Q1.
The electric field in a plane electromagnetic wave is given by
$$E_z = 60 \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{V/m}.$$
Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
- A. $B_y = 60 \sin\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$
- B. $B_y = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$ ✓
- C. $B_x = 2 \times 10^{-7} \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$
- D. $B_z = 60 \cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$
Solution: In an electromagnetic wave $E$ and $B$ are in the same phase and $B_0 = \dfrac{E_0}{c}$; their planes are perpendicular to each other.
$$\therefore\ B_y = \frac{60}{c}\cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$$
$$= \frac{60}{3 \times 10^{8}}\cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$$
$$B_y = 2 \times 10^{-7}\cos\left(5x + 1.5 \times 10^{9} t\right)\ \text{T}$$
Q2.
A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
- A. $2f$
- B. $\dfrac{f}{2}$
- C. $f$ ✓
- D. $\dfrac{3f}{2}$
Solution: Fundamental frequency of a pipe open at both ends
$$f = \frac{v}{2L} \qquad \ldots(i)$$
Now immersed in water, the open pipe behaves as a closed pipe of length $\dfrac{L}{2}$:
$$f' = \frac{v}{4\left(\dfrac{L}{2}\right)} = \frac{v}{2L} \qquad \ldots(ii)$$
Comparing (i) and (ii), $f = f'$.
Q3.
An electron (mass $9 \times 10^{-31}$ kg and charge $1.6 \times 10^{-19}$ C) moving with speed $c/100$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4}$ T perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c = 3 \times 10^{8}\ \text{ms}^{-1}$)
- A. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{4}\ \text{V m}^{-1}$
- B. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{4}\ \text{V m}^{-1}$
- C. $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^{2}\ \text{V m}^{-1}$ ✓
- D. $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^{2}\ \text{V m}^{-1}$
Solution: For no deflection of the electron, $\vec{F}_B = \vec{F}_E$.
$$-e\left(\vec{v} \times \vec{B}\right) = -e\vec{E}$$
$$\Rightarrow\ \vec{E} = \vec{v} \times \vec{B} \quad \Rightarrow\ \vec{E} \perp \vec{B}$$
$$E = vB = \frac{c}{100} \times 9 \times 10^{-4} = \frac{3 \times 10^{8}}{100} \times 9 \times 10^{-4}$$
$$= 27 \times 10^{2}\ \text{V m}^{-1}$$
Q4.
In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power ($p$) and magnification ($m$) for each lens will be, respectively
- A. $p^4$ and $m^4$
- B. $4p$ and $4m$
- C. $p^4$ and $4m$
- D. $4p$ and $m^4$ ✓
Solution: For a series combination of lenses in contact the powers add and the magnifications multiply:
$$p_{\text{eff}} = p_1 + p_2 + p_3 + p_4 = 4p$$
$$m_{\text{eff}} = m_1 \times m_2 \times m_3 \times m_4 = m^4$$
Q5.
A 2 amp current is flowing through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be
- A. 4 : 1
- B. 1 : 4 ✓
- C. 1 : 2
- D. 2 : 1
Solution: Magnetic moment of a current carrying circular loop is $M = IA$.
Since the current is the same in both coils, $M \propto A$.
$$\frac{M_1}{M_2} = \frac{A_1}{A_2} = \frac{\pi r_1^{2}}{\pi r_2^{2}} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}$$
Q6.
Two gases $A$ and $B$ are filled at the same pressure in separate cylinders with movable pistons of radius $r_A$ and $r_B$, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas $A$ and $B$ are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio $\dfrac{r_A}{r_B}$ is equal to
- A. $\dfrac{\sqrt{3}}{2}$
- B. $\dfrac{4}{3}$
- C. $\dfrac{3}{4}$ ✓
- D. $\dfrac{2}{\sqrt{3}}$
Solution: Using the first law of thermodynamics,
$$\Delta Q = \Delta U + P\Delta V$$
$\Delta Q$ is the same and $\Delta U$ is also the same, so $W_A = W_B$.
$$\therefore\ (P\Delta V)_A = (P\Delta V)_B$$
$P$ is also the same, so $A_A d_A = A_B d_B$:
$$\pi r_A^{2} d_A = \pi r_B^{2} d_B$$
$$\frac{r_A}{r_B} = \left(\frac{d_B}{d_A}\right)^{\frac{1}{2}} = \left(\frac{9}{16}\right)^{\frac{1}{2}} = \frac{3}{4}$$
Q7.
A container has two chambers of volumes $V_1 = 2$ litres and $V_2 = 3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_1 = 5$ and $n_2 = 4$ moles of ideal gas at pressures $p_1 = 1$ atm and $p_2 = 2$ atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of
- A. 1.8 atm
- B. 1.3 atm
- C. 1.6 atm ✓
- D. 1.4 atm
Solution: $$P_1V_1 + P_2V_2 = P\left(V_1 + V_2\right)$$
$$1(2) + 2(3) = P(2 + 3)$$
$$\frac{8}{5} = P$$
$$\Rightarrow\ P = 1.6\ \text{atm}$$
Q8.
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
- A. 124 earth days
- B. 88 earth days ✓
- C. 225 earth days
- D. 172 earth days
Solution: Applying Kepler's third law: $T^{2} \propto R^{3}$.
Radius of the Martian orbit, $R' = 4R$.
$$\left(\frac{T'}{T}\right)^{2} = \left(\frac{R'}{R}\right)^{3} = \left(\frac{4R}{R}\right)^{3} = 4^{3} = 64 \quad \Rightarrow \quad \frac{T'}{T} = 8$$
$$\therefore\ \text{Length of 1 year on Mercury} = T = \frac{T'}{8} = \frac{687}{8} = 85.88\ \text{days}$$
Q9.
To an ac power supply of 220 V at 50 Hz, a resistor of 20 $\Omega$, a capacitor of reactance 25 $\Omega$ and an inductor of reactance 45 $\Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively
- A. 15.6 A and 45°
- B. 7.8 A and 30°
- C. 7.8 A and 45° ✓
- D. 15.6 A and 30°
Solution: $X_L = 45\ \Omega$, $X_C = 25\ \Omega$, $R = 20\ \Omega$
$$I = \frac{220}{\sqrt{\left(X_L - X_C\right)^{2} + R^{2}}} = \frac{220}{\sqrt{(45 - 25)^{2} + 20^{2}}}$$
$$= \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} = 7.779\ \text{A}$$
$$\tan\phi = \frac{X_L - X_C}{R} = \frac{45 - 25}{20} = 1$$
$$\phi = 45^{\circ}$$
Q10.
A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
- A. $\dfrac{R}{8}$
- B. $\dfrac{R}{64}$
- C. $\dfrac{R}{32}$
- D. $\dfrac{R}{16}$ ✓
Solution: After being cut into 8 equal pieces,
$$\Rightarrow\ \text{Resistance of each piece} = R' = \frac{R}{8}$$
Each set has 4 pieces in parallel combination
$$\Rightarrow\ \text{Resistance of each set} = R'' = \frac{R'}{4} = \frac{R}{32}$$
Both sets are connected in series
$$\therefore\ R_{\text{eq}} = R'' + R'' = 2 \times \frac{R}{32} = \frac{R}{16}$$
Q11.
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
- A. $\dfrac{3F}{8}$ ✓
- B. $\dfrac{3F}{5}$
- C. $\dfrac{2F}{3}$
- D. $\dfrac{F}{2}$
Solution: Initially each sphere carries $q$ at separation $r$:
$$F = \frac{Kqq}{r^{2}}$$
The third (uncharged) sphere touches $A$: charge on $A$ becomes $\dfrac{q}{2}$ and the third sphere carries $\dfrac{q}{2}$.
It then touches $B$: the pair $\left(\dfrac{q}{2},\, q\right)$ shares to give $B$ a charge $\dfrac{3q}{4}$.
$$F' = \frac{K \cdot \dfrac{q}{2} \cdot \dfrac{3q}{4}}{r^{2}}$$
$$F' = \frac{3F}{8}$$
Q12.
Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at $x = 0.1$ cm when the jaws of Vernier callipers are closed.
If the main scale reading for the diameter is $M = 5$ cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is
- A. 5.00 cm
- B. 5.18 cm
- C. 5.08 cm
- D. 4.98 cm ✓
Solution: $$\text{Least count} = 1\,\text{MSD} - 1\,\text{VSD}$$
$$= 1\,\text{MSD} - \frac{9}{10}\text{MSD} = \frac{1}{10}\text{MSD}$$
$$= \frac{1}{10} \times 0.1\ \text{cm} = 0.01\ \text{cm}$$
Zero error $= +0.1$ cm
Main scale reading $= 5$ cm
Vernier scale reading $= 8 \times 0.01 = 0.08$ cm
Final measurement of diameter
$$= 5 + 0.08 - 0.1 = 4.98\ \text{cm}$$
Q13.
In some appropriate units, time ($t$) and position ($x$) relation of a moving particle is given by $t = x^{2} + x$. The acceleration of the particle is
- A. $+\dfrac{2}{2x + 1}$
- B. $-\dfrac{2}{\left(x + 2\right)^{3}}$
- C. $-\dfrac{2}{\left(2x + 1\right)^{3}}$ ✓
- D. $+\dfrac{2}{\left(x + 1\right)^{3}}$
Solution: $$t = x^{2} + x$$
$$\frac{dt}{dx} = 2x + 1$$
$$v = \frac{dx}{dt} = \frac{1}{\left(2x + 1\right)}$$
$$\frac{dv}{dx} = \frac{-2}{\left(2x + 1\right)^{2}}$$
$$a = v\frac{dv}{dx} = \frac{1}{\left(2x + 1\right)}\left[\frac{-2}{\left(2x + 1\right)^{2}}\right]$$
$$= -\frac{2}{\left(2x + 1\right)^{3}}$$
Q14.
Which of the following options represent the variation of photoelectric current with property of light shown on the $x$-axis?
A. Photoelectric current against intensity of light — a straight line through the origin
B. Photoelectric current against intensity of light — a horizontal line
C. Photoelectric current against frequency of light — a straight line through the origin
D. Photoelectric current against frequency of light — a straight line meeting the axis at a positive frequency
- A. B and D
- B. A only ✓
- C. A and C
- D. A and D
Solution: Photoelectric current is directly proportional to the intensity of light, so the correct graph is the straight line through the origin against intensity — graph A only.
Q15.
A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the $n^{\text{th}}$ orbit and the particle's speed $\nu$ in the orbit depend on $n$ as
- A. $r \propto n^{4/3}$; $\nu \propto n^{-1/3}$
- B. $r \propto n^{1/3}$; $\nu \propto n^{1/3}$
- C. $r \propto n^{1/3}$; $\nu \propto n^{2/3}$
- D. $r \propto n^{2/3}$; $\nu \propto n^{1/3}$ ✓
Solution: Given, the force is constant:
$$F = \frac{mv^{2}}{r}$$
$$\Rightarrow\ \frac{v^{2}}{r} = \text{constant}$$
$$\Rightarrow\ r \propto v^{2} \qquad \ldots(1)$$
and from Bohr's quantisation condition,
$$L = mvr = \frac{nh}{2\pi} \qquad \ldots(2)$$
Solving (1) and (2):
$$\nu \propto n^{1/3} \quad \text{and} \quad r \propto n^{2/3}$$
Q16.
A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:
- A. $\left(\dfrac{\sin\theta}{2 + 3\sin\theta}\right)^{\frac{1}{2}}$ ✓
- B. $\left(\sin\theta\right)^{\frac{1}{2}}$
- C. $\left(\dfrac{1}{2 + 3\sin\theta}\right)^{\frac{1}{2}}$
- D. $\left(\dfrac{\cos\theta}{2 + 3\sin\theta}\right)^{\frac{1}{2}}$
Solution: At point $P$, the string is slack, so the weight component along the string supplies the centripetal force:
$$mg\sin\theta = \frac{mv^{2}}{l} \qquad \ldots(1)$$
By conservation of mechanical energy between $Q$ and $P$:
$$\frac{1}{2}mv_0^{2} = \frac{1}{2}mv^{2} + mg\left(l + l\sin\theta\right)$$
$$\frac{v_0^{2}}{2} = \frac{v^{2}}{2} + gl\left(1 + \sin\theta\right)$$
Put $gl = \dfrac{v^{2}}{\sin\theta}$ using (1):
$$\frac{v_0^{2}}{2} = \frac{v^{2}}{2} + \frac{v^{2}}{\sin\theta}\left(1 + \sin\theta\right)$$
$$\frac{v_0^{2}}{2} = \frac{v^{2}}{2} + \frac{v^{2}}{\sin\theta} + v^{2}$$
$$v_0^{2} = v^{2}\left[3 + \frac{2}{\sin\theta}\right]$$
$$\frac{v}{v_0} = \left(\frac{\sin\theta}{3\sin\theta + 2}\right)^{\frac{1}{2}}$$
Q17.
A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is filled in it) has small area $A$. It is filled with a gas of density $\rho$ and takes a spherical shape of radius $R$. When the gas is allowed to flow freely out of it, its radius $r$ changes from $R$ to 0 (zero) in time $T$. If the speed $v(r)$ of gas coming out of the balloon depends on $r$ as $r^{a}$ and $T \propto S^{\alpha} A^{\beta} \rho^{\gamma} R^{\delta}$ then
- A. $a = \dfrac{1}{2}, \alpha = \dfrac{1}{2}, \beta = -\dfrac{1}{2}, \gamma = \dfrac{1}{2}, \delta = \dfrac{7}{2}$
- B. $a = \dfrac{1}{2}, \alpha = \dfrac{1}{2}, \beta = -1, \gamma = +1, \delta = \dfrac{3}{2}$
- C. $a = -\dfrac{1}{2}, \alpha = -\dfrac{1}{2}, \beta = -1, \gamma = -\dfrac{1}{2}, \delta = \dfrac{5}{2}$
- D. $a = -\dfrac{1}{2}, \alpha = -\dfrac{1}{2}, \beta = -1, \gamma = \dfrac{1}{2}, \delta = \dfrac{7}{2}$ ✓
Solution: $$T \propto S^{\alpha} A^{\beta} \rho^{\gamma} R^{\delta}$$
$$\text{M}^{0}\text{L}^{0}\text{T}^{1} = K\left(\text{MT}^{-2}\right)^{\alpha}\left(\text{L}^{2}\right)^{\beta}\left(\text{ML}^{-3}\right)^{\gamma}\text{L}^{\delta}$$
$$\text{M}^{0}\text{L}^{0}\text{T}^{1} = K\left[\text{M}^{\alpha + \gamma}\,\text{L}^{2\beta - 3\gamma + \delta}\,\text{T}^{-2\alpha}\right]$$
Comparing powers:
$$-2\alpha = 1 \ \Rightarrow\ \alpha = -\frac{1}{2}$$
$$\alpha + \gamma = 0 \ \Rightarrow\ \gamma = \frac{1}{2}$$
$$2\beta - 3\gamma + \delta = 0 \ \Rightarrow\ 2\beta - 3\left(\frac{1}{2}\right) + \delta = 0$$
Taking $\beta = -1$ from the options,
$$2(-1) - \frac{3}{2} + \delta = 0 \qquad \therefore\ \delta = \frac{7}{2}$$
Q18.
A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is
- A. 250
- B. 100
- C. 125 ✓
- D. 150
Solution: $$m = \frac{L}{f_o} \times \frac{D}{f_e}$$
$$= \frac{40}{2} \times \frac{25}{4}$$
$$m = 125$$
Q19.
Two identical point masses $P$ and $Q$, suspended from two separate massless springs of spring constants $k_1$ and $k_2$, respectively, oscillate vertically. If their maximum speeds are the same, the ratio $\left(A_Q/A_P\right)$ of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ is
- A. $\sqrt{\dfrac{k_1}{k_2}}$ ✓
- B. $\dfrac{k_2}{k_1}$
- C. $\dfrac{k_1}{k_2}$
- D. $\sqrt{\dfrac{k_2}{k_1}}$
Solution: Maximum velocity $V = A\omega$.
