eamcet 13

Physics1. A wire of length L and linear density m is

stretched by a force T and the frequency isn1. Another wire of same material of length2L and same linear density is stretched by aforce 9T and its frequency is n2 . Then, thevalue of ( / )n n2 1 is,(a) 4 : 1 (b) 1 : 3

(c) 3 : 2 (d) 1 : 2

2. A pipe closed at one end and open at theother end resonates with a sound offrequency 135 Hz and also with 165 Hz, butnot at any other frequency intermediatebetween these two. Then, the frequency ofthe fundamental note of the pipe is(a) 15 Hz (b) 60 Hz

(c) 7.5 Hz (d) 30 Hz

3. A small angled prism of refractive index 1.4is combined with another small angledprism of refractive index 1.6 to producedispersion without deviation. If the angle offirst prism is 6°, the angle of the secondprism is(a) 8° (b) 6°

(c) 4° (d) 2°

4. The magnifying power of the astronomicaltelescope for normal adjustment is 50. Thefocal length of the eyepiece is 2 cm. Therequired length of the telescope for normaladjustment is(a) 102 cm (b) 100 cm

(c) 98 cm (d) 25 cm

5. Light of wavelenght λ from a point sourcefalls on a small circular obstacle of diameterd. Dark and bright circular rings around acentral bright spot are formed on a screenbeyond the obstacle. The distance betweenthe screen and obstacle is D. Then, thecondition for the formation of rings is,

(a) λ ≈ d

D4(b) λ ≈ d

D

2

4

(c) dD

≈ λ2

(d) λ ≈ D

4

6. A bar magnet of moment of inertia I isvibrated in a magnetic field of induction 0 4 10 4. × − T. The time period of vibration is12 s. The magnetic moment of the magnet is120 Am2 . The moment of inertia of themagnet is (in kgm2) approximately(a) 172.8 × −10 4

(b) 2.1 π2 210−

(c) 1.57 × −10 2

(d) 1728 × −10 2

7. A short magnet of magnetic moment M, isplaced on a straight line. The ratio ofmagnetic induction fields B , B , B1 2 3 valueson this line at points which are at distances30 cm, 60 cm and 90 cm respectively fromthe centre of the magnet is(a) 27 : 3.37 : 1

(b) 37.3 : 1 : 27

(c) 27 : 8 : 3.37

(d) 1 : 2 : 3

Solved Paper 2013

EAMCETMedical Entrance Exam

8. Two capacitors having capacitances C1 and C2 are charged with 120 V and 200 Vbatteries respectively. When they areconnected in parallel now, it is found thatthe potential on each one of them is zero.Then,(a) 5 31 2C C= (b) 8 51 2C C=(c) 9 51 2C C= (d) 3 51 2C C=

9. A parallel plate capacitor with a dielectricslab of dielectric constant 3 filling the spacebetween the plates is charged to a potentialV. The battery is then disconnected and thedielectric slab is withdrawn. It is thenreplaced by another dielectric slab ofdielectric constant 2. If the energies storedin the capacitor before and after thedielectric slab is changed are E1 and E2 , then E E1 2/ is(a)

9

5 (b)

4

9(c)

2

3 (d)

3

2

10. Copper and carbon wires are connected inseries and the combined resistor is kept at 0° C. Assuming the combined resistance does not vary with temperature, the ratio of theresistances of carbon and copper wires at 0° C is (Temperature coefficients ofresistivity of copper and carbon respectivelyare 4 10 3× °− / C and − × °−0.5 C)10 3 /(a) 4 (b) 8

(c) 6 (d) 2

11. In the equation 1p

yk TBβ

= , where p is the

pressure, y is the distance, kB is Boltzmannconstant and T is the temperature.Dimensions of β are(a) [ ]M L T−1 1 2 (b) [ ]M L T0 2 0

(c) [ ]M L T1 1 2− − (d) [ ]M L T0 0 0

12. A person reaches a point directly opposite on the other bank of a river. The velocity of thewater in the river is 4 1ms− and the velocityof the person in still water is 5 1ms− . If thewidth of the river is 84.6 m, time taken tocross the river in seconds is(a) 28.2 (b) 9.4

(c) 2 (d) 84.6

13. A body is thrown vertically upward from apoint A 125 m above the ground. It goes up to a maximum height of 250 m above theground and passes through A on itsdownward journey. The velocity of the bodywhen it is at a height of 70 m above theground is ( )g = −10 2ms(a) 50 1ms− (b) 60 1ms−

(c) 80 1ms− (d) 20 1ms−

14. A body of mass 300 kg is moved through10 m along a smooth inclined plane of angle 30°. The work done in moving in joules is ( )g = −9.8 ms 2

(a) 4900 (b) 9800

(c) 14,700 (d) 2450

15. A balloon starting from rest ascendsvertically with uniform acceleration to aheight of 100 m in 10 s. The force on thebottom of the balloon by a mass of 50 kg is ( )g = −10 2ms(a) 100 N (b) 300 N

(c) 600 N (d) 400 N

16. Three particles, each of mass m, are placedat the vertices of a right angled triangle asshown in figure. The position vector of thecentre of mass of the system is (O is theorigin and $, $, $i j k are unit vectors)

(a) 1

3( $ $ )a bi j−

(b) 2

3( $ $ )a bi j−

(c) 2

3( $ $ )a i j− b

(d) 1

3( $ $ )a bi + j

2 | EAMCET (Medical) l Solved Paper 2013

XO m a A

mB

Y

m

17. A ball of mass m moving with a horizontalvelocity v strikes the bob of a pendulum atrest. Mass of the bob of the pendulum is alsom. During this collision the ball sticks withthe bob of the pendulum. The height towhich the combined mass rises(g = acceleration due to gravity)

(a) v

g

2

4(b)

v

g

2

8 (c)

v

g

2

(b) v

g

2

2

18. A 3 kg block is placed over a 10 kg block andboth are placed on a smooth horizontalsurface. The coefficient of friction betweenthe blocks is 0.2. If a horizontal force of 20 Nis applied to 3 kg block, accelerations of thetwo blocks in ms−2 are ( )g = −10 2ms

(a) 13

4, 0.6 (b)

14

43,

(c) 13

43, (d)

14

3, 0.6

19. A stone tied to a rope is rotated in a verticalcircle with uniform speed. If the differencebetween the maximum and minimumtensions in the rope is 20 N, mass of thestone in kg is ( )g