MECEE-BL Entrance Model Set 1 with Answers

1. a200 m
2. b20 m
3. c100 m
4. d50 m

Answer (c): $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(10)^2 = 100$ m.

Revise: Kinematics (Class 11 notes)

1. a$ML^{-1}T^{-2}$
2. b$MLT^{-2}$
3. c$ML^2T^{-2}$
4. d$ML^{-2}T^{-2}$

Answer (a): Pressure = force/area = $MLT^{-2}/L^2 = ML^{-1}T^{-2}$.

Revise: Physical Quantities (Class 11 notes)

1. a7 N
2. b5 N
3. c1 N
4. d12 N

Answer (b): $R = \sqrt{3^2 + 4^2} = 5$ N.

Revise: Vectors (Class 11 notes)

1. a$6\ \mathrm{m/s^2}$
2. b$8\ \mathrm{m/s^2}$
3. c$2\ \mathrm{m/s^2}$
4. d$4\ \mathrm{m/s^2}$

Answer (d): Net force $= 30 - 10 = 20$ N, so $a = F/m = 20/5 = 4\ \mathrm{m/s^2}$.

Revise: Dynamics (Class 11 notes)

1. a10 J
2. b50 J
3. c100 J
4. d25 J

Answer (c): $W = mgh = 2 \times 10 \times 5 = 100$ J.

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

1. a$g/2$
2. b$g/9$
3. c$g/3$
4. d$g/4$

Answer (d): $g' = GM/(2R)^2 = g/4$, since $g$ varies as $1/r^2$.

Revise: Gravitation (Class 11 notes)

1. a1 N
2. b8 N
3. c16 N
4. d4 N

Answer (d): $F = mv^2/r = 0.5 \times 16 / 2 = 4$ N.

Revise: Circular Motion (Class 11 notes)

1. a$10^{8}\ \mathrm{N/m^2}$
2. b$10^{10}\ \mathrm{N/m^2}$
3. c$10^{11}\ \mathrm{N/m^2}$
4. d$10^{12}\ \mathrm{N/m^2}$

Answer (c): Stress $= 100/10^{-6} = 10^8$ Pa; $Y = \text{stress}/\text{strain} = 10^8/10^{-3} = 10^{11}$ N/m$^2$.

Revise: Elasticity (Class 11 notes)

1. a100 J
2. b200 J
3. c20 J
4. d50 J

Answer (a): $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(2)(100) = 100$ J.

Revise: Rotational Dynamic (Class 12 notes)

1. abecomes half
2. bbecomes four times
3. cremains unchanged
4. dbecomes double

Answer (d): $T = 2\pi\sqrt{L/g}$, so $T \propto \sqrt{L}$; four times $L$ gives $\sqrt{4} = 2$ times $T$.

Revise: Periodic Motion (Class 12 notes)

1. a8400 J
2. b42000 J
3. c84000 J
4. d840000 J

Answer (c): $Q = mc\Delta T = 2 \times 4200 \times 10 = 84000$ J.

Revise: Quantity of Heat (Class 11 notes)

1. a$40^\circ$
2. b$-40^\circ$
3. c$0^\circ$
4. d$-273^\circ$

Answer (b): Put $C = F$ in $F = 1.8C + 32$: $C = 1.8C + 32$, so $C = -40$.

Revise: Heat and Temperature (Class 11 notes)

1. a1.2 mm
2. b0.12 mm
3. c12 mm
4. d2.4 mm

Answer (a): $\Delta L = L\alpha\Delta T = 2 \times 1.2\times10^{-5} \times 50 = 1.2\times10^{-3}$ m $= 1.2$ mm.

Revise: Thermal Expansion (Class 11 notes)

1. a700 J
2. b300 J
3. c200 J
4. d-300 J

Answer (b): $\Delta U = Q - W = 500 - 200 = 300$ J.

Revise: First Law of Thermodynamics (Class 12 notes)

1. a75%
2. b33%
3. c12.5%
4. d25%

Answer (d): $\eta = 1 - T_2/T_1 = 1 - 300/400 = 0.25$.

Revise: Second Law of Thermodynamics (Class 12 notes)

1. aheat exchanged is zero
2. bpressure stays constant
3. cinternal energy does not change
4. dvolume stays constant

Answer (c): Internal energy of an ideal gas depends only on temperature, so it is unchanged when temperature is constant.

Revise: First Law of Thermodynamics (Class 12 notes)

1. abecomes double
2. bbecomes four times
3. cremains the same
4. dbecomes half

Answer (d): $H = kA\Delta T/d$, so $H \propto 1/d$.

Revise: Rate of Heat Flow (Class 11 notes)

1. a$200\ \mathrm{m/s}$
2. b$12.5\ \mathrm{m/s}$
3. c$54\ \mathrm{m/s}$
4. d$100\ \mathrm{m/s}$

Answer (a): $v = f\lambda = 50 \times 4 = 200$ m/s.

Revise: Wave Motion (Class 12 notes)

1. a340 Hz
2. b680 Hz
3. c85 Hz
4. d170 Hz

Answer (d): $f = v/4L = 340/(4 \times 0.5) = 170$ Hz.

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

1. a2
2. b4
3. c8
4. d516

Answer (b): Beat frequency $= |260 - 256| = 4$.

Revise: Acoustic Phenomena (Class 12 notes)

1. a7.5 cm
2. b15 cm
3. c30 cm
4. d20 cm

Answer (b): $1/v = 1/f - 1/u = 1/10 - 1/30 = 2/30$, so $v = 15$ cm.

Revise: Reflection at Curved Mirrors (Class 11 notes)

1. a$2\times10^8\ \mathrm{m/s}$
2. b$4.5\times10^8\ \mathrm{m/s}$
3. c$1.5\times10^8\ \mathrm{m/s}$
4. d$3\times10^8\ \mathrm{m/s}$

Answer (a): $v = c/\mu = 3\times10^8/1.5 = 2\times10^8$ m/s.

