MECEE-BL Entrance Model Set 2 with Answers

1. a10 m
2. b30 m
3. c20 m
4. d40 m

Answer (c): $h = \frac{u^2}{2g} = \frac{400}{20} = 20$ m.

Revise: Kinematics (Class 11 notes)

1. aequal
2. bgreater for $30^\circ$
3. cin the ratio $1:\sqrt{3}$
4. dgreater for $60^\circ$

Answer (a): Range $R = \frac{u^2\sin 2\theta}{g}$ and $\sin 60^\circ = \sin 120^\circ$, so complementary angles give equal ranges.

Revise: Kinematics (Class 11 notes)

1. a5 m/s
2. b15 m/s
3. c60 m/s
4. d12 m/s

Answer (b): $a = F/m = 5\ \mathrm{m/s^2}$, so $v = at = 5 \times 3 = 15$ m/s.

Revise: Dynamics (Class 11 notes)

1. a2 N s
2. b8 N s
3. c0 N s
4. d4 N s

Answer (d): Impulse $= m(v - (-v)) = 0.2 \times 20 = 4$ N s.

Revise: Dynamics (Class 11 notes)

1. a50 W
2. b200 W
3. c500 W
4. d1000 W

Answer (c): $P = \frac{mgh}{t} = \frac{100 \times 10 \times 10}{20} = 500$ W.

Revise: Work, Energy and Power (Class 11 notes)

1. a20 m/s
2. b100 m/s
3. c5 m/s
4. d10 m/s

Answer (d): Friction supplies the centripetal force: $v_{max} = \sqrt{\mu r g} = \sqrt{0.5 \times 20 \times 10} = 10$ m/s.

Revise: Circular Motion (Class 11 notes)

1. a5.6 km/s
2. b15.8 km/s
3. c11.2 km/s
4. d7.9 km/s

Answer (d): $v_o = v_e/\sqrt{2} = 11.2/1.414 \approx 7.9$ km/s.

Revise: Gravitation (Class 11 notes)

1. a$2\times10^{9}\ \mathrm{J/m^3}$
2. b$2\times10^{3}\ \mathrm{J/m^3}$
3. c$1\times10^{3}\ \mathrm{J/m^3}$
4. d$5\times10^{2}\ \mathrm{J/m^3}$

Answer (c): Energy density $= \frac{1}{2}\times\text{stress}\times\text{strain} = \frac{1}{2}(2\times10^{6})(10^{-3}) = 10^{3}\ \mathrm{J/m^3}$.

Revise: Elasticity (Class 11 notes)

1. a0.6
2. b1.0
3. c0.5
4. d0.4

Answer (a): Floating means weight equals buoyant force: $\rho_{block}V = \rho_{water}V_{sub}$, so $V_{sub}/V = 600/1000 = 0.6$.

Revise: Fluid Statics (Class 12 notes)

1. a1.57 s
2. b6.3 s
3. c0.063 s
4. d0.63 s

Answer (d): $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{1/100} = 2\pi/10 \approx 0.63$ s.

Revise: Periodic Motion (Class 12 notes)

1. a$100^\circ$C
2. b$60^\circ$C
3. c$50^\circ$C
4. d$40^\circ$C

Answer (c): Equal masses of the same liquid give the mean: $(80+20)/2 = 50^\circ$C.

Revise: Quantity of Heat (Class 11 notes)

1. a6.72 kJ
2. b67.2 kJ
3. c168 kJ
4. d33.6 kJ

Answer (b): $Q = mL = 200 \times 336 = 67200$ J $= 67.2$ kJ.

Revise: Quantity of Heat (Class 11 notes)

1. a$4^\circ$C
2. b$100^\circ$C
3. c$0^\circ$C
4. d$10^\circ$C

Answer (a): Water shows anomalous expansion: below $4^\circ$C it expands on cooling, so its density is greatest at $4^\circ$C.

Revise: Thermal Expansion (Class 11 notes)

1. a1.5 times
2. b2 times
3. c4 times
4. d12 times

Answer (b): Charles' law: $V \propto T$. $T_1 = 300$ K and $T_2 = 600$ K, so $V_2/V_1 = 2$.

Revise: Ideal Gas (Class 11 notes)

1. a300 J
2. b1000 J
3. c60 J
4. d600 J

Answer (d): $W = P\Delta V = 2\times10^{5}\times 3\times10^{-3} = 600$ J.

Revise: First Law of Thermodynamics (Class 12 notes)

1. a6
2. b1.2
3. c5
4. d0.2

Answer (c): $\mathrm{COP} = \frac{T_c}{T_h - T_c} = \frac{250}{50} = 5$.

Revise: Second Law of Thermodynamics (Class 12 notes)

1. a2 times
2. b4 times
3. c8 times
4. d16 times

Answer (d): Stefan's law: $P \propto T^4$, so doubling $T$ gives $2^4 = 16$ times.

Revise: Rate of Heat Flow (Class 11 notes)

1. a200 m/s
2. b100 m/s
3. c800 m/s
4. d402 m/s

Answer (a): Compare with $y = A\sin(\omega t - kx)$: $\omega = 400$, $k = 2$, so $v = \omega/k = 200$ m/s.

Revise: Wave Motion (Class 12 notes)

1. a50 Hz
2. b25 Hz
3. c200 Hz
4. d100 Hz

Answer (d): $f_1 = \frac{v}{2L} = \frac{100}{2\times0.5} = 100$ Hz.

Revise: Wave in Pipes and Strings (Class 12 notes)

1. a500 Hz
2. b550 Hz
3. c534 Hz
4. d450 Hz

Answer (b): $f' = f,\frac{v + v_o}{v} = 500\times\frac{374}{340} = 550$ Hz.

Revise: Acoustic Phenomena (Class 12 notes)

1. a10 dB
2. b20 dB
3. c100 dB
4. d2 dB

Answer (b): $\Delta\beta = 10\log_{10}(100) = 20$ dB.

Revise: Acoustic Phenomena (Class 12 notes)

1. a10 cm
2. b25 cm
3. c2.4 cm
4. d50 cm

Answer (a): $P = 4 + 6 = 10$ D, so $f = 1/P = 0.1$ m $= 10$ cm.

Revise: Lenses (Class 11 notes)

1. a$30^\circ$
2. b$60^\circ$
3. c$45^\circ$
4. d$15^\circ$

Answer (a): $\mu = \frac{\sin\frac{A+D}{2}}{\sin\frac{A}{2}}$ gives $\sin\frac{60+D}{2} = \sqrt{2}\times0.5 = \sin45^\circ$, so $D = 30^\circ$.

Revise: Refraction Through Prisms (Class 11 notes)

1. a$2.5\times10^{-4}$ rad
2. b$1\times10^{-3}$ rad
3. c$2.5\times10^{-3}$ rad
4. d$4\times10^{-3}$ rad

Answer (c): $a\sin\theta = \lambda$, so $\theta \approx \frac{500\times10^{-9}}{2\times10^{-4}} = 2.5\times10^{-3}$ rad.