Given $v_P = v_Q$:
$$A_P\omega_P = A_Q\omega_Q$$
$$\frac{A_Q}{A_P} = \frac{\omega_P}{\omega_Q} \qquad \left(\omega = \sqrt{\frac{k}{m}}\right)$$
$$= \sqrt{\frac{k_P}{m_P} \cdot \frac{m_Q}{k_Q}}$$
The masses are identical, so
$$= \sqrt{\frac{k_1}{k_2}}$$
Q20.
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
- A. Zero between the plates and non-zero outside
- B. Zero at all places
- C. Constant between the plates and zero outside the plates
- D. Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates ✓
Solution: Let the surface charge density be $\sigma = \dfrac{q}{A}$.
Given $\dfrac{d\sigma}{dt} = \text{constant}$
$$\therefore\ \frac{d}{dt}\left(\frac{q}{A}\right) = \text{constant} \ \Rightarrow\ \frac{1}{A}i = \text{constant}$$
It means the displacement current is constant.
This system will act like a cylindrical wire: the magnetic field grows with $r$ inside the plate region and falls off outside, so $B$ is non-zero everywhere and is maximum at the imaginary cylindrical surface joining the peripheries of the plates.
Q21.
An electric dipole with dipole moment $5 \times 10^{-6}$ Cm is aligned with the direction of a uniform electric field of magnitude $4 \times 10^{5}$ N/C. The dipole is then rotated through an angle of $60^{\circ}$ with respect to the electric field. The change in the potential energy of the dipole is:
- A. 1.5 J
- B. 0.8 J
- C. 1.0 J ✓
- D. 1.2 J
Solution: Given
$$\left|\vec{P}\right| = 5 \times 10^{-6}\ \text{C m}, \qquad \left|\vec{E}\right| = 4 \times 10^{5}\ \text{N/C}$$
$$\theta_i = 0^{\circ} \quad \text{and} \quad \theta_f = 60^{\circ}$$
$$\Delta U = U_f - U_i = -PE\cos\theta_f + PE\cos\theta_i = PE\left[\cos\theta_i - \cos\theta_f\right]$$
$$= 5 \times 10^{-6} \times 4 \times 10^{5}\left[1 - \frac{1}{2}\right]$$
$$= 10 \times 10^{-6} \times 10^{5} = 1\ \text{J}$$
Q22.
There are two inclined surfaces of equal length ($L$) and same angle of inclination $45^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction ($\mu_k$) between the object and the rough surface is close to
- A. 0.75 ✓
- B. 0.25
- C. 0.40
- D. 0.5
Solution: $$t_{\text{rough}} = 2\,t_{\text{smooth}}$$
$$a_{\text{smooth}} = g\sin\theta$$
$$t \propto \frac{1}{\sqrt{a}} \ \Rightarrow\ t_{\text{smooth}} \propto \frac{1}{\sqrt{g\sin\theta}}$$
$$a_{\text{rough}} = g\sin\theta - \mu_k g\cos\theta$$
$$\frac{t_{\text{rough}}}{t_{\text{smooth}}} = \frac{\sqrt{\sin\theta}}{\sqrt{\sin\theta - \mu_k\cos\theta}} = 2$$
Squaring both sides,
$$\frac{\sin\theta}{\sin\theta - \mu_k\cos\theta} = 4 \ \Rightarrow\ \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}} - \mu_k \times \dfrac{1}{\sqrt{2}}} = 4$$
$$\Rightarrow\ 1 - \mu_k = \frac{1}{4}$$
$$\mu_k = \frac{3}{4} = 0.75$$
Q23.
De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052$ nm)
- A. 2.67 nm
- B. 0.067 nm
- C. 0.67 nm ✓
- D. 1.67 nm
Solution: $$r = 0.052\,n^{2}$$
For $n = 2$:
$$r = 0.052 \times 4 = 0.208\ \text{nm}$$
From Bohr's quantisation condition,
$$Mvr = \frac{nh}{2\pi}$$
$$\lambda = \frac{h}{Mv} = \pi r$$
$$= 3.14 \times 0.208\ \text{nm} = 0.65317\ \text{nm} \approx 0.67\ \text{nm}$$
Q24.
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
- A. 108 days ✓
- B. 100 days
- C. 105 days
- D. 115 days
Solution: Assuming the Sun to be a solid sphere, $I = \dfrac{2}{5}mR^{2}$.
Using conservation of angular momentum, $I'\omega' = I\omega$:
$$\Rightarrow\ \frac{2}{5}m(2R)^{2} \times \frac{2\pi}{T'} = \frac{2}{5}mR^{2} \times \frac{2\pi}{T}$$
$$\Rightarrow\ T' = 4T = 4 \times 27 = 108\ \text{days}$$
Q25.
A physical quantity $P$ is related to four observations $a$, $b$, $c$ and $d$ as follows:
$$P = a^{3}b^{2}\,/\,c\sqrt{d}$$
The percentage errors of measurement in $a$, $b$, $c$ and $d$ are 1%, 3%, 2%, and 4% respectively. The percentage error in the quantity $P$ is
- A. 15%
- B. 10%
- C. 2%
- D. 13% ✓
Solution: Maximum percentage error in $P$:
$$\frac{\Delta P}{P} \times 100 = 3\left(\frac{\Delta a}{a} \times 100\right) + 2\left(\frac{\Delta b}{b} \times 100\right) + \left(\frac{\Delta c}{c} \times 100\right) + \frac{1}{2}\left(\frac{\Delta d}{d} \times 100\right)$$
$$= 3 \times (1) + 2 \times (3) + (2) + \frac{1}{2} \times (4)$$
$$= 13\%$$
Q26.
The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\dfrac{3}{8}d$ and $\dfrac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.
If $K_1 = 1.25\,K_2$, the value of $K_1$ is:
- A. 1.33
- B. 2.66 ✓
- C. 2.33
- D. 1.60
Solution: Using
$$C_{eq} = \frac{\varepsilon_0 A}{\dfrac{t_1}{K_1} + \dfrac{t_2}{K_2} + \dfrac{t_3}{K_3}}$$
here $C_0 = \dfrac{\varepsilon_0 A}{d}$, $t_1 = \dfrac{3d}{8}$, $t_2 = \dfrac{d}{2}$, $t_3 = \dfrac{d}{8}$
$$K_1 = K_1, \qquad K_2 = \frac{K_1}{1.25} \quad \text{and} \quad K_3 = 1$$
Given $C_{eq} = 2C_0$:
$$\Rightarrow\ 2C_0 = \frac{\varepsilon_0 A}{\dfrac{3d}{8K_1} + \dfrac{d \times 1.25}{2K_1} + \dfrac{d}{8}}$$
$$\Rightarrow\ \frac{2\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{\dfrac{3d}{8K_1} + \dfrac{d}{2K_1} \times \dfrac{5}{4} + \dfrac{d}{8}}$$
$$\Rightarrow\ 2 = \frac{1}{\dfrac{3}{8K_1} + \dfrac{5}{8K_1} + \dfrac{1}{8}} \ \Rightarrow\ K_1 = \frac{8}{3} = 2.66$$
Q27.
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8$ m/s$^{2}$)
- A. 84 N s
- B. 21 N s ✓
- C. 7 N s
- D. 0
Solution: Speed just before impact:
$$v_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 40} = \sqrt{784} = 28\ \text{m s}^{-1}$$
Speed just after impact:
$$v_2 = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14\ \text{m s}^{-1}$$
$$\text{Impulse} = \Delta\vec{p} = m\left(\vec{v}_f - \vec{v}_i\right) = m\left(\vec{v}_2 - \vec{v}_1\right)$$
$$= \frac{1}{2}\left(14 - (-28)\right) = 21\ \text{N s}$$
Q28.
Two cities $X$ and $Y$ are connected by a regular bus service with a bus leaving in either direction every $T$ min. A girl is driving scooty with a speed of 60 km/h in the direction $X$ to $Y$ notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed (assumed constant) of the buses.
- A. 15 min, 120 km/h ✓
- B. 9 min, 40 km/h
- C. 25 min, 100 km/h
- D. 10 min, 90 km/h
Solution: Let the velocity of the bus be $v$ km/hr.
For $X \to Y$ (same direction), the relative velocity of the bus w.r.t. the scooty is $(v - 60)$, and the distance between 2 consecutive buses is $vT$:
$$(v - 60)\,30 = vT \qquad \ldots(i)$$
For $Y \to X$ (opposite direction):
$$(v + 60)\,10 = vT \qquad \ldots(ii)$$
Equating (i) and (ii):
$$(v - 60)\,30 = (v + 60)\,10$$
$$\therefore\ v = 120\ \text{km/hr}$$
$$T = 15\ \text{min}$$
Q29.
An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27$^{\circ}$C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:
[Given, $R = \dfrac{100}{12}\ \text{J mol}^{-1}\text{K}^{-1}$, and molecular mass of O$_2$ = 32, 1 atm pressure $= 1.01 \times 10^{5}$ N/m$^{2}$]
- A. 0.156 kg
- B. 0.125 kg
- C. 0.144 kg
- D. 0.116 kg ✓
Solution: Number of moles left (the absolute pressure is $11 + 1 = 12$ atm):
$$n = \frac{PV}{RT} = \frac{12 \times 1.01 \times 10^{5}\ \text{N/m}^{2} \times 30 \times 10^{-3}\ \text{m}^{3}}{\dfrac{100}{12} \times 300}$$
$$n = \frac{12 \times 1.01 \times 12}{10} = 14.54\ \text{moles}$$
Moles removed $= 18.2 - 14.54 = 3.656$ moles
Mass removed $= 3.656 \times 32 = 116.99\ \text{g} = 0.116\ \text{kg}$
Q30.
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
- A. $\omega(t)$ decreasing with $t$; $A(t)$ decreasing with $t$
- B. $\omega(t)$ increasing with $t$; $A(t)$ constant with $t$
- C. $\omega(t)$ increasing with $t$; $A(t)$ decreasing with $t$ ✓
- D. $\omega(t)$ increasing with $t$; $A(t)$ increasing with $t$
Solution: At any point of time, the time period is given by
$$T = 2\pi\sqrt{\frac{m}{k}}$$
Here $m$ is decreasing, so the time period $T$ will be decreasing.
Since $\omega = \dfrac{2\pi}{T}$, as mass leaks $\omega$ will increase.
Now, at any instant $mg = kx_0$, so the equilibrium length is
$$x_0 = \frac{mg}{k}$$
where $m$ is decreasing. So the equilibrium length will decrease, and the amplitude also goes on decreasing.
Q31.
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron)
- A. $\dfrac{heB}{2\pi m}$
- B. $\dfrac{he}{\pi m}$
- C. $\dfrac{he}{2\pi m}$ ✓
- D. $\dfrac{heB}{\pi m}$
Solution: The magnetic force provides the centripetal force:
$$evB = \frac{mv^{2}}{r} \ \Rightarrow\ v = \frac{eBr}{m}$$
The flux condition $\phi = BA$ gives
$$\frac{nh}{e} = B\pi r^{2} \ \Rightarrow\ Br^{2} = \frac{nh}{e\pi}$$
The magnetic moment is
$$\mu = IA = \frac{e}{T}\pi r^{2} = \frac{e \times v}{2\pi r}\pi r^{2} = \frac{evr}{2}$$
$$= \frac{1}{2}e \times \frac{eBr}{m} \times r = \frac{1}{2}e^{2}\frac{Br^{2}}{m}$$
$$\mu = \frac{1}{2}e^{2}\frac{nh}{e\pi m} = \frac{neh}{2\pi m}$$
For $n = 1$:
$$\mu = \frac{eh}{2\pi m}$$
Q32.
A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :
- A. 36 N
- B. 16 N
- C. 27 N ✓
- D. 32 N
Solution: $W = mg$ and $g = \dfrac{GM}{R^{2}}$, $g_h = \dfrac{GM}{(R + h)^{2}}$
$$\Rightarrow\ \frac{W_h}{W} = \frac{mg_h}{mg} = \frac{g_h}{g} = \frac{R^{2}}{(R + h)^{2}} \qquad \left(h = \frac{R}{3}\right)$$
$$\Rightarrow\ \frac{W_h}{W} = \frac{R^{2}}{\left(R + \dfrac{R}{3}\right)^{2}} = \frac{R^{2}}{\left(\dfrac{4R}{3}\right)^{2}} = \frac{9}{16}$$
$$\Rightarrow\ W_h = \frac{9}{16}W = \frac{9}{16} \times 48 = 27\ \text{N}$$
Q33.
The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at $22.5^{\circ}$ from the polarization axis of one of the polaroids, is ($I_0$ is the intensity of polarised light after passing through the first polaroid):
- A. $\dfrac{I_0}{16}$
- B. $\dfrac{I_0}{2}$
- C. $\dfrac{I_0}{4}$
- D. $\dfrac{I_0}{8}$ ✓
Solution: After the middle polaroid, set at $\dfrac{45}{2}$ degrees:
$$I_1 = I_0\cos^{2}\left(\frac{45}{2}\right)$$
The third polaroid is crossed with the first, so it makes $\left(90 - \dfrac{45}{2}\right)$ with the middle one:
$$I_2 = I_1\cos^{2}\left(90 - \frac{45}{2}\right) = I_0\cos^{2}\left(\frac{45}{2}\right)\sin^{2}\left(\frac{45}{2}\right)$$
$$= \frac{I_0}{4}\left(4\cos^{2}\left(\frac{45}{2}\right)\sin^{2}\left(\frac{45}{2}\right)\right)$$
$$= \frac{I_0}{4}\sin^{2}45^{\circ} = \frac{I_0}{8}$$
Q34.
A photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\lambda_{\text{photon}}/\lambda_{\text{electron}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)
- A. $\dfrac{1}{c}\sqrt{\dfrac{E}{2m}}$
- B. $\sqrt{\dfrac{E}{2m}}$
- C. $c\sqrt{2mE}$
- D. $c\sqrt{\dfrac{2m}{E}}$ ✓
Solution: For the photon,
$$E = \frac{hc}{\lambda_{\text{Ph}}} \ \Rightarrow\ \lambda_{\text{Ph}} = \frac{hc}{E}$$
For the electron, with $p$ the momentum,
$$E = \frac{p^{2}}{2m} = \left(\frac{h}{\lambda_e}\right)^{2} \times \frac{1}{2m}$$
$$\Rightarrow\ \lambda_e = \frac{h}{\sqrt{2mE}}$$
$$\therefore\ \frac{\lambda_{\text{Ph}}}{\lambda_e} = \frac{\dfrac{hc}{E}}{\dfrac{h}{\sqrt{2mE}}} = c\sqrt{\frac{2m}{E}}$$
Q35.
An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then
- A. Transmitted light is completely polarized with angle of refraction close to $30^{\circ}$
- B. Reflected light is completely polarized and the angle of reflection is close to $60^{\circ}$ ✓
- C. Reflected light is partially polarized and the angle of reflection is close to $30^{\circ}$
- D. Both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to $60^{\circ}$ and $30^{\circ}$, respectively
Solution: Using Brewster's law,
$$\mu = \tan\theta_P$$
$$\Rightarrow\ 1.73 = \tan\theta_P$$
$$\Rightarrow\ \sqrt{3} = \tan\theta_P$$
$$\Rightarrow\ \theta_P = 60^{\circ}$$
At this polarising angle, reflected light is perfectly polarized and transmitted light is partially polarised.
Q36.
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take $g = 10$ m/s$^{2}$)
- A. $200\sqrt{3}$ N
- B. 100 N
- C. $100\sqrt{3}$ N ✓
- D. 200 N
Solution: For translational equilibrium,
$$N_1 = Mg, \qquad N_2 = f$$
For rotational equilibrium, taking torque about $A$:
$$Mg\frac{L}{2}\cos\theta = N_2 L\sin\theta$$
$$\frac{Mg}{2}\cot\theta = N_2 = f$$
The rod makes $60^{\circ}$ with the wall, so it makes $30^{\circ}$ with the floor:
$$\frac{Mg}{2}\cot 30^{\circ} = f$$
$$\frac{Mg}{2}\sqrt{3} = N_2$$
$$100\sqrt{3} = f$$
Q37.