Revise: Refraction at Plane Surface (Class 11 notes)

1. a+4 D
2. b-4 D
3. c+25 D
4. d+0.04 D

Answer (a): $P = 1/f(\mathrm{m}) = 1/0.25 = +4$ D, positive for a converging lens.

Revise: Lenses (Class 11 notes)

1. a0.6 mm
2. b10 mm
3. c1 mm
4. d0.1 mm

Answer (c): $\beta = \lambda D/d = (600\times10^{-9} \times 1)/(0.6\times10^{-3}) = 10^{-3}$ m.

Revise: Interference (Class 12 notes)

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

Answer (b): Brewster's law: $\tan\theta_p = \mu = \sqrt{3}$, so $\theta_p = 60^\circ$.

Revise: Polarization of Light (Class 12 notes)

1. a$2\ \Omega$
2. b$9\ \Omega$
3. c$4.5\ \Omega$
4. d$0.5\ \Omega$

Answer (a): $R = (6\times3)/(6+3) = 2\ \Omega$.

Revise: DC Circuit (Class 11 notes)

1. a8 A
2. b80 A
3. c0.125 A
4. d12.5 A

Answer (a): $I = P/V = 2000/250 = 8$ A.

Revise: DC Circuit (Class 11 notes)

1. a$3.6\times10^3$ J
2. b$3.6\times10^5$ J
3. c$10^3$ J
4. d$3.6\times10^6$ J

Answer (d): $1\ \mathrm{kWh} = 1000\ \mathrm{W} \times 3600\ \mathrm{s} = 3.6\times10^6$ J.

Revise: DC Circuit (Class 11 notes)

1. a10 V
2. b12 V
3. c2 V
4. d11 V

Answer (a): $I = 12/(5+1) = 2$ A; $V = IR = 2 \times 5 = 10$ V.

Revise: Electrical Circuit One (Class 12 notes)

1. a0.25 N
2. b2 N
3. c1 N
4. d4 N

Answer (c): $F = BIL = 0.5 \times 4 \times 0.5 = 1$ N.

Revise: Magnetic Field (Class 12 notes)

1. a2 V
2. b0.4 V
3. c3 V
4. d0.08 V

Answer (a): $\varepsilon = \Delta\Phi/\Delta t = 0.4/0.2 = 2$ V.

Revise: Electromagnetic Induction (Class 12 notes)

1. a4 V
2. b100 V
3. c20 V
4. d500 V

Answer (b): $V_s/V_p = N_s/N_p$, so $V_s = 20 \times 500/100 = 100$ V.

Revise: Electromagnetic Induction (Class 12 notes)

1. a100 V
2. b141.4 V
3. c70.7 V
4. d50 V

Answer (c): $V_{rms} = V_0/\sqrt{2} = 100/1.414 \approx 70.7$ V.

Revise: Alternating Currents (Class 12 notes)

1. aparamagnetic
2. bdiamagnetic
3. cferromagnetic
4. dantiferromagnetic

Answer (b): Diamagnetic substances develop a magnetisation opposite to the applied field, so they are weakly repelled.

Revise: Magnetic Properties of Materials (Class 12 notes)

1. a0.06 N
2. b0.6 N
3. c6 N
4. d1.8 N

Answer (b): $F = kq_1q_2/r^2 = 9\times10^9 \times 6\times10^{-12}/0.09 = 0.6$ N.

Revise: Electric Charges (Class 11 notes)

1. ahalf
2. bdouble
3. cfour times
4. done quarter

Answer (d): $E \propto 1/r^2$, so doubling $r$ reduces $E$ to $1/4$.

Revise: Electric Field (Class 11 notes)

1. a$9\ \mu$F
2. b$1\ \mu$F
3. c$2\ \mu$F
4. d$4.5\ \mu$F

Answer (c): $C = (6\times3)/(6+3) = 2\ \mu$F.

Revise: Capacitor (Class 11 notes)

1. a0.5 J
2. b0.1 J
3. c0.05 J
4. d$5\times10^{-3}$ J

Answer (c): $U = \frac{1}{2}CV^2 = \frac{1}{2}(10^{-5})(10^4) = 0.05$ J.

Revise: Capacitor (Class 11 notes)

1. a5 eV
2. b2 eV
3. c1 eV
4. d3 eV

Answer (c): $K_{max} = h\nu - \phi = 3 - 2 = 1$ eV.

Revise: Photons (Class 12 notes)

1. a4 eV
2. b1 eV
3. c0.5 eV
4. d2 eV

Answer (d): $E = 1240/\lambda\ (\mathrm{nm})$ eV $= 1240/620 = 2$ eV.

Revise: Photons (Class 12 notes)

1. abecomes double
2. bremains the same
3. cbecomes half
4. dbecomes four times

Answer (c): $\lambda = h/p$, so $\lambda \propto 1/p$.

Revise: Quantization of Energy (Class 12 notes)

1. a-13.6 eV
2. b-3.4 eV
3. c-1.51 eV
4. d-6.8 eV

Answer (b): $E_n = -13.6/n^2 = -13.6/4 = -3.4$ eV.

Revise: Quantization of Energy (Class 12 notes)

1. a1/3
2. b1/6
3. c1/16
4. d1/8

Answer (d): 30 days is 3 half-lives, so fraction $= (1/2)^3 = 1/8$.

Revise: Radioactivity and Nuclear Reaction (Class 12 notes)

1. a146
2. b92
3. c238
4. d330

Answer (a): $N = A - Z = 238 - 92 = 146$.

Revise: Nuclear Physics (Class 11 notes)

1. a$3.1\ \text{\AA}$
2. b$0.31\ \text{\AA}$
3. c$0.031\ \text{\AA}$
4. d$1.24\ \text{\AA}$

Answer (b): $\lambda_{min} = 12400/V\ \text{\AA} = 12400/40000 = 0.31\ \text{\AA}$.

Revise: Electrons (Class 12 notes)

1. aphosphorus
2. bboron
3. caluminium
4. dgallium

Answer (a): Phosphorus is pentavalent and donates a free electron; the others are trivalent acceptors.

