 4\pi\ \text{V}$$ $$\therefore\ \text{Maximum induced current in the coil}, I_{\max} = \frac{e_{\max}}{R} = \frac{4\pi}{12.56} = 1\ \text{A}$$
Match List-I with List-II. List-I: (A) Gravitational constant ($G$) (B) Gravitational potential energy (C) Gravitational potential (D) Gravitational intensity List-II: (i) $[\text{L}^{2}\text{T}^{-2}]$ (ii) $[\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]$ (iii) $[\text{LT}^{-2}]$ (iv) $[\text{ML}^{2}\text{T}^{-2}]$ Choose the correct answer from the options given below
Solution: A. According to Newton's law of gravitation, $F = G\dfrac{m_1m_2}{r^{2}}$, so $G = \dfrac{Fr^{2}}{m_1m_2}$: $$[G] = \frac{[F][r^{2}]}{[m_1][m_2]} = \frac{[\text{MLT}^{-2}][\text{L}^{2}]}{[\text{M}][\text{M}]} = [\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]$$ B. Gravitational potential energy, $U = mgh$: $$[U] = [\text{M}][\text{LT}^{-2}][\text{L}] = [\text{ML}^{2}\text{T}^{-2}]$$ C. Gravitational potential, $V = -\dfrac{Gm}{r}$: $$[V] = \frac{[\text{M}^{-1}\text{L}^{3}\text{T}^{-2}][\text{M}]}{[\text{L}]} = [\text{L}^{2}\text{T}^{-2}]$$ D. Gravitational intensity, $I = \dfrac{GM}{r^{2}}$: $$[I] = \frac{[\text{M}^{-1}\text{L}^{3}\text{T}^{-2}][\text{M}]}{[\text{L}^{2}]} = [\text{LT}^{-2}]$$ Hence the correct matching is (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii).
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R) A coil spring of copper has more tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: When a coil spring is stretched, neither its length nor its volume changes, but there is only the change in its shape. Therefore, the stretching of a coil spring is determined by the shear modulus. So the Assertion is true. Young's modulus of a steel spring is more than Young's modulus of a copper spring, therefore the tensile strength of a steel spring is more than the tensile strength of a copper spring — the Reason states the opposite. Hence, Assertion is true but Reason is false.
The area of a rectangular field (in m$^{2}$) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is
Solution: Area $=$ Length $\times$ Breadth $$= 55.3 \times 25 = 1382.5$$ $$= 13.825 \times 10^{2} = 13.83 \times 10^{2}$$ The breadth, 25 m, carries only two significant figures, so the product must be rounded to two significant figures: $$= 13.8 \times 10^{2} = 14 \times 10^{2}$$
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is
Solution: Molecular weight of water $= 18$ g $= 1.8 \times 10^{-2}$ kg. Mass of water, $M = 4.5$ kg. $$\therefore\ \text{number of moles of water}, n = \frac{\text{Mass of water}}{\text{Molecular weight of water}} = \frac{4.5}{1.8 \times 10^{-2}} = 250$$ At STP, $T = 273$ K, $p = 10^{5}$ N/m$^{2}$. Using the equation $pV = nRT$, $$\Rightarrow\ V = \frac{nRT}{p} = \frac{250 \times 8.3 \times 273}{10^{5}} = 5.6\ \text{m}^{3}$$
A wheatstone bridge is used to determine the value of unknown resistance $X$ by adjusting the variable resistance $Y$ as shown in the figure. For the most precise measurement of $X$, the resistances $P$ and $Q$
Solution: For the most precise measurement of the unknown resistance $X$, the resistances $P$ and $Q$ should be approximately equal, and should be small so that it could decrease the error in the experiment. With $P \approx Q$ the bridge is most sensitive, and keeping them small keeps the currents large enough for the galvanometer to register a small imbalance.
A series $L$-$C$-$R$ circuit with inductance 10 H, capacitance 10 $\mu$F, resistance 50 $\Omega$ is connected to an AC source of voltage, $V = 200\sin(100\,t)$ volt. If the resonant frequency of the LCR circuit is $\nu_0$ and the frequency of the AC source is $\nu$, then
Solution: Given, in the series $L$-$C$-$R$ circuit, $L = 10$ H, $C = 10\ \mu$F $= 10^{-5}$ F, $R = 50\ \Omega$, and source voltage $V = 200\sin(100\,t)$ volt. $$\omega = 100\ \text{rad/s}$$ $$\therefore\ 2\pi\nu = 100 \ \Rightarrow\ \nu = \frac{100}{2\pi} = \frac{50}{\pi}\ \text{Hz}$$ Resonance frequency, $$\nu_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10 \times 10^{-5}}} = \frac{1}{2\pi \times 10^{-2}} = \frac{50}{\pi}\ \text{Hz}$$ Hence, $\nu = \nu_0 = \dfrac{50}{\pi}$ Hz.
A capacitor of capacitance $C = 900$ pF is charged fully by 100 V battery $B$ as shown in figure (a). Then, it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C = 900$ pF as shown in figure (b). The electrostatic energy stored by the system (b) is
Solution: After the charged capacitor is connected across the uncharged one, the common potential is $$V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{C \times 100 + C \times 0}{C + C} = 50\ \text{V}$$ $$\therefore\ \text{Electrostatic stored potential energy}, U = 2 \times \frac{1}{2}CV^{2} = CV^{2}$$ $$= 9 \times 10^{-10} \times 50^{2}$$ $$= 2.25 \times 10^{-6}\ \text{J}$$
Two point charges $-q$ and $+q$ are placed at a distance of $L$, as shown in the figure. The magnitude of electric field intensity at a distance $R\ (R \gg L)$ on the axis of the dipole varies as
Solution: The arrangement of charges shown in the figure is an electric dipole. Hence, the electric field at point $P$ at a distance $R\ (R \gg L)$ from the dipole is equal to the electric field due to an electric dipole on the axial position, i.e. $$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{R^{3}}$$ where $p = qL$ (electric dipole moment). $$\therefore\ E \propto \frac{1}{R^{3}}$$
A ball is projected with a velocity, 10 ms$^{-1}$, at an angle of 60$^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be
Solution: Initial speed of ball, $u = 10$ ms$^{-1}$, and the angle of projection is 60$^{\circ}$. At the highest point of the trajectory of the ball, its vertical component of velocity becomes zero and only the horizontal component of velocity exists, which is equal to the horizontal component of velocity at the time of projection. i.e. speed of ball at highest point $$= u\cos\theta = 10\cos 60^{\circ} = 10 \times \frac{1}{2} = 5\ \text{ms}^{-1}$$
From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is :
Solution: According to Ampere's circuital law, the magnetic field inside a current carrying long straight wire of circular cross-section is a linearly increasing function of distance $r\ (r < R)$ up to the boundary of the wire, and the magnetic field outside the wire at distance $r\ (r > R)$ from its axis is a decreasing one with $\dfrac{1}{r}$ dependence, $$B \propto r \quad (r < R), \qquad B \propto \frac{1}{r} \quad (r > R)$$ where $R$ is the radius of the circular cross-section of the wire.
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is
Solution: Given, length of first pendulum, $l_1 = 121$ cm $= 1.21$ m; length of second pendulum, $l_2 = 100$ cm $= 1$ m. According to the given situation, both pendulums are again in phase if the number of vibrations made by the shorter pendulum is known. Let $T_1$ and $T_2$ be the time periods of the longer and shorter pendulum respectively. Since $T = 2\pi\sqrt{\dfrac{l}{g}}$, $$\Rightarrow\ \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} = \sqrt{\frac{1.21}{1}} = 1.1 = \frac{11}{10}$$ $$\Rightarrow\ 10\,T_1 = 11\,T_2$$ $$\therefore\ 10\ \text{vibrations of the longer pendulum} = 11\ \text{vibrations of the shorter pendulum}$$ $$\therefore\ \text{Minimum number of vibrations of the shorter pendulum} = 11$$
The given graph is a representation of kinetics of a reaction, drawn at constant temperature $T$ as a horizontal line against the $x$ axis. The $y$ and $x$ axes for zero and first order reactions, respectively are
Solution: The graph is a horizontal line, so the plotted quantity is independent of the quantity on the $x$ axis. For a zero order reaction, the reaction rate is independent of concentration and remains constant. So the graph is of zero order if the $y$-axis represents rate and $x$ represents concentration. For a first order reaction, the half-life $t_{1/2}$ is independent of concentration. So the same horizontal graph represents first order if $y$ is half-life ($t_{1/2}$) and $x$ is concentration. Hence zero order ($y$ = rate and $x$ = concentration), first order ($y = t_{1/2}$ and $x$ = concentration).
Which compound amongst the following is not an aromatic compound?
Solution: The cycloheptatriene structure follows Huckel's $(4n + 2)\pi$ electron rule, as it has 6$\pi$ electrons. But the given structure is NON-PLANAR due to the presence of an $sp^{3}$ hybridised carbon atom in the ring. Hence, it is a non-aromatic compound. The other three options are planar, fully conjugated and follow Huckel's rule, hence they are aromatic.
$$R\text{Mg}X + \text{CO}_2 \xrightarrow{\ \text{Dry ether}\ } Y \xrightarrow{\ \text{H}_3\text{O}^{+}\ } R\text{COOH}$$ What is $Y$ in the above reaction ?
Solution: The alkyl group of the Grignard reagent has a partial negative charge while magnesium has a partial positive charge. The alkyl group attacks the carbon atom of CO$_2$ and forms the salt of the carboxylic acid, which on hydrolysis gives the carboxylic acid. $$R\text{Mg}X + \text{O=C=O} \xrightarrow{\ \text{Dry ether}\ } R\text{COO}^{-}\text{Mg}^{+}X\ (Y) \xrightarrow{\ \text{H}_3\text{O}^{+}\ } R\text{COOH}$$ Hence '$Y$' in the reaction is $R\text{COO}^{-}\text{Mg}^{+}X$.
Which of the following statement is not correct about diborane?
Solution: The incorrect statement about diborane is that both the boron atoms are $sp^{2}$-hybridised. In diborane, boron has $sp^{3}$-hybridisation and the molecule as a whole is non-planar: the four terminal hydrogens and the two borons lie in one plane, while the two bridging hydrogens lie above and below that plane. The other three statements are correct — the four terminal B—H bonds are ordinary 2-centre-2-electron bonds, and the two bridges are 3-centre-2-electron bonds.
Gadolinium has a low value of third ionisation enthalpy because of
Solution: The electronic configuration of gadolinium is $$\text{Gd} = [\text{Xe}]4f^{7}5d^{1}6s^{2}$$ $$\text{Gd}^{3+} = [\text{Xe}]4f^{7}$$ After removal of the 3rd electron, it will attain a half-filled electronic configuration, due to which $4f$ will have high exchange energy. So, Gd has a low value of third ionisation energy.