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1/T_2$ is
- A. $\dfrac{5}{4}$
- B. $\dfrac{3}{2}$
- C. $\dfrac{4}{3}$
- D. $\dfrac{5}{3}$ ✓
Solution: In series, $R_{\text{eq.}} = R_1 + R_2 + R_3$
$$= \frac{l}{2KA} + \frac{l}{KA} + \frac{l}{2KA} = \frac{4l}{2KA}$$
$$R_{\text{eq.}} = \frac{2l}{KA}$$
In series the rate of heat flow is the same:
$$\therefore\ \frac{3T - T_1}{R_1} = \frac{3T - T}{R_{\text{eq.}}}$$
$$\frac{(3T - T_1)2KA}{l} = \frac{(2T)KA}{2l}$$
$$\Rightarrow\ 6T - 2T_1 = T \ \Rightarrow\ 2T_1 = 5T \ \Rightarrow\ T_1 = \frac{5T}{2} \qquad \ldots(1)$$
Now equate the heat flow rate in the 3rd section and the total section:
$$\frac{T_2 - T}{R_3} = \frac{3T - T}{R_{\text{eq.}}}$$
$$\Rightarrow\ \frac{(T_2 - T)(2KA)}{l} = \frac{2T(KA)}{2l}$$
$$\Rightarrow\ 2T_2 - 2T = T \ \Rightarrow\ T_2 = \frac{3T}{2} \qquad \ldots(2)$$
By equations (1) and (2):
$$\frac{T_1}{T_2} = \frac{5T \times 2}{2 \times 3T} = \frac{5}{3}$$
Q38.
The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car $A$ stops after 1000 m and car $B$ stops after 1500 m. If $F_A$ and $F_B$ are the forces applied by the breaks on cars $A$ and $B$ respectively, then the ratio of $\dfrac{F_A}{F_B}$ is
- A. $\dfrac{1}{2}$
- B. $\dfrac{3}{2}$
- C. $\dfrac{2}{3}$ ✓
- D. $\dfrac{1}{3}$
Solution: By the work-energy theorem,
$$FS = \Delta K.E$$
$$\Rightarrow\ -FS = k_f - k_i \ \Rightarrow\ FS = k_i - k_f$$
Each car is brought to rest, so $k_f = 0$:
$$\Rightarrow\ \frac{F_A}{F_B} = \frac{k_A}{k_B} \times \frac{S_B}{S_A}$$
$$= \frac{100}{225} \times \frac{1500}{1000}$$
$$= \frac{150}{225} = \frac{2}{3}$$
Q39.
If the molar conductivity ($\Lambda_m$) of a 0.050 mol L$^{-1}$ solution of a monobasic weak acid is 90 S cm$^{2}$ mol$^{-1}$, its extent (degree) of dissociation will be
[Assume $\Lambda^{\circ}_{+} = 349.6$ S cm$^{2}$ mol$^{-1}$ and $\Lambda^{\circ}_{-} = 50.4$ S cm$^{2}$ mol$^{-1}$.]
- A. 0.215
- B. 0.115
- C. 0.125
- D. 0.225 ✓
Solution: Degree of dissociation ($\alpha$) is given as
$$\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m}$$
$$\Lambda^{\circ}_m = \Lambda^{\circ}_{+} + \Lambda^{\circ}_{-}$$
$$= 349.6 + 50.4 = 400\ \text{S cm}^{2}\ \text{mol}^{-1}$$
$$\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{90}{400} = 0.225$$
Q40.
Given below are two statements :
Statement I : A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II : As bond order increases, the bond length increases.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false ✓
- D. Statement I is true but Statement II is false
Solution: A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule, so Statement I is false.
When bond order increases, the bond length decreases, so Statement II is false as well.
Q41.
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n = 2 \rightarrow n = 3$ and $n = 4 \rightarrow n = 6$ transitions, respectively, is
- A. $\dfrac{1}{4}$ ✓
- B. $\dfrac{1}{36}$
- C. $\dfrac{1}{16}$
- D. $\dfrac{1}{9}$
Solution: $$\Delta E = \frac{hc}{\lambda} = E_{\text{final}} - E_{\text{initial}} \qquad \left(E_n = \frac{-R_H}{n^{2}}\right)$$
For $n = 2 \rightarrow 3$:
$$\Delta E_{2 \rightarrow 3} = \frac{hc}{\lambda_{2 \rightarrow 3}} = E_3 - E_2 = \frac{-R_H}{3^{2}} - \left(\frac{-R_H}{2^{2}}\right) = R_H\left(\frac{1}{4} - \frac{1}{9}\right) = R_H \times \frac{5}{36}$$
$$\therefore\ \lambda_{2 \rightarrow 3} = \frac{hc \cdot 36}{R_H \cdot 5}$$
For $n = 4 \rightarrow 6$:
$$\Delta E_{4 \rightarrow 6} = E_6 - E_4 = \frac{-R_H}{36} + \frac{R_H}{16} = \frac{R_H \times 20}{36 \times 16}$$
$$\lambda_{4 \rightarrow 6} = \frac{hc \times 36 \times 16}{R_H \cdot 20}$$
$$\frac{\lambda_{2 \rightarrow 3}}{\lambda_{4 \rightarrow 6}} = \frac{\dfrac{hc \cdot 36}{R_H \cdot 5}}{\dfrac{hc \times 36 \times 16}{R_H \cdot 20}} = \frac{1}{4}$$
Q42.
The correct order of the wavelength of light absorbed by the following complexes is,
A. $[\text{Co}(\text{NH}_3)_6]^{3+}$
B. $[\text{Co}(\text{CN})_6]^{3-}$
C. $[\text{Cu}(\text{H}_2\text{O})_4]^{2+}$
D. $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$
Choose the correct answer from the options given below:
- A. C < A < D < B
- B. B < D < A < C
- C. B < A < D < C ✓
- D. C < D < A < B
Solution: $$\lambda \propto \frac{1}{\text{strength of ligand}}, \qquad \lambda \propto \frac{1}{\text{splitting}}$$
The absorption maxima are
- A. $[\text{Co}(\text{NH}_3)_6]^{3+}$ — 475 nm
- B. $[\text{Co}(\text{CN})_6]^{3-}$ — 310 nm
- C. $[\text{Cu}(\text{H}_2\text{O})_4]^{2+}$ — 600 nm
- D. $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ — 498 nm
Order of $\lambda$ = C > D > A > B, i.e. in increasing order B < A < D < C.
Q43.
If the rate constant of a reaction is 0.03 s$^{-1}$, how much time does it take for 7.2 mol L$^{-1}$ concentration of the reactant to get reduced to 0.9 mol L$^{-1}$?
(Given: $\log 2 = 0.301$)
- A. 21.0 s
- B. 69.3 s ✓
- C. 23.1 s
- D. 210 s
Solution: $k = 0.03$ s$^{-1}$, and for a first order reaction
$$t = \frac{2.303}{k}\log\frac{a}{a - x}$$
$$= \frac{2.303}{0.03}\log\frac{7.2}{0.9}$$
$$= \frac{2.303}{0.03}\log 8$$
$$= \frac{2.303}{0.03} \times 3 \times \log 2$$
$$= \frac{2.303}{0.03} \times 3 \times 0.301$$
$$= 69.3\ \text{s}$$
Q44.
Match List I with List II
List-I (Mixture):
A. CHCl$_3$ + C$_6$H$_5$NH$_2$
B. Crude oil in petroleum industry
C. Glycerol from spent-lye
D. Aniline - water
List-II (Method of separation):
(I) Distillation under reduced pressure
(II) Steam distillation
(III) Fractional distillation
(IV) Simple distillation
Choose the correct answer from the options given below:
- A. A-III, B-IV, C-II, D-I
- B. A-IV, B-III, C-I, D-II ✓
- C. A-IV, B-III, C-II, D-I
- D. A-III, B-IV, C-I, D-II
Solution: Matching each mixture to the method that separates it:
A. CHCl$_3$ + C$_6$H$_5$NH$_2$ — Simple distillation (their boiling points differ widely enough)
B. Crude oil in petroleum industry — Fractional distillation
C. Glycerol from spent-lye — Distillation under reduced pressure (glycerol decomposes at its normal boiling point)
D. Aniline - water — Steam distillation
So A-IV, B-III, C-I, D-II.
Q45.
The major product of the following reaction is
3-benzoylpropanenitrile (C$_6$H$_5$COCH$_2$CH$_2$CN) treated with (i) CH$_3$MgBr (excess), then (ii) H$_3$O$^{+}$
- A. 1-phenylpentane-1,4-dione, C$_6$H$_5$COCH$_2$CH$_2$COCH$_3$
- B. 4-hydroxy-4-phenylpentanenitrile, C$_6$H$_5$C(CH$_3$)(OH)CH$_2$CH$_2$CN
- C. 5-hydroxy-5-phenylhexan-2-one, C$_6$H$_5$C(CH$_3$)(OH)CH$_2$CH$_2$COCH$_3$ ✓
- D. 2-methyl-5-phenylhexane-2,5-diol, C$_6$H$_5$C(CH$_3$)(OH)CH$_2$CH$_2$C(CH$_3$)$_2$OH
Solution: Excess CH$_3$MgBr attacks both the ketone and the nitrile.
At the ketone carbonyl it gives a bromomagnesium alkoxide, and at the nitrile it gives a ketimine salt (C=NMgBr).
On acidic work-up with H$_3$O$^{+}$, the alkoxide becomes a tertiary alcohol and the ketimine hydrolyses to a ketone.
The major product is therefore C$_6$H$_5$C(CH$_3$)(OH)CH$_2$CH$_2$COCH$_3$.
Q46.
Which one of the following compounds can exist as cis-trans isomers?
- A. 1, 2-Dimethylcyclohexane ✓
- B. Pent-1-ene
- C. 2-Methylhex-2-ene
- D. 1, 1-Dimethylcyclopropane
Solution: Cis-trans isomerism needs restricted rotation and two different groups on each of the two centres.
1,1-Dimethylcyclopropane carries both methyls on the same carbon, so there is no cis-trans pair.
Pent-1-ene, CH$_3$CH$_2$CH$_2$CH=CH$_2$, has two hydrogens on the terminal carbon — no cis-trans.
2-Methylhex-2-ene, CH$_3$CH$_2$CH$_2$CH=C(CH$_3$)CH$_3$, has two identical methyls on one doubly-bonded carbon — no cis-trans.
1,2-Dimethylcyclohexane has one methyl on each of two adjacent ring carbons, and the ring restricts rotation, so it exists as cis and trans forms.
Q47.
Among the following, choose the ones with equal number of atoms.
A. 212 g of Na$_2$CO$_3$(s) [molar mass = 106 g]
B. 248 g of Na$_2$O(s) [molar mass = 62 g]
C. 240 g of NaOH(s) [molar mass = 40 g]
D. 12 g of H$_2$(g) [molar mass = 2 g]
E. 220 g of CO$_2$(g) [molar mass = 44 g]
Choose the correct answer from the options given below :
- A. B, D, and E only
- B. A, B, and C only
- C. A, B, and D only ✓
- D. B, C, and D only
Solution: $$\text{Number of atoms} = \frac{\text{given mass}}{\text{molar mass}} \times \text{atomicity} \times N_A$$
A. $\dfrac{212}{106} \times 6 \times N_A = 12\,N_A$
B. $\dfrac{248}{62} \times 3 \times N_A = 12\,N_A$
C. $\dfrac{240}{40} \times 3 \times N_A = 18\,N_A$
D. $\dfrac{12}{2} \times N_A \times 2 = 12\,N_A$
E. $\dfrac{220}{44} \times N_A \times 3 = 15\,N_A$
A, B and D have the same number of atoms.
Q48.
Among the given compounds I-III, the correct order of bond dissociation energy of C–H bond marked with * is :
I. the ring C–H of benzene
II. the terminal alkyne C–H of phenylacetylene, C$_6$H$_5$C$\equiv$C–H
III. a CH$_2$ hydrogen of cyclopropene's saturated carbon
- A. II > III > I
- B. II > I > III ✓
- C. I > II > III
- D. III > II > I
Solution: The hybridisation of the carbon bearing the marked hydrogen decides the strength:
I. In benzene, the carbon of this bond is $sp^{2}$ hybridised.
II. In phenylacetylene, the carbon of this bond is $sp$ hybridised.
III. In the cyclopropene, the carbon of this bond is $sp^{3}$ hybridised.
Higher the percentage s character, stronger is the C–H bond. Correct order of bond dissociation energy of C–H bond:
II > I > III
Q49.
The standard heat of formation, in kcal/mol of Ba$^{2+}$ is :
[Given : standard heat of formation of SO$_4^{2-}$ ion (aq) = $-216$ kcal/mol, standard heat of crystallisation of BaSO$_4$(s) = $-4.5$ kcal/mol, standard heat of formation of BaSO$_4$(s) = $-349$ kcal/mol]
- A. + 220.5
- B. $-$ 128.5 ✓
- C. $-$ 133.0
- D. + 133.0
Solution: $$\text{S} + 2\text{O}_2 + 2e^{-} \longrightarrow \text{SO}_4^{2-} \qquad \Delta H_f = -216\ \text{kcal/mol} \qquad \ldots(1)$$
$$\text{Ba}^{2+}(g) + \text{SO}_4^{2-}(g) \longrightarrow \text{BaSO}_4(s) \qquad \Delta H_{\text{crystallisation}} = -4.5\ \text{kcal/mol} \qquad \ldots(2)$$
$$\text{Ba} + \text{S} + 2\text{O}_2 \longrightarrow \text{BaSO}_4(s) \qquad \Delta H_{f(\text{BaSO}_4)} = -349\ \text{kcal/mol} \qquad \ldots(3)$$
$$\text{Ba}(s) \longrightarrow \text{Ba}^{2+}(g) + 2e^{-} \qquad \ldots(4)$$
From equations (1), (2) and (3) we get equation (4). Applying $(3) - (1) - (2)$:
$$-349 - (-4.5) - (-216)$$
$$= -349 + 4.5 + 216$$
$$= -349 + 220.5$$
$$= -128.5\ \text{kcal/mol}$$
Q50.
Consider the following compounds :
KO$_2$, H$_2$O$_2$ and H$_2$SO$_4$
The oxidation state of the underlined elements (K in KO$_2$, O in H$_2$O$_2$, S in H$_2$SO$_4$) in them are, respectively,
- A. +4, $-$4, and +6
- B. +1, $-$1, and +6 ✓
- C. +2, $-$2, and +6
- D. +1, $-$2, and +4
Solution: KO$_2$ — an alkali metal always shows the $+1$ oxidation state, therefore the oxidation state of K is $+1$.
H$_2$O$_2$ — the structure is H–O–O–H, with each oxygen bonded to one hydrogen and one oxygen, so the oxidation state of oxygen in H$_2$O$_2$ is $-1$.
H$_2$SO$_4$ — with two S–OH bonds and two S=O bonds, the oxidation state of sulphur in H$_2$SO$_4$ is $+6$.
Q51.
Which one of the following reactions does NOT give benzene as the product?
- A. Benzenediazonium chloride, C$_6$H$_5$N$_2^{+}$Cl$^{-}$, with H$_2$O on warming ✓
- B. Sodium benzoate, C$_6$H$_5$COO$^{-}$Na$^{+}$, heated with sodalime
- C. n-Hexane over Mo$_2$O$_3$ at 773 K and 10–20 atm
- D. Ethyne, H–C$\equiv$C–H, passed through a red hot iron tube at 873 K
Solution: Benzenediazonium chloride warmed with water gives phenol, not benzene:
$$\text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-} + \text{H}_2\text{O} \xrightarrow{\ \text{warm}\ } \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{HCl}$$
The other three do give benzene:
Sodium benzoate with sodalime on heating undergoes decarboxylation to benzene.
n-Hexane over Mo$_2$O$_3$ at 773 K and 10–20 atm aromatises to benzene.