Revise: Semiconductor Devices (Class 12 notes)

1. a50 Hz
2. b100 Hz
3. c25 Hz
4. d200 Hz

Answer (b): Both half cycles are used, so the output has twice the input frequency.

Revise: Semiconductor Devices (Class 12 notes)

1. aAND
2. bNAND
3. cOR
4. dXOR

Answer (b): Any other gate can be built from NAND gates alone, so NAND is universal.

Revise: Semiconductor Devices (Class 12 notes)

1. a0.8 mA
2. b20 mA
3. c$2\ \mu$A
4. d2 mA

Answer (d): $I_C = \beta I_B = 50 \times 40\ \mu\mathrm{A} = 2000\ \mu\mathrm{A} = 2$ mA.

Revise: Semiconductor Devices (Class 12 notes)

1. a18.63 MeV
2. b1.863 MeV
3. c186.3 MeV
4. d46.6 MeV

Answer (a): $E = 0.02 \times 931.5 = 18.63$ MeV.

Revise: Radioactivity and Nuclear Reaction (Class 12 notes)

1. a$3.0\times10^{23}$
2. b$6.02\times10^{23}$
3. c$1.5\times10^{23}$
4. d$2.4\times10^{24}$

Answer (c): Moles $= 11/44 = 0.25$; molecules $= 0.25 \times 6.02\times10^{23} \approx 1.5\times10^{23}$.

Revise: Stoichiometry (Class 11 notes)

1. a4 g
2. b0.2 g
3. c2 g
4. d20 g

Answer (c): Moles $= 0.1 \times 0.5 = 0.05$; mass $= 0.05 \times 40 = 2$ g.

Revise: Stoichiometry (Class 11 notes)

1. a8
2. b9
3. c18
4. d32

Answer (c): Maximum $= 2n^2 = 2 \times 9 = 18$.

Revise: Atomic Structure (Class 11 notes)

1. a3
2. b4
3. c2
4. d5

Answer (d): $\mathrm{Mn^{2+}}$ is $[\mathrm{Ar}]3d^5$; each of the five d orbitals holds one unpaired electron.

Revise: Atomic Structure (Class 11 notes)

1. a(3, 2, -2)
2. b(2, 1, 0)
3. c(1, 0, 0)
4. d(2, 2, 1)

Answer (d): $l$ must be less than $n$, so $n = 2$ cannot have $l = 2$.

Revise: Atomic Structure (Class 11 notes)

1. atrigonal planar
2. btetrahedral
3. cbent
4. dtrigonal pyramidal

Answer (d): Three bond pairs and one lone pair on N give a pyramidal shape.

Revise: Chemical Bonding (Class 11 notes)

1. a$sp^3d^2$
2. b$sp^3d$
3. c$sp^3$
4. d$sp^2$

Answer (a): Six bond pairs and no lone pair mean six hybrid orbitals, so $sp^3d^2$.

Revise: Chemical Bonding (Class 11 notes)

1. a4 atm
2. b0.25 atm
3. c2 atm
4. d1 atm

Answer (a): $P_1V_1 = P_2V_2$: $P_2 = 1 \times 2/0.5 = 4$ atm.

Revise: States of Matter (Class 11 notes)

1. a16 : 1
2. b1 : 4
3. c4 : 1
4. d2 : 1

Answer (c): Graham's law: $r_1/r_2 = \sqrt{M_2/M_1} = \sqrt{32/2} = 4$.

Revise: States of Matter (Class 11 notes)

1. atowards the product side
2. btowards the reactant side
3. cnowhere, it has no effect
4. dtowards the side with more moles

Answer (a): There are fewer moles of gas on the product side (2 against 4), so higher pressure favours the forward reaction.

Revise: Chemical Equilibrium (Class 11 notes)

1. a1
2. b12
3. c2
4. d0.01

Answer (c): $\mathrm{pH} = -\log[H^+] = -\log(10^{-2}) = 2$.

Revise: Ionic Equilibrium (Class 12 notes)

1. a$\mathrm{NH_4OH + NH_4Cl}$
2. b$\mathrm{CH_3COOH + CH_3COONa}$
3. c$\mathrm{HCl + NaCl}$
4. d$\mathrm{NaOH + NaCl}$

Answer (b): An acidic buffer is a weak acid with its salt of a strong base; the second pair is a basic buffer and the others are not buffers.

Revise: Ionic Equilibrium (Class 12 notes)

1. a-676 kJ
2. b-110 kJ
3. c+110 kJ
4. d-283 kJ

Answer (b): Hess's law: subtract the second equation from the first: $-393 - (-283) = -110$ kJ.

Revise: Chemical Thermodynamic (Class 12 notes)

1. a1.27 g
2. b0.635 g
3. c0.3175 g
4. d6.35 g

Answer (b): $Q = 1930$ C $= 0.02$ F; $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$ gives $0.01$ mol $= 0.635$ g.

Revise: Electrochemistry (Class 12 notes)

1. a0.42 V
2. b1.10 V
3. c1.52 V
4. d0.76 V

Answer (b): $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10$ V.

Revise: Electrochemistry (Class 12 notes)

1. a10 min
2. b1 min
3. c100 min
4. d6.93 min

Answer (a): $t_{1/2} = 0.693/k = 0.693/0.0693 = 10$ min.

Revise: Chemical Kinetics (Class 12 notes)

1. adouble
2. bunchanged
3. ceight times
4. dfour times

Answer (d): Rate $\propto [A]^2$, so $2^2 = 4$ times.

Revise: Chemical Kinetics (Class 12 notes)

1. achlorine
2. bfluorine
3. coxygen
4. dnitrogen

Answer (b): Electronegativity increases across a period and up a group; fluorine has the highest value (4.0).

Revise: Classification of Elements (Class 11 notes)

1. aNa
2. bMg
3. cNe
4. dK

Answer (c): Noble gases have the highest ionisation energies in their period, and Ne is smaller than the others listed.

Revise: Classification of Elements (Class 11 notes)

1. apotassium
2. blithium
3. ccaesium
4. dsodium

Answer (d): Na gives golden yellow, K lilac, Li crimson red.