Which one is not correct mathematical equation for Dalton's law of partial pressure? Here $p$ = total pressure of gaseous mixture
Solution: According to Dalton's law of partial pressure, the partial pressure of a gas equals the mole fraction of that gas in the gaseous mixture multiplied by the TOTAL pressure of the gaseous mixture. So, $$p_1 = x_1 p, \qquad p_2 = x_2 p, \qquad p_3 = x_3 p$$ Then, the total pressure is $p = p_1 + p_2 + p_3$. The expression $p_i = x_i p_i^{\circ}$, using the pressure of the pure gas, is Raoult's law rather than Dalton's law. Hence statement (c) is incorrect.
Given below are two statements. Statement I The boiling points of the following hydrides of group 16 elements increases in the order H$_2$O < H$_2$S < H$_2$Se < H$_2$Te Statement II The boiling points of these hydride increases with increase in molar mass. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: The correct order of boiling points of group 16 hydrides is $$\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te} < \text{H}_2\text{O}$$ so Statement I, which puts H$_2$O lowest, is incorrect. Boiling point does increase with molar mass down the group, but H$_2$O has an anomalously high boiling point due to the presence of hydrogen bonding, so molar mass alone does not govern the order. Hence Statement II as a general claim about these hydrides is also incorrect. Hence, both statement I and statement II are incorrect.
Given below are two statements. Statement I The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group. Statement II $o$-nitrophenol, $m$-nitrophenol and $p$-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct: the electron withdrawing nitro group increases the acidity of phenol. Statement II is incorrect, because the order of acidic strength is $$p\text{-nitrophenol} > o\text{-nitrophenol} > m\text{-nitrophenol} > \text{phenol}$$ The NO$_2$ group at the ortho and para positions withdraws electrons of the O—H bond towards itself by the stronger $-R$ effect, while the NO$_2$ group at the meta position withdraws electrons of the O—H bond by the weaker $-I$ effect. Thus, ortho and para-nitrophenol are more acidic than meta-nitrophenol. Among ortho and para-nitrophenols, $o$-nitrophenol is less acidic due to intramolecular hydrogen bonding, which makes loss of a proton a little more difficult.
The Kjeldahl's method for the estimation of nitrogen can be used to estimate the amount of nitrogen in which one of the following compounds?
Solution: Kjeldahl's method is not applicable to compounds containing nitrogen as nitro and azo groups, and nitrogen present in the ring (pyridine). It is because nitrogen of these compounds does not change into ammonium sulphate. Hence, it is applicable only to aniline, where the —NH$_2$ group can be converted into ammonium sulphate.
Match List-I with List-II. List-I (Drug class): (A) Antacids (B) Antihistamines (C) Analgesics (D) Antimicrobials List-II (Drug molecule): (i) Salvarsan (ii) Morphine (iii) Cimetidine (iv) Seldane Choose the correct answer from the options given below
Solution: Cimetidine is an antacid, used to prevent the interaction of histamine with the receptors present in the stomach wall, due to which less amount of acid is released. Seldane is an antihistamine. It interferes with the natural action of histamine by competing with histamine for binding sites of the receptor where histamine exerts its effect. Morphine is an analgesic. They are chiefly used for the relief of pain, cardiac pain etc. Salvarsan is an antimicrobial. It mainly affects the bacteria spirochete which causes syphilis in human beings. So (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i).
The IUPAC name of an element with atomic number 119 is
Solution: In the IUPAC systematic nomenclature for elements with atomic number above 100, each digit is replaced by its numerical root and the name ends in -ium. The roots for 1, 1 and 9 are un, un and enn respectively. Hence the symbol and name respectively are Uue and ununennium.
Amongst the following which one will have maximum 'lone pair - lone pair' electron repulsions?
Solution: Counting the lone pairs on the central atom in each species: SF$_4$ contains 1 lone pair. XeF$_2$ contains 3 lone pairs. ClF$_3$ contains 2 lone pairs. IF$_5$ contains 1 lone pair. Among all the options XeF$_2$ has the maximum number of lone pairs, i.e. 3. Hence, it will show maximum 'lone pair - lone pair' repulsion.
Which of the following sequence of reactions is suitable to synthesise chlorobenzene?
Solution: To synthesise chlorobenzene, benzene reacts with Cl$_2$ in the presence of FeCl$_3$. The reaction taking place is as follows: $$\text{C}_6\text{H}_6 \xrightarrow[\text{Anhydrous FeCl}_3]{+\text{Cl}_2} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}$$ This is an electrophilic substitution; anhydrous FeCl$_3$ acts as the Lewis acid catalyst that generates the chloronium electrophile.
Which of the following $p$-$V$ curve represents maximum work done? Each option shows an isothermal curve on $p$-$V$ axes: (a) a curve starting at high pressure and running far out to large volume, sweeping out a large area; (b) a nearly vertical line at small volume; (c) a curve that falls steeply and stops at a small volume; (d) a closed elliptical loop.
Solution: In a $p$-$V$ graph, the area under the curve represents the magnitude of work. The maximum area is covered in graph (a). Hence, it represents maximum work done.
Given below are half-cell reactions $$\text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}, \qquad E^{\circ}_{\text{Mn}^{2+}/\text{MnO}_4^{-}} = -1.510\ \text{V}$$ $$\frac{1}{2}\text{O}_2 + 2\text{H}^{+} + 2e^{-} \longrightarrow \text{H}_2\text{O}, \qquad E^{\circ}_{\text{O}_2/\text{H}_2\text{O}} = +1.223\ \text{V}$$ Will the permanganate ion, MnO$_4^{-}$ liberate O$_2$ from water in the presence of an acid?
Solution: Reaction at cathode (reduction) $$2\text{MnO}_4^{-} + 16\text{H}^{+} + 10e^{-} \longrightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}, \qquad E^{\circ}_{\text{RP}} = 1.510\ \text{V}$$ Reaction at anode (oxidation) $$5\text{H}_2\text{O} \longrightarrow \frac{5}{2}\text{O}_2 + 10\text{H}^{+} + 10e^{-}, \qquad E^{\circ}_{\text{OP}} = -1.223\ \text{V}$$ The overall reaction is $$2\text{MnO}_4^{-} + 6\text{H}^{+} \longrightarrow 2\text{Mn}^{2+} + \frac{5}{2}\text{O}_2 + 3\text{H}_2\text{O}$$ $$E^{\circ} = E^{\circ}_{\text{MnO}_4^{-}/\text{Mn}^{2+}} - E^{\circ}_{\text{O}_2/\text{H}_2\text{O}} = 1.510 - 1.223 = 0.287\ \text{V}$$ Here, since the value of $E^{\circ}$ is positive, the reaction is spontaneous. Hence, MnO$_4^{-}$ will liberate O$_2$.
The IUPAC name of the complex [Ag(H$_2$O)$_2$][Ag(CN)$_2$] is
Solution: The complex [Ag(H$_2$O)$_2$][Ag(CN)$_2$] has coordination number 2 and the oxidation state is Ag$^{+}$ in both the cationic and the anionic parts. The cation is named first with the metal's ordinary name, and the anion second with the -ate suffix. Hence, its IUPAC name is diaquasilver (I) dicyanidoargentate (I).
Identify the incorrect statement from the following.
Solution: The oxidation number of K in KO$_2$ is $+1$, not $+4$, because in this compound oxygen is present as the superoxide ion: $$\text{K}^{+}\text{O}_2^{-} \quad (\text{O}_2^{-}\ \text{superoxide ion})$$ An alkali metal can only ever show the $+1$ oxidation state, so $+4$ is impossible. The other three statements are correct.
What mass of 95% pure CaCO$_3$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? $$\text{CaCO}_3(s) + 2\text{HCl}(aq) \longrightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$$ [Calculate upto second place of decimal point]
Solution: $$\text{CaCO}_3(s) + 2\text{HCl}(aq) \longrightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$$ Number of moles of CaCO$_3$ (pure) $= \dfrac{1}{2} \times$ number of moles of HCl $$[\text{Moles} = \text{Molarity} \times \text{Volume (in L)}]$$ $$= \frac{1}{2} \times 0.5 \times \frac{50}{1000} = 0.0125$$ Weight of CaCO$_3$ (pure) $=$ Mole $\times$ Molecular weight $= 0.0125 \times 100 = 1.25$ g $$\%\ \text{Purity} = \frac{\text{Weight of pure substance}}{\text{Weight of impure sample}} \times 100$$ $$95 = \frac{1.25}{\text{Weight of impure sample}} \times 100$$ $$\text{Weight of impure sample} = \frac{1.25 \times 100}{95} = 1.32\ \text{g}$$
Given below are two statements. Statement I Primary aliphatic amines react with HNO$_2$ to give unstable diazonium salts. Statement II Primary aromatic amines react with HNO$_2$ to form diazonium salts which are stable even above 300 K. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Primary aliphatic amines react with HNO$_2$ to give alkyl diazonium ions, which are unstable — so Statement I is correct. $$R\text{—NH}_2 \xrightarrow{\ \text{HNO}_2\ } R\text{—}\overset{\oplus}{\text{N}}_2$$ Primary aromatic amines react with HNO$_2$ to form aryl diazonium ions, which are stable only at LOW temperature, i.e. 0-5$^{\circ}$C, not above 300 K — so Statement II is incorrect. Hence, statement I is correct but statement II is incorrect.
Which statement regarding polymers is not correct?
Solution: The incorrect statement regarding polymers is that thermosetting polymers are reusable. It is because thermosetting polymers are NOT reusable due to the presence of cross linking in them. This cross-linking makes the thermosetting polymer more resistant to high temperature and prevents remelting, while all the other options are correct.
The incorrect statement regarding chirality is
Solution: The incorrect statement regarding chirality is that enantiomers are superimposable mirror images on each other. It is because enantiomers are NON-superimposable mirror images of each other — that non-superimposability is the definition of chirality. The other three statements are correct.
Given below are two statements. Statement I In the coagulation of a negative sol, the flocculating power of the three given ions is in the order Al$^{3+}$ > Ba$^{2+}$ > Na$^{+}$ Statement II In the coagulation of a positive sol, the flocculating power of the three given salts is in the order NaCl > Na$_2$SO$_4$ > Na$_3$PO$_4$ In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: According to the Hardy-Schulze rule, for negative sol coagulation, the greater the charge on the cation, the greater is its flocculation power. Hence, statement I is correct. For positive sol coagulation, the greater the charge on the anion, the greater is its flocculating power. Hence the correct order is $$\text{NaCl} < \text{Na}_2\text{SO}_4 < \text{Na}_3\text{PO}_4$$ So statement II, which reverses this, is incorrect.