Three molecules of ethyne cyclically polymerise in a red hot iron tube at 873 K to benzene.
Q52.
Which of the following are paramagnetic?
A. [NiCl$_4$]$^{2-}$
B. Ni(CO)$_4$
C. [Ni(CN)$_4$]$^{2-}$
D. [Ni(H$_2$O)$_6$]$^{2+}$
E. Ni(PPh$_3$)$_4$
Choose the correct answer from the options given below :
- A. A, D and E only
- B. A and C only
- C. B and E only
- D. A and D only ✓
Solution: A. [NiCl$_4$]$^{2-}$; Ni$^{+2}$; 3$d^{8}$; $sp^{3}$ hybridisation; 2 unpaired electrons; paramagnetic
B. Ni(CO)$_4$; Ni; 3$d^{8}$4$s^{2}$; $sp^{3}$ hybridisation; zero unpaired electrons; diamagnetic
C. [Ni(CN)$_4$]$^{2-}$; Ni$^{+2}$; 3$d^{8}$; $dsp^{2}$ hybridisation; zero unpaired electrons; diamagnetic
D. [Ni(H$_2$O)$_6$]$^{2+}$; Ni$^{2+}$; 3$d^{8}$; $sp^{3}d^{2}$ hybridisation; two unpaired electrons; paramagnetic
E. Ni(PPh$_3$)$_4$; Ni; 3$d^{8}$4$s^{2}$; $sp^{3}$ hybridisation; zero unpaired electrons; diamagnetic
So only A and D are paramagnetic.
Q53.
Which one of the following compounds does not decolourize bromine water?
- A. Aniline, C$_6$H$_5$NH$_2$
- B. Cyclohexane ✓
- C. Phenol, C$_6$H$_5$OH
- D. Styrene, C$_6$H$_5$CH=CH$_2$
Solution: Bromine water is a test for unsaturation: the reddish orange colour of the bromine solution discharges when bromine adds to an unsaturation site, or when it substitutes into a strongly activated ring.
Styrene: C$_6$H$_5$CH=CH$_2$ + Br$_2$ $\rightarrow$ C$_6$H$_5$CHBr–CH$_2$Br — colour discharged.
Aniline: bromination gives 2,4,6-tribromoaniline — colour discharged.
Phenol: bromination gives 2,4,6-tribromophenol — colour discharged.
Cyclohexane is a saturated hydrocarbon: no reaction, and no discharge of the orange colour.
Q54.
Match List-I with List-II.
List-I:
A. Haber process
B. Wacker oxidation
C. Wilkinson catalyst
D. Ziegler catalyst
List-II:
I. Fe catalyst
II. PdCl$_2$
III. [(PPh$_3$)$_3$RhCl]
IV. TiCl$_4$ with Al(CH$_3$)$_3$
Choose the correct answer from the options given below :
- A. A-I, B-IV, C-III, D-II
- B. A-I, B-II, C-IV, D-III
- C. A-II, B-III, C-I, D-IV
- D. A-I, B-II, C-III, D-IV ✓
Solution: Matching each process with the catalyst it uses:
A. Haber process — Fe catalyst
B. Wacker oxidation — PdCl$_2$
C. Wilkinson catalyst — [(PPh$_3$)$_3$RhCl]
D. Ziegler catalyst — TiCl$_4$ with Al(CH$_3$)$_3$
So A-I, B-II, C-III, D-IV.
Q55.
Match List-I with List-II.
List-I (Name of Vitamin):
A. Vitamin B$_{12}$
B. Vitamin D
C. Vitamin B$_2$
D. Vitamin B$_6$
List-II (Deficiency disease):
I. Cheilosis
II. Convulsions
III. Rickets
IV. Pernicious anaemia
Choose the correct answer from the options given below:
- A. A-IV, B-III, C-II, D-I
- B. A-I, B-III, C-II, D-IV
- C. A-IV, B-III, C-I, D-II ✓
- D. A-II, B-III, C-I, D-IV
Solution: Matching each vitamin with the disease its deficiency causes:
A. Vitamin B$_{12}$ — Pernicious anaemia
B. Vitamin D — Rickets
C. Vitamin B$_2$ — Cheilosis
D. Vitamin B$_6$ — Convulsions
So A-IV, B-III, C-I, D-II.
Q56.
Given below are two statements :
Statement I : Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II : The number of unpaired electrons in a Cr$^{2+}$ ion (Z = 24) is the same as that of a Nd$^{3+}$ ion (Z = 60).
In the light of the above statements, choose the correct answer from the options given below :
- A. Statement I is false but Statement II is true
- B. Both Statement I and Statement II are true
- C. Both Statement I and Statement II are false
- D. Statement I is true but Statement II is false ✓
Solution: Substances which are attracted very strongly in an applied magnetic field are termed ferromagnetic. In fact, ferromagnetism is an extreme form of paramagnetism. Hence Statement I is correct.
$$\text{Cr}^{2+} = 3d^{4}4s^{0} \quad \Rightarrow \quad \text{unpaired electrons} = 4$$
$$\text{Nd}^{3+} = 4f^{3}6s^{0} \quad \Rightarrow \quad \text{unpaired electrons} = 3$$
Hence Statement II is incorrect.
Q57.
If the half-life ($t_{1/2}$) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to :
- A. 10 minutes ✓
- B. 2 minutes
- C. 4 minutes
- D. 5 minutes
Solution: For a first order reaction
$$kt = 2.303\log\frac{A_0}{A_t}$$
where $A_0$ is the initial concentration and $A_t$ the final concentration.
For 99.9% completion,
$$t_{99.9\%} = 10\,t_{1/2}$$
$$t_{99.9\%} = 10 \times 1\ \text{minute} = 10\ \text{minutes}$$
Q58.
The correct order of decreasing basic strength of the given amines is:
- A. benzenamine > ethanamine > N-methylaniline > N-ethylethanamine
- B. N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
- C. N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
- D. N-ethylethanamine > ethanamine > N-methylaniline > benzenamine ✓
Solution: Lower the value of p$K_b$, higher is the basicity. Also, aliphatic amines are stronger bases than aromatic amines.
p$K_b$ : Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine
Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine
Q59.
Match List I with List II
List-I (Ion):
A. Co$^{2+}$
B. Mg$^{2+}$
C. Pb$^{2+}$
D. Al$^{3+}$
List-II (Group Number in Cation Analysis):
I. Group-I
II. Group-III
III. Group-IV
IV. Group-VI
Choose the correct answer from the options given below :
- A. A-III, B-II, C-I, D-IV
- B. A-III, B-IV, C-II, D-I
- C. A-III, B-IV, C-I, D-II ✓
- D. A-III, B-II, C-IV, D-I
Solution: Placing each ion in its analytical group:
A. Co$^{2+}$ — Group-IV
B. Mg$^{2+}$ — Group-VI
C. Pb$^{2+}$ — Group-I
D. Al$^{3+}$ — Group-III
So A-III, B-IV, C-I, D-II.
Q60.
Phosphoric acid ionizes in three steps with their ionization constant values $K_{a_1}$, $K_{a_2}$ and $K_{a_3}$, respectively, while K is the overall ionization constant. Which of the following statements are true?
A. $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$
B. H$_3$PO$_4$ is a stronger acid than H$_2$PO$_4^{-}$ and HPO$_4^{2-}$
C. $K_{a_1} > K_{a_2} > K_{a_3}$
D. $K_{a_1} = \dfrac{K_{a_3} + K_{a_2}}{2}$
Choose the correct answer from the options given below :
- A. A, B and C only ✓
- B. A and B only
- C. A and C only
- D. B, C and D only
Solution: H$_3$PO$_4$ is a stronger acid than H$_2$PO$_4^{-}$ and HPO$_4^{2-}$:
$$\text{H}_3\text{PO}_4(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{H}_2\text{PO}_4^{-}(aq) \qquad K_{a_1} = 7.5 \times 10^{-3}$$
$$\text{H}_2\text{PO}_4^{-}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{HPO}_4^{2-}(aq) \qquad K_{a_2} = 6.2 \times 10^{-8}$$
$$\text{HPO}_4^{2-}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{PO}_4^{3-}(aq) \qquad K_{a_3} = 1.7 \times 10^{-12}$$
So $K_{a_1} > K_{a_2} > K_{a_3}$.
The overall constant is the product of the stepwise constants, $K = K_{a_1}K_{a_2}K_{a_3}$, so
$$\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$$
Statement D is not a relation between ionization constants at all.
Hence (A), (B) and (C) only.
Q61.
Which of the following statements are true?
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and Cl are not the same.
C. Ar, K$^{+}$, Cl$^{-}$, Ca$^{2+}$, and S$^{2-}$ are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below :
- A. A, C, and E only
- B. A, B, and E only
- C. C and E only ✓
- D. C and D only
Solution: A. Both Ga and Cs have low melting points — Ga 303 K, Cs 302 K — so A is false.
B. On the Pauling scale the electronegativity values of N and Cl are the same (3.0), so B is false.
C. Ar, K$^{+}$, Cl$^{-}$, Ca$^{2+}$ and S$^{2-}$ all have 18 electrons, so these are isoelectronic species — C is true.
D. The correct order of first ionization enthalpy is Si > Mg > Al > Na, not Si > Al > Mg > Na, so D is false. The first ionisation enthalpy of Mg is higher than Al because the penetration of a 3s-electron to the nucleus is more than that of a 2p-electron.
E. Generally down the group atomic radii increase — Li 152 pm, Rb 244 pm, Cs 262 pm — so E is true.
Hence C and E only.
Q62.
Given below are two statements :
Statement I : Like nitrogen that can form ammonia, arsenic can form arsine.
Statement II : Antimony cannot form antimony pentoxide.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are correct
- C. Both Statement I and Statement II are incorrect
- D. Statement I is correct but Statement II is incorrect ✓
Solution: All the elements of group 15 form hydrides of EH$_3$ type. Nitrogen forms ammonia (NH$_3$) while arsenic forms arsine (AsH$_3$), so Statement I is correct.
All the elements of group 15 form two types of oxides: E$_2$O$_3$ and E$_2$O$_5$. Antimony forms antimony pentoxide, Sb$_2$O$_5$.
Hence Statement I is correct and Statement II is incorrect.
Q63.
Which of the following aqueous solution will exhibit highest boiling point?
- A. 0.015M C$_6$H$_{12}$O$_6$
- B. 0.01M Urea
- C. 0.01M KNO$_3$
- D. 0.01M Na$_2$SO$_4$ ✓
Solution: $$\Delta T_b = iK_b \times m \quad \Rightarrow \quad \Delta T_b \propto i \times m$$
By considering molarity same as molality:
0.015 M C$_6$H$_{12}$O$_6$ : $i \times m = 1 \times 0.015 = 0.015$
0.01 M Urea : $i \times m = 1 \times 0.01 = 0.01$
0.01 M KNO$_3$ : $i \times m = 2 \times 0.01 = 0.02$
0.01 M Na$_2$SO$_4$ : $i \times m = 3 \times 0.01 = 0.03$
Since $T_b' = T_b^{*} + \Delta T_b$, the higher the value of $(i \times m)$ the higher will be the boiling point — so 0.01 M Na$_2$SO$_4$.
Q64.
Given below are two statements :
Statement-I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273 – 278 K. It decomposes easily in the dry state.
Statement-II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but Statement II is correct
- B. Both Statement I and Statement II are correct ✓
- C. Both Statement I and Statement II are incorrect
- D. Statement I is correct but Statement II is incorrect
Solution: Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid at 273–278 K. Nitrous acid is produced in the reaction mixture by reaction of NaNO$_2$ with HCl:
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\ \text{K}} \text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-} + \text{NaCl} + 2\text{H}_2\text{O}$$
Benzene diazonium chloride decomposes easily in the dry state, so Statement I is correct.
Iodobenzene is prepared by shaking benzene diazonium salt with KI because direct insertion of iodine into the benzene ring is difficult:
$$\text{C}_6\text{H}_5\text{N}_2^{+}\text{X}^{-} \xrightarrow{\ \text{KI}\ } \text{C}_6\text{H}_5\text{I} + \text{N}_2$$
So Statement II is correct as well.
Q65.
Identify the suitable reagent for the following conversion.
Methyl benzoate, C$_6$H$_5$COOCH$_3$, is converted to benzaldehyde, C$_6$H$_5$CHO.
- A. H$_2$/Pd-BaSO$_4$
- B. (i) LiAlH$_4$, (ii) H$^{+}$/H$_2$O
- C. (i) AlH(iBu)$_2$, (ii) H$_2$O ✓
- D. (i) NaBH$_4$, (ii) H$^{+}$/H$_2$O
Solution: Esters are reduced to aldehydes with DIBAL-H, diisobutylaluminium hydride, AlH(iBu)$_2$:
$$\text{C}_6\text{H}_5\text{COOCH}_3 \xrightarrow[\ (ii)\ \text{H}_2\text{O}\ ]{\ (i)\ \text{DIBAL-H}\ } \text{C}_6\text{H}_5\text{CHO}$$
LiAlH$_4$ would carry the reduction on to the primary alcohol, NaBH$_4$ does not reduce esters, and H$_2$/Pd-BaSO$_4$ (Rosenmund) reduces acid chlorides, not esters.
Q66.
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The primary alkyl iodide shown undergoes S$_N$2 reaction faster than the corresponding primary alkyl chloride of the same carbon skeleton.
Reason (R) : Iodine is a better leaving group because of its large size.
In the light of the above statements, choose the correct answer from the options given below:
- A. A is false but R is true
- B. Both A and R are true and R is the correct explanation of A ✓
- C. Both A and R are true but R is not the correct explanation of A
- D. A is true but R is false
Solution: The rate of the S$_N$2 reaction of the alkyl iodide is faster than that of the alkyl chloride, because iodide is a good leaving group due to the large size of iodine, which stabilises the I$^{-}$ ion.
So both A and R are true, and R is the correct explanation of A.
Q67.
The correct order of decreasing acidity of the following aliphatic acids is
- A. HCOOH > (CH$_3$)$_3$CCOOH > (CH$_3$)$_2$CHCOOH > CH$_3$COOH
- B. (CH$_3$)$_3$CCOOH > (CH$_3$)$_2$CHCOOH > CH$_3$COOH > HCOOH
- C. CH$_3$COOH > (CH$_3$)$_2$CHCOOH > (CH$_3$)$_3$CCOOH > HCOOH
- D. HCOOH > CH$_3$COOH > (CH$_3$)$_2$CHCOOH > (CH$_3$)$_3$CCOOH ✓
Solution: An electron donating group decreases the acidity of carboxylic acids, and the donating effect grows as hydrogens are replaced by methyl groups.
So the correct order is
HCOOH > CH$_3$COOH > (CH$_3$)$_2$CHCOOH > (CH$_3$)$_3$CCOOH
Q68.
Which one of the following reactions does NOT belong to "Lassaigne's test"?
- A. $2\text{CuO} + \text{C} \xrightarrow{\Delta} 2\text{Cu} + \text{CO}_2$ ✓
- B. $\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN}$
- C. $2\text{Na} + \text{S} \xrightarrow{\Delta} \text{Na}_2\text{S}$
- D. $\text{Na} + \text{X} \xrightarrow{\Delta} \text{NaX}$
Solution: Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by "Lassaigne's test", in which the element is converted to its sodium salt by fusion with sodium metal:
$$\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN}$$
$$2\text{Na} + \text{S} \xrightarrow{\Delta} \text{Na}_2\text{S}$$
$$\text{Na} + \text{X} \xrightarrow{\Delta} \text{NaX} \qquad (\text{X} = \text{Cl, Br, I})$$
The reaction of copper(II) oxide with carbon is the estimation of carbon, not part of Lassaigne's test.
Q69.
How many products (including stereoisomers) are expected from monochlorination of the following compound?