Revise: Chemistry of Metals (Class 11 notes)

1. a$\mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O}$
2. b$\mathrm{CaSO_4\cdot2H_2O}$
3. c$\mathrm{CaSO_4}$
4. d$\mathrm{CaSO_4\cdot H_2O}$

Answer (a): Gypsum on heating at about 120 C loses part of its water to give the hemihydrate.

Revise: Chemistry of Metals (Class 11 notes)

1. aoxide ores
2. bcarbonate ores
3. chalide ores
4. dsulphide ores

Answer (d): Sulphide ore particles are wetted by oil and carried up in the froth, leaving gangue behind.

Revise: Chemistry of Metals (Class 11 notes)

1. aFe
2. bNi
3. c$\mathrm{Al_2O_3}$
4. d$\mathrm{V_2O_5}$

Answer (d): $\mathrm{V_2O_5}$ catalyses $2SO_2 + O_2 \to 2SO_3$; Fe is the Haber process catalyst.

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

1. a$\mathrm{NO_2}$
2. b$\mathrm{N_2O}$
3. c$\mathrm{H_2}$
4. dNO

Answer (d): $3Cu + 8HNO_3(dil) \to 3Cu(NO_3)_2 + 2NO + 4H_2O$.

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

1. abromine
2. bchlorine
3. ciodine
4. dfluorine

Answer (a): Bromine is a red-brown volatile liquid; Cl and F are gases and iodine is a solid.

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

1. a$\mathrm{Zn^{2+}}$
2. b$\mathrm{Cu^{2+}}$
3. c$\mathrm{Ni^{2+}}$
4. d$\mathrm{Co^{2+}}$

Answer (a): $\mathrm{Zn^{2+}}$ is $3d^{10}$, with no d-d transition possible, so it is colourless.

Revise: Transition Metals (Class 12 notes)

1. a+3
2. b+7
3. c+12
4. d+6

Answer (d): $2(+1) + 2x + 7(-2) = 0$, so $x = +6$.

Revise: Oxidation and Reduction (Class 11 notes)

1. a3-methylbutane
2. b2-methylbutane
3. c2-ethylpropane
4. d1,1-dimethylpropane

Answer (b): The longest chain has four carbons, and numbering from the end nearest the branch puts the methyl on C-2.

Revise: Fundamental Principles (Class 11 notes)

1. a3
2. b4
3. c1
4. d2

Answer (d): Butane and 2-methylpropane (isobutane).

Revise: Fundamental Principles (Class 11 notes)

1. a1,2-dibromoethane
2. b1,1-dibromoethane
3. cbromoethane
4. dethyne

Answer (a): Bromine adds across the double bond, putting one Br on each carbon.

Revise: Hydrocarbons (Class 11 notes)

1. a1-bromopropane
2. b1,2-dibromopropane
3. c2-bromopropane
4. d2-bromopropene

Answer (c): The H goes to the carbon with more hydrogens, so Br attaches to the middle carbon.

Revise: Hydrocarbons (Class 11 notes)

1. adil. $\mathrm{HNO_3}$ only
2. b$\mathrm{NaNO_2}$ + HCl
3. cconc. $\mathrm{HNO_3}$ + conc. $\mathrm{H_2SO_4}$
4. dconc. HCl + conc. $\mathrm{HNO_3}$

Answer (c): The nitrating mixture produces the electrophile $\mathrm{NO_2^+}$.

Revise: Aromatic Hydrocarbons (Class 11 notes)

1. aethene
2. bethane
3. cethanol
4. ddiethyl ether

Answer (c): Aqueous KOH causes nucleophilic substitution; alcoholic KOH would cause elimination to ethene.

Revise: Haloalkanes (Class 12 notes)

1. abutan-1-ol
2. b2-methylpropan-2-ol
3. cbutan-2-ol
4. dethanol

Answer (b): Tertiary alcohols give the most stable carbocation and react at once.

Revise: Alcohols (Class 12 notes)

1. ablood red
2. byellow
3. cviolet
4. dgreen

Answer (c): Phenol forms a violet iron(III) phenoxide complex, which is a standard test.

Revise: Phenols (Class 12 notes)

1. a$140^\circ$C
2. b$170^\circ$C
3. c$100^\circ$C
4. d$200^\circ$C

Answer (a): At $140^\circ$C, ether forms by dehydration of two ethanol molecules; at $170^\circ$C ethene forms.

Revise: Ethers (Class 12 notes)

1. aethanal
2. bpropanone
3. cethanol
4. ddiethyl ether

Answer (a): Aldehydes reduce the silver ammonia complex to a silver mirror; ketones do not.

Revise: Aldehydes and Ketones (Class 12 notes)

1. a$\mathrm{CH_3COOH}$
2. b$\mathrm{HCOOH}$
3. c$\mathrm{CH_3CH_2COOH}$
4. d$\mathrm{(CH_3)_2CHCOOH}$

Answer (b): Alkyl groups donate electrons and weaken the acid; formic acid has none.

Revise: Carboxylic Acids (Class 12 notes)

1. adimethylamine
2. baniline
3. cammonia
4. dtrimethylamine

Answer (a): Two alkyl groups raise the electron density on N, and the secondary amine is also well solvated; aniline is weakened by resonance.

Revise: Amines (Class 12 notes)

1. asucrose
2. bglucose
3. cmaltose
4. dlactose

Answer (a): In sucrose the anomeric carbons of glucose and fructose are joined through the glycosidic bond, leaving no free aldehyde or ketone group.

1. a$\mathrm{-COO-}$
2. b$\mathrm{-NH-NH-}$
3. c$\mathrm{-CO-NH-}$
4. d$\mathrm{-CO-O-CO-}$

Answer (c): A peptide bond is an amide bond between the carboxyl group of one amino acid and the amino group of the next.

1. aethylene glycol and terephthalic acid
2. bhexamethylenediamine and adipic acid
3. cphenol and formaldehyde
4. dcaprolactam alone

Answer (b): Diamine plus diacid gives a polyamide with loss of water.