Match List-I with List-II. List-I (Hydrides): (A) MgH$_2$ (B) GeH$_4$ (C) B$_2$H$_6$ (D) HF List-II (Nature): (i) Electron precise (ii) Electron deficient (iii) Electron rich (iv) Ionic Choose the correct answer from the options given below
Solution: An electron deficient hydride contains less than 8 electrons around the central atom. An electron precise hydride contains 8 electrons without a lone pair. An electron rich hydride contains 8 electrons along with lone pair(s). MgH$_2$ is an ionic hydride, GeH$_4$ is electron precise, B$_2$H$_6$ is electron deficient and HF is electron rich. So (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii).
Match List-I with List-II. List-I: (A) Li (B) Na (C) KOH (D) Cs List-II: (i) Absorbent for carbon dioxide (ii) Electrochemical cells (iii) Coolant in fast breeder reactors (iv) Photoelectric cell Choose the correct answer from the options given below
Solution: Lithium is used in electrochemical cells. These cells generate electrical energy from chemical reactions. Coolants are used to fill the engine's cooling system, to act as a heat exchange fluid. Sodium is one of its constituents, used in fast breeder reactors. KOH is a strong base and CO$_2$ is slightly acidic. It absorbs CO$_2$ readily at room temperature. Caesium is used in photoelectric cells. So (A)-(ii), (B)-(iii), (C)-(i), (D)-(iv).
Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells. Reason (R) In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Assertion (A) is correct, because in point defects of ionic solids, electrical neutrality is an essential condition. Reason (R) is also correct, because the cation dislocates from the lattice site to an interstitial position in a Frenkel defect. But the Assertion describes cations MISSING from the unit cell, which is the Schottky defect, while the Reason describes the Frenkel defect. Hence, both (A) and (R) are correct but (R) is not the correct explanation of (A).
The incorrect statement regarding enzymes is
Solution: The incorrect statement about enzymes is that enzymes are polysaccharides. Enzymes are PROTEIN in nature and comprised of amino acids that are linked together in one or more polypeptide chains. The rest of the given options are correct.
Match List-I with List-II. List-I (Products formed): (A) Cyanohydrin (B) Acetal (C) Schiff's base (D) Oxime List-II (Reaction of carbonyl compound with): (i) NH$_2$OH (ii) $R$NH$_2$ (iii) Alcohol (iv) HCN Choose the correct answer from the options given below
Solution: The reactions involved are as follows: A carbonyl compound with HCN gives the cyanohydrin, which carries an OH and a CN on the same carbon. A carbonyl compound with two molecules of an alcohol gives the acetal, which carries two OR groups on the same carbon. A carbonyl compound with a primary amine $R$NH$_2$ gives the Schiff's base, C=N—$R$. A carbonyl compound with NH$_2$OH gives the oxime, C=N—OH. So (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i).
In one molal solution that contains 0.5 mole of a solute, there is
Solution: We know that, $$\text{Molality} = \frac{\text{number of moles of solute}}{\text{weight of the solvent (g)}} \times 1000$$ Given that molality $= 1$ and moles of solute $= 0.5$: $$1 = \frac{0.5}{\text{Weight of the solvent (g)}} \times 1000$$ $$\text{Weight of the solvent} = 500\ \text{g of solvent}$$
Choose the correct statement.
Solution: In diamond, each carbon is bonded with four other carbon atoms, so the hybridisation of the carbon atom is $sp^{3}$. In graphite, each carbon atom is bonded with three other carbon atoms, so the hybridisation of the carbon atom in graphite is $sp^{2}$. Hence diamond is $sp^{3}$-hybridised and graphite is $sp^{2}$-hybridised. The rest of the options are incorrect: only graphite is used as a dry lubricant, diamond has a 3-D network structure, and both diamond and graphite are covalent.
Which amongst the following is incorrect statement?
Solution: Among the given options, the incorrect statement is that O$_2^{+}$ is diamagnetic. The molecular orbital configuration of O$_2^{+}$ is $$\sigma 1s^{2}\,\sigma^{*}1s^{2}\,\sigma 2s^{2}\,\sigma^{*}2s^{2}\,\sigma 2p_z^{2}\,\pi 2p_x^{2}\,\pi 2p_y^{2}\,\pi^{*}2p^{1}$$ O$_2^{+}$ ions have 15 electrons, so it contains one unpaired electron. Hence, it is paramagnetic in nature, not diamagnetic.
Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) ICl is more reactive than I$_2$. Reason (R) I—Cl bond is weaker than I—I bond. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: In an interhalogen compound of group 17, ICl is more reactive than I$_2$ due to polar bonds. Also, the $X$—$X'$ bond is weaker than the $X$—$X$ bond except in F$_2$. So the I—Cl bond is weaker than the I—I bond. Hence, both A and R are correct and R is the correct explanation of A.
Given below are two statements. Statement I The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole-dipole interactions. Statement II The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: The boiling point of comparable molecular mass molecules is as follows: $$R\text{—OH} > \text{aldehyde-ketone} > \text{alkane}$$ The order followed above is due to presence of hydrogen bonding (strong molecular association) in alcohol. In aldehyde-ketone, dipole-dipole interactions are present (weak molecular association), while an alkane is non-polar hence has the lowest boiling point among all three. Hence, both statement I and statement II are correct.
The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given p$K_a$ of CH$_3$COOH = 4.57]
Solution: Given, p$K_a$ of CH$_3$COOH $= 4.57$, concentration of acetic acid (CH$_3$COOH) $= 0.01$ M and concentration of sodium acetate (CH$_3$COONa) $= 0.10$ M. A weak acid (CH$_3$COOH) and the salt of the weak acid with a strong base (CH$_3$COONa) form an acidic buffer. The pH of an acidic buffer solution is given by $$\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$$ $$= 4.57 + \log\left(\frac{0.1}{0.01}\right) = 4.57 + 1 = 5.57$$ Hence, the given solution has a pH of 5.57.
Identify the incorrect statement from the following.
Solution: The incorrect statement is the third one, because the shapes of the first four orbitals, i.e. $d_{xy}$, $d_{yz}$, $d_{zx}$ and $d_{x^{2}-y^{2}}$ are similar to each other — all are the double dumb-bell shape — whereas the fifth one, i.e. $d_{z^{2}}$, is different from the others.
At 298 K, the standard electrode potentials of Cu$^{2+}$/Cu, Zn$^{2+}$/Zn, Fe$^{2+}$/Fe and Ag$^{+}$/Ag are 0.34 V, $-0.76$ V, $-0.44$ V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction can not occur?
Solution: The standard reduction potential values can be arranged as follows: $$E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} < E^{\circ}_{\text{Fe}^{2+}/\text{Fe}} < E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} < E^{\circ}_{\text{Ag}^{+}/\text{Ag}}$$ So the reactivity order will be Zn > Fe > Cu > Ag. The reactions given in the options are displacement reactions. In a displacement reaction, more reactive metals (i.e. low SRP) can displace less reactive metals (higher SRP) from their salt solution. So, CuSO$_4$(aq) $+$ 2Ag(s) $\longrightarrow$ Cu(s) $+$ Ag$_2$SO$_4$(aq) is not possible, as Ag is less reactive compared to Cu.
The order of energy absorbed which is responsible for the colour of complexes (A) [Ni(H$_2$O)$_2$(en)$_2$]$^{2+}$ (B) [Ni(H$_2$O)$_4$(en)]$^{2+}$ and (C) [Ni(en)$_3$]$^{2+}$ is
Solution: 'en' is a strong field ligand. As the number of en (strong field ligand) increases, splitting also increases. So, $\Delta_o$ increases. Hence, [Ni(en)$_3$]$^{2+}$ will have the highest energy due to the presence of 3 en ligands, followed by [Ni(H$_2$O)$_2$(en)$_2$]$^{2+}$ and [Ni(H$_2$O)$_4$(en)]$^{2+}$. So, the correct order is (C) > (A) > (B).
A 10.0 L flask contains 64 g of oxygen at 27$^{\circ}$C. (Assume O$_2$ gas is behaving ideally). The pressure inside the flask in bar is (Given, $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$)
Solution: Given, $V = 10$ L, $W_{\text{O}_2} = 64$ g, $T = 27^{\circ}$C $= 27 + 273 = 300$ K, $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$. $$n_{\text{O}_2} = \frac{64}{32} = 2$$ Using the ideal gas equation $pV = nRT$, $$\Rightarrow\ p = \frac{nRT}{V}$$ Substituting the values, $$p = \frac{2 \times 0.0831 \times 300}{10}$$ $$p = 4.986 \approx 4.9\ \text{bar}$$
Which one of the following is not formed when acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating?
Solution: When acetone reacts with 2-pentanone in the presence of dilute NaOH followed by heating, the products will be formed according to aldol condensation. Self aldol of acetone with acetone gives 4-methylpent-3-en-2-one. Cross aldol between acetone and 2-pentanone gives the two mixed condensation products. The structure drawn in option (a), which carries an ethyl branch on the alkene carbon, cannot arise from any of these enolate combinations, so it will not form.
The product formed from the following reaction sequence is Cyanobenzene, C$_6$H$_5$CN, treated with (i) LiAlH$_4$, H$_2$O; (ii) NaNO$_2$, HCl; (iii) H$_2$O.
Solution: In this reaction cyanobenzene reacts with a strong reducing agent, i.e. LiAlH$_4$, to yield benzyl amine. $$\text{C}_6\text{H}_5\text{CN} \xrightarrow{\ \text{LiAlH}_4,\ \text{H}_2\text{O}\ } \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$$ The benzyl amine on further reaction with NaNO$_2$ + HCl will give the diazonium salt, which is unstable as it is an aliphatic (benzylic) diazonium ion. This salt on reaction with H$_2$O will yield benzyl alcohol. $$\text{C}_6\text{H}_5\text{CH}_2\overset{\oplus}{\text{N}}_2\text{Cl}^{-} \xrightarrow{\ \text{H}_2\text{O}\ } \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{N}_2 + \text{HCl}$$
Copper crystallises in fcc unit cell with cell edge length of $3.608 \times 10^{-8}$ cm. The density of copper is 8.92 g cm$^{-3}$. Calculate the atomic mass of copper.