(CH$_3$)$_2$CH–CH$_2$–CH$_3$
Solution: Possible monochlorination products:
(CH$_3$)$_2$CH–CH$_2$–CH$_2$Cl — one isomer
(CH$_3$)$_2$CH–CHCl–CH$_3$ — two isomers, due to the presence of a chiral carbon
(CH$_3$)$_2$CCl–CH$_2$–CH$_3$ — one isomer
ClCH$_2$(CH$_3$)CH–CH$_2$–CH$_3$ — two isomers, due to a chiral carbon
Total 6 isomers.
Q70.
Sugar 'X'
A. is found in honey
B. is a keto sugar
C. exists in $\alpha$ and $\beta$ – anomeric forms.
D. Is laevorotatory.
'X' is :
- A. Sucrose
- B. D-Glucose
- C. D-Fructose ✓
- D. Maltose
Solution: D-Fructose is found in honey and is a keto sugar.
It cyclises to the furanose form, which exists as $\alpha$-D-(–)-fructofuranose and $\beta$-D-(–)-fructofuranose, the two anomers.
It is laevorotatory, which is why it is written D-(–)-fructose.
Q71.
Dalton's Atomic theory could not explain which of the following?
- A. Law of gaseous volume ✓
- B. Law of conservation of mass
- C. Law of constant proportion
- D. Law of multiple proportion
Solution: Dalton's theory could explain the laws of chemical combination — conservation of mass, constant proportion and multiple proportion. However, it could not explain the law of gaseous volumes.
Q72.
Higher yield of NO in $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$ can be obtained at
[$\Delta$H of the reaction = +180.7 kJ mol$^{-1}$]
A. Higher temperature
B. Lower temperature
C. Higher concentration of N$_2$
D. Higher concentration of O$_2$
Choose the correct answer from the options given below :
- A. A, C, D only ✓
- B. A, D only
- C. B, C only
- D. B, C, D only
Solution: Yield of the product generally depends on temperature, on the concentration of reactants and products, and on pressure.
As this is an endothermic reaction ($\Delta$H = +180.7 kJ mol$^{-1}$), an increase in temperature will shift the equilibrium in the forward direction and increase the yield of NO.
An increase in the concentration of the reactants (N$_2$ and O$_2$) also shifts the equilibrium in the forward direction and increases the yield of NO.
Hence (A), (C) and (D) only will increase the yield of NO.
Q73.
Match List-I with List-II
List-I:
A. XeO$_3$
B. XeF$_2$
C. XeOF$_4$
D. XeF$_6$
List-II:
(I) $sp^{3}d$; linear
(II) $sp^{3}$; pyramidal
(III) $sp^{3}d^{3}$; distorted octahedral
(IV) $sp^{3}d^{2}$; square pyramidal
Choose the correct answer from the options given below :
- A. A-IV, B-II, C-I, D-III
- B. A-II, B-I, C-IV, D-III ✓
- C. A-II, B-I, C-III, D-IV
- D. A-IV, B-II, C-III, D-I
Solution: Taking each xenon compound with its hybridisation and shape:
XeO$_3$ — $sp^{3}$; pyramidal (three bond pairs and one lone pair)
XeF$_2$ — $sp^{3}d$; linear (two bond pairs and three lone pairs)
XeOF$_4$ — $sp^{3}d^{2}$; square pyramidal (five bond pairs and one lone pair)
XeF$_6$ — $sp^{3}d^{3}$; distorted octahedral (six bond pairs and one lone pair)
So A-II, B-I, C-IV, D-III.
Q74.
Match List-I with List-II
List-I (Example):
A. Humidity
B. Alloys
C. Amalgams
D. Smoke
List-II (Type of Solution):
I. Solid in solid
II. Liquid in gas
III. Solid in gas
IV. Liquid in solid
Choose the correct answer from the options given below:
- A. A-III, B-II, C-I, D-IV
- B. A-II, B-IV, C-I, D-III
- C. A-II, B-I, C-IV, D-III ✓
- D. A-III, B-I, C-IV, D-II
Solution: Humidity is a solution of liquid in gas.
Alloy is a solution of solid in solid.
Amalgam is a solution of liquid in solid.
Smoke is a solution of solid in gas.
So A-II, B-I, C-IV, D-III.
Q75.
Energy and radius of first Bohr orbit of He$^{+}$ and Li$^{2+}$ are
[Given R$_H$ = $2.18 \times 10^{-18}$ J, $a_0$ = 52.9 pm]
- A. E$_n$(Li$^{2+}$) = $-8.72 \times 10^{-16}$ J; r$_n$(Li$^{2+}$) = 17.6 pm; E$_n$(He$^{+}$) = $-19.62 \times 10^{-16}$ J; r$_n$(He$^{+}$) = 17.6 pm
- B. E$_n$(Li$^{2+}$) = $-19.62 \times 10^{-18}$ J; r$_n$(Li$^{2+}$) = 17.6 pm; E$_n$(He$^{+}$) = $-8.72 \times 10^{-18}$ J; r$_n$(He$^{+}$) = 26.4 pm ✓
- C. E$_n$(Li$^{2+}$) = $-8.72 \times 10^{-18}$ J; r$_n$(Li$^{2+}$) = 26.4 pm; E$_n$(He$^{+}$) = $-19.62 \times 10^{-18}$ J; r$_n$(He$^{+}$) = 17.6 pm
- D. E$_n$(Li$^{2+}$) = $-19.62 \times 10^{-16}$ J; r$_n$(Li$^{2+}$) = 17.6 pm; E$_n$(He$^{+}$) = $-8.72 \times 10^{-16}$ J; r$_n$(He$^{+}$) = 26.4 pm
Solution: $$E_n = \frac{-2.18 \times 10^{-18} \times z^{2}}{n^{2}}\ \text{J}; \qquad r_n = \frac{52.9 \times n^{2}}{z}\ \text{pm}$$
For He$^{+}$ ($z = 2$, $n = 1$):
$$E_{\text{He}^{+}} = -2.18 \times 10^{-18} \times 4 = -8.72 \times 10^{-18}\ \text{J}$$
$$r_{\text{He}^{+}} = \frac{52.9 \times 1}{2} = 26.45\ \text{pm}$$
For Li$^{2+}$ ($z = 3$, $n = 1$):
$$E_{\text{Li}^{2+}} = -2.18 \times 10^{-18} \times 9 = -19.62 \times 10^{-18}\ \text{J}$$
$$r_{\text{Li}^{2+}} = \frac{52.9 \times 1}{3} = 17.63\ \text{pm}$$
Q76.
Which among the following electronic configurations belong to main group elements?
A. [Ne]3s$^{1}$
B. [Ar]3d$^{3}$4s$^{2}$
C. [Kr]4d$^{10}$5s$^{2}$5p$^{5}$
D. [Ar]3d$^{10}$4s$^{1}$
E. [Rn]5f$^{0}$6d$^{2}$7s$^{2}$
Choose the correct answer from the option given below :
- A. A, C and D only
- B. B and E only
- C. A and C only ✓
- D. D and E only
Solution: Identifying each element and its block:
(A) [Ne]3s$^{1}$ — Na (s-block)
(B) [Ar]3d$^{3}$4s$^{2}$ — V (d-block)
(C) [Kr]4d$^{10}$5s$^{2}$5p$^{5}$ — I (p-block)
(D) [Ar]3d$^{10}$4s$^{1}$ — Cu (d-block)
(E) [Rn]5f$^{0}$6d$^{2}$7s$^{2}$ — Th (f-block)
Main group elements are the s- and p-block ones, so A and C only.
Q77.
Predict the major product 'P' in the following sequence of reactions-
1-Methylcyclopentene, treated with (i) HBr, benzoyl peroxide; (ii) KCN; (iii) Na(Hg)/C$_2$H$_5$OH, gives P (Major).
- A. 1-Cyano-1-methylcyclopentane — NC and CH$_3$ on the same ring carbon
- B. (2-Methylcyclopentyl)methanamine — CH$_3$ on one ring carbon and CH$_2$NH$_2$ on the adjacent one ✓
- C. (1-Methylcyclopentyl)methanamine — CH$_3$ and CH$_2$NH$_2$ on the same ring carbon
- D. 2-Methylcyclopentanecarbonitrile — CH$_3$ on one ring carbon and NC on the adjacent one
Solution: HBr with benzoyl peroxide adds anti-Markovnikov across the double bond of 1-methylcyclopentene, putting the bromine on the carbon adjacent to the methyl-bearing carbon.
KCN then displaces the bromide by nucleophilic substitution, giving the nitrile with CN on that adjacent carbon.
Na(Hg)/C$_2$H$_5$OH reduces the nitrile to a primary amine, so the CN becomes CH$_2$NH$_2$.
P is therefore the cyclopentane carrying CH$_3$ on one carbon and CH$_2$NH$_2$ on the neighbouring carbon.
Q78.
Identify the correct orders against the property mentioned
A. H$_2$O > NH$_3$ > CHCl$_3$ – dipole moment
B. XeF$_4$ > XeO$_3$ > XeF$_2$ – number of lone pairs on central atom
C. O–H > C–H > N–O – bond length
D. N$_2$ > O$_2$ > H$_2$ – bond enthalpy
Choose the correct answer from the options given below:
- A. B, C only
- B. A, D only ✓
- C. B, D only
- D. A, C only
Solution: A. Dipole moments $\mu$(D): H$_2$O 1.85, NH$_3$ 1.47, CHCl$_3$ 1.04 — so H$_2$O > NH$_3$ > CHCl$_3$ is correct.
B. Lone pairs on the central atom: XeF$_4$ has 2, XeO$_3$ has 1, XeF$_2$ has 3 — so the given order is wrong.
C. The order of bond length is N–O > C–H > O–H, so the given order is wrong.
D. Bond orders are N$_2$ = 3, O$_2$ = 2, H$_2$ = 1, so bond enthalpy follows N$_2$ > O$_2$ > H$_2$ — correct.
Hence A and D only.
Q79.
Total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C$_4$H$_8$O is :
Solution: For a cyclic ether the oxygen must be in the ring. Counting each skeleton and its stereoisomers:
2-Methyloxetane — the ring carbon bearing the methyl is chiral, so 2 (d, l pair)
Tetrahydrofuran (oxolane) — 1
cis-2,3-Dimethyloxirane — a meso compound, 1
3-Methyloxetane — 1
2-Ethyloxirane — chiral, 2 (d, l pair)
2,2-Dimethyloxirane — 1
trans-2,3-Dimethyloxirane — chiral, 2 (d, l pair)
Total number of isomers $= 2 + 1 + 1 + 1 + 2 + 1 + 2 = 10$
Q80.
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given : R = 0.0831 L atm mol$^{-1}$ K$^{-1}$]
K$_P$ for the reaction at 1000 K is
- A. 0.021
- B. 83.1
- C. $2.077 \times 10^{5}$
- D. 0.033 ✓
Solution: $$K_C = \frac{k_f}{k_b} = \frac{1}{2500}$$
$$K_P = K_C(RT)^{\Delta n_g} \qquad (\Delta n_g = 2 - 1 = 1)$$
$$= \frac{1}{2500} \times 0.0831 \times 1000$$
$$= 0.033$$
Q81.
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
- A. The solution has volume greater than the sum of individual volumes.
- B. The solution shows positive deviation.
- C. The solution shows negative deviation. ✓
- D. The solution is ideal.
Solution: $$P_{\text{total}} = X_x P_x^{\circ} + X_y P_y^{\circ}$$
$$= \frac{5}{15} \times 63 + \frac{10}{15} \times 78$$
$$= 21 + 52 = 73\ \text{torr}$$
The observed total pressure of the solution is 70 torr. It is less than the calculated total pressure, hence it shows negative deviation.
Q82.
Which of the following is the unit of productivity of an Ecosystem?
- A. (KCal m$^{-2}$)yr$^{-1}$ ✓
- B. gm$^{-2}$
- C. KCal m$^{-2}$
- D. KCal m$^{-3}$
Solution: The rate of biomass production is called productivity. It is expressed in terms of g m$^{-2}$ yr$^{-1}$ or (KCal m$^{-2}$) yr$^{-1}$ to compare the productivity of different ecosystems.
The rate is what matters, so the unit must carry a per-year term.
Q83.
The first menstruation is called :
- A. Ovulation
- B. Menopause
- C. Menarche ✓
- D. Diapause
Solution: The first menstruation begins at puberty and is called menarche.
Ovulation is the process that deals with the release of the secondary oocyte from the mature Graafian follicle.
In human beings, menstrual cycles cease around 50 years of age; that is termed menopause.
Diapause is a state of dormancy or developmental arrest in an organism.
Q84.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : All vertebrates are chordates but all chordates are not vertebrate.
Reason (R) : The members of subphylum vertebrata possess notochord during the embryonic period, the notochord is replaced by cartilaginous or bony vertebral column in adults.
In the light of the above statements, choose the correct answer from the options given below:
- A. (A) is false but (R) is true
- B. Both (A) and (R) are true and (R) is the correct explanation of (A) ✓
- C. Both (A) and (R) are true but (R) is not the correct explanation of (A)
- D. (A) is true but (R) is false
Solution: Both (A) and (R) are true and (R) is the correct explanation of (A).
The members of subphylum Vertebrata possess a notochord during the embryonic period. The notochord is replaced by a cartilaginous or bony vertebral column in the adult.
Thus, all vertebrates are chordates but all chordates are not vertebrates.
Q85.
Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?
- A. Phenotypic ratio - 9 : 7
- B. Phenotypic ratio - 1 : 2 : 1
- C. Phenotypic ratio - 3 : 1
- D. Phenotypic ratio - 9 : 3 : 3 : 1 ✓
Solution: This is a classical dihybrid cross of the kind Mendel performed: a cross made between a pure round yellow seeded pea plant (RRYY) and a wrinkled green seeded plant (rryy). Yellow colour is dominant over green, and round seed shape over wrinkled seed shape.
Since the two genes assort independently, the phenotypic ratio in the F2 generation is
9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green
that is 9 : 3 : 3 : 1.
Q86.
Given below are two statements :
Statement I : The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA.
Statement II : Smaller size DNA fragments are observed near anode while larger fragments are found near the wells in an agarose gel.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but statement II is correct
- B. Both statement I and statement II are correct ✓
- C. Both statement I and statement II are incorrect
- D. Statement I is correct but statement II is incorrect
Solution: The cutting of DNA by restriction endonucleases results in fragments of DNA. These fragments can be separated by a technique known as gel electrophoresis.
The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution. The DNA fragments purified in this way are used in constructing rDNA by joining them with cloning vectors — so Statement I is correct.
In gel electrophoresis, the DNA fragments separate (resolve) according to their size through the sieving effect provided by the agarose gel. Hence the smaller the fragment size, the farther it moves from the cathode towards the anode — so Statement II is correct too.
Q87.
What is the main function of the spindle fibers during mitosis?
- A. To regulate cell growth
- B. To separate the chromosomes ✓
- C. To synthesize new DNA
- D. To repair damaged DNA
Solution: During mitosis, spindle fibres get attached to the kinetochores of the chromosomes and help in the separation of the chromosomes.
Q88.
How many meiotic and mitotic divisions need to occur for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm plant?
- A. No Meiosis and 2 Mitosis
- B. 2 Meiosis and 3 Mitosis
- C. 1 Meiosis and 2 Mitosis
- D. 1 Meiosis and 3 Mitosis ✓
Solution: Development of a mature female gametophyte, i.e. the embryo sac, from a megaspore mother cell in an angiosperm plant requires 1 meiotic and 3 mitotic divisions.
The megaspore mother cell divides meiotically to give four megaspores, of which one remains functional; that megaspore then undergoes three successive mitotic divisions to give the 8-nucleate, 7-celled embryo sac.
Q89.
Identify the statement that is NOT correct.
- A. Constant region of heavy and light chains are located at C-terminus of antibody molecules
- B. Each antibody has two light and two heavy chains.
- C. The heavy and light chains are held together by disulfide bonds.
- D. Antigen binding site is located at C-terminal region of antibody molecules. ✓
Solution: Each antibody molecule has four peptide chains, two small called light chains and two longer called heavy chains. Hence, an antibody is represented as H$_2$L$_2$.