1. abutan-2-ol
2. bpropan-2-ol
3. cethanol
4. dbutan-1-ol

Answer (a): Butan-2-ol has a carbon attached to four different groups (H, OH, CH3, C2H5).

Revise: Fundamental Principles (Class 11 notes)

1. adiethyl ether
2. bethanal
3. cethyl ethanamide
4. dethyl ethanoate

Answer (d): This is esterification, giving an ester and water.

Revise: Carboxylic Acids (Class 12 notes)

1. aspeed up the setting
2. bincrease its colour
3. cincrease its alkalinity
4. dslow down the setting

Answer (d): Without gypsum, cement sets almost instantly; gypsum controls the setting time.

Revise: Cement (Class 12 notes)

1. acalcium salt of a mineral acid
2. bsodium salt of a long-chain fatty acid
3. cester of glycerol
4. dsodium salt of a benzene sulphonic acid

Answer (b): Saponification of fats with NaOH gives sodium salts of fatty acids plus glycerol.

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

1. a35%
2. b46.7%
3. c28%
4. d64%

Answer (b): Molar mass $= 60$; N $= 28$; $28/60 \times 100 = 46.7%$.

Revise: Modern Chemical Manufactures (Class 11 notes)

1. aphenolphthalein
2. bmethyl orange
3. cthymol blue
4. dno indicator is needed

Answer (b): The equivalence point is acidic (pH below 7), which is matched by methyl orange's range of 3.1 to 4.4.

Revise: Volumetric Analysis (Class 12 notes)

1. a0.05 M
2. b0.1 M
3. c0.2 M
4. d0.4 M

Answer (c): $M_1V_1 = M_2V_2$: $0.1 \times 20 = M \times 10$, so $M = 0.2$ M.

Revise: Volumetric Analysis (Class 12 notes)

1. aapple green
2. bbrick red
3. ccrimson red
4. dgolden yellow

Answer (b): Ca gives brick red, Ba apple green, Sr crimson red and Na golden yellow.

1. aDarwin
2. bDe Vries
3. cLamarck
4. dWeismann

Answer (c): Lamarck's theory of inheritance of acquired characters (use and disuse) explains the giraffe neck this way. Darwin used natural selection and De Vries mutation.

Revise: Theories of Organic Evolution (Class 11 notes)

1. aanalogous organs
2. bvestigial organs
3. catavistic organs
4. dhomologous organs

Answer (d): Both share the same basic bone plan (same origin) but are modified for different functions, which defines homologous organs and shows divergent evolution.

Revise: Evidences of Evolution (Class 11 notes)

1. anatural selection
2. binheritance of acquired characters
3. cgenetic drift
4. dartificial selection

Answer (a): Dark moths were better camouflaged on soot-covered trees, so they survived and reproduced more: selection by the environment.

Revise: Theories of Organic Evolution (Class 11 notes)

1. aCnidaria
2. bPlatyhelminthes
3. cPorifera
4. dAnnelida

Answer (c): Sponges (Porifera) have flagellated collar cells lining the spongocoel that create water current and trap food.

Revise: Kingdom Animalia (Class 11 notes)

1. aPlanaria
2. bEarthworm
3. cStarfish
4. dHydra

Answer (d): Cnidoblasts (nematocyst-bearing cells) are the defining feature of phylum Cnidaria, to which Hydra belongs.

Revise: Kingdom Animalia (Class 11 notes)

1. aEchinodermata
2. bMollusca
3. cArthropoda
4. dNematoda

Answer (a): Starfish and other echinoderms move by tube feet operated by the water vascular system.

Revise: Kingdom Animalia (Class 11 notes)

1. aTaenia
2. bAscaris
3. cHydra
4. dPheretima

Answer (b): Ascaris (Aschelminthes) has a pseudocoelom. Taenia is acoelomate with no gut, Hydra is diploblastic and Pheretima is a true coelomate and segmented.

Revise: Kingdom Animalia (Class 11 notes)

1. aSimple columnar epithelium
2. bStratified squamous epithelium
3. cCuboidal epithelium
4. dSimple squamous epithelium

Answer (d): A single layer of thin flat cells gives the shortest diffusion path.

Revise: Animal Tissues (Class 12 notes)

1. aosteoblasts
2. bosteoclasts
3. cchondrocytes
4. dfibroblasts

Answer (a): Osteoblasts secrete bone matrix; osteoclasts resorb bone and chondrocytes are cartilage cells.

Revise: Animal Tissues (Class 12 notes)

1. askeletal muscle
2. bcardiac muscle
3. csmooth muscle
4. dnervous tissue

Answer (b): Intercalated discs join cardiac muscle fibres and let impulses spread quickly for synchronized contraction.

Revise: Animal Tissues (Class 12 notes)

1. aaxon
2. bmyelin sheath
3. cdendrite
4. dnode of Ranvier

Answer (c): Dendrites receive stimuli; the axon carries the impulse away from the cell body.

Revise: Animal Tissues (Class 12 notes)

1. anephridia
2. bMalpighian tubules
3. cflame cells
4. dgreen glands

Answer (a): Earthworm has segmentally arranged nephridia. Malpighian tubules occur in cockroach and flame cells in flatworms.

Revise: Detailed Study of Earthworm (Class 11 notes)

1. aone atrium and one ventricle
2. btwo atria and one ventricle
3. ctwo atria and two ventricles
4. done atrium and two ventricles

Answer (b): The frog heart is three-chambered, so oxygenated and deoxygenated blood mix partly in the single ventricle.

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

1. abook lungs
2. bgills
3. ctracheae opening by spiracles
4. dskin

Answer (c): Insects have a tracheal system; air enters through paired spiracles on the thorax and abdomen.

Revise: Kingdom Animalia (Class 11 notes)

1. amerozoite
2. bgametocyte
3. ctrophozoite
4. dsporozoite

Answer (d): Sporozoites from the mosquito's salivary glands are injected into human blood and go to the liver.

Revise: Parasite (Class 11 notes)

1. anephridia
2. bMalpighian tubules
3. ckidneys
4. dgreen glands

Answer (b): Malpighian tubules collect uric acid from the haemolymph and pass it into the hindgut.