Solution: Given $a = 3.608 \times 10^{-8}$ cm and $d = 8.92$ g/cm$^{3}$. For an fcc unit cell, $Z = 4$. Using the formula of density, $$d = \frac{Z \times M}{N_A \times a^{3}}$$ where $d$ = density, $M$ = atomic mass, $a$ = edge length, $N_A$ = Avogadro's number and $Z$ = number of atoms in unit cell. Now, substituting the values in the above formula, $$8.92 = \frac{4 \times M}{6.022 \times 10^{23} \times (3.608 \times 10^{-8})^{3}}$$ $$M = 63.1\ \text{g/mol}$$ Or atomic mass $(M) = 63.1$ u.
$$3\text{O}_2(g) \rightleftharpoons 2\text{O}_3(g)$$ For the above reaction at 298 K, $K_C$ is found to be $3.0 \times 10^{-59}$. If the concentration of O$_2$ at equilibrium is 0.040 M then concentration of O$_3$ in M is
Solution: Given $K_C = 3 \times 10^{-59}$ and [O$_2$] $= 0.040$ M. $$K_C = \frac{[\text{O}_3]^{2}}{[\text{O}_2]^{3}}$$ Substituting the values, $$3 \times 10^{-59} = \frac{[\text{O}_3]^{2}}{(4 \times 10^{-2})^{3}}$$ $$[\text{O}_3]^{2} = 3 \times 10^{-59} \times (4 \times 10^{-2})^{3}$$ $$[\text{O}_3]^{2} = 19.2 \times 10^{-64}$$ $$[\text{O}_3] = 4.38 \times 10^{-32}\ \text{M}$$
Given below are two statements. Statement I In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnCl$_2$, known as Lucas reagent. Statement II Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas reagent. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: 1$^{\circ}$, 2$^{\circ}$ and 3$^{\circ}$ alcohols can be distinguished by using Lucas reagent (conc. HCl + anhy. ZnCl$_2$) on the basis of the time taken for turbidity to appear, so Statement I is correct. A primary alcohol gives turbidity only on heating; a secondary alcohol gives turbidity in about 5 minutes; and a tertiary alcohol gives turbidity immediately. Hence the order of reactivity of alcohols towards Lucas reagent is 3$^{\circ}$ > 2$^{\circ}$ > 1$^{\circ}$, so Statement II, which claims primary alcohols are most reactive, is incorrect.
In the neutral or faintly alkaline medium, KMnO$_4$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from
Solution: The reaction that takes place in the given condition is as follows: $$\overset{+7}{\text{KMnO}_4} + \text{I}^{-} \xrightarrow[\text{Weak alkaline medium}]{\text{Neutral}} \overset{+4}{\text{MnO}_2} + \text{IO}_3^{-}$$ So, the change in oxidation state of manganese in this reaction is from $+7$ to $+4$.
Match List-I with List-II. List-I (Ores): (A) Haematite (B) Magnetite (C) Calamine (D) Kaolinite List-II (Composition): (i) Fe$_3$O$_4$ (ii) ZnCO$_3$ (iii) Fe$_2$O$_3$ (iv) [Al$_2$(OH)$_4$Si$_2$O$_5$] Choose the correct answer from the options given below
Solution: Haematite is an ore of iron; its composition is Fe$_2$O$_3$. Magnetite also is an ore of iron; its composition is Fe$_3$O$_4$. Calamine is an ore of zinc; its composition is ZnCO$_3$. Kaolinite is a clay mineral with chemical formula [Al$_2$(OH)$_4$Si$_2$O$_5$]. So (A)-(iii), (B)-(i), (C)-(ii), (D)-(iv).
The correct IUPAC name of the compound drawn as a six-carbon chain carrying Cl on one carbon, OH on the next-but-one, a methyl branch between them, and Br at the far end is
Solution: Numbering the chain so that the principal functional group, the alcohol, gets the lowest possible locant, the bromine falls on C-1, the hydroxyl on C-3, the methyl branch on C-4 and the chlorine on C-5. The substituents are then cited alphabetically: bromo, chloro, methyl. The IUPAC name of the given compound is 1-bromo-5-chloro-4-methylhexan-3-ol.
If radius of second Bohr orbit of the He$^{+}$ ion is 105.8 pm, what is the radius of third Bohr orbit of Li$^{2+}$ ion?
Solution: According to Bohr's atomic model, $$r \propto \frac{n^{2}}{Z}$$ For the 2nd orbit of He$^{+}$: $n_2 = 2$, $z_2 = 2$. For the 3rd orbit of Li$^{2+}$: $n_1 = 3$, $z_1 = 3$. From the above equation, $$\frac{(r_3)_{\text{Li}^{2+}}}{(r_2)_{\text{He}^{+}}} = \frac{n_1^{2} \times z_2}{n_2^{2} \times z_1}$$ $$\frac{(r_3)_{\text{Li}^{2+}}}{105.8\ \text{pm}} = \frac{3 \times 3 \times 2/3}{2 \times 2}$$ $$(r_3)_{\text{Li}^{2+}} = \frac{3 \times 3 \times 2 \times 105.8}{4 \times 3} = 158.7\ \text{pm}$$
Compound $X$ on reaction with O$_3$ followed by Zn/H$_2$O gives formaldehyde and 2-methyl propanal as products. The compound $X$ is
Solution: The compound '$X$' that reacts with O$_3$ followed by reaction with Zn/H$_2$O to give formaldehyde and 2-methyl propanal is 3-methylbut-1-ene. The ozonolysis reaction involved is as follows: $$\text{CH}_3\text{—CH(CH}_3)\text{—CH=CH}_2 \xrightarrow[\text{(ii) Zn + H}_2\text{O}]{\text{(i) O}_3} (\text{CH}_3)_2\text{CH—CHO} + \text{H—CHO}$$ The terminal =CH$_2$ becomes formaldehyde, and the other fragment, carrying the isopropyl group, becomes 2-methyl propanal.
For a first order reaction $A \rightarrow$ products, initial concentration of $A$ is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min$^{-1}$ is
Solution: Given, initial concentration $[A_0] = 0.1$ M, concentration after 5 min $[A_t] = 0.001$ M and $t = 5$ minutes. Using the rate constant formula for a 1st order reaction, $$K = \frac{2.303}{t}\log\left(\frac{A_0}{A_t}\right)$$ $$= \frac{2.303}{5}\log\left(\frac{0.1}{0.001}\right)$$ $$K = \frac{2.303 \times 2}{5} = 0.9212\ \text{min}^{-1}$$
The pollution due to oxides of sulphur gets enhanced due to the presence of (I) particulate matter (II) ozone (III) hydrocarbons (IV) hydrogen peroxide Choose the most appropriate answer from the options given below
Solution: The pollution due to oxides of sulphur gets enhanced due to the presence of particulate matter, ozone and hydrogen peroxide. The particulate matter present in polluted air catalyses the oxidation of sulphur dioxide to sulphur trioxide: $$2\text{SO}_2(g) + \text{O}_2(g) \longrightarrow 2\text{SO}_3(g)$$ Similarly, the reaction is also promoted by ozone and hydrogen peroxide: $$\text{SO}_2(g) + \text{O}_3(g) \longrightarrow \text{SO}_3(g) + \text{O}_2(g)$$ $$\text{SO}_2(g) + \text{H}_2\text{O}_2(l) \longrightarrow \text{H}_2\text{SO}_4(aq)$$
Which of the following is not a method of $ex$ $situ$ conservation?
Solution: National parks are not a method of $ex$-$situ$ conservation. These are sites of protecting an endangered plant or animal species in its natural habitat, i.e. $in$-$situ$ conservation. $Ex$-$situ$ conservation literally means "off-site conservation". It is the process of protecting an endangered species, variety or breed, of plant or animal outside of its natural habitat, for example by removing part of the population from a threatened habitat and placing it in a new location. Cryopreservation, micropropagation and $in$-$vitro$ fertilisation are all $ex$-$situ$ methods.
Given below are two statements Statement I The primary CO$_2$ acceptor in C$_4$ plants is phosphoenolpyruvate and is found in the mesophyll cells. Statement II Mesophyll cells of C$_4$ plants lack RuBisCO enzyme. In the light of the above statements, choose the correct answer from the options given below.
Solution: Both statements I and II are correct. The primary CO$_2$ acceptor is a 3-carbon molecule, phosphoenolpyruvate (PEP), in C$_4$ plants and is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase. Also, mesophyll cells lack the RuBisCO enzyme. The C$_4$ acid is OAA, which is formed in the mesophyll cells. It then forms other 4-carbon compounds like malic acid or aspartic acid in the mesophyll cells itself, which are transported to the bundle sheath cells. In the bundle sheath cells these C$_4$ acids are broken down to release CO$_2$ and a 3-carbon molecule.
XO type of sex-determination can be found in
Solution: The XO type of sex-determination can be found in grasshoppers, crickets and some other insects. In this type of insects, the females are homogametic, which means they produce only one single type of chromosome, that is XX, whereas the males have only one X chromosome.
In old trees, the greater part of secondary xylem is dark brown and resistant to insect attack due to I. secretion of secondary metabolites and their deposition in the lumen of vessels. II. deposition of organic compounds like tannins and resins in the central layers of stem. III. deposition of suberin and aromatic substances in the outer layer of stem. IV. deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem. V. presence of parenchyma cells, functionally active xylem elements and essential oils. Choose the correct answer from the options given below.
Solution: Options (I) and (II) are correct. In older trees the greater part of secondary xylem is dark brown due to the deposition of tannins, resins, oil, gums, aromatic substances and essential oil in the central or the innermost layer of the stem. It also has secretion of secondary metabolites and their deposition in the lumen of vessels. These substances make it hard, durable and resistant to the attacks of microorganisms and insects.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) Polymerase chain reaction is used in DNA amplification. Reason (R) The ampicillin resistant gene is used as a selectable marker to check transformation. In the light of the above statements, choose the correct answer from the options given below.
Solution: Both (A) and (R) are correct, but (R) is not the correct explanation of (A). Polymerase chain reaction is a method of amplifying the fragments of DNA. This method can make multiple copies of even a single DNA fragment or the gene of interest in a test tube. If we spread the transformed cells on agar plates containing ampicillin, only transformants will grow and untransformed recipient cells will die. Thus, due to the ampicillin resistance gene, one is able to select a transformed cell in the presence of ampicillin. The ampicillin resistance gene in this case is called a selectable marker. The two describe different techniques, so R does not explain A.
Which one of the following statement is not true regarding gel electrophoresis technique?
Solution: The statement about a chromogenic substrate giving blue coloured DNA bands is not related to gel electrophoresis — that describes the blue-white screening of recombinant colonies. DNA fragments can be separated by a technique known as gel electrophoresis. Since DNA fragments are negatively charged molecules, they can be separated by forcing them to move towards the anode under an electric field through a medium/matrix. The most commonly used matrix is agarose, a natural polymer extracted from sea weeds. The DNA fragments are separated according to their size through the sieving effect provided by the agarose gel — the smaller the fragment size, the farther it moves. The separated DNA fragments can be visualised as bright orange coloured bands after staining the DNA with ethidium bromide followed by exposure to UV radiation. The separated bands are cut out from the agarose gel and extracted from the gel piece; this step is known as elution.
The gaseous plant growth regulator is used in plants to
Solution: Ethylene is a simple gaseous PGR. It is synthesised in large amounts by tissues undergoing senescence and ripening fruits. It promotes root growth and root hair formation, thus helping the plants to increase their absorption surface. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds, sprouting of potato tubers and promotes rapid internode/petiole elongation in deep water rice plants. It also helps leaves and upper parts of the shoot to remain above water.