Revise: Diffraction of Light (Class 12 notes)

1. a$90^\circ$
2. b$30^\circ$
3. c$60^\circ$
4. d$45^\circ$

Answer (b): $\sin C = 1/\mu = 1/2$, so $C = 30^\circ$.

Revise: Refraction at Plane Surface (Class 11 notes)

1. a$40\ \Omega$
2. b$10\ \Omega$
3. c$5\ \Omega$
4. d$20\ \Omega$

Answer (a): Volume is fixed, so area halves while length doubles: $R \propto l^2$, giving $4\times10 = 40\ \Omega$.

Revise: DC Circuit (Class 11 notes)

1. a$60\ \Omega$
2. b$20\ \Omega$
3. c$15\ \Omega$
4. d$40\ \Omega$

Answer (a): Balance: $P/Q = R/S$, so $S = RQ/P = 30\times20/10 = 60\ \Omega$.

Revise: Electrical Circuit One (Class 12 notes)

1. a12 V
2. b0.3 V
3. c3.0 V
4. d1.2 V

Answer (d): $E = (\text{gradient})\times l = 0.2\times6 = 1.2$ V.

Revise: Electrical Circuit One (Class 12 notes)

1. amaximum
2. bminimum
3. cnegative
4. dzero

Answer (a): The emf rises to a maximum at the neutral temperature (where $dE/dT = 0$) and then falls.

Revise: Thermoelectric Effects (Class 12 notes)

1. a$1.26\times10^{-5}$ T
2. b$2.5\times10^{-4}$ T
3. c$1.26\times10^{-4}$ T
4. d$6.3\times10^{-4}$ T

Answer (c): $B = \frac{\mu_0 N I}{2r} = \frac{4\pi\times10^{-7}\times10\times2}{0.2} = 4\pi\times10^{-5} \approx 1.26\times10^{-4}$ T.

Revise: Magnetic Field (Class 12 notes)

1. a$2\times10^{-4}$ N/m, attractive
2. b$2\times10^{-4}$ N/m, repulsive
3. c$2\times10^{-5}$ N/m, attractive
4. d$4\times10^{-4}$ N/m, repulsive

Answer (a): $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2\times10^{-7}\times50}{0.05} = 2\times10^{-4}$ N/m; parallel currents attract.

Revise: Magnetic Field (Class 12 notes)

1. aremains the same
2. bbecomes double
3. cbecomes four times
4. dbecomes half

Answer (b): $r = \frac{mv}{qB}$, so $r \propto v$ and doubling the speed doubles the radius.

Revise: Magnetic Field (Class 12 notes)

1. a0.8 V
2. b8 V
3. c80 V
4. d20 V

Answer (c): $\varepsilon = L\frac{\Delta I}{\Delta t} = 2\times\frac{4}{0.1} = 80$ V.

Revise: Electromagnetic Induction (Class 12 notes)

1. a1590 Hz
2. b159 Hz
3. c628 Hz
4. d15.9 Hz

Answer (b): $f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-6}}} = \frac{1000}{2\pi} \approx 159$ Hz.

Revise: Alternating Currents (Class 12 notes)

1. a900 V
2. b90 V
3. c9 V
4. d45 V

Answer (b): $V = \frac{kq}{r} = \frac{9\times10^{9}\times5\times10^{-9}}{0.5} = 90$ V.

Revise: Potential Difference and Potential Energy (Class 11 notes)

1. a$2\times10^{-2}$ J
2. b$2\times10^{-6}$ J
3. c$5\times10^{7}$ J
4. d$2\times10^{-4}$ J

Answer (d): $W = qV = 2\times10^{-6}\times100 = 2\times10^{-4}$ J.

Revise: Potential Difference and Potential Energy (Class 11 notes)

1. a1/4 times
2. b16 times
3. c4 times
4. dunchanged

Answer (c): $C = K C_0$, so the capacitance is multiplied by 4.

Revise: Capacitor (Class 11 notes)

1. athe same as at the surface
2. buniform and non-zero
3. czero
4. dmaximum at the centre

Answer (c): Charge resides on the outer surface of a conductor, so by Gauss' law the field inside the cavity is zero.

Revise: Electric Field (Class 11 notes)

1. a1.24 eV
2. b0.62 eV
3. c2.48 eV
4. d4.96 eV

Answer (c): $\phi = \frac{hc}{\lambda_0} = \frac{1240}{500} = 2.48$ eV.

Revise: Photons (Class 12 notes)

1. abecomes four times
2. bbecomes half
3. cdoubles
4. dremains the same

Answer (d): Maximum kinetic energy depends on frequency and work function, not on intensity. Intensity only changes the number of photoelectrons.

Revise: Photons (Class 12 notes)

1. a1.89 eV
2. b13.6 eV
3. c10.2 eV
4. d3.4 eV

Answer (c): $E = 13.6\left(1 - \frac{1}{4}\right) = 10.2$ eV.

Revise: Quantization of Energy (Class 12 notes)

1. athe intensity of the beam
2. bthe material of the target
3. cthe accelerating voltage only
4. dthe filament current

Answer (b): Characteristic lines come from electron transitions in the target atoms (Moseley's law), so they depend on the target element.

Revise: Electrons (Class 12 notes)

1. a10 Å
2. b100 Å
3. c0.1 Å
4. d1 Å

Answer (d): $\lambda = \frac{12.27}{\sqrt{V}}\ \text{Å} = \frac{12.27}{\sqrt{150}} \approx 1$ Å.

Revise: Quantization of Energy (Class 12 notes)

1. a5 min
2. b7.5 min
3. c3 min
4. d15 min

Answer (a): $800\to100$ is a factor of 8 $= 2^3$, so three half-lives in 15 min, giving $T_{1/2} = 5$ min.

Revise: Radioactivity and Nuclear Reaction (Class 12 notes)

1. aan isobar of the original nucleus
2. ban isotope of the original nucleus
3. can isotone of the original nucleus
4. dthe same nucleus

Answer (b): Alpha: $Z-2$, $A-4$. Two beta-minus: $Z+2$. The net $Z$ is unchanged while $A$ drops by 4, so it is an isotope.

Revise: Radioactivity and Nuclear Reaction (Class 12 notes)

1. anuclear fusion of hydrogen
2. bradioactive decay of thorium
3. cchemical burning of carbon
4. dnuclear fission of uranium

Answer (a): In the Sun, hydrogen nuclei fuse to form helium, releasing energy.

Revise: Nuclear Physics (Class 11 notes)

1. aincreases
2. bdecreases
3. cremains unchanged
4. dbecomes infinite

Answer (b): Forward bias pushes majority carriers towards the junction and narrows the depletion layer.