Revise: Kingdom Animalia (Class 11 notes)

1. a1 to 3
2. b5 to 7
3. c14 to 16
4. d20 to 22

Answer (c): The glandular clitellum forms a girdle in segments 14, 15 and 16 and secretes the cocoon.

Revise: Detailed Study of Earthworm (Class 11 notes)

1. apepsin
2. blipase
3. ctrypsin
4. dsalivary amylase

Answer (d): Salivary amylase (ptyalin) hydrolyses starch to maltose in the mouth.

Revise: Digestive System (Class 12 notes)

1. abeta cells of the islets of Langerhans
2. balpha cells of the islets of Langerhans
3. cacinar cells of pancreas
4. dadrenal cortex

Answer (a): Beta cells make insulin (lowers blood glucose); alpha cells make glucagon.

Revise: Endocrinology (Class 12 notes)

1. aatrioventricular node
2. bbundle of His
3. csinoatrial node
4. dPurkinje fibres

Answer (c): The SA node in the right atrium generates the impulse that initiates each heartbeat.

Revise: Circulatory System (Class 12 notes)

1. aneuron
2. balveolus
3. cvillus
4. dnephron

Answer (d): A nephron consists of a renal corpuscle and a tubule and forms urine.

Revise: Excretory System (Class 12 notes)

1. a500 mL
2. b1200 mL
3. c3500 mL
4. d5000 mL

Answer (a): Normal quiet breathing moves about 500 mL of air per breath.

Revise: Respiratory System (Class 12 notes)

1. aAB positive
2. bO negative
3. cA negative
4. dB positive

Answer (b): Group O red cells carry no A or B antigens and Rh negative lacks the D antigen, so they are accepted by all.

Revise: Circulatory System (Class 12 notes)

1. acerebrum
2. bmedulla oblongata
3. cthalamus
4. dcerebellum

Answer (d): The cerebellum integrates sensory input to maintain equilibrium and smooth muscle movement.

Revise: Nervous System (Class 12 notes)

1. arod cells
2. bcone cells
3. cbipolar cells
4. dganglion cells

Answer (a): Rods contain rhodopsin and are very sensitive to low light but do not give colour vision.

Revise: Sense Organs (Class 12 notes)

1. aanterior pituitary
2. bposterior pituitary
3. cthyroid
4. dadrenal medulla

Answer (b): Oxytocin is made in the hypothalamus and stored and released by the posterior pituitary.

Revise: Endocrinology (Class 12 notes)

1. auterus
2. bvagina
3. campulla of the fallopian tube
4. dovary

Answer (c): The egg meets sperm in the upper part (ampulla) of the oviduct, and the zygote then moves to the uterus.

Revise: Reproductive System (Class 12 notes)

1. aO
2. bB
3. cAB
4. dNone of the groups

Answer (a): Group A plasma has anti-B antibodies, so B or AB cells would clump. O cells have no A or B antigens.

Revise: Circulatory System (Class 12 notes)

1. ais alkaline due to bile
2. bis strongly acidic due to HCl
3. ccontains lipase
4. dis neutral

Answer (b): HCl provides pH about 2 and converts inactive pepsinogen to pepsin, which digests proteins.

Revise: Digestive System (Class 12 notes)

1. aloop of Henle
2. bcollecting duct
3. cglomerulus
4. ddistal convoluted tubule

Answer (c): Healthy glomerular filtration holds back large proteins. Damage lets them leak into the filtrate.

Revise: Excretory System (Class 12 notes)

1. acretinism
2. bacromegaly
3. cAddison's disease
4. dmyxoedema

Answer (d): Hypothyroidism in adults gives myxoedema. In infants it gives cretinism.

Revise: Endocrinology (Class 12 notes)

1. adissolved oxygen in plasma
2. boxyhaemoglobin
3. ccarbaminohaemoglobin
4. dbicarbonate

Answer (b): About 97 percent binds to haemoglobin as oxyhaemoglobin and only about 3 percent dissolves in plasma.

Revise: Respiratory System (Class 12 notes)

1. aVibrio cholerae
2. bMycobacterium tuberculosis
3. cSalmonella typhi
4. dPlasmodium vivax

Answer (c): Salmonella typhi is a bacterium spread by contaminated food and water.

Revise: Microbial Diseases (Class 12 notes)

1. amacrophages
2. bmast cells
3. cplatelets
4. dplasma cells

Answer (d): B lymphocytes differentiate into plasma cells, which release antibodies.

Revise: Microbial Diseases (Class 12 notes)

1. apassive immunity
2. bactive natural immunity
3. cactive artificial immunity
4. dno immunity

Answer (a): Immunity from antibodies supplied from outside is passive. Active immunity requires the body to make its own antibodies, as after a vaccine.

Revise: Microbial Diseases (Class 12 notes)

1. atyphoid
2. bpolio
3. ctuberculosis
4. dtetanus

Answer (c): BCG is a weakened Mycobacterium bovis strain that protects against tuberculosis.

Revise: Microbial Diseases (Class 12 notes)

1. aX-ray radiography
2. bCT scan
3. cradioactive tracer scan
4. dultrasonography

Answer (d): Ultrasound uses high-frequency sound waves, safe for the foetus.

Revise: Human Health and Disorders (Class 12 notes)

1. aPCR
2. bgel electrophoresis
3. cELISA
4. dblotting

Answer (a): Polymerase chain reaction cycles denaturation, annealing and extension with Taq polymerase to amplify DNA.

Revise: Biotechnology (Class 12 notes)

1. aex situ conservation
2. bin situ conservation
3. ccaptive breeding
4. dcryopreservation

Answer (b): In situ means in the original place. Zoos and seed banks are ex situ.

Revise: Conservation Biology (Class 11 notes)

1. aWildlife Protection Act
2. bCITES appendix only
3. cRamsar list
4. dIUCN Red Data Book

Answer (d): The IUCN Red List (Red Data Book) classifies species by extinction risk.