What amount of energy is released from glucose during lactic acid fermentation?
Solution: In lactic acid fermentation not much energy is released, i.e. less than 7 per cent of the energy in glucose is released, and not all of it is trapped as high energy bonds of ATP. Also, the processes are hazardous — either acid or alcohol is produced.
Read the following statements about the vascular bundles. I. In roots, xylem and phloem in a vascular bundle are arranged in an alternate manner along the different radii. II. Conjoint closed vascular bundles do not possess cambium. III. In open vascular bundles, cambium is present in between xylem and phloem. IV. The vascular bundles of dicotyledonous stem possess endarch protoxylem. V. In monocotyledonous root, usually there are more than six xylem bundles present Choose the correct answer from the options given below
Solution: All statements (I), (II), (III), (IV) and (V) are correct. When xylem and phloem within a vascular bundle are arranged in an alternate manner along the different radii, the arrangement is called radial, such as in roots. Conjoint closed vascular bundles do not possess cambium, and in open vascular bundles cambium is present in between xylem and phloem. The vascular bundles of dicotyledonous stem possess endarch protoxylem. In monocotyledonous root, usually there are more than six xylem bundles present, i.e. the polyarch condition.
Identify the correct set of statements. I. The leaflets are modified into pointed hard thorns in $Citrus$ and $Bougainvillea$. II. Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin. III. Stem is flattened and fleshy in $Opuntia$ and modified to perform the function of leaves. IV. $Rhizophora$ shows vertically upward growing roots that help to get oxygen for respiration. V. Subaerially growing stems in grasses and strawberry help in vegetative propagation. Choose the correct answer from the options given below.
Solution: Statements (II), (III), (IV) and (V) are correct. Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin. Stem is flattened and fleshy in $Opuntia$ and modified to perform the function of leaves. $Rhizophora$ shows vertically upward growing roots that help to get oxygen for respiration. Subaerially growing stems in grasses and strawberry help in vegetative propagation. Incorrect statement (I) can be corrected as: in $Citrus$ and $Bougainvillea$ the AXILLARY BUDS get modified into pointed hard thorns, not the leaflets.
Which one of the following plants does not show plasticity?
Solution: Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity. For example, in cotton, buttercup, coriander and larkspur plants, the leaves of the juvenile phase are different in shape from those in the mature phase — this is known as heterophylly. Maize does not show this.
Given below are two statements, Statement I Cleistogamous flowers are invariably autogamous. Statement II Cleistogamy is disadvantageous as there is no chance for cross-pollination. In the light of the above statements, choose the correct answer from the options given below.
Solution: Both statement I and statement II are correct. Cleistogamous flowers are flowers which do not open at all. In such flowers, the anthers and stigma lie close to each other. When anthers dehisce in the flower buds, pollen grains come in contact with the stigma for effective pollination. Thus, cleistogamous flowers are invariably autogamous, as there is no chance of cross-pollen landing on the stigma. It is a big disadvantage of being closed always. Cleistogamous flowers also produce assured seed-set even in the absence of pollinators.
'Girdling Experiment' was performed by plant physiologists to identify the plant tissue through which
Solution: The "Girdling Experiment" was performed by plant physiologists to identify the plant tissue through which food is transported. In this experiment, on the trunk of a tree a ring of bark up to a depth of the phloem layer is carefully removed. In the absence of downward movement of food, the portion of the bark above the ring on the stem becomes swollen after a few weeks. This experiment shows that phloem is the tissue which is responsible for translocation of food and that transport takes place in one direction, i.e. towards the roots.
Hydrocolloid carrageen is obtained from
Solution: Hydrocolloid carrageen is obtained from Rhodophyceae (red algae) only. Certain marine brown and red algae produce large amounts of hydrocolloids (water holding substances), e.g. algin (brown algae) and carrageen (red algae), which are used commercially.
Given below are two statements, Statement I Mendel studied seven pairs of contrasting traits in pea plants and proposed the law of inheritance. Statement II Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height. In the light of the above statements, choose the correct answer from the options given below
Solution: Both statement I and statement II are correct. Gregor Johann Mendel conducted hybridisation experiments on garden peas for seven years (1856-1863) and proposed the laws of inheritance in living organisms. For his experiments, he studied seven pairs of contrasting traits in pea plants. These were seed shape and colour, flower colour, pod shape and colour, flower position and stem height.
Production of cucumber has increased manifold in recent years. Application of which of the following phytohormones has resulted in this increased yield as the hormone is known to produce female flowers in the plants?
Solution: Ethylene is a simple gaseous PGR. It is synthesised in large amounts by tissues undergoing senescence and ripening fruits. It promotes female flowers in cucumbers, thereby increasing the yield.
The process of translation of $m$RNA to proteins begins as soon as
Solution: The process of translation of $m$RNA to proteins begins as soon as the small subunit of the ribosome encounters $m$RNA. Translation refers to the process of polymerisation of amino acids to form a polypeptide. The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits, a large subunit and a small subunit. When the small subunit encounters an $m$RNA, the process of translation of the $m$RNA to protein begins.
Habitat loss and fragmentation, overexploitation, alien-species invasion and co-extinction are causes for
Solution: Habitat loss and fragmentation, over exploitation, alien-species invasion and co-extinction are causes for biodiversity loss. These are the "Evil Quartet" of biodiversity loss. In general, loss of biodiversity in a region may lead to decline in plant production, lowered resistance to environmental perturbations such as drought and increased variability in certain ecosystem processes such as plant productivity, water use, and pest and disease cycles.
Read the following statements and choose the set of correct statements. I. Euchromatin is loosely packed chromatin. II. Heterochromatin is transcriptionally active. III. Histone octamer is wrapped by negatively charged DNA in nucleosome. IV. Histones are rich in lysine and arginine. V. A typical nucleosome contains 400 bp of DNA helix. Choose the correct answer from the options given below.
Solution: The set of correct statements are (I), (III) and (IV). The incorrect statements can be corrected as follows: heterochromatin is more densely packed and stains dark. Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive. A typical nucleosome contains 200 bp of DNA helix, not 400 bp.
Which of the following is incorrectly matched?
Solution: Option (a) is incorrectly matched. It can be corrected as $Ulothrix$ belongs to class Chlorophyceae and the stored food material in them is starch. Mannitol is present in members of class Phaeophyceae.
The device which can remove particulate matter present in the exhaust from a thermal power plant is
Solution: The device which can remove particulate matter present in the exhaust from a thermal power plant is the electrostatic precipitator. It can remove over 99 per cent of the particulate matter present in the exhaust from a thermal power plant. It has electrode wires that are maintained at several thousand volts, which produce a corona that releases electrons. These electrons attach to dust particles giving them a net negative charge. The collecting plates are grounded and attract the charged dust particles.
DNA polymorphism forms the basis of
Solution: DNA polymorphism forms the basis of both genetic mapping and DNA fingerprinting. DNA fingerprinting is a technique to find out variation in individuals of a population at DNA level. It works on the principle of polymorphism in DNA sequences. It has immense applications in the field of forensic science, genetic biodiversity and evolutionary biology. Gene mapping refers to the process of determining the location of genes on chromosomes. It involves sequencing a genome and then using computer programs to analyse the sequence to identify the location of genes.
Which one of the following statements cannot be connected to predation?
Solution: The statement that both the interacting species are negatively impacted cannot be connected to predation. In predation only one species is negatively impacted and the other is completely benefitted. Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. Predation might lead to extinction of a species, but it is necessitated by nature to maintain the ecological balance.
Which one of the following never occurs during mitotic cell division?
Solution: Pairing of homologous chromosomes never occurs during mitotic cell division — it is a feature of prophase I of meiosis. Mitosis, or the equational division, is usually restricted to the diploid cells. In it, spindle fibres attach to kinetochores of chromosomes, movement of centrioles towards opposite poles occurs and coiling and condensation of the chromatids occurs.
Exoskeleton of arthropods is composed of
Solution: The exoskeleton of arthropods is composed of chitin. Arthropods are the most successful group of animals. Their success is due to the unique chitinous cuticle. The exoskeleton is light weight, tough and composed of the structural polysaccharide chitin. It is strengthened with proteins and calcium carbonate occurring on the outside.
What is the net gain of ATP when each molecule of glucose is converted to two molecules of pyruvic acid?
Solution: The net gain of ATP when each molecule of glucose is converted to two molecules of pyruvic acid is 2 ATP. Glycolysis consumes 2 ATP in the preparatory phase and produces 4 ATP in the pay-off phase, so the net gain is 2 ATP.
Which one of the following plants shows vexillary aestivation and diadelphous stamens?
Solution: Vexillary aestivation and diadelphous dithecous anthers are seen in $Pisum$ $sativum$. In them, there are five petals: the largest (standard) overlaps the two lateral petals (wings), which in turn overlap the two smallest anterior petals (keel). This type of aestivation is known as vexillary. When the stamens are united into two bundles, it is diadelphous.
Identify the incorrect statement related to pollination.
Solution: Statement (c) is incorrect. It can be corrected as honeybees are the most dominant pollinating agents among animals. The rest of the statements are correct: wind pollination is the more common of the abiotic modes, some flowers do produce foul odours to attract flies and beetles, and water pollination is quite rare in flowering plants.
Given below are two statements. Statement I Decomposition is a process in which the detritus is degraded into simpler substances by microbes. Statement II Decomposition is faster if the detritus is rich in lignin and chitin. In the light of the above statements, choose the correct answer from the options given below.
Solution: Statement I is correct, but statement II is incorrect. The incorrect statement can be corrected as: decomposition is largely an oxygen-requiring process. The rate of decomposition is controlled by the chemical composition of detritus and climatic factors. In a particular climatic condition, the decomposition rate is SLOWER if detritus is rich in lignin and chitin, and is QUICKER if detritus is rich in nitrogen and water-soluble substances like sugars.
The appearance of recombination nodules on homologous chromosomes during meiosis characterises
Solution: The appearance of recombination nodules on homologous chromosomes during meiosis characterises sites at which crossing over occurs between non-sister chromatids of the homologous chromosomes. Crossing over is the exchange of genetic material between two homologous chromosomes.
Which one of the following produces nitrogen-fixing nodules on the roots of $Alnus$ ?
Solution: $Frankia$ produces nitrogen-fixing nodules on the roots of non-leguminous plants like $Alnus$. $Rhizobium$ forms nodules on the roots of legumes, while $Rhodospirillum$ and $Beijerinckia$ are free-living nitrogen fixers.