Revise: Semiconductor Devices (Class 12 notes)

1. aat least one input is 0
2. bboth inputs are 1
3. cat least one input is 1
4. dboth inputs are 0

Answer (b): NAND is the complement of AND, and AND is 1 only when both inputs are 1, so NAND is 0 only then.

Revise: Semiconductor Devices (Class 12 notes)

1. a0.02
2. b50
3. c98
4. d49

Answer (d): $\beta = \frac{\alpha}{1-\alpha} = \frac{0.98}{0.02} = 49$.

Revise: Semiconductor Devices (Class 12 notes)

1. areverse breakdown
2. bthe cut-off of a transistor
3. cforward bias
4. dthe saturation region only

Answer (a): In reverse breakdown the voltage across a Zener stays almost constant over a wide current range, which regulates the supply.

Revise: Semiconductor Devices (Class 12 notes)

1. a$6.02\times10^{23}$
2. b$1.2\times10^{23}$
3. c$1.8\times10^{24}$
4. d$3.6\times10^{24}$

Answer (c): $18$ g is 1 mol of $\mathrm{H_2O}$, which is $6.02\times10^{23}$ molecules with 3 atoms each, giving $1.8\times10^{24}$ atoms.

Revise: Stoichiometry (Class 11 notes)

1. a$\mathrm{CH_4O}$
2. b$\mathrm{CHO}$
3. c$\mathrm{CH_2O}$
4. d$\mathrm{C_2H_4O}$

Answer (c): Moles: C $= 40/12 = 3.33$, H $= 6.7/1 = 6.7$, O $= 53.3/16 = 3.33$. The ratio is $1:2:1$.

Revise: Stoichiometry (Class 11 notes)

1. a72 g
2. b4 g
3. c36 g
4. d18 g

Answer (c): $4$ g $\mathrm{H_2}$ is 2 mol, giving 2 mol of water $= 2\times18 = 36$ g.

Revise: Stoichiometry (Class 11 notes)

1. a2
2. b3
3. c0
4. d1

Answer (d): Radial nodes $= n - l - 1 = 3 - 1 - 1 = 1$.

Revise: Atomic Structure (Class 11 notes)

1. a$[\mathrm{Ar}],3d^{4}4s^{2}$
2. b$[\mathrm{Ar}],3d^{6}$
3. c$[\mathrm{Ar}],4s^{2}4p^{4}$
4. d$[\mathrm{Ar}],3d^{5}4s^{1}$

Answer (d): A half-filled $d$ subshell is extra stable, so one $4s$ electron moves to $3d$ giving $3d^54s^1$.

Revise: Atomic Structure (Class 11 notes)

1. a$\mathrm{Al^{3+}}$
2. b$\mathrm{Mg^{2+}}$
3. c$\mathrm{Na^+}$
4. d$\mathrm{O^{2-}}$

Answer (d): All have 10 electrons. The smallest nuclear charge ($Z = 8$) holds them least tightly, so $\mathrm{O^{2-}}$ is the largest.

Revise: Classification of Elements (Class 11 notes)

1. a2.5
2. b1.5
3. c2
4. d3

Answer (a): $\mathrm{O_2}$ has bond order 2. Removing one antibonding electron gives $(10-5)/2 = 2.5$.

Revise: Chemical Bonding (Class 11 notes)

1. a$\mathrm{BF_3}$
2. b$\mathrm{NH_3}$
3. c$\mathrm{SO_2}$
4. d$\mathrm{H_2O}$

Answer (a): $\mathrm{BF_3}$ is trigonal planar, so the three bond dipoles cancel. The others are bent or pyramidal.

Revise: Chemical Bonding (Class 11 notes)

1. a$\mathrm{Cu}$
2. b$\mathrm{Zn}$
3. c$\mathrm{CuSO_4}$
4. d$\mathrm{ZnSO_4}$

Answer (c): $\mathrm{Cu^{2+}}$ gains electrons (is reduced), so $\mathrm{CuSO_4}$ is the oxidising agent.

Revise: Oxidation and Reduction (Class 11 notes)

1. a32 g/mol
2. b16 g/mol
3. c64 g/mol
4. d8 g/mol

Answer (a): Moles $= 5.6/22.4 = 0.25$, so molar mass $= 8/0.25 = 32$ g/mol.

Revise: States of Matter (Class 11 notes)

1. a3 atm
2. b5 atm
3. c2 atm
4. d2.5 atm

Answer (c): $p_{N_2} = x_{N_2}P = \frac{2}{5}\times5 = 2$ atm.

Revise: States of Matter (Class 11 notes)

1. a$K_p = K_c(RT)^{2}$
2. b$K_p = K_c(RT)$
3. c$K_p = K_c(RT)^{-1}$
4. d$K_p = K_c$

Answer (b): $K_p = K_c(RT)^{\Delta n}$ with $\Delta n = 2 - 1 = 1$.

Revise: Chemical Equilibrium (Class 11 notes)

1. a1.3
2. b1
3. c2
4. d0.05

Answer (b): $[\mathrm{H^+}] = 2\times0.05 = 0.1$ M, so pH $= -\log0.1 = 1$.

Revise: Ionic Equilibrium (Class 12 notes)

1. a$1\times10^{-10}$ M
2. b$1\times10^{-5}$ M
3. c$1\times10^{-3}$ M
4. d$2\times10^{-5}$ M

Answer (b): $K_{sp} = s^2$, so $s = \sqrt{10^{-10}} = 10^{-5}$ M.

Revise: Ionic Equilibrium (Class 12 notes)

1. a$-4RT$
2. b$-2RT$
3. c$0$
4. d$+2RT$

Answer (b): $\Delta H - \Delta U = \Delta n_g RT$ with $\Delta n_g = 2 - 4 = -2$.

Revise: Chemical Thermodynamic (Class 12 notes)

1. a$-212.3$ kJ/mol
2. b$-106.2$ kJ/mol
3. c$+212.3$ kJ/mol
4. d$-2.12$ kJ/mol

Answer (a): $\Delta G^\circ = -nFE^\circ = -2\times96500\times1.10 = -212300$ J $= -212.3$ kJ.

Revise: Electrochemistry (Class 12 notes)

1. a$\mathrm{s^{-1}}$
2. b$\mathrm{mol,L^{-1},s^{-1}}$
3. c$\mathrm{L^2,mol^{-2},s^{-1}}$
4. d$\mathrm{L,mol^{-1},s^{-1}}$

Answer (d): Rate $= k[A]^2$, so $k = \frac{\mathrm{mol,L^{-1},s^{-1}}}{(\mathrm{mol,L^{-1}})^2} = \mathrm{L,mol^{-1},s^{-1}}$.