Revise: Conservation Biology (Class 11 notes)

1. aamino acids
2. bnucleotides
3. cmonosaccharides
4. dfatty acids

Answer (a): Amino acids join through peptide bonds to form polypeptides.

Revise: Biomolecules (Class 11 notes)

1. astarch
2. bglycogen
3. ccellulose
4. dchitin

Answer (b): Glycogen is stored in liver and muscle. Starch is the plant reserve.

Revise: Biomolecules (Class 11 notes)

1. aChlamydomonas
2. bVolvox
3. cSpirogyra
4. dUlothrix

Answer (c): In Spirogyra the contents of one filament cell pass through a conjugation tube into the cell of another filament.

Revise: Algae (Class 11 notes)

1. achitin
2. bcellulose
3. cpeptidoglycan
4. dlignin

Answer (a): Fungal walls contain chitin, unlike plant walls (cellulose) and bacterial walls (peptidoglycan).

Revise: Mycota (Fungi) (Class 11 notes)

1. asporophyte
2. bgametophyte
3. czygote
4. dsporangium

Answer (b): The leafy or thalloid plant is haploid gametophyte, and the sporophyte depends on it.

Revise: Bryophyta (Class 11 notes)

1. aprotonema
2. bstrobilus
3. cprothallus
4. dsorus

Answer (c): A fern spore germinates into a small heart-shaped prothallus bearing antheridia and archegonia.

Revise: Pteridophyta (Class 11 notes)

1. ahave no seeds
2. bhave no flowers or seeds
3. clack vascular tissue
4. dhave naked seeds

Answer (d): Their ovules are not enclosed in an ovary, so seeds are exposed on cone scales.

Revise: Gymnosperms (Class 11 notes)

1. adouble-stranded DNA
2. bsingle-stranded RNA
3. csingle-stranded DNA
4. ddouble-stranded RNA

Answer (b): TMV has a helical protein coat around a single-stranded RNA.

Revise: Introductory Microbiology (Class 11 notes)

1. apink
2. bcolourless
3. cpurple
4. dgreen

Answer (c): Thick peptidoglycan retains the crystal violet stain, so the cells appear purple. Gram-negative cells appear pink.

Revise: Introductory Microbiology (Class 11 notes)

1. ahaploid
2. bdiploid
3. ctetraploid
4. dtriploid

Answer (d): One sperm fuses with the two polar nuclei in double fertilization, so endosperm is 3n.

Revise: Angiosperms (Class 11 notes)

1. aan alga or cyanobacterium and a fungus
2. ba moss and a fern
3. ca bacterium and a virus
4. da fungus and a root

Answer (a): The photobiont makes food and the mycobiont gives protection and water.

Revise: Lichens (Class 11 notes)

1. a1000 J
2. b10 J
3. c100 J
4. d10000 J

Answer (c): Producers 10000 J, herbivores 1000 J, secondary consumers 100 J.

Revise: Ecology (Class 11 notes)

1. ashrubs
2. btrees
3. cgrasses
4. dlichens

Answer (d): Crustose lichens secrete acids that weather rock and begin soil formation.

Revise: Ecology (Class 11 notes)

1. aShorea robusta
2. bPinus roxburghii
3. cRhododendron arboreum
4. dAbies spectabilis

Answer (a): Sal (Shorea robusta) forms the main forest of the Terai and Siwalik region.

Revise: Vegetation (Class 11 notes)

1. axerophytes
2. bhydrophytes
3. chalophytes
4. depiphytes

Answer (b): Aerenchyma gives buoyancy and stores air in aquatic plants.

Revise: Ecology (Class 11 notes)

1. aribosome
2. bGolgi body
3. clysosome
4. dmitochondrion

Answer (d): ATP is produced by oxidative phosphorylation in the mitochondria.

Revise: Cell (Class 11 notes)

1. aPachytene of prophase I
2. bMetaphase I
3. cAnaphase II
4. dLeptotene of prophase I

Answer (a): Non-sister chromatids of synapsed homologues exchange segments in pachytene.

Revise: Cell Division (Class 11 notes)

1. a46 chromosomes
2. b23 chromosomes
3. c92 chromosomes
4. d12 chromosomes

Answer (b): Meiosis halves the chromosome number, so each gamete is haploid with 23.

Revise: Cell Division (Class 11 notes)

1. aWatson and Crick
2. bSchleiden and Schwann
3. cSinger and Nicolson
4. dRobertson

Answer (c): Singer and Nicolson (1972) described proteins floating in a fluid lipid bilayer.

Revise: Cell (Class 11 notes)

1. a28
2. b14
3. c7
4. d56

Answer (a): Sister chromatids separate, so the cell momentarily has 28 chromosomes, 14 at each pole.

Revise: Cell Division (Class 11 notes)

1. a1:1
2. b3:1
3. c9:3:3:1
4. d1:2:1

Answer (b): Tt x Tt gives 3 dominant to 1 recessive phenotype.

Revise: Mendelian Genetics (Class 12 notes)

1. aa back cross to the dominant parent
2. ba reciprocal cross
3. ca test cross
4. da dihybrid cross

Answer (c): A test cross reveals whether the dominant phenotype is homozygous or heterozygous.

Revise: Mendelian Genetics (Class 12 notes)

1. a3:1
2. b1:2:1
3. call tall
4. d1:1

Answer (d): Tt gives T and t gametes and tt gives only t, so offspring are 1 Tt (tall) to 1 tt (dwarf).

Revise: Mendelian Genetics (Class 12 notes)

1. a30
2. b20
3. c40
4. d70

Answer (b): A = T = 30 percent, leaving 40 percent for G + C, so G = 20 percent (Chargaff's rule).

Revise: Genetic Materials (Class 12 notes)

1. aethylene
2. bauxin
3. ccolchicine
4. dpenicillin

Answer (c): Colchicine blocks spindle formation, so chromosomes fail to separate and the number doubles.

Revise: Mutation and Polyploidy (Class 12 notes)

1. asister chromatids of one chromosome
2. bnon-homologous chromosomes
3. ctwo daughter cells
4. dnon-sister chromatids of homologous chromosomes

Answer (d): This recombination breaks linkage and produces new allele combinations.