Which one of the following is not true regarding the release of energy during ATP synthesis through chemiosmosis? It involves
Solution: Statement (a) is not related to chemiosmosis. In chemiosmosis there is breakdown of a PROTON gradient, not an electron gradient. The proton gradient is important because it is the breakdown of this gradient that leads to the synthesis of ATP. The gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of the CF$_0$ of the ATP synthase. The NADP reductase enzyme is located on the stroma side of the membrane.
The flowers are zygomorphic in I. Mustard II. Gulmohar III. $Cassia$ IV. $Datura$ V. Chilly Choose the correct answer from the options given below.
Solution: The flowers are zygomorphic in gulmohar and $Cassia$. Zygomorphic flowers are those flowers that have their floral parts unequal in size or form, so that the flower is capable of division into essentially symmetrical halves by only one longitudinal plane passing through the axis. Mustard, $Datura$ and chilly bear actinomorphic flowers.
Match List-I with List-II. List-I: A. Manganese B. Magnesium C. Boron D. Iron List-II: (i) Activates the enzyme catalase (ii) Required for pollen germination (iii) Activates enzymes of respiration (iv) Functions in splitting of water during photosynthesis Choose the correct answer from the options given below.
Solution: Manganese functions in splitting of water during photosynthesis. Magnesium activates enzymes of respiration. Boron is required for pollen germination. Iron activates the enzyme catalase. So (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i).
Addition of more solutes in a given solution will
Solution: Addition of more solutes in a given solution will lower its water potential. Water potential is the chemical potential of water and depends on the number of water molecules. When a solute has been dissolved in water, there is a relative decrease in the number of solvent or water molecules. Hence, water potential decreases. The water potential of pure water has been fixed as zero. The water potential of all solutions is negative.
Read the following statements on lipids and find out correct set of statements I. Lecithin found in the plasma membrane is a glycolipid. II. Saturated fatty acids possess one or more C=C bonds. III. Gingelly oil has lower melting point, hence remains as oil in winter. IV. Lipids are generally insoluble in water, but soluble in some organic solvents. V. When fatty acid is esterified with glycerol, monoglycerides are formed. Choose the correct answer from the options given below.
Solution: Statements III, IV and V are correct. The incorrect statements can be corrected as: some lipids have phosphorus and a phosphorylated organic compound in them. These are PHOSPHOLIPIDS, not glycolipids. They are found in the cell membrane, like lecithin. Fatty acids could be saturated (without double bond) or unsaturated (with one or more C=C double bonds), so statement II reverses the definition.
What is the role of large bundle sheath cells found around the vascular bundles in C$_4$ plants?
Solution: The role of large bundle sheath cells found around the vascular bundles in C$_4$ plants is to increase the number of chloroplasts for the operation of the Calvin cycle. The leaves which have such anatomy are said to have 'Kranz' anatomy. 'Kranz' means 'wreath' and is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.
The entire fleet of buses in Delhi were converted to CNG from diesel. In reference to this, which one of the following statements is false?
Solution: Statement (a) is false. All the buses of Delhi were converted to run on CNG from diesel by the end of 2002. This was done because CNG is better than diesel. CNG burns most efficiently, unlike petrol or diesel, in automobiles and very little of it is left unburnt. Moreover, CNG is cheaper than petrol or diesel, cannot be siphoned off by thieves and cannot be adulterated like petrol or diesel. The main problem with switching over to CNG is the difficulty of laying down pipelines to deliver CNG through distribution points/pumps and ensuring uninterrupted supply.
Transposons can be used during which one of the following ?
Solution: Transposons can be used during gene silencing. Transposons or mobile genetic elements in viruses are the sources of the complementary $ds$RNA, which in turn binds to specific $m$RNA and causes RNA interference (RNA$i$). RNA$i$ takes place in all eukaryotic organisms as a method of cellular defence.
Match List-I with List-II. List-I: A. Metacentric chromosome B. Acrocentric chromosome C. Sub-metacentric chromosome D. Telocentric chromosome List-II: (i) Centromere situated close to the end forming one extremely short and one very long arms (ii) Centromere at the terminal end (iii) Centromere in the middle forming two equal arms of chromosomes (iv) Centromere slightly away from the middle forming one shorter arm and one longer arm Choose the correct answer from the options given below.
Solution: Metacentric chromosome — centromere in the middle forming two equal arms of chromosomes. Acrocentric chromosome — centromere situated close to the end forming one extremely short and one very long arm. Sub-metacentric chromosome — centromere slightly away from the middle forming one shorter arm and one longer arm. Telocentric chromosome — centromere at the terminal end. So (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii).
Match the plant with the kind of life cycle it exhibits. List-I: A. $Spirogyra$ B. Fern C. $Funaria$ D. $Cycas$ List-II: (i) Dominant diploid sporophyte vascular plant, with highly reduced male or female gametophyte (ii) Dominant haploid free-living gametophyte (iii) Dominant diploid sporophyte alternating with reduced gametophyte called prothallus (iv) Dominant haploid leafy gametophyte alternating with partially dependent multicellular sporophyte Choose the correct answer from the options given below.
Solution: $Spirogyra$ — dominant haploid free-living gametophyte. Fern — dominant diploid sporophyte alternating with reduced gametophyte called prothallus. $Funaria$ — dominant haploid leafy gametophyte alternating with partially dependent multicellular sporophyte. $Cycas$ — dominant diploid sporophyte vascular plant, with highly reduced male or female gametophyte. So (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i).
Which of the following occurs due to the presence of autosome linked dominant trait?
Solution: Myotonic dystrophy occurs due to the presence of an autosome linked dominant trait. This means one copy of the altered gene in each cell is sufficient to cause the disorder. In most cases, an affected person has one parent with the condition. Haemophilia is X-linked recessive, while thalassemia and sickle cell anaemia are autosomal recessive.
The anatomy of spring wood shows some peculiar features. Identify the correct set of statement about springwood. I. It is also called as the early wood. II. In spring season cambium produces xylem elements with narrow vessels. III. It is light in colour. IV. The springwood along with autumnwood shows alternate concentric rings forming annual rings. V. It has lower density. Choose the correct answer from the options given below.
Solution: Statements I, III, IV and V are related to spring wood. In the spring season, cambium is very active and produces a large number of xylary elements having vessels with WIDER cavities, so statement II is wrong. The wood formed during this season is called spring wood or early wood. The spring wood is lighter in colour and has a lower density, whereas the autumn wood is darker and has a higher density. Together they form the alternate concentric rings known as annual rings.
Which part of the fruit, labelled in the given figure makes it a false fruit?
Solution: The part labelled as C (thalamus) in the figure makes it a false fruit. A false fruit is derived from the floral parts other than the ovary, like the thalamus. False fruit is also called pseudocarp or accessory fruit or spurious fruit.
In the following palindromic base sequences of DNA, which one can be cut easily by particular restriction enzyme?
Solution: The sequence 5$'$ G A A T T C 3$'$ / 3$'$ C T T A A G 5$'$ is a palindromic sequence of DNA, which can be cut easily by a particular restriction enzyme. The palindrome in DNA is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same.
While explaining interspecific interaction of population, (+) sign is assigned for beneficial interaction, ($-$) sign is assigned for detrimental interaction and (0) for neutral interaction. Which of the following interactions can be assigned (+) for one species and ($-$) for another species involved in the interaction ?
Solution: Predation is an interaction in which we can assign (+) for one species and ($-$) for another species involved. The other options can be explained as: in competition both the species are in loss. The interaction where one species is benefitted and the other is neither benefitted nor harmed is called commensalism. In amensalism, on the other hand, one species is harmed whereas the other is unaffected.
Which one of the following will accelerate phosphrus cycle?
Solution: Weathering of rocks will accelerate the phosphorus cycle. The natural reservoir of phosphorus is rock, which contains phosphorus in the form of phosphates. When rocks are weathered, minute amounts of these phosphates dissolve in soil solution and are absorbed by the roots of the plants. Phosphorus is a major constituent of biological membranes, nucleic acids and cellular energy transfer systems. Many animals also need large quantities of this element to make shells, bones and teeth.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) Mendel's law of independent assortment does not hold good for the genes that are located closely on the same chromosome. Reason (R) Closely located genes assort independently. In the light of the above statements, choose the correct answer from the options given below.
Solution: (A) is correct, but (R) is not correct. Mendel's law of independent assortment does not hold good for the genes that are located closely on the same chromosome. These genes are called 'linked genes'. Closely located genes assort TOGETHER, and distantly located genes, due to recombination, assort independently. Linkage maps therefore correspond to the arrangement of genes on a chromosome.
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by him is called as,
Solution: Sequence annotation will be used by a geneticist as a blind approach for sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions. For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes and cloned in a suitable host using specialised vectors. The cloning resulted into amplification of each piece of DNA fragment so that it subsequently could be sequenced with ease.
Given below are two statements. Statement I Autoimmune disorder is a condition where body defence mechanism recognises its own cells as foreign bodies. Statement II Rheumatoid arthritis is a condition where body does not attack self cells. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct, but statement II is incorrect. Autoimmune disorders are abnormal immune responses in which the immune system fails to properly differentiate between self and non-self cells and destroys the body's own protein. Rheumatoid arthritis is an autoimmune disorder that primarily affects joints — so the body DOES attack self cells. Thus, joints become painful, stiff and swollen.
Given below are two statements. Statement I The coagulum is formed of network of threads called thrombins. Statement II Spleen is the graveyard of erythrocytes. In the light of the above statements, choose the most appropriate answer from the options given below
Solution: Statement I is incorrect, but statement II is correct. A blood clot or coagulum is formed mainly of a network of threads called FIBRINS, not thrombins, in which dead and damaged formed elements of blood are trapped. Spleen is the graveyard of RBCs because after completion of lifespan, old and damaged RBCs or erythrocytes are destroyed in the spleen where they are ingested by free macrophages.
Identify the asexual reproductive structure associated with $Penicillium$.
Solution: Conidia is the asexual reproductive structure associated with $Penicillium$. Asexual reproduction refers to the production of offspring by a single parent without the fusion of gametes. Conidia are non-motile spores that are produced exogenously.
Given below are two statements. Statement I The release of sperms into the seminiferous tubules is called spermiation. Statement II Spermiogenesis is the process of formation of sperms from spermatogonia. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct, but statement II is incorrect. Spermiation is the process of releasing mature spermatozoa from the Sertoli cells into the lumen of seminiferous tubules. Spermiogenesis is the process of transformation of SPERMATIDS to spermatozoa or sperms, not of spermatogonia to sperms.
Under normal physiological condition in human being every 100 mL of oxygenated blood can deliver ......... mL of O$_2$ to the tissues.
Solution: About 20 mL of oxygen is present in 100 mL of blood; when it reaches the tissues only 5 mL of oxygen is delivered and 15 mL returns back. Thus, under normal physiological conditions in human beings every 100 mL of oxygenated blood can deliver 5 mL of O$_2$ to the tissues.