In an antibody molecule, the antigen binding site is located at the N-terminal region, not the C-terminal region — so that statement is not correct.
Q90.
Consider the following :
A. The reductive division for the human female gametogenesis starts earlier than that of the male gametogenesis.
B. The gap between the first meiotic division and the second meiotic division is much shorter for males compared to females.
C. The first polar body is associated with the formation of the primary oocyte.
D. Luteinizing Hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding.
Choose the correct answer from the options given below:
- A. B and C are true
- B. A and B are true ✓
- C. A and C are true
- D. B and D are true
Solution: Statements A and B are true while statements C and D are false.
The first polar body is associated with the formation of the secondary oocyte, not the primary oocyte.
The LH surge leads to ovulation. It is the decreased level of progesterone during the late luteal phase that leads to degeneration of the endometrium and the onset of menstrual bleeding.
Q91.
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.
Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells.
In light of the above statements, choose the most appropriate answer from the options given below:
- A. A is false but R is true
- B. Both A and R are true and R is the correct explanation of A
- C. Both A and R are true but R is NOT the correct explanation of A
- D. A is true but R is false ✓
Solution: Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus, so the Assertion is true.
The presence of more than one nucleus in the tapetal cells increases the efficiency of nourishing the developing pollen grains — not the microspore mother cells as the Reason states. So the Reason is false.
Q92.
The blue and white selectable markers have been developed which differentiate recombinant colonies from non-recombinant colonies on the basis of their ability to produce colour in the presence of a chromogenic substrate.
Given below are two statements about this method:
Statement I : The blue coloured colonies have DNA insert in the plasmid and they are identified as recombinant colonies.
Statement II : The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but Statement II is correct ✓
- B. Both Statement I and Statement II are correct
- C. Both Statement I and Statement II are incorrect
- D. Statement I is correct but Statement II is incorrect
Solution: Statement I is incorrect but Statement II is correct.
A recombinant DNA is inserted within the coding sequence of an enzyme, $\beta$-galactosidase. This results in inactivation of the gene for synthesis of this enzyme.
Thus, the presence of the insert results in insertional inactivation of the $\beta$-galactosidase gene, and the colonies do not produce any colour — these are identified as recombinant colonies. Whereas non-recombinant transformants will produce blue colour in the presence of the chromogenic substrate.
Q93.
In bryophytes, the gemmae help in which one of the following?
- A. Gaseous exchange
- B. Sexual reproduction
- C. Asexual reproduction ✓
- D. Nutrient absorption
Solution: Gemmae are green, multicellular, asexual buds which develop in small receptacles called gemma cups and help in asexual reproduction in bryophytes.
Q94.
Match List I with List II:
List-I:
A. Adenosine
B. Adenylic acid
C. Adenine
D. Alanine
List-II:
I. Nitrogen base
II. Nucleotide
III. Nucleoside
IV. Amino acid
Choose the option with all correct matches.
- A. A-II, B-III, C-I, D-IV
- B. A-III, B-IV, C-II, D-I
- C. A-III, B-II, C-IV, D-I
- D. A-III, B-II, C-I, D-IV ✓
Solution: The correct answer is A-III, B-II, C-I, D-IV.
Adenosine — it is a nucleoside, which is composed of a nitrogen base and sugar only.
Adenylic acid — it is a nucleotide, composed of a nitrogen base and sugar, with a phosphate group esterified to the sugar.
Adenine — a nitrogen base (purine).
Alanine — an amino acid that contains a methyl group as the 'R' group.
Q95.
Consider the following statements regarding function of adrenal medullary hormones :
(A) It causes pupilary constriction.
(B) It is a hyperglycemic hormone.
(C) It causes piloerection.
(D) It increases strength of heart contraction.
Choose the correct answer from the options given below :
- A. D only
- B. C and D only
- C. B, C and D only ✓
- D. A, C and D only
Solution: The adrenal medullary hormones are adrenaline and noradrenaline, the catecholamines of the fight-or-flight response.
They are hyperglycemic hormones: they stimulate glycogenolysis and raise blood glucose, so (B) is correct.
They cause piloerection, so (C) is correct.
They increase the strength of heart contraction as well as the rate, so (D) is correct.
They cause pupillary dilation, not constriction, so (A) is incorrect.
Hence B, C and D only.
Q96.
Which of the following is an example of a zygomorphic flower?
- A. Chilli
- B. Petunia
- C. Datura
- D. Pea ✓
Solution: Zygomorphic flowers can be divided into two equal halves by only a single vertical plane and show bilateral symmetry.
Pea possesses zygomorphic flowers.
Chilli, Petunia and Datura possess actinomorphic flowers.
Q97.
Who proposed that the genetic code for amino acids should be made up of three nucleotides?
- A. Franklin Stahl
- B. George Gamow ✓
- C. Francis Crick
- D. Jacque Monod
Solution: George Gamow, a physicist, proposed that the genetic code for amino acids should be made up of three nucleotides.
His argument was combinatorial: with only four bases, a code of two letters would give 4$^2$ = 16 combinations, too few for 20 amino acids, while three letters give 4$^3$ = 64, which is enough.
Q98.
Given below are two statements:
Statement I: In ecosystem, there is unidirectional flow of energy of sun from producers to consumers.
Statement II: Ecosystems are exempted from 2$^{\text{nd}}$ law of thermodynamics.
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but statement II is correct
- B. Both statement I and statement II are correct
- C. Both statement I and statement II are incorrect
- D. Statement I is correct but statement II is incorrect ✓
Solution: Sun is the only source of energy for all ecosystems on Earth, except for the deep sea hydro-thermal ecosystem. The energy flow is unidirectional from the sun to producers and then to consumers, so Statement I is correct.
Ecosystems are not exempted from the second law of thermodynamics. They need a constant supply of energy to synthesise the molecules they require, to counteract the universal tendency towards increasing disorderliness. So Statement II is incorrect.
Q99.
Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain the evolution.
- A. Analogy, divergent
- B. Analogy, convergent ✓
- C. Homology, divergent
- D. Homology, convergent
Solution: Sweet potato is a root modification while potato is a stem modification, but both of them have the same function.
Analogous structures are not anatomically similar structures though they perform similar functions. Analogous structures are the result of convergent evolution.
Homologous organs are anatomically similar but they do not perform similar functions; homologous organs are the result of divergent evolution.
Q100.
All living members of the class Cyclostomata are :
- A. Ectoparasite ✓
- B. Free living
- C. Endoparasite
- D. Symbiotic
Solution: All living members of class Cyclostomata are ectoparasites — they attach to the outside of their fish hosts with their circular, jawless sucking mouth.
Q101.
Histones are enriched with -
- A. Phenylalanine & Arginine
- B. Lysine & Arginine ✓
- C. Leucine & Lysine
- D. Phenylalanine & Leucine
Solution: In eukaryotes, packaging of DNA is much more complex. There is a set of positively charged, basic proteins called histones.
Histones are organised to form a unit of eight molecules called the histone octamer.
They are rich in the basic amino acid residues lysine and arginine, whose positive charge lets them bind the negatively charged DNA.
Q102.
Which one of the following equations represents the Verhulst-Pearl Logistic Growth of population?
- A. $\dfrac{dN}{dt} = N\left(\dfrac{r - K}{K}\right)$
- B. $\dfrac{dN}{dt} = r\left(\dfrac{K - N}{K}\right)$
- C. $\dfrac{dN}{dt} = rN\left(\dfrac{K - N}{K}\right)$ ✓
- D. $\dfrac{dN}{dt} = rN\left(\dfrac{N - K}{N}\right)$
Solution: Logistic growth is described by the Verhulst-Pearl logistic growth equation
$$\frac{dN}{dt} = rN\left(\frac{K - N}{K}\right)$$
where $N$ is the population size, $r$ the intrinsic rate of natural increase and $K$ the carrying capacity.
Q103.
Given below are two statements : one is labelled as Assertion (A), and the other is labelled as Reason (R).
Assertion (A) : The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell.
Reason (R) : Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus.
In the light of the above statements, choose the correct answer from the options given below :
- A. A is false but R is true
- B. Both A and R are true and R is the correct explanation of A
- C. Both A and R are true but R is not the correct explanation of A ✓
- D. A is true but R is false
Solution: The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell — this statement is correct, and the reason statement is also correct.
Golgi apparatus remains in close association with the endoplasmic reticulum. Here, assertion and reason statements are both correct but the reason is not correctly explaining the assertion.
Q104.
Which of the following statements about RuBisCO is true?
- A. It catalyzes the carboxylation of RuBP ✓
- B. It is active only in the dark
- C. It has higher affinity for oxygen than carbon dioxide
- D. It is an enzyme involved in the photolysis of water
Solution: Carboxylation is the most crucial step of the Calvin cycle, where CO$_2$ is utilised for the carboxylation of RuBP. This reaction is catalysed by the enzyme RuBP carboxylase.
Since this enzyme also has an oxygenase activity, it is named RuBisCO; but it has a higher affinity for carbon dioxide than for oxygen.
Q105.
Match List-I with List-II.
List-I:
A. Progesterone
B. Relaxin
C. Melanocyte stimulating hormone
D. Catecholamines
List-II:
I. Pars intermedia
II. Ovary
III. Adrenal Medulla
IV. Corpus luteum
Choose the correct answer from the options given below :
- A. A-III, B-II, C-IV, D-I
- B. A-IV, B-II, C-I, D-III ✓
- C. A-IV, B-II, C-III, D-I
- D. A-II, B-IV, C-I, D-III
Solution: Matching each hormone with the structure that secretes it:
A. Progesterone — Corpus luteum
B. Relaxin — Ovary
C. Melanocyte stimulating hormone — Pars intermedia of the pituitary
D. Catecholamines — Adrenal Medulla
So A-IV, B-II, C-I, D-III.
Q106.
The protein portion of an enzyme is called:
- A. Prosthetic group
- B. Cofactor
- C. Coenzyme
- D. Apoenzyme ✓
Solution: There are a number of cases in which non-protein constituents called co-factors are bound to the enzyme to make the enzyme catalytically active.
In these instances, the protein portion of the enzyme is called the apoenzyme.
Three kinds of co-factors are identified: prosthetic groups, co-enzymes and metal ions. Prosthetic groups are organic compounds and they are tightly bound with the apoenzyme. Co-enzymes are also organic compounds but their association with the apoenzyme is only transient.
Q107.
Which of the following enzyme(s) are NOT essential for gene cloning?
A. Restriction enzymes
B. DNA ligase
C. DNA mutase
D. DNA recombinase
E. DNA polymerase
Choose the correct answer from the options given below:
- A. B and C only
- B. C and D only ✓
- C. A and B only
- D. D and E only
Solution: Gene cloning is a process where a specific gene or DNA sequence is isolated and replicated, creating multiple identical copies.
In gene cloning, restriction enzymes, DNA ligase and DNA polymerase are primarily used.
DNA mutase and DNA recombinase are not required for it.
Q108.
Which of the following type of immunity is present at the time of birth and is a non-specific type of defence in the human body?
- A. Humoral Immunity
- B. Acquired Immunity
- C. Innate Immunity ✓
- D. Cell-mediated Immunity
Solution: Innate immunity is a non-specific type of defence that is present at the time of birth. This is accomplished by providing different types of barriers to the entry of foreign agents into our body.
Acquired immunity is pathogen specific, characterised by memory cells.
The immune response mediated by B-lymphocytes is humoral immunity, and the other immune response mediated by T-lymphocytes is called cell-mediated immunity.
Q109.
Which factor is important for termination of transcription?
- A. $\gamma$ (gamma)
- B. $\alpha$ (alpha)
- C. $\sigma$ (sigma)
- D. $\rho$ (rho) ✓
Solution: In prokaryotes the RNA polymerase is only capable of catalysing the process of elongation. It associates transiently with the initiation factor ($\sigma$) and the termination factor ($\rho$) to initiate and terminate the transcription respectively.
Q110.
Which of the following hormones released from the pituitary is actually synthesized in the hypothalamus?
- A. Adrenocorticotropic hormone (ACTH)
- B. Luteinizing hormone (LH)
- C. Anti-diuretic hormone (ADH) ✓
- D. Follicle-stimulating hormone (FSH)
Solution: The neurohypophysis, i.e. the posterior pituitary (pars nervosa), stores and releases two hormones called oxytocin and vasopressin (also called ADH, i.e. antidiuretic hormone), which are actually synthesised by the hypothalamus and are transported axonally to the neurohypophysis.
The pars distalis (anterior pituitary) produces follicle stimulating hormone (FSH), adrenocorticotropic hormone (ACTH) and luteinizing hormone (LH).
Q111.
Which of the following microbes is NOT involved in the preparation of household products?
A. Aspergillus niger
B. Lactobacillus
C. Trichoderma polysporum
D. Saccharomyces cerevisiae
E. Propionibacterium sharmanii
Choose the correct answer from the options given below:
- A. C and E only
- B. A and B only
- C. A and C only ✓
- D. C and D only
Solution: Lactobacillus is used for production of curd.
Saccharomyces cerevisiae is used for the fermentation of palm sap to obtain toddy drink.
Propionibacterium sharmanii is used for production of swiss cheese.
Aspergillus niger is used for the commercial production of citric acid.
Trichoderma polysporum is used for the production of cyclosporin A and also acts as a biocontrol agent.
A and C are used in industrial production of citric acid and cyclosporin-A, not in household products.
Q112.
Given below are two statements :
Statement I : Fig fruit is a non-vegetarian fruit as it has enclosed fig wasps in it.
Statement II : Fig wasp and fig tree exhibit mutual relationship as fig wasp completes its life cycle in fig fruit and fig fruit gets pollinated by fig wasp.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but statement II is correct
- B. Both statement I and statement II are correct
- C. Both statement I and statement II are incorrect ✓
- D. Statement I is correct but statement II is incorrect
Solution: Fig fruit is a vegetarian fruit, as it only gets pollinated by the wasp. Fig tree and fig wasps show mutualism in which both species are benefitted. So Statement I is incorrect.
Statement II is also not correct, as it is the fig inflorescence/flower that gets pollinated by the fig wasp, not the fruit.
Q113.
Role of the water vascular system in Echinoderms is :
A. Respiration and Locomotion
B. Excretion and Locomotion
C. Capture and transport of food
D. Digestion and Respiration
E. Digestion and Excretion
Choose the correct answer from the options given below :
- A. B, D and E Only
- B. A and B Only
- C. A and C Only ✓
- D. B and C Only
Solution: The water vascular system in Echinoderms helps in locomotion, capture and transport of food, and respiration.
An excretory system is absent in echinoderms; excretion takes place through the general body surface.
So A and C only.
Q114.
After maturation, in primary lymphoid organs, the lymphocytes migrate for interaction with antigens to secondary lymphoid organ(s) / tissue(s) like
A. thymus
B. bone marrow
C. spleen
D. lymph nodes
E. Peyer's patches
Choose the correct answer from the options given below
- A. C, D, E only ✓
- B. B, C, D only
- C. A, B, C only
- D. E, A, B only
Solution: The primary lymphoid organs are bone marrow and thymus, where immature lymphocytes differentiate into antigen-sensitive lymphocytes.
After maturation, the lymphocytes migrate into secondary lymphoid organs like spleen, lymph nodes, Peyer's patches of the small intestine and appendix.
These secondary lymphoid organs provide the sites for interaction of lymphocytes with the antigen.
Q115.
Match List I with List II:
List I:
A. The Evil Quartet
B. Ex situ conservation
C. Lantana camara
D. Dodo
List II:
I. Cryopreservation
II. Alien species invasion
III. Causes of biodiversity losses
IV. Extinction
Choose the option with all correct matches.
- A. A-III, B-II, C-IV, D-I
- B. A-III, B-II, C-I, D-IV
- C. A-III, B-I, C-II, D-IV ✓
- D. A-III, B-IV, C-II, D-I
Solution: The Evil Quartet — Causes of biodiversity losses
Ex situ conservation — Cryopreservation
Lantana camara — Alien species invasion
Dodo — Extinction
So A-III, B-I, C-II, D-IV.
Q116.
Read the following statements on plant growth and development.
(A) Parthenocarpy can be induced by auxins.
(B) Plant growth regulators can be involved in promotion as well as inhibition of growth.
(C) Dedifferentiation is a pre-requisite for re-differentiation.
(D) Abscisic acid is a plant growth promoter.
(E) Apical dominance promotes the growth of lateral buds.
Choose the option with all correct statements.
- A. B, D, E only
- B. A, B, C only ✓
- C. A, C, E only
- D. A, D, E only
Solution: ABA is a plant growth inhibitor and an inhibitor of plant metabolism, so statement (D) is wrong.
Apical dominance promotes growth of the apical bud and suppresses the lateral buds, so statement (E) is wrong.
Statements A, B and C are correct.
Q117.
Match List-I with List-II.
List-I:
A. Pteridophyte
B. Bryophyte
C. Angiosperm
D. Gymnosperm
List-II:
(I) Salvia
(II) Ginkgo
(III) Polytrichum
(IV) Salvinia
Choose the option with all correct matches.
- A. A-IV, B-III, C-II, D-I
- B. A-III, B-IV, C-II, D-I
- C. A-IV, B-III, C-I, D-II ✓
- D. A-III, B-IV, C-I, D-II
Solution: Pteridophyte — Salvinia
Bryophyte — Polytrichum
Angiosperm — Salvia
Gymnosperm — Ginkgo
So A-IV, B-III, C-I, D-II.
Q118.
Why can't insulin be given orally to diabetic patients?
- A. Its bioavailability will be increased
- B. Human body will elicit strong immune response
- C. It will be digested in Gastro-Intestinal (GI) tract ✓
- D. Because of structural variation
Solution: Insulin can't be administered orally to diabetic patients as, being a proteinaceous molecule, it will be digested in the gastro-intestinal tract.
Q119.
Which one of the following is the characteristic feature of gymnosperms?
- A. Gymnosperms have flowers for reproduction
- B. Seeds are enclosed in fruits
- C. Seeds are naked ✓
- D. Seeds are absent
Solution: The gymnosperms (Gymnos : naked, sperma : seed) are plants in which the ovules are not enclosed by an ovary wall and remain exposed, both before and after fertilization. The seeds that develop post-fertilization are not covered, i.e. naked.
Q120.
Frogs respire in water by skin and buccal cavity and on land by skin, buccal cavity and lungs.
Choose the correct answer from the following :
- A. The statement is false for both the environment
- B. The statement is true for water but false for land
- C. The statement is true for both the environment
- D. The statement is false for water but true for land ✓
Solution: In water, frogs respire through skin and not through the buccal cavity, i.e. they undergo cutaneous respiration only. So the statement is false for water.
On land, the buccal cavity, skin and lungs act as respiratory organs, i.e. they undergo buccopharyngeal, cutaneous and pulmonary respiration. So the statement is true for land.
Q121.
Silencing of specific mRNA is possible via RNAi because of
- A. Non-complementary ssRNA
- B. Complementary dsRNA ✓
- C. Inhibitory ssRNA
- D. Complementary tRNA
Solution: RNAi (RNA interference) takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA.
Q122.
Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?
- A. They have 75% identical genetic content.
- B. They are monozygotic twins.
- C. They are fraternal twins. ✓
- D. They were conceived through in vitro fertilization.
Solution: Fraternal twins, or dizygotic twins, come from 2 separate fertilized eggs; they usually develop 2 separate amniotic sacs, placentas and supporting structures.
Monozygotic twins come from one zygote and so are always of the same sex. If twins are a boy and a girl, this indicates they are fraternal twins.
Q123.
Match List I with List II :
List I:
A. Scutellum
B. Non-albuminous seed
C. Epiblast
D. Perisperm
List II:
I. Persistent nucellus
II. Cotyledon of Monocot seed
III. Groundnut
IV. Rudimentary cotyledon
Choose the option with all correct matches.
- A. A-II, B-IV, C-III, D-I
- B. A-II, B-III, C-IV, D-I ✓
- C. A-IV, B-III, C-II, D-I
- D. A-IV, B-III, C-I, D-II
Solution: Scutellum is the cotyledon of a monocot seed.
Groundnut seed is a non-albuminous seed.
Epiblast is the rudimentary cotyledon in a monocot seed.
Perisperm is persistent nucellus.
So A-II, B-III, C-IV, D-I.
Q124.
In frog, the Renal portal system is a special venous connection that acts to link :
- A. Kidney and lower part of body ✓
- B. Liver and intestine
- C. Liver and kidney
- D. Kidney and intestine
Solution: In frogs, special venous connections between the liver and intestine as well as between the kidney and lower parts of the body are present. The former is called the hepatic portal system and the latter is called the renal portal system.
Q125.
Match List-I with List-II.
List-I:
A. Heart
B. Kidney
C. Gastro-intestinal tract
D. Adrenal Cortex
List-II:
(I) Erythropoietin
(II) Aldosterone
(III) Atrial natriuretic factor
(IV) Secretin
Choose the correct answer from the options given below :
- A. A-III, B-I, C-IV, D-II ✓
- B. A-II, B-I, C-III, D-IV
- C. A-IV, B-III, C-II, D-I
- D. A-I, B-III, C-IV, D-II
Solution: Matching each organ with the hormone it secretes:
Heart — Atrial natriuretic factor
Kidney — Erythropoietin
Gastro-intestinal tract — Secretin
Adrenal cortex — Aldosterone
So A-III, B-I, C-IV, D-II.
Q126.
Cardiac activities of the heart are regulated by:
A. Nodal tissue
B. A special neural centre in the medulla oblongata
C. Adrenal medullary hormones
D. Adrenal cortical hormones
Choose the correct answer from the options given below :
- A. A, B and D Only
- B. A, B and C Only ✓
- C. A, B, C and D
- D. A, C and D Only
Solution: Normal cardiac activities of the heart are regulated intrinsically, i.e. auto regulated by specialised muscles (nodal tissue), hence the heart is called myogenic. A special neural centre in the medulla oblongata can moderate the cardiac function through the autonomic nervous system.
The sympathetic nervous system can increase the rate of heartbeat, ventricular contraction and thereby cardiac output.
Parasympathetic neural signals decrease the rate of heartbeat, speed of conduction of action potential and thereby the cardiac output. Adrenal medullary hormones can also increase the cardiac output.
So A, B and C only.
Q127.
Streptokinase produced by bacterium Streptococcus is used for
- A. Removing clots from blood vessels ✓
- B. Curd production
- C. Ethanol production
- D. Liver disease treatment
Solution: Streptokinase produced by the bacterium Streptococcus and modified by genetic engineering is used as a 'clot buster' for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to heart attack.
Curd production is done by Lactobacillus and ethanol production is done by Saccharomyces.
Q128.
Who is known as the father of Ecology in India?
- A. Birbal Sahni
- B. S.R. Kashyap
- C. Ramdeo Misra ✓
- D. Ram Udar
Solution: Ramdeo Misra is known as the father of Ecology in India.
Q129.
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : A typical unfertilised, angiosperm embryo sac at maturity is 8 nucleate and 7-celled.
Reason (R) : The egg apparatus has 2 polar nuclei.
In the light of the above statements, choose the correct answer from the options given below :
- A. A is false but R is true
- B. Both A and R are true and R is the correct explanation of A
- C. Both A and R are true but R is NOT the correct explanation of A
- D. A is true but R is false ✓
Solution: A typical angiosperm embryo sac at maturity is 7-celled and 8 nucleate, so the Assertion is true.
The polar nuclei are situated below the egg apparatus in the large central cell — they are not part of the egg apparatus. Three cells are grouped together at the micropylar end and constitute the egg apparatus, namely one egg cell and two synergids.
Hence A is true but R is false.
Q130.
Neoplastic characteristics of cells refer to :
A. A mass of proliferating cell
B. Rapid growth of cells
C. Invasion and damage to the surrounding tissue
D. Those confined to original location
Choose the correct answer from the options given below:
- A. B, C, D only
- B. A, B only
- C. A, B, C only ✓
- D. A, B, D only
Solution: The correct answer will include A, B and C only.
A neoplasm is a general term for any abnormal growth of tissue. Neoplastic characteristics of cells refer to a mass of proliferating cells, rapid growth of cells, and invasion and damage to the surrounding tissue.
Cancer specifically refers to malignant neoplasms, which are cancerous and invasive.
Benign tumours remain confined to their original location. Thus, D is not included in the answer.
The malignant tumours, on the other hand, are a mass of proliferating cells called neoplastic or tumour cells. These cells grow very rapidly, invading and damaging the surrounding normal tissues.
Q131.
Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence.
A. Prothallus stage
B. Meiosis in spore mother cells
C. Fertilisation
D. Formation of archegonia and antheridia in gametophyte.
E. Transfer of antherozoids to the archegonia in presence of water.
Choose the correct answer from the options given below:
- A. E, D, C, B, A
- B. B, A, D, E, C ✓
- C. B, A, E, C, D
- D. D, E, C, A, B
Solution: In a pteridophyte's life cycle, the correct sequence of stages is as follows:
B — Meiosis in spore mother cells
A — Prothallus stage
D — Formation of archegonia and antheridia in gametophyte
E — Transfer of antherozoids to the archegonia in presence of water
C — Fertilisation will occur
So the correct sequence is B, A, D, E, C.
Q132.
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Both wind and water pollinated flowers are not very colourful and do not produce nectar.
Reason (R) : The flowers produce enormous amount of pollen grains in wind and water pollinated flowers.
In the light of the above statements, choose the correct answer from the options given below:
- A. A is false but R is true
- B. Both A and R are true and R is the correct explanation of A
- C. Both A and R are true but R is NOT the correct explanation of A ✓
- D. A is true but R is false
Solution: Both statements are correct. Wind and water pollinated flowers are not very colourful and do not produce nectar, because they do not need to attract animal pollinators.
They also produce enormous amounts of pollen grains, because pollination by wind or water is a chance event and most of the pollen is lost.
But the abundance of pollen is not the reason for the absence of colour and nectar — both are separate consequences of abiotic pollination. So R is not the correct explanation of A.
Q133.
Which one of the following enzymes contains 'Haem' as the prosthetic group?
- A. Catalase ✓
- B. RuBisCo
- C. Carbonic anhydrase
- D. Succinate dehydrogenase
Solution: In peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is part of the active site of the enzymes.
Zinc is the cofactor in the enzyme carbonic anhydrase.
RuBisCo is the most abundant protein in the whole of the biosphere.
Succinate is the substrate of the enzyme succinic dehydrogenase.
Q134.
Match List-I with List-II.
List-I:
A. Emphysema
B. Angina Pectoris
C. Glomerulonephritis
D. Tetany
List-II:
I. Rapid spasms in muscle due to low Ca$^{++}$ in body fluid
II. Damaged alveolar walls and decreased respiratory surface
III. Acute chest pain when not enough oxygen is reaching to heart muscle
IV. Inflammation of glomeruli of kidney
Choose the correct answer from the options given below :
- A. A-II, B-III, C-IV, D-I ✓
- B. A-III, B-I, C-IV, D-II
- C. A-III, B-I, C-II, D-IV
- D. A-II, B-IV, C-III, D-I
Solution: Emphysema — Damaged alveolar walls and decreased respiratory surface
Angina pectoris — Acute chest pain when not enough oxygen is reaching the heart muscle
Glomerulonephritis — Inflammation of glomeruli of kidney
Tetany — Rapid spasms in muscle due to low Ca$^{++}$ in body fluid
So A-II, B-III, C-IV, D-I.
Q135.
Find the statement that is NOT correct with regard to the structure of monocot stem.
- A. Phloem parenchyma is absent.
- B. Hypodermis is parenchymatous. ✓
- C. Vascular bundles are scattered.
- D. Vascular bundles are conjoint and closed.
Solution: In a monocot stem, the hypodermis is sclerenchymatous, not parenchymatous — so that statement is not correct.
The other three are true of a monocot stem: phloem parenchyma is absent, the vascular bundles are scattered in the ground tissue, and they are conjoint and closed (no cambium).
Q136.
Which of the following statement is correct about location of the male frog copulatory pad?
- A. First digit of the fore limb ✓
- B. First and Second digit of fore limb
- C. First digit of hind limb
- D. Second digit of fore limb
Solution: In male frogs, a copulatory pad is present on the first digit of the forelimbs, which is absent in female frogs.
Q137.
Given below are two statements:
Statement I : The primary source of energy in an ecosystem is solar energy.
Statement II : The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP).
In the light of the above statements, choose the most appropriate answer from the options given below:
- A. Statement I is incorrect but statement II is correct
- B. Both statement I and statement II are correct
- C. Both statement I and statement II are incorrect
- D. Statement I is correct but statement II is incorrect ✓
Solution: The primary source of energy in the ecosystem is solar energy, so Statement I is correct.
Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. Net primary productivity is what is left after the respiratory losses of the producers are subtracted. Hence Statement II is incorrect.
Q138.
Polymerase chain reaction (PCR) amplifies DNA following the equation.
- A. $2N^{2}$
- B. $N^{2}$
- C. $2^{n}$ ✓
- D. $2n + 1$
Solution: PCR, i.e. polymerase chain reaction, amplifies DNA as per the equation $2^{n}$, where '$n$' refers to the number of cycles.
Thus, say, if 3 PCR cycles are run, then $2^{3}$, i.e. $2 \times 2 \times 2 \Rightarrow 8$ DNA fragments will be formed.
Q139.
Match List - I with List - II.
List - I:
A. Head
B. Middle piece
C. Acrosome
D. Tail
List - II:
(I) Enzymes
(II) Sperm motility
(III) Energy
(IV) Genetic material
Choose the correct answer from the options given below :
- A. A-III, B-II, C-I, D-IV
- B. A-IV, B-III, C-I, D-II ✓
- C. A-IV, B-III, C-II, D-I
- D. A-III, B-IV, C-II, D-I
Solution: The sperm head contains an elongated nucleus which possesses the genetic material.
The middle piece possesses numerous mitochondria, which produce energy for movement.
Acrosome is a cap-like structure filled with enzymes that help in fertilization of the ovum.
The tail of the sperm facilitates sperm motility essential for fertilisation.
So A-IV, B-III, C-I, D-II.
Q140.
Given below are two statements:
Statement I: In a floral formula $\oplus$ stands for zygomorphic nature of the flower, and $\underline{\text{G}}$ stands for inferior ovary.
Statement II: In a floral formula $\oplus$ stands for actinomorphic nature of the flower and $\underline{\text{G}}$ stands for superior ovary.
In the light of the above statements, choose the correct answer from the options given below:
- A. Statement I is incorrect but Statement II is correct ✓
- B. Both Statement I and Statement II are correct
- C. Both Statement I and Statement II are incorrect
- D. Statement I is correct but Statement II is incorrect
Solution: The floral formula symbol $\oplus$ is used for an actinomorphic flower, while % is used for a zygomorphic flower.
The symbol G represents the gynoecium, and G with a line underneath represents a superior ovary, while an inferior ovary is represented by G with a line above it.
Thus, Statement I is incorrect and Statement II is correct.
Q141.
From the statements given below choose the correct option :
A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S.
B. Each ribosome has two sub-units.
C. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S.
D. The two sub-units of 80S ribosome are 60S and 20S and that of 70S are 50S and 20S.
E. The two sub-units of 80S are 60S and 30S and that of 70S are 50S and 30S.
- A. B, D, E are true
- B. A, B, C are true ✓
- C. A, B, D are true
- D. A, B, E are true
Solution: The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S type.
Each ribosome has two sub-units.
The two sub-units of the 80S ribosome are 60S and 40S while those of the 70S are 50S and 30S.
So A, B and C are true.
Q142.
Each of the following characteristics represent a Kingdom proposed by Whittaker. Arrange the following in increasing order of complexity of body organization.
A. Multicellular heterotrophs with cell wall made of chitin.
B. Heterotrophs with tissue/organ/organ system level of body organization.
C. Prokaryotes with cell wall made of polysaccharides and amino acids.
D. Eukaryotic autotrophs with tissue/organ level of body organization.
E. Eukaryotes with cellular body organization.
Choose the correct answer from the options given below :
- A. C, E, A, B, D
- B. A, C, E, B, D
- C. C, E, A, D, B ✓
- D. A, C, E, D, B
Solution: Increasing order of complexity of body organisation in the kingdoms given by R.H. Whittaker is as follows:
C. Monera — Prokaryotes with cell wall made up of polysaccharide
E. Protista — Unicellular eukaryotes
A. Fungi — Multicellular heterotrophic with cell wall made up of chitin
D. Plantae — Eukaryotic autotrophs with tissue body organisation
B. Animalia — Heterotrophs with tissue/organ/organ system body organisation
Correct sequence is C, E, A, D, B.
Q143.
The correct sequence of events in the life cycle of bryophytes is
A. Fusion of antherozoid with egg.
B. Attachment of gametophyte to substratum.
C. Reduction division to produce haploid spores.
D. Formation of sporophyte.
E. Release of antherozoids into water.
Choose the correct answer from the options given below :
- A. D, E, A, B, C
- B. D, E, A, C, B
- C. B, E, A, C, D
- D. B, E, A, D, C ✓
Solution: The correct sequence of events in the life cycle of bryophytes is:
Attachment of gametophyte to substratum.
Release of antherozoids into water.
Fusion of antherozoid with egg.
Formation of sporophyte.
Reduction division to produce haploid spores.
So B, E, A, D, C.
Q144.
Which are correct:
A. Computed tomography and magnetic resonance imaging detect cancers of internal organs.
B. Chemotheraputics drugs are used to kill non-cancerous cells.
C. $\alpha$-interferon activate the cancer patients' immune system and helps in destroying the tumour.
D. Chemotherapeutic drugs are biological response modifiers.
E. In the case of leukaemia blood cell counts are decreased.
Choose the correct answer from the options given below:
- A. A and C only ✓
- B. B and D only
- C. D and E only
- D. C and D only
Solution: Statements A and C are correct while statements B, D and E are incorrect.
Chemotherapeutic drugs are used to kill cancerous cells, not non-cancerous ones.
In case of leukaemia, blood cell counts are increased.
$\alpha$-interferons are biological response modifiers, not the chemotherapeutic drugs.
Q145.
Name the class of enzyme that usually catalyze the following reaction :
$$\text{S} - \text{G} + \text{S}^{\#} \rightarrow \text{S} + \text{S}^{\#} - \text{G}$$
Where, G $\rightarrow$ a group other than hydrogen; S $\rightarrow$ a substrate; S$^{\#}$ $\rightarrow$ another substrate
- A. Ligase
- B. Hydrolase
- C. Lyase
- D. Transferase ✓
Solution: Enzymes catalysing a transfer of a G group (other than hydrogen) between a pair of substrates, S and S', are known as transferases.
$$\text{S} - \text{G} + \text{S}^{\#} \rightarrow \text{S} + \text{S}^{\#} - \text{G}$$
Ligases catalyse the linking together of 2 compounds such as C–O, C–S, C–N bonds etc.
Lyases catalyse removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds.
Hydrolases are enzymes that catalyse hydrolysis of ester, ether, peptide, glycosidic, C–C, C–halide or P–N bonds.
Q146.
Find the correct statement :
(A) In human pregnancy, the major organ systems are formed at the end of 12 weeks.
(B) In human pregnancy the major organ systems are formed at the end of 8 weeks.
(C) In human pregnancy heart is formed after one month of gestation.
(D) In human pregnancy, limbs and digits develop by the end of second month.
(E) In human pregnancy the appearance of hair is usually observed in the fifth month.
Choose the correct answer from the options given below :
- A. A, C, D and E only ✓
- B. A and E only
- C. B and C only
- D. B, C, D and E only
Solution: In a human female's pregnancy:
By the end of 12 weeks (1st trimester), most of the major organ systems are formed — not by the end of 8 weeks.
After one month of pregnancy, the embryo's heart is formed.
By the end of the second month of pregnancy, the foetus develops limbs and digits.
The first movements of the foetus and appearance of hair on the head are usually observed during the fifth month.
So A, C, D and E only.
Q147.
Which of the following is an example of non-distilled alcoholic beverage produced by yeast?
- A. Rum
- B. Whisky
- C. Brandy
- D. Beer ✓
Solution: Wine and beer are produced without distillation, whereas whisky, brandy and rum are produced by distillation of the fermented broth.
Q148.
Given below are two statements :
Statement I : In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable.
Statement II : DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but statement II is correct
- B. Both statement I and statement II are correct ✓
- C. Both statement I and statement II are incorrect
- D. Statement I is correct but statement II is incorrect
Solution: In the RNA world, RNA was the first genetic material, as there is enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.) evolved around RNA. RNA used to act as a genetic material as well as a catalyst — there are some important biochemical reactions in living systems that are catalysed by RNA catalysts, not by protein enzymes. So Statement I is correct.
Statement II is also correct, as DNA, being double stranded and having complementary strands, further resists changes by evolving a process of repair.
Q149.
Given below are two statements :
Statement I : Transfer RNAs and ribosomal RNA do not interact with mRNA.
Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.
In the light of the above statements, choose the most appropriate answer from the options given below :
- A. Statement I is incorrect but statement II is correct ✓
- B. Both statement I and statement II are correct
- C. Both statement I and statement II are incorrect
- D. Statement I is correct but statement II is incorrect
Solution: Both transfer RNAs and ribosomal RNA interact with mRNA — tRNA reads the codons on mRNA and rRNA forms the ribosome on which mRNA is translated. So Statement I is incorrect.
RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence, so Statement II is correct.
Q150.
What is the pattern of inheritance for polygenic trait?
- A. X-linked recessive inheritance pattern
- B. Mendelian inheritance pattern
- C. Non-mendelian inheritance pattern ✓
- D. Autosomal dominant pattern
Solution: Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes.
When human disorders are determined by mutation in a single gene then they are transmitted to the offspring as per the Mendelian principle. A polygenic trait shows a non-Mendelian inheritance pattern.
Q151.
In the seeds of cereals, the outer covering of endosperm separates the embryo by a protein-rich layer called :
- A. Aleurone layer ✓
- B. Coleoptile
- C. Coleorhiza
- D. Integument
Solution: In monocot seeds, the outer covering of endosperm separates the embryo by a proteinous layer called the aleurone layer.
Q152.
Match List I with List II:
List-I:
A. Chlorophyll a
B. Chlorophyll b
C. Xanthophylls
D. Carotenoids
List-II:
(I) Yellow-green
(II) Yellow
(III) Blue-green
(IV) Yellow to Yellow-orange
Choose the option with all correct matches.
- A. A-I, B-IV, C-III, D-II
- B. A-III, B-IV, C-II, D-I
- C. A-III, B-I, C-II, D-IV ✓
- D. A-I, B-II, C-IV, D-III
Solution: A chromatographic separation of the leaf pigments shows that the colour that we see in leaves is not due to a single pigment but due to four pigments:
Chlorophyll a — bright or blue-green in the chromatogram
Chlorophyll b — yellow-green
Xanthophylls — yellow
Carotenoids — yellow to yellow-orange
So A-III, B-I, C-II, D-IV.
Q153.
Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?
- A. Phage
- B. Bacterium ✓
- C. Yeast
- D. Virus
Solution: The correct answer is bacterium.
In 1983, Eli Lilly, an American company, prepared two DNA sequences corresponding to 'A' and 'B' chains of human insulin and introduced them into plasmids of E. coli (a gram negative bacterium) to produce the insulin chains.
Q154.
Which of the following are the post-transcriptional events in an eukaryotic cell?
A. Transport of pre-mRNA to cytoplasm prior to splicing.
B. Removal of introns and joining of exons.
C. Addition of methyl group at 5' end of hnRNA.
D. Addition of adenine residues at 3' end of hnRNA.
E. Base pairing of two complementary RNAs.
Choose the correct answer from the options given below :
- A. C, D, E only
- B. A, B, C only
- C. B, C, D only ✓
- D. B, C, E only
Solution: The primary transcript is converted into functional mRNA after post-transcriptional processing, which involves 3 steps:
Modification of the 5' end by capping (addition of a methyl group),
Tailing (addition of adenine residues at the 3' end),
Splicing (removal of introns and joining of exons).
Transport of pre-mRNA to the cytoplasm prior to splicing is not a post-transcriptional processing event, and base pairing of two complementary RNAs is not one either.
Hence statements B, C and D are the post-transcriptional modification events in a eukaryotic cell.
Q155.
Match List-I with List-II.
List-I:
A. Centromere
B. Cilium
C. Cristae
D. Cell membrane
List-II:
I. Mitochondrion
II. Cell division
III. Cell movement
IV. Phospholipid Bilayer
Choose the correct answer from the options given below :
- A. A-II, B-III, C-I, D-IV ✓
- B. A-I, B-II, C-III, D-IV
- C. A-II, B-I, C-IV, D-III
- D. A-IV, B-II, C-III, D-I
Solution: Centromere — helps in cell division
Cilium — helps in cell movement
Cristae — finger-like infoldings of the inner membrane of mitochondria
Cell membrane — is a phospholipid bilayer
So A-II, B-III, C-I, D-IV.
Q156.
Match List I with List II:
List-I:
A. Alfred Hershey and Martha Chase
B. Euchromatin
C. Frederick Griffith
D. Heterochromatin
List-II:
I. Streptococcus pneumoniae
II. Densely packed and dark-stained
III. Loosely packed and light-stained
IV. DNA as genetic material confirmation
Choose the correct answer from the options given below:
- A. A-III, B-II, C-IV, D-I
- B. A-II, B-IV, C-I, D-III
- C. A-IV, B-II, C-I, D-III
- D. A-IV, B-III, C-I, D-II ✓
Solution: The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase.
Euchromatin is the lightly stained region with loosely packed chromatin fibre.
Frederick Griffith performed a series of experiments by selecting the different strains of Streptococcus pneumoniae.
Heterochromatin is the darkly stained region with tightly packed chromatin fibre.
So A-IV, B-III, C-I, D-II.
Q157.
Which chromosome in the human genome has the highest number of genes?
- A. Chromosome 10
- B. Chromosome X
- C. Chromosome Y
- D. Chromosome 1 ✓
Solution: In the human genome, Chromosome 1 has the highest number of genes, i.e. 2968.
Q158.
What are the potential drawbacks in adoption of the IVF method?
A. High fatality risk to mother
B. Expensive instruments and reagents
C. Husband/wife necessary for being donors
D. Less adoption of orphans
E. Not available in India
F. Possibility that the early embryo does not survive
Choose the correct answer from the options given below:
- A. A, B, C, E, F only
- B. B, D, F only ✓
- C. A, C, D, F only
- D. A, B, C, D only
Solution: Statements B, D and F are correct while statements A, C and E are incorrect.
Husband/wife is not necessary for being donors, and IVF is available in India.
The genuine drawbacks are the expense of the instruments and reagents, the fall in adoption of orphans, and the possibility that the early embryo does not survive.
Q159.
Which one of the following is an example of ex-situ conservation?
- A. Protected areas
- B. National Park
- C. Wildlife Sanctuary
- D. Zoos and botanical gardens ✓
Solution: Zoological parks (zoos), botanical gardens and wildlife safari parks are examples of ex-situ conservation.
Sacred groves, biosphere reserves, national parks and wildlife sanctuaries are examples of in-situ conservation.
Q160.
A specialised membranous structure in a prokaryotic cell which helps in cell wall formation, DNA replication and respiration is
- A. Endoplasmic Reticulum
- B. Mesosome ✓
- C. Chromatophores
- D. Cristae
Solution: Mesosome is a membranous extension in the bacterial cell that helps in cell wall formation and DNA replication, and contains enzymes for respiration.
Q161.
The plasmid shown carries an origin of replication (Ori), an ampicillin resistance gene (amp), a tetracycline resistance gene (Tet R), a BamH1 site, and an EcoRI site that lies within the $\beta$-galactosidase gene.
In the above represented plasmid, an alien piece of DNA is inserted at EcoRI site. Which of the following strategies will be chosen to select the recombinant colonies?
- A. Blue color colonies grown on ampicillin plates can be selected.
- B. Using ampicillin & tetracycline containing medium plate.
- C. Blue color colonies will be selected.
- D. White color colonies will be selected. ✓
Solution: The correct answer is that white-coloured colonies will be selected.
Since an alien piece of DNA is being inserted at the EcoRI site, the gene $\beta$-galactosidase present here will undergo insertional inactivation.
This gene is responsible for producing blue-coloured colonies, but since it has been insertionally inactivated, white coloured colonies will be produced.
Ampicillin and tetracycline resistance genes present in the given DNA will remain intact. Thus, the given DNA will show amp$^{R}$ and tet$^{R}$, so neither antibiotic distinguishes recombinants from non-recombinants.
Q162.
What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?
- A. Vena cava ✓
- B. Aorta
- C. Pulmonary artery
- D. Pulmonary vein
Solution: Frog's heart is a muscular structure with three chambers. It receives deoxygenated blood from body parts through the major veins called vena cava. Vena cava carries deoxygenated blood.
Aorta and pulmonary vein carry oxygenated blood, whereas the pulmonary artery carries deoxygenated blood towards the lungs.
Q163.
Which of following organisms cannot fix nitrogen?
A. Azotobacter
B. Oscillatoria
C. Anabaena
D. Volvox
E. Nostoc
Choose the correct answer from the options given below:
- A. E only
- B. A only
- C. D only ✓
- D. B only
Solution: Azotobacter, Oscillatoria, Anabaena and Nostoc can fix nitrogen, but Volvox cannot fix nitrogen — it is a colonial green alga, not a nitrogen-fixing prokaryote.
Q164.
While trying to find out the characteristic of a newly found animal, a researcher did the histology of adult animal and observed a cavity with presence of mesodermal tissue towards the body wall but no mesodermal tissue was observed towards the alimentary canal. What could be the possible coelome of that animal?
- A. Spongocoelomate
- B. Acoelomate
- C. Pseudocoelomate ✓
- D. Schizocoelomate
Solution: In pseudocoelomates, the body cavity is not entirely lined with mesoderm; instead, mesodermal tissue is present along the body wall but not towards the gut.
Schizocoelomates are animals whose coelom or body cavity develops from a split in the mesoderm, the middle germ layer of the embryo.
In acoelomates, the coelom is absent.
Spongocoel is a central cavity found in sponges.
Q165.
Which one of the following statements refers to Reductionist Biology?
- A. Behavioural approach to study and understand living organisms
- B. Physico-chemical approach to study and understand living organisms ✓
- C. Physiological approach to study and understand living organisms
- D. Chemical approach to study and understand living organisms
Solution: The physico-chemical approach to study and understand living organisms is called 'Reductionist Biology'.
Q166.
Epiphytes that are growing on a mango branch is an example of which of the following?
- A. Amensalism
- B. Commensalism ✓
- C. Mutualism
- D. Predation
Solution: Commensalism is the type of interaction in which one species benefits and another is neither harmed nor benefited.
An orchid growing as an epiphyte on a mango branch is an example of commensalism: the orchid gains support and light, while the mango tree is unaffected.
Q167.
Which one of the following phytohormones promotes nutrient mobilization which helps in the delay of leaf senescence in plants?
- A. Cytokinin ✓
- B. Ethylene
- C. Abscisic acid
- D. Gibberellin
Solution: Cytokinins help to overcome apical dominance. They promote nutrient mobilisation, which helps in the delay of leaf senescence.
Q168.
The complex II of mitochondrial electron transport chain is also known as
- A. NADH dehydrogenase
- B. Cytochrome bc$_1$
- C. Succinate dehydrogenase ✓
- D. Cytochrome c oxidase
Solution: Complex II of the mitochondrial electron transport chain is also known as succinate dehydrogenase.
The others are: NADH dehydrogenase (complex I), cytochrome bc$_1$ (complex III) and cytochrome c oxidase (complex IV).