Revise: Chemical Kinetics (Class 12 notes)

1. a$\mathrm{MgO}$
2. b$\mathrm{Al_2O_3}$
3. c$\mathrm{Na_2O}$
4. d$\mathrm{CO_2}$

Answer (b): $\mathrm{Al_2O_3}$ reacts with both acids and bases, so it is amphoteric. $\mathrm{Na_2O}$ and $\mathrm{MgO}$ are basic and $\mathrm{CO_2}$ is acidic.

Revise: Classification of Elements (Class 11 notes)

1. anickel
2. bvanadium pentoxide
3. cfinely divided iron
4. dplatinum

Answer (c): Haber's process uses iron as the catalyst with molybdenum as a promoter. Platinum is used in the Ostwald process.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. alimestone
2. bcaustic soda
3. cquicklime solution
4. ddry slaked lime

Answer (d): $\mathrm{Ca(OH)_2 + Cl_2 \to CaOCl_2 + H_2O}$ with dry slaked lime.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. a$\mathrm{H_3PO_3}$
2. b$\mathrm{H_3PO_4}$
3. c$\mathrm{HNO_3}$
4. d$\mathrm{H_3PO_2}$

Answer (a): $\mathrm{H_3PO_3}$ has two ionisable $-\mathrm{OH}$ hydrogens (one H is bonded to P), so it is dibasic.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. atetrahedral
2. boctahedral
3. csee-saw
4. dsquare planar

Answer (d): Xe has 4 bond pairs and 2 lone pairs (octahedral electron geometry). The lone pairs sit opposite each other, leaving a square planar shape.

Revise: Chemistry of Non-Metals (Class 11 notes)

1. aremove silica
2. bact as a reducing agent
3. coxidise the aluminium
4. dlower the melting point and improve conductivity

Answer (d): Pure alumina melts near 2050 degrees C. Cryolite lowers the fusion temperature and makes the melt conduct.

Revise: Chemistry of Metals (Class 11 notes)

1. acalcium and magnesium chlorides
2. bcalcium sulphate
3. csodium carbonate
4. dcalcium and magnesium bicarbonates

Answer (d): Bicarbonates of Ca and Mg are removed on boiling, which is why the hardness is temporary.

Revise: Chemistry of Metals (Class 11 notes)

1. a3.87 BM
2. b2.83 BM
3. c4.90 BM
4. d5.92 BM

Answer (a): $\mathrm{Cr^{3+}}$ is $3d^3$ with 3 unpaired electrons: $\mu = \sqrt{n(n+2)} = \sqrt{15} = 3.87$ BM.

Revise: Transition Metals (Class 12 notes)

1. awhite anhydrous $\mathrm{CuSO_4}$
2. bblack CuO at once
3. cgreen $\mathrm{CuSO_4\cdot3H_2O}$
4. dred $\mathrm{Cu_2O}$

Answer (a): The water of crystallisation is driven off, leaving white anhydrous copper sulphate (decomposition to CuO needs a much higher temperature).

Revise: Copper (Class 12 notes)

1. alead
2. barsenic
3. ccadmium
4. dmercury

Answer (d): Minamata disease in Japan was caused by methylmercury in contaminated fish.

Revise: Study of Heavy Metals (Class 12 notes)

1. a2-pentyne
2. bpent-2-ene
3. cpentane
4. dpent-3-ene

Answer (b): Five carbons with a double bond starting at C-2 when numbered from the nearer end: pent-2-ene.

Revise: Fundamental Principles (Class 11 notes)

1. aethane
2. bmethanol
3. cacetaldehyde
4. ddimethyl ether

Answer (d): Both have the formula $\mathrm{C_2H_6O}$: $\mathrm{CH_3CH_2OH}$ and $\mathrm{CH_3OCH_3}$ differ in functional group.

Revise: Fundamental Principles (Class 11 notes)

1. a$(\mathrm{CH_3})_3\mathrm{C^+}$
2. b$\mathrm{CH_3CH_2^+}$
3. c$\mathrm{CH_3^+}$
4. d$(\mathrm{CH_3})_2\mathrm{CH^+}$

Answer (a): Stability rises with the number of alkyl groups (inductive effect and hyperconjugation), so the tertiary ion is most stable.

Revise: Fundamental Principles (Class 11 notes)

1. abutane
2. bmethane
3. cethane
4. dpropane

Answer (c): Wurtz reaction: $\mathrm{2CH_3Br + 2Na \to CH_3{-}CH_3 + 2NaBr}$.

Revise: Hydrocarbons (Class 11 notes)

1. abut-2-yne
2. bbutane
3. cbut-1-yne
4. dbut-1-ene

Answer (c): Only a terminal alkyne has an acidic $\equiv\mathrm{C{-}H}$ that forms a silver acetylide precipitate.

Revise: Hydrocarbons (Class 11 notes)

1. a$-\mathrm{Cl}$
2. b$-\mathrm{CH_3}$
3. c$-\mathrm{NO_2}$
4. d$-\mathrm{OH}$

Answer (c): $-\mathrm{NO_2}$ withdraws electrons by resonance and induction and directs the incoming group to the meta position.

Revise: Aromatic Hydrocarbons (Class 11 notes)

1. aphenol
2. btoluene
3. cbenzaldehyde
4. dchlorobenzene

Answer (b): Friedel-Crafts alkylation introduces a methyl group on the ring, giving toluene.

Revise: Aromatic Hydrocarbons (Class 11 notes)

1. amethyl bromide
2. bethyl bromide
3. ctert-butyl bromide
4. dn-propyl bromide

Answer (c): $S_N1$ proceeds through a carbocation, and the tertiary carbocation is the most stable.

Revise: Haloalkanes (Class 12 notes)

1. athe C-Cl bond has partial double bond character due to resonance
2. bit has no halogen atom
3. cbenzene ring is non-planar
4. dchlorine is not electronegative

Answer (a): The lone pair on chlorine delocalises into the ring, strengthening and shortening the C-Cl bond.

Revise: Haloarenes (Class 12 notes)

1. aethanoic acid
2. bethanal only
3. cethyl ethanoate
4. dethene

Answer (a): A primary alcohol is oxidised first to the aldehyde and then to the carboxylic acid when the oxidant is in excess.

Revise: Alcohols (Class 12 notes)

1. apropan-2-ol
2. b2-methylpropan-2-ol
3. cbutan-2-ol
4. d2-methylpropan-1-ol

Answer (b): A Grignard reagent adds to a ketone to give a tertiary alcohol: $\mathrm{(CH_3)_2C{=}O + CH_3MgBr \to (CH_3)_3COH}$.

Revise: Aldehydes and Ketones (Class 12 notes)

1. a2,4,6-tribromophenol
2. b2-bromophenol
3. c4-bromophenol
4. dbromobenzene

Answer (a): The $-\mathrm{OH}$ group strongly activates the ortho and para positions, so all three are brominated.