Revise: Linkage and Crossing Over (Class 12 notes)

1. axylem
2. bphloem
3. ccambium
4. dcollenchyma

Answer (a): Xylem vessels and tracheids carry water from roots to leaves. Phloem carries food.

Revise: Plant Anatomy (Class 12 notes)

1. aapical meristem
2. bintercalary meristem
3. cvascular cambium
4. dphellem only

Answer (c): Cambium adds secondary xylem and phloem and so increases thickness (secondary growth).

Revise: Plant Anatomy (Class 12 notes)

1. amonocot
2. bgymnosperm only
3. cpteridophyte
4. ddicot

Answer (d): Open bundles with cambium in a ring are typical of dicot stems. Monocots have scattered closed bundles.

Revise: Plant Anatomy (Class 12 notes)

1. athylakoid membranes of grana
2. bstroma
3. cmatrix of mitochondria
4. dcytoplasm

Answer (a): Photosystems and electron carriers are in thylakoid membranes. The Calvin cycle occurs in stroma.

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

1. a4
2. b2
3. c8
4. d38

Answer (b): 4 ATP are formed and 2 are used, so the net gain is 2 ATP.

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

1. aauxin
2. bgibberellin
3. ccytokinin
4. dethylene

Answer (d): Ethylene is a gaseous hormone that triggers ripening and senescence.

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

1. aabscisic acid
2. bgibberellic acid
3. cindole acetic acid
4. dzeatin

Answer (a): ABA makes guard cells lose turgor, so stomata close and water loss falls.

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

1. acarbon dioxide
2. bwater
3. cglucose
4. dATP

Answer (b): Photolysis of water in photosystem II releases O2, as shown by isotopic labelling.

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

1. alenticels
2. bcuticle
3. cstomata
4. droot hairs

Answer (c): Most water vapour leaves through the stomatal pores of the leaves.

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

1. a7-celled and 8-nucleate
2. b8-celled and 8-nucleate
3. c4-celled and 4-nucleate
4. d3-celled and 3-nucleate

Answer (a): It has 3 antipodals, 2 synergids, 1 egg and a central cell with 2 polar nuclei: 7 cells, 8 nuclei.

Revise: Angiosperms (Class 11 notes)

1. aendothecium
2. btapetum
3. cepidermis
4. dmiddle layer

Answer (b): The innermost tapetum supplies food to microspore mother cells and pollen.

Revise: Angiosperms (Class 11 notes)

1. aapical dominance
2. bvernalization
3. ctotipotency
4. dphotoperiodism

Answer (c): Totipotency is the ability of a living plant cell to give rise to a complete plant.

Revise: Biotechnology (Class 12 notes)

1. afungal wilt
2. bdrought
3. cherbicides only
4. dbollworm

Answer (d): It carries a gene from Bacillus thuringiensis for a toxin lethal to bollworm larvae.

Revise: Biotechnology (Class 12 notes)

1. aAzotobacter
2. bRhizobium
3. cNostoc
4. dMycorrhiza

Answer (b): Rhizobium lives symbiotically in legume nodules and fixes atmospheric nitrogen.

Revise: Applied Biology (Class 12 notes)

1. acooking
2. bcutting
3. ceating
4. dkitchen

Answer (c): A book is used for reading, a fork is used for eating (tool to action).

1. amango
2. bapple
3. cbanana
4. dpotato

Answer (d): Mango, apple and banana are fruits that grow above ground, potato is a tuber (vegetable).

1. aabundant
2. brare
3. climited
4. dmeagre

Answer (a): Scarce means in short supply; abundant is the opposite. The others are synonyms.

1. asparrow
2. beagle
3. cbat
4. dparrot

Answer (c): Bat is a mammal. The others are birds.

1. acruel
2. bcareless
3. cselfish
4. dkind

Answer (d): Benevolent means well-meaning and kindly.

1. a42
2. b40
3. c44
4. d36

Answer (a): Differences are 4, 6, 8, 10, so the next is 12 and 30 + 12 = 42.

1. a225
2. b300
3. c450
4. d675

Answer (b): Number = 45 x 100 / 15 = 300.

1. a24
2. b32
3. c48
4. d40

Answer (d): One part = 64 / 8 = 8, so the larger is 5 x 8 = 40.

1. a25
2. b20
3. c100
4. d15

Answer (a): Profit = 100, and 100 / 400 x 100 = 25 percent.

1. a10
2. b12
3. c14
4. d15

Answer (b): Total = 60, the four sum to 48, so the fifth is 12.

1. aT
2. bV
3. cU
4. dS

Answer (c): Positions are 1, 3, 6, 10, 15 with gaps +2, +3, +4, +5, so the next gap is +6: 21, which is U.

1. a38
2. b36
3. c40
4. d35

Answer (a): Differences are 3, 5, 7, 9, so the next is 11 and 27 + 11 = 38.

1. aseed
2. bflower
3. cplant
4. dfruit

Answer (b): The order is seed, then plant, then flower, then fruit.

1. a15
2. b19
3. c17
4. d16

Answer (c): These are consecutive prime numbers and the one after 13 is 17.

1. aThursday
2. bSaturday
3. cWednesday
4. dFriday

Answer (d): 100 = 14 weeks + 2 days, so two days after Wednesday is Friday.

1. a5 km
2. b3 km
3. c8 km
4. d13 km

Answer (b): He goes 5 km north, 3 km east, then 5 km south, so he is back level with the start, 3 km east of it.

1. a8
2. b6
3. c12
4. d1

Answer (c): Edge cubes (not corner) have two painted faces: 12 edges x 1 cube each = 12. Corners (8) have three, face centres (6) have one.

1. ap
2. bq
3. cb
4. dd

Answer (d): A vertical mirror reverses left and right, so b appears as d.

1. a8 o'clock
2. b6 o'clock
3. c7 o'clock
4. d4 o'clock

Answer (a): Mirror time = 12 - 4 = 8.

1. a4
2. b6
3. c5
4. d3

Answer (c): 7 - 2 = 5.