$In$-$situ$ conservation refers to
Solution: $In$-$situ$ conservation involves the protection of species in their natural habitat. It is the conservation and protection of the whole ecosystem and its biodiversity at all levels in order to protect the threatened species. For example, National parks, Wildlife sanctuaries, etc.
Natural selection where more individuals acquire specific character value other than the mean character value, leads to
Solution: Natural selection where more individuals acquire a specific character value other than the mean character value leads to directional change. This operates in response to gradual changes in environmental condition, like during the industrial revolution in peppered moths.
Breeding crops with higher levels of vitamins and minerals or higher proteins and healthier fats is called
Solution: Biofortification is the method of breeding crops with a higher level of nutrients to improve public health. The objective of breeding for improved nutritional quality is to enhance protein and oil content, vitamin content and micronutrient and mineral content. For example, Atlas-66 is a protein rich wheat variety.
Which of the following is present between the adjacent bones of the vertebral column?
Solution: Cartilage is a flexible connective tissue present between the adjacent bones of the vertebral column. It provides a cushioning effect and reduces friction between the bones during movement.
Nitrogenous waste is excreted in the form of pellet or paste by
Solution: $Pavo$ belongs to class Aves and thus excretes nitrogenous waste as uric acid in the form of a pellet or paste with minimum loss of water. This mode of excretion is called uricotelism.
Which of the following statements with respect to endoplasmic reticulum is incorrect?
Solution: Prokaryotes are organisms that do not have a nuclear membrane and other membrane bound organelles. Thus, prokaryotes do not have RER or SER at all, so the statement that only RER are present in prokaryotes is incorrect.
Which of the following statements are true for spermatogenesis but do not hold true for oogenesis? I. It results in the formation of haploid gametes. II. Differentiation of gamete occurs after the completion of meiosis. III. Meiosis occurs continuously in a mitotically dividing stem cell population. IV. It is controlled by the Luteinising hormone (LH) and Follicle Stimulating Hormone (FSH) secreted by the anterior pituitary. V. It is initiated at puberty. Choose the most appropriate answer from the options given below.
Solution: Statements (II), (III) and (V) are true for spermatogenesis but incorrect for oogenesis. Oogenesis is the process of formation of the mature female gamete. Differentiation of the gamete will not occur after the completion of meiosis. Meiosis occurs continuously in a mitotically dividing egg mother cell or oogonia — this is not so in oogenesis, where the oogonia stop dividing well before birth. It is initiated even before birth, i.e. during the embryonic development stage, but is completed only after fertilisation — so it is not initiated at puberty. Statements (I) and (IV) hold for both processes.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) All vertebrates are chordates but all chordates are not vertebrates. Reason (R) Notochord is replaced by vertebral column in the adult vertebrates. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both (A) and (R) are correct and (R) is the correct explanation of (A). The characteristic feature of phylum Chordata includes the presence of a notochord and paired pharyngeal gill slits. In sub-phylum Vertebrata, the notochord present in the embryo gets replaced by a cartilaginous or bony vertebral column in adults. But in cephalochordates, notochord persists throughout life as such, and in urochordates notochord is present only in larval stages and absent in adults. Thus, it can be said that all vertebrates are chordates but all chordates are not vertebrates.
In which of the following animals, digestive tract has additional chambers like crop and gizzard?
Solution: The digestive tract of animals like $Pavo$, $Psittacula$ and $Corvus$ has additional chambers like crop and gizzard. These animals belong to class Aves. Animals of this class (birds) store their food in the crop after swallowing. Then food is passed to the true stomach or proventriculus. The food then gets passed to the gizzard and then back to the proventriculus. The function of the gizzard is to chew and grind the food.
Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) Osteoporosis is characterised by decreased bone mass and increased chances of fractures. Reason (R) Common cause of osteoporosis is increased levels of oestrogen. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: (A) is correct, but (R) is not correct. Osteoporosis is characterised by decreased bone mass and increased chances of fractures. It results in weakness of skeletal muscles. It is caused due to excessive reabsorption of calcium and phosphorus from bone. The DECREASED level of oestrogen is the common cause of this, not an increased level.
In gene therapy of Adenosine Deaminase (ADA) deficiency, the patient requires periodic infusion of genetically engineered lymphocytes because
Solution: ADA (Adenosine deaminase) deficiency is a rare inherited disorder of purine metabolism, characterised by the accumulation of metabolic substrates that lead to hampering of the functioning of the immune system and cause Severe Combined Immunodeficiency (SCID). Gene therapy is the replacement of a defective or faulty gene with a normal or healthy gene to correct a genetic disorder. In gene therapy of ADA deficiency, the patient requires periodic infusion of genetically engineered lymphocytes because these cells are not immortal.
Select the incorrect statement with reference to mitosis.
Solution: Statement (a) is incorrect with reference to mitosis, because spindle fibres attach to the KINETOCHORES of chromosomes in metaphase, not directly to the centromere. The other three statements are correct descriptions of mitosis.
In the taxonomic categories, which hierarchical arrangement in ascending order is correct in case of animals ?
Solution: The hierarchical arrangement given in option (d) is correct in case of animals. Thus, the correct arrangement is Kingdom $\rightarrow$ Phylum $\rightarrow$ Class $\rightarrow$ Order $\rightarrow$ Family $\rightarrow$ Genus $\rightarrow$ Species
Given below are two statements. Statement I Restriction endonucleases recognise specific sequence to cut DNA known as palindromic nucleotide sequence. Statement II Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both statement I and statement II are correct. Each restriction endonuclease functions by 'inspecting' the length of a DNA sequence. Once it finds its specific recognition sequence, it will bind to the DNA and cut each of the two strands of the double helix at specific points in their sugar-phosphate backbones. Restriction enzymes cut the strand of DNA a little away from the centre of the palindromic site, but between the same two bases on the opposite strands. This leaves single stranded portions at the ends, known as sticky ends.
Given below are two statements. Statement I $Mycoplasma$ can pass through less than 1 micron filter size. Statement II $Mycoplasma$ are bacteria with cell wall. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is correct but statement II is incorrect. $Mycoplasma$ are organisms that completely LACK a cell wall. These organisms are often called PPLOs (Pleuropneumonia like Organisms). These are the smallest free-living microorganisms (0.1 – 0.5 $\mu$m) and are so small that they can easily pass through bacteria-proof filters.
Which of the following is not a connective tissue?
Solution: Neuroglia cells are part of NEURAL tissue, not connective tissue. These are the supporting cells which form a packing around the neurons in the brain, spinal cord and ganglia. Neuroglia makes up more than one half of the volume of neural tissue in our body. Adipose tissue, cartilage and blood are all connective tissues.
Lippes loop is a type of contraceptive used as
Solution: Lippes loop is a type of contraceptive used as a non-medicated IUD. IUDs (Intra Uterine Devices) contain either copper or progesterone and are inserted by doctors in the uterus through the vagina; the Lippes loop carries neither, so it is the non-medicated type.
At which stage of life, the oogenesis process is initiated?
Solution: Oogenesis is the process of formation of a mature female gamete. The production of eggs in the female begins before birth, that is during the embryonic development stage, but is completed only after fertilisation.
Identify the microorganism which is responsible for the production of an immunosuppressive molecule cyclosporin-A.
Solution: Cyclosporin-A is a bioactive molecule that is produced by the microorganism $Trichoderma$ $polysporum$ (a fungus). Cyclosporin-A is used as an immunosuppressive agent in organ transplant patients.
Given below are two statements. Statement I Fatty acids and glycerols cannot be absorbed into the blood. Statement II Specialised lymphatic capillaries called lacteals carry chylomicrons into lymphatic vessels and ultimately into the blood. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Both statement I and statement II are correct. Fatty acids and glycerol, being insoluble in water, cannot be absorbed into the blood directly. They are first modified into small droplets called micelles which move into intestinal mucosal cells. Lacteal is a lymphatic vessel of the small intestine which absorbs digested fat. Lacteals carry chylomicrons into lymphatic vessels and ultimately into the blood.
In an $E.coli$ strain, $i$ gene gets mutated and its product cannot bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome?
Solution: The $i$ gene synthesises the repressor of the operon. The repressor binds to the operator region of the operon, preventing RNA polymerase from transcribing the operon, and is inactivated by interaction with the inducer such as lactose and allolactose. Now, if the $i$ gene gets mutated and its product cannot bind the inducer molecule, then the repressor will remain activated, thus blocking transcription. The repressor protein binds to the operator region of the operon and as a result prevents RNA polymerase from binding to the operon. Hence, $z$, $y$ and $a$ genes will not be translated.
Tegmina in cockroach, arises from
Solution: The first pair of wings arises from the mesothorax and the second pair arises from the metathorax. The first pair of wings are called tegmina, which are opaque, dark and leathery. Tegmina encloses the hindwings, which are transparent, membranous and are used in flight.
Which of the following is not the function of conducting part of respiratory system?
Solution: The conducting part of the respiratory system is responsible for providing passage for air. It conditions the incoming air by warming, moistening and cleaning it. It consists of nasopharynx, larynx, trachea, bronchi, bronchioles and terminal bronchioles. The respiratory portion, on the other hand, is responsible for providing surface for diffusion of O$_2$ and CO$_2$ — so that is not a function of the conducting part.
Which of the following is a correct match for disease and its symptoms?
Solution: Arthritis is a disease causing painful inflammation and stiffness of the joints, so (d) is correctly matched. The other three match pairs can be corrected as: Tetany — LOW Ca$^{2+}$ level causing rapid spasms. Myasthenia gravis — AUTOIMMUNE disorder resulting in weakening and paralysis of skeletal muscle. Muscular dystrophy — a GENETIC disorder causing progressive degeneration of skeletal muscle.
Regarding meiosis, which of the statements is incorrect?
Solution: Meiosis is a type of cell division that produces haploid gametes from diploid cells. DNA replication occurs in the S-phase of interphase BEFORE meiosis-I begins, to produce identical sister chromatids — there is no S-phase before meiosis-II. So that statement is incorrect. The other three statements are correct.
Detritivores breakdown detritus into smaller particles. This process is called
Solution: Fragmentation is a process that causes breakdown of detritus into smaller particles. It is an initial stage of decomposition. In this process, detritivores feed on the dead organic matter, turning it into smaller fragments.
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs?
Solution: Given, length of DNA molecule $= 1.1$ metres. As we know, the length of the DNA in a human diploid cell is 2.2 metres, and the number of base pairs in a human genome (diploid cell) is $6.6 \times 10^{9}$ bp. Thus, 2.2 m of DNA $= 6.6 \times 10^{9}$ bp. Hence, 1.1 m of DNA $$= \frac{6.6 \times 10^{9}}{2} = 3.3 \times 10^{9}\ \text{bp}$$
Which of the following functions is not performed by secretions from salivary glands?