Revise: Phenols (Class 12 notes)

1. aacetaldehyde
2. bformaldehyde
3. cbenzaldehyde
4. d2,2-dimethylpropanal

Answer (a): An aldol reaction needs an alpha hydrogen. Only acetaldehyde has one among the options.

Revise: Aldehydes and Ketones (Class 12 notes)

1. aacetone
2. bchloroethane
3. cchloroacetic acid
4. dacetyl chloride

Answer (c): Hell-Volhard-Zelinsky reaction replaces an alpha hydrogen with halogen.

Revise: Carboxylic Acids (Class 12 notes)

1. aacetyl chloride
2. bbenzenesulphonyl chloride
3. cnitrous acid
4. dchloroform and KOH

Answer (b): Benzenesulphonyl chloride separates primary, secondary and tertiary amines.

Revise: Amines (Class 12 notes)

1. amethylamine
2. btrimethylamine
3. cdimethylamine
4. dN-methylaniline

Answer (a): Only primary amines react with chloroform and alcoholic KOH to form foul-smelling isocyanides.

Revise: Amines (Class 12 notes)

1. atwo molecules of glucose
2. bglucose and galactose
3. cfructose and galactose
4. dglucose and fructose

Answer (d): Sucrose is a disaccharide of alpha-D-glucose and beta-D-fructose; the mixture is called invert sugar.

1. astarch
2. bprotein
3. clignin
4. dcellulose

Answer (d): Paper is made from wood pulp, which is mostly cellulose after the lignin is removed.

Revise: Paper and Pulp (Class 12 notes)

1. a$\mathrm{CaSO_4}$
2. b$\mathrm{Ca(H_2PO_4)_2}$
3. c$\mathrm{(NH_4)_2SO_4}$
4. d$\mathrm{Ca_3(PO_4)_2}$

Answer (b): Triple superphosphate is mainly calcium dihydrogen phosphate, a soluble phosphate source.

Revise: Modern Chemical Manufactures (Class 11 notes)

1. aparacetamol
2. bacetylsalicylic acid
3. cbenzoic acid
4. dsalicylamide

Answer (b): Aspirin is the acetyl ester of salicylic acid, used as an analgesic and antipyretic.

Revise: Chemistry in the Service to Mankind (Class 12 notes)

1. a0.14 g
2. b0.28 g
3. c1.12 g
4. d0.056 g

Answer (b): Moles $\mathrm{MnO_4^-} = 0.05\times0.02 = 10^{-3}$. Each takes 5 $\mathrm{Fe^{2+}}$, giving $5\times10^{-3}$ mol $\times56 = 0.28$ g.

Revise: Volumetric Analysis (Class 12 notes)

1. achloride ions
2. bcarbonate ions
3. cnitrate ions
4. dsulphate ions

Answer (c): Nitrate with $\mathrm{FeSO_4}$ and conc. $\mathrm{H_2SO_4}$ added down the side gives a brown ring of $[\mathrm{Fe(NO)}]^{2+}$.

Revise: Practical Work (Class 12 notes)

1. adeep blue
2. bblood red
3. cwhite precipitate
4. dyellow precipitate

Answer (b): $\mathrm{Fe^{3+}}$ forms the blood-red complex $[\mathrm{Fe(SCN)}]^{2+}$.

Revise: Practical Work (Class 12 notes)

1. a0.24
2. b0.36
3. c0.16
4. d0.48

Answer (d): By the Hardy-Weinberg equation, heterozygotes $= 2pq = 2 \times 0.6 \times 0.4 = 0.48$.

Revise: Evolutionary Biology (Class 11 notes)

1. agene flow
2. bnatural selection
3. cgenetic drift
4. dmutation pressure

Answer (c): Genetic drift is chance fluctuation of allele frequencies and has a larger effect when the population is small. Gene flow is movement of alleles between populations.

Revise: Evolutionary Biology (Class 11 notes)

1. aconvergent evolution
2. bindustrial melanism
3. cvestigial organs
4. dadaptive radiation

Answer (d): A single ancestral stock diverged into many forms adapted to different niches, which is adaptive radiation.

Revise: Evidences of Evolution (Class 11 notes)

1. aHydra
2. bTaenia (flatworm)
3. cAscaris
4. dPheretima

Answer (b): Flame (protonephridial) cells occur in phylum Platyhelminthes such as Taenia and Planaria. Ascaris belongs to Nematoda and Pheretima to Annelida.

Revise: Kingdom Animalia (Class 11 notes)

1. aAnnelida
2. bMollusca
3. cArthropoda
4. dEchinodermata

Answer (b): Mantle, muscular foot and radula are molluscan features, as in Pila (apple snail).

Revise: Kingdom Animalia (Class 11 notes)

1. aPetromyzon (lamprey)
2. bScoliodon (dogfish)
3. cLabeo (rohu)
4. dHippocampus (sea horse)

Answer (a): Lampreys (class Cyclostomata) have a sucker-like mouth and no jaws. The others are jawed fishes.

Revise: Kingdom Animalia (Class 11 notes)

1. aCephalochordata
2. bUrochordata (tunicates)
3. cVertebrata
4. dHemichordata

Answer (b): In tunicates such as Herdmania the tadpole-like larva has a notochord in its tail that disappears at metamorphosis. In Cephalochordata it persists throughout life.

Revise: Kingdom Animalia (Class 11 notes)

1. aurinary bladder
2. bfallopian tube (oviduct)
3. cstomach
4. durethra

Answer (b): Cilia in the oviduct lining beat to push the ovum toward the uterus. The stomach has non-ciliated columnar epithelium.

Revise: Animal Tissues (Class 12 notes)

1. amuscle to bone
2. bbone to bone
3. cmuscle to muscle
4. dskin to muscle

Answer (b): Ligaments join bone to bone, whereas tendons join muscle to bone.

Revise: Animal Tissues (Class 12 notes)

1. abiceps of the arm
2. bwall of the intestine
3. cheart
4. dskeletal muscle of the thigh

Answer (b): Smooth muscle lines hollow viscera such as the intestine. Cardiac muscle is striated and skeletal muscle is voluntary.

Revise: Animal Tissues (Class 12 notes)

1. aneurons
2. bosteocytes
3. cneuroglia
4. dfibroblasts

Answer (c): Neuroglia support, insulate and nourish neurons; neurons are the conducting cells.

Revise: Animal Tissues (Class 12 notes)

1. aincrease the absorptive surface
2. bsecrete bile
3. cgrind food
4. dstore urea

Answer (a): The typhlosole is a median dorsal fold of the intestine that increases the area for absorption.

Revise: Detailed Study of Earthworm (Class 11 notes)

1. ainside red blood cells
2. bonly in the coelomic fluid
3. cdissolved in the blood plasma
4. din the muscle fibres only

Answer (c): Earthworm blood is red because haemoglobin is dissolved in plasma; its blood corpuscles are colourless.