Solution: The saliva secreted into the oral cavity contains electrolytes and two enzymes, salivary amylase and lysozyme. Starch is hydrolysed in the oral cavity by the action of salivary amylase into a DISACCHARIDE, i.e. maltose — so saliva produces disaccharides rather than digesting them. Lysozyme acts as an antibacterial agent and kills bacteria, thus preventing infection.
A dehydration reaction links two glucose molecules to produce maltose. If the formula for glucose is C$_6$H$_{12}$O$_6$ then what is the formula for maltose?
Solution: In a dehydration reaction, when two molecules are linked together, one molecule of water (H$_2$O) is removed. $$\text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6 \longrightarrow \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O}$$ So the formula for maltose is C$_{12}$H$_{22}$O$_{11}$.
If '8' $Drosophila$ in a laboratory population of '80' died during a week, the death rate in the population is ............ individuals per $Drosophila$ per week.
Solution: Number of $Drosophila$ in laboratory $= 80$. Number of $Drosophila$ died in a week $= 08$. $$\text{Death rate} = \frac{\text{Number of individuals died}}{\text{Total number of individuals}} = \frac{8}{80} = 0.1$$ The death rate in the population is 0.1 individuals per $Drosophila$ per week.
Given below are two statements. Statement I In a scrubber, the exhaust from the thermal plant is passed through the electric wires to charge the dust particles. Statement II Particulate matter (Pm 2.5) cannot be removed by scrubber but can be removed by an electrostatic precipitator. In the light of the above statements, choose the most appropriate answer from the options given below.
Solution: Statement I is incorrect, but statement II is correct. In a scrubber the exhaust from the thermal plant is passed through a spray of lime or water, not through electric wires. Water dissolves the gases and lime reacts with SO$_2$ to form a precipitate of calcium sulphate and sulphide. Charging dust particles with electrode wires is what an electrostatic precipitator does, and that is the device which removes fine particulate matter.
Which of the following is a correct statement?
Solution: Cyanobacteria or blue-green algae are Gram positive autotrophic prokaryotes which perform oxygenic photosynthesis. Cyanobacteria are classified under kingdom Monera. The other three statements can be corrected as: bacteria are not exclusively heterotrophic, they can be photosynthetic or chemosynthetic autotrophs; slime moulds are saprophytic organisms classified under kingdom Protista; and $Mycoplasma$ have DNA and ribosomes but LACK a cell wall.
Ten $E.coli$ cells with $^{15}$N-$ds$DNA are incubated in medium containing $^{14}$N nucleotide. After 60 minutes, how many $E.coli$ cells will have DNA totally free from $^{15}$N?
Solution: Each division of $E.coli$ takes 20 minutes. Every 20 minutes the 2 strands of DNA get separated and a new DNA strand is synthesised using the $^{14}$N medium. After 40 minutes, we have 2 hybrid density DNA and 2 light density DNA. After another 20 minutes, i.e. 60 minutes, each of these four DNA will synthesise a new DNA strand. Hence, after 60 minutes we have 2/8 DNA of hybrid density ($^{15}$N – $^{14}$N) and the rest 6/8 DNA of light density ($^{14}$N – $^{14}$N). Thus, 1 $E.coli$ will form 6 $E.coli$ cells that are totally free from $^{15}$N. Thus, 10 $E.coli$ will produce $10 \times 6 = 60$ $E.coli$ cells free from $^{15}$N DNA.
Select the incorrect statement with respect to acquired immunity.
Solution: The statement given in option (c) is incorrect and can be corrected as: INNATE immunity is the non-specific type of defence mechanism present at the time of birth. Acquired immunity is also called specific immunity. It is a type of immunity that develops when a person's immune system responds to a foreign substance or microbe, or that occurs after a person receives antibodies from another source.
Statements related to human insulin are given below. Which statement(s) is/are correct about genetically engineered insulin? I. Pro-hormone insulin contains extra stretch of C-peptide. II. A-peptide and B-peptide chains of insulin were produced separately in $E.coli$, extracted and combined by creating disulphide bond between them. III. Insulin used for treating diabetes was extracted from cattles and pigs. IV. Pro-hormone insulin needs to be processed for converting into a mature and functional hormone. V. Some patients develop allergic reactions to the foreign insulin. Choose the most appropriate answer from the options given below.
Solution: The statement given in option (II) is most appropriately correct, as insulin consists of two short polypeptide chains, i.e. chain 'A' and 'B', that are linked together by a disulphide bridge. In the genetically engineered route, the A and B chains were produced separately in $E.coli$, extracted and then combined by creating the disulphide bond between them.
Which of the following are not the effects of parathyroid hormone? I. Stimulates the process of bone resorption. II. Decreases Ca$^{2+}$ level in blood. III. Reabsorption of Ca$^{2+}$ by renal tubules. IV. Decreases the absorption of Ca$^{2+}$ from digested food. V. Increases metabolism of carbohydrates. Choose the most appropriate answer from the options given below.
Solution: Parathyroid hormone increases serum calcium by increasing bone reabsorption of calcium into blood. It increases renal reabsorption of calcium. Thus, PTH helps to prevent low calcium levels by acting on bones, intestine and kidneys. So statements II, IV and V — decreasing blood Ca$^{2+}$, decreasing absorption of Ca$^{2+}$ from digested food, and increasing carbohydrate metabolism — are not effects of parathyroid hormone.
Select the incorrect statement regarding synapses.
Solution: Statement (c) is incorrect and can be corrected as: impulse transmission across an ELECTRICAL synapse is always faster than that across a chemical synapse. In an electrical synapse the membranes are in close proximity and current flows directly, so there is no synaptic delay; a chemical synapse has to release and diffuse a neurotransmitter, which takes longer.
Match List-I with List-II. List-I: A. Bronchioles B. Goblet cell C. Tendons D. Adipose tissue List-II: (i) Dense regular connective tissue (ii) Loose connective tissue (iii) Glandular tissue (iv) Ciliated epithelium Choose the correct answer from the options given below.
Solution: If the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and Fallopian tubes. Columnar or cuboidal cells get specialised for secretion and are called glandular epithelium. They are mainly of two types: unicellular, consisting of isolated glandular cells (goblet cells), and multicellular, consisting of a cluster of cells (salivary gland). Tendons, which attach skeletal muscles to bones, are an example of dense regular connective tissue. Adipose tissue is a type of loose connective tissue located beneath the skin. So (A)-(iv), (B)-(iii), (C)-(i), (D)-(ii).
The recombination frequency between the genes $a$ and $c$ is 5%, $b$ and $c$ is 15%, $b$ and $d$ is 9%, $a$ and $b$ is 20%, $c$ and $d$ is 24% and $a$ and $d$ is 29%. What will be the sequence of these genes on a linear chromosome?
Solution: Recombination frequency is proportional to the map distance between two genes, so the distances must add up along the chromosome. Placing $a$ first, $c$ lies 5% away. $b$ lies 15% beyond $c$, which puts it 20% from $a$ — matching the given $a$-$b$ distance. $d$ lies 9% beyond $b$, which puts it 24% from $c$ and 29% from $a$ — again matching the given values. Hence the sequence of genes on the linear chromosome is $a$, $c$, $b$, $d$.
If a colourblind female marries a man whose mother was also colourblind, what are the chances of her progeny having colour blindness?
Solution: Colour blindness is an X-linked recessive condition. This disease mainly affects male offspring. For a female to be born colour blind, the mother has to be at least a carrier and the father has to be colourblind. A colourblind female is denoted by the genotype X$^{c}$X$^{c}$. A man whose mother was colour blind will also be colourblind, X$^{c}$Y. Thus the cross is X$^{c}$X$^{c}$ $\times$ X$^{c}$Y, which gives X$^{c}$X$^{c}$ (colour blind female) and X$^{c}$Y (colour blind male) only. Hence 100% of the progeny will be colour blind.
Match List-I with List-II. List-I (Biological molecules): A. Glycogen B. Globulin C. Steroids D. Thrombin List-II (Biological functions): (i) Hormone (ii) Biocatalyst (iii) Antibody (iv) Storage product Choose the correct answer from the options given below.
Solution: Glycogen is a polysaccharide that serves as a form of stored energy in animals and fungi. Antibodies are immunoglobulins or gamma globulins that are large, Y-shaped proteins produced in response to foreign cells or antigens by plasma cells. Steroids are hormones that are produced by the adrenal cortex and testes. These are important for proper body functioning. Enzymes such as thrombin are biocatalytic proteins that catalyse biological reactions and are indispensable for metabolism of living beings. So (A)-(iv), (B)-(iii), (C)-(i), (D)-(ii).
Which of the following is not a desirable feature of a cloning vector?
Solution: There are three important characteristics of cloning vectors: (a) Origin of replication (b) Selectable markers (c) One or more unique restriction sites into which a DNA fragment can be inserted The site must be UNIQUE. If a vector carries two or more recognition sites for the same enzyme, the cut would generate several fragments and complicate the gene cloning, so that is not a desirable feature.
Which of the following statements is not true?
Solution: The statement in option (c) is incorrect and can be corrected as: flippers of penguins and dolphins are a pair of ANALOGOUS organs. They have similar function, but different origin. Analogous structures arise by convergent evolution, while homologous structures indicate common ancestry and arise by divergent evolution.
Which one of the following statements is correct?
Solution: Statement (b) is correct: during joint diastole, when all the chambers are relaxed, blood moves freely from the atrium into the ventricle. The rest can be corrected as follows. As the blood fills the chamber, the atria contract to push the blood through the valve, and this is called atrial systole — it is the stage when both bicuspid and tricuspid valves open. Increased ventricular pressure causes OPENING of the semilunar valves. The SINO-ATRIAL node generates an action potential to stimulate atrial contraction, not the AVN.
Match List-I with List-II with respect to methods of contraception and their respective actions. List-I: A. Diaphragms B. Contraceptive pills C. Intra Uterine Devices D. Lactational amenorrhea List-II: (i) Inhibit ovulation and implantation (ii) Increase phagocytosis of sperm within uterus (iii) Absence of menstrual cycle and ovulation following parturition (iv) They cover the cervix blocking the entry of sperms Choose the correct answer from the options given below.
Solution: Diaphragms are made up of rubber and are inserted into the female reproductive tract to cover the cervix during intercourse. They prevent conception by blocking entry of sperms through the cervix. Contraceptive pills alter or inhibit ovulation and implantation. They also modify the quality of cervical mucus to prevent or retard the entry of sperms. IUDs increase phagocytosis of sperms within the uterus and can also suppress sperm motility. Lactational amenorrhea is the absence of menstruation during the period of intense lactation following parturition. So (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii).
Full-length mock tests, an AI study workbook and a free rank predictor — built for NEET, JEE & boards.