Revise: Detailed Study of Earthworm (Class 11 notes)

1. alungs from birth
2. bexternal gills at first, later internal gills
3. cbook lungs
4. dtracheae

Answer (b): The tadpole is aquatic and respires by gills, which are replaced by lungs at metamorphosis.

Revise: Frog (Rana tigrina) (Class 11 notes)

1. auricotelic
2. bammonotelic
3. cguanotelic
4. dureotelic

Answer (d): Adult frogs convert ammonia to urea in the liver and excrete it, so they are ureotelic. Tadpoles are ammonotelic.

Revise: Frog (Rana tigrina) (Class 11 notes)

1. aforegut and midgut
2. bcrop and gizzard
3. cmidgut and hindgut
4. dhindgut and anus

Answer (c): Malpighian tubules arise at the midgut-hindgut junction and remove nitrogenous waste from the haemolymph.

Revise: Kingdom Animalia (Class 11 notes)

1. aliver of man
2. bred blood cells of man
3. cstomach of female Anopheles
4. dsalivary gland of mosquito

Answer (c): Gametocytes taken up by the mosquito form gametes that fuse in its stomach. Sporozoites later move to the salivary glands.

Revise: Parasite (Class 11 notes)

1. apepsin
2. benterokinase
3. camylase
4. dlipase

Answer (b): Enterokinase from the intestinal mucosa converts inactive trypsinogen to trypsin.

Revise: Digestive System (Class 12 notes)

1. abreaking proteins into peptides
2. bhydrolysing starch
3. cemulsifying fats
4. dneutralising enzymes

Answer (c): Bile salts break large fat globules into tiny droplets, increasing the surface for lipase.

Revise: Digestive System (Class 12 notes)

1. a2.5 L per minute
2. b7 L per minute
3. c10 L per minute
4. d5 L per minute

Answer (d): Cardiac output $= 72 \times 70 = 5040$ mL, which is about 5 L per minute.

Revise: Circulatory System (Class 12 notes)

1. apulmonary artery
2. bsuperior vena cava
3. chepatic portal vein
4. dpulmonary vein

Answer (d): Pulmonary veins bring oxygenated blood from the lungs to the left atrium.

Revise: Circulatory System (Class 12 notes)

1. athymus
2. bpancreas
3. cspleen
4. dthyroid

Answer (c): The spleen is called the graveyard of RBCs, where macrophages remove worn-out cells.

Revise: Circulatory System (Class 12 notes)

1. acarbaminohaemoglobin
2. bdissolved gas
3. cbicarbonate ions in plasma
4. dcarbonic acid crystals

Answer (c): About 70 percent of CO2 is carried as bicarbonate formed by carbonic anhydrase in RBCs.

Revise: Respiratory System (Class 12 notes)

1. arelaxes and becomes dome-shaped
2. bcontracts and rises
3. cdoes not change shape
4. dcontracts and flattens

Answer (d): A flattened diaphragm increases thoracic volume and lowers the pressure, so air flows in.

Revise: Respiratory System (Class 12 notes)

1. adiabetes mellitus
2. bdiabetes insipidus
3. cgoitre
4. dAddison's disease

Answer (b): Without ADH the collecting duct reabsorbs little water, so a large volume of dilute urine is passed.

Revise: Excretory System (Class 12 notes)

1. aSchwann cells
2. boligodendrocytes
3. castrocytes
4. dosteoblasts

Answer (a): In the peripheral nervous system Schwann cells wrap the axon. Oligodendrocytes do this in the central nervous system.

Revise: Nervous System (Class 12 notes)

1. a3 Na+ out and 2 K+ into the cell
2. b2 Na+ out and 3 K+ into the cell
3. c3 Na+ in and 2 K+ out
4. d1 Na+ out and 1 K+ in

Answer (a): Each cycle uses ATP to send 3 sodium ions out and bring 2 potassium ions in, maintaining the resting potential.

Revise: Nervous System (Class 12 notes)

1. aconvex lens
2. bcylindrical lens
3. cbifocal convex lens
4. dconcave lens

Answer (d): In myopia the image forms in front of the retina; a diverging concave lens shifts it back onto the retina.

Revise: Sense Organs (Class 12 notes)

1. aProgesterone
2. bProlactin
3. cOxytocin
4. dLH (luteinizing hormone)

Answer (d): A mid-cycle LH surge causes the Graafian follicle to rupture and release the ovum, around day 14.

Revise: Reproductive System (Class 12 notes)

1. aFSH
2. brelaxin
3. chuman chorionic gonadotropin (hCG)
4. dcalcitonin

Answer (c): hCG keeps the corpus luteum producing progesterone, and it is the basis of pregnancy tests.

Revise: Development Biology (Class 12 notes)

1. aparathormone
2. bcalcitonin
3. cthyroxine
4. dglucagon

Answer (b): Calcitonin from the thyroid parafollicular cells lowers blood calcium; parathormone raises it.

Revise: Endocrinology (Class 12 notes)

1. agigantism
2. bdwarfism
3. ccretinism
4. dmyxoedema

Answer (a): Excess growth hormone before the growth plates close causes gigantism; in adults it causes acromegaly.

Revise: Endocrinology (Class 12 notes)

1. aSalmonella typhi
2. bVibrio cholerae
3. cMycobacterium tuberculosis
4. dTreponema pallidum

Answer (c): TB is caused by the acid-fast bacterium Mycobacterium tuberculosis, which mainly affects the lungs.

Revise: Microbial Diseases (Class 12 notes)

1. acholera
2. btyphoid
3. cdysentery
4. dhepatitis A

Answer (a): Vibrio cholerae releases a toxin that causes massive fluid loss; treatment is oral rehydration.

Revise: Microbial Diseases (Class 12 notes)

1. aB memory cells only
2. bred blood cells
3. cplatelets
4. dCD4 helper T cells

Answer (d): HIV infects CD4 helper T lymphocytes, weakening both humoral and cell-mediated immunity.

Revise: Microbial Diseases (Class 12 notes)

1. aAnopheles mosquito
2. bCulex mosquito
3. ctsetse fly
4. dAedes mosquito

Answer (d): Dengue virus is carried by Aedes aegypti. Anopheles carries malaria and Culex carries filariasis.

Revise: Microbial Diseases (Class 12 notes)

1. aventricular depolarization
2. batrial depolarization
3. cventricular repolarization
4. datrial repolarization

Answer (a): The P wave is atrial depolarization, QRS is ventricular depolarization and the T wave is ventricular repolarization.

Revise: Human Health and Disorders (Class 12 notes)

1. akidneys
2. bliver
3. clungs
4. dpancreas

Answer (a): A dialysis machine filters urea and excess salts from the blood when the kidneys fail.

Revise: Human Health and Disorders (Class 12 notes)

1. aex situ conservation
2. bin situ conservation
3. cbiosphere reserve zoning
4. dsacred grove protection

Answer (a): Ex situ means off-site. National parks and sanctuaries are in situ methods.

Revise: Conservation Biology (Class 11 notes)

1. aChitwan National Park
2. bSagarmatha National Park
3. cBardiya National Park
4. dLangtang National Park

Answer (a): Chitwan National Park (1973) protects the one-horned rhinoceros and Bengal tiger.

Revise: Conservation Biology (Class 11 notes)

1. aglycosidic bond
2. bester bond
3. cpeptide bond
4. dphosphodiester bond

Answer (c): A peptide bond forms between the carboxyl group of one amino acid and the amino group of the next, releasing water.

Revise: Biomolecules (Class 11 notes)

1. aactive site
2. ballosteric lock
3. ccofactor shell
4. dprosthetic tail

Answer (a): The substrate fits the active site, forming an enzyme-substrate complex.

Revise: Biomolecules (Class 11 notes)

1. aphycoerythrin
2. bchlorophyll b only
3. canthocyanin
4. dfucoxanthin

Answer (d): Fucoxanthin gives brown algae their colour. Laminarin is their reserve carbohydrate.

Revise: Algae (Class 11 notes)

1. aconidia only
2. bzoospores
3. cbinary fission only
4. dbudding

Answer (d): A yeast cell forms a small outgrowth, the bud, that detaches as a new cell.

Revise: Mycota (Fungi) (Class 11 notes)

1. ais the dominant free-living phase
2. bhas true roots and vascular tissue
3. cis attached to and depends on the gametophyte
4. dproduces seeds

Answer (c): The bryophyte sporophyte stays on the gametophyte for nutrition and bears the capsule with spores.

Revise: Bryophyta (Class 11 notes)

1. aa moss
2. bPinus
3. ca green alga
4. da fern

Answer (d): Sori are characteristic of ferns (Pteridophyta), e.g. Dryopteris.

Revise: Pteridophyta (Class 11 notes)

1. afemale cones
2. bmale cones
3. cflowers
4. dfruits

Answer (b): Male cones bear microsporophylls with pollen sacs. Wings on the pollen help wind dispersal.

Revise: Gymnosperms (Class 11 notes)

1. amango (a dicot)
2. bwheat (a monocot)
3. cpea (a dicot)
4. dPinus

Answer (b): Fibrous roots, parallel venation and scattered bundles are monocot features.

Revise: Angiosperms (Class 11 notes)

1. asoil salinity
2. bwater hardness
3. cozone hole
4. dair pollution (sulphur dioxide)

Answer (d): Lichens absorb water and gases directly and disappear where SO2 is high.

Revise: Lichens (Class 11 notes)

1. aAristotle
2. bRobert Whittaker
3. cCharles Darwin
4. dCarolus Linnaeus

Answer (d): Linnaeus gave each species a genus name and a species epithet, e.g. Mangifera indica.

Revise: Biodiversity (Class 11 notes)

1. aviroid
2. bprion
3. cbacteriophage
4. dmycoplasma

Answer (c): Bacteriophages attach to a bacterial cell and inject their nucleic acid.

Revise: Introductory Microbiology (Class 11 notes)

1. ainverted
2. bupright
3. cspindle-shaped
4. dupright only in forests

Answer (b): Energy is lost at each level, so the base (producers) always holds the most energy.

Revise: Ecology (Class 11 notes)

1. aRhizobium
2. bNitrosomonas
3. cdenitrifying bacteria such as Pseudomonas
4. dAzotobacter

Answer (c): Denitrification returns N2 to the atmosphere. Rhizobium and Azotobacter fix nitrogen.

Revise: Ecology (Class 11 notes)

1. aRhododendron arboreum (lali guras)
2. bRosa indica
3. cNelumbo nucifera
4. dTagetes erecta

Answer (a): Rhododendron arboreum, locally called lali guras, is the national flower.

Revise: Vegetation (Class 11 notes)

1. ahydrophytes
2. bmesophytes in rain forest
3. cepiphytes only
4. dxerophytes

Answer (d): Xerophytes reduce water loss in dry habitats with such features.

Revise: Ecology (Class 11 notes)

1. aG1 phase
2. bG2 phase
3. cS phase
4. dM phase

Answer (c): The S (synthesis) phase doubles the DNA content before division.

Revise: Cell Division (Class 11 notes)

1. acontain hydrolytic enzymes that can digest the cell itself
2. bcontain DNA for self-destruction
3. cstore toxins
4. dproduce ATP quickly

Answer (a): If the lysosome membrane bursts, its enzymes digest the cell.

Revise: Cell (Class 11 notes)

1. aleptotene stage
2. bzygotene stage
3. cdiplotene stage
4. ddiakinesis

Answer (b): Zygotene of prophase I is when homologues pair to form bivalents.

Revise: Cell Division (Class 11 notes)

1. acell plate formation
2. bcleavage furrow
3. cbudding
4. damitosis

Answer (a): A cell plate grows from the centre outward to form the new wall. Animal cells divide by a furrow.

Revise: Cell Division (Class 11 notes)

1. a80S
2. b70S
3. c60S
4. d100S

Answer (a): Eukaryotic cytoplasmic ribosomes are 80S (60S + 40S); prokaryotic ones are 70S.

Revise: Cell (Class 11 notes)

1. a1/4
2. b3/16
3. c1/16
4. d9/16

Answer (c): Each locus gives aa with probability 1/4, so aabb $= \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.

Revise: Mendelian Genetics (Class 12 notes)

1. a3 red : 1 white
2. b1 red : 2 pink : 1 white
3. c1 red : 1 white
4. d9 : 3 : 3 : 1

Answer (b): In incomplete dominance the heterozygote is intermediate, and the F2 phenotypic ratio equals the genotypic ratio 1:2:1.

Revise: Mendelian Genetics (Class 12 notes)

1. aAB or O
2. bA, B, AB or O
3. cO only
4. dA or B only

Answer (d): IAIB x ii gives IAi (group A) and IBi (group B).

Revise: Mendelian Genetics (Class 12 notes)

1. a25 percent
2. b50 percent
3. c100 percent
4. d0 percent

Answer (b): A son gets his X from the mother, who gives the affected X in half of cases.

Revise: Mendelian Genetics (Class 12 notes)

1. asemi-conservative
2. bconservative
3. cdispersive
4. drandom

Answer (a): Using heavy nitrogen they found each new DNA has one old and one new strand.

Revise: Genetic Materials (Class 12 notes)

1. aUAA
2. bAUG
3. cUUU
4. dGCA

Answer (a): UAA, UAG and UGA are stop codons. AUG is the start codon and codes for methionine.

Revise: Genetic Materials (Class 12 notes)

1. acompanion cells
2. btracheids
3. csclereids
4. dguard cells

Answer (a): Companion cells control the sieve tube elements, which lack a nucleus.

Revise: Plant Anatomy (Class 12 notes)

1. aendodermis of the root
2. bepidermis of the leaf
3. ccortex of the stem
4. dpith of the monocot stem

Answer (a): Casparian strips of suberin in the endodermis force water through the cell membrane.

Revise: Plant Anatomy (Class 12 notes)

1. ain a ring and open
2. bscattered and closed (no cambium)
3. cradial
4. dabsent

Answer (b): Monocot stems have scattered closed bundles, hence no secondary growth.

Revise: Plant Anatomy (Class 12 notes)

1. apea
2. bmaize (C4 plant)
3. cHydrilla
4. dsunflower

Answer (b): C4 plants such as maize and sugarcane have a wreath of bundle sheath cells.

Revise: Plant Physiology: Photosynthesis and Respiration (Class 12 notes)

1. aoxaloacetic acid
2. bglucose
3. c3-phosphoglyceric acid (3-PGA)
4. dpyruvic acid

Answer (c): RuBisCO fixes CO2 to RuBP, forming two molecules of 3-PGA.

Revise: Plant Physiology: Photosynthesis and Respiration (Class 12 notes)

1. acytoplasm
2. binner membrane
3. cmitochondrial matrix
4. dthylakoid

Answer (c): Krebs cycle enzymes are found in the matrix; the electron transport chain is on the inner membrane.

Revise: Plant Physiology: Photosynthesis and Respiration (Class 12 notes)

1. a0.7
2. b1
3. c0.9
4. dmore than 1

Answer (b): $C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$ gives $RQ = 6/6 = 1$.

Revise: Plant Physiology: Photosynthesis and Respiration (Class 12 notes)

1. aauxin
2. bcytokinin
3. cethylene
4. dabscisic acid

Answer (a): Auxin from the shoot tip inhibits lateral buds; removing the tip releases them.

Revise: Plant Physiology: Plant Hormones and Growth (Class 12 notes)

1. aethylene
2. babscisic acid
3. cauxin only
4. dgibberellin

Answer (d): Gibberellins promote stem elongation, bolting and flowering in rosette plants.

Revise: Plant Physiology: Plant Hormones and Growth (Class 12 notes)

1. achalaza
2. bfuniculus
3. cmicropyle
4. dhilum

Answer (c): Entry through the micropyle is called porogamy.

Revise: Angiosperms (Class 11 notes)

1. aapomixis
2. bpolyembryony
3. cvivipary
4. dparthenocarpy

Answer (d): Parthenocarpic fruits, such as banana, are seedless.

Revise: Angiosperms (Class 11 notes)

1. aGregor Mendel
2. bHargobind Khorana
3. cNorman Borlaug
4. dHugo de Vries

Answer (c): Norman Borlaug's semi-dwarf wheat varieties raised yields greatly.

Revise: Applied Biology (Class 12 notes)

1. avitamin C
2. bbeta-carotene (provitamin A)
3. ciron only
4. dvitamin K

Answer (b): Genes for beta-carotene synthesis were added to rice to fight vitamin A deficiency.

Revise: Biotechnology (Class 12 notes)

1. acarbon dioxide
2. blight
3. cnitrogen gas
4. dphosphorus and water

Answer (d): The fungal hyphae enlarge the absorbing area, aiding phosphorus and water uptake.

Revise: Applied Biology (Class 12 notes)

1. astudent
2. bschool
3. cbook
4. dclassroom

Answer (a): A doctor treats a patient, and a teacher teaches a student.

1. acareless
2. bhasty
3. cloud
4. dcareful

Answer (d): Meticulous means very careful and precise about details.

1. aextend
2. bgrow
3. ccontract
4. dinflate

Answer (c): To expand is to become larger; the opposite is to contract.

1. aliver
2. bfemur
3. ckidney
4. dlung

Answer (b): Liver, kidney and lung are soft organs; the femur is a bone.

1. ahumidity
2. bwind speed
3. crainfall
4. datmospheric pressure

Answer (d): A barometer measures atmospheric pressure just as a thermometer measures temperature.

1. a5 hours
2. b4 hours
3. c6 hours
4. d7 hours

Answer (a): Speed $= 180/3 = 60$ km/h, so time $= 300/60 = 5$ hours.

1. aRs 400
2. bRs 1000
3. cRs 800
4. dRs 640

Answer (c): $SI = \frac{5000 \times 8 \times 2}{100} = 800$.

1. a4 days
2. b6 days
3. c9 days
4. d3 days

Answer (a): Together they do $\frac{1}{12} + \frac{1}{6} = \frac{1}{4}$ of the job per day, so 4 days.

1. a13
2. b14
3. c12
4. d15

Answer (a): Total is 120; remaining total is $120 - 29 = 91$, and $91/7 = 13$.

1. aRs 650
2. bRs 600
3. cRs 575
4. dRs 700

Answer (b): Discount $= 0.25 \times 800 = 200$, so price $= 800 - 200 = 600$.

1. a45
2. b48
3. c46
4. d50

Answer (b): The terms are $2^2-1, 3^2-1, 4^2-1, 5^2-1, 6^2-1$, so the next is $7^2-1 = 48$.

1. aP
2. bR
3. cQ
4. dO

Answer (c): Each letter moves 3 places ahead, and N + 3 = Q.

1. a1, 3, 4, 2
2. b3, 4, 1, 2
3. c3, 1, 2, 4
4. d3, 1, 4, 2

Answer (d): Symptoms appear first, then diagnosis, then treatment, then recovery.

1. aOnly I follows
2. bOnly II follows
3. cBoth I and II follow
4. dNeither I nor II follows

Answer (b): II is the first statement restated. I does not follow because the teachers among graduates need not be doctors.

1. a36
2. b34
3. c35
4. d32

Answer (c): The terms are $11^2-1, 10^2-1, 9^2-1, 8^2-1, 7^2-1$, so next is $6^2 - 1 = 35$.

1. a90 degrees
2. b60 degrees
3. c105 degrees
4. d75 degrees

Answer (d): At 3:30 the hour hand is at $3.5 \times 30 = 105$ degrees and the minute hand at 180 degrees, so the angle is 75 degrees.

1. a27
2. b36
3. c64
4. d9

Answer (a): The unpainted cubes form the inner $3\times3\times3$ block, so $3^3 = 27$.

1. anorth
2. bsouth
3. ceast
4. dwest

Answer (b): North, clockwise 90 gives east; turning 180 gives west; anticlockwise 90 from west gives south.

1. a2
2. b3
3. c4
4. d8

Answer (c): Two folds make four layers, and each layer gets one hole.

1. a4
2. b6
3. c8
4. d5

Answer (d): There are 4 small squares and 1 large square, giving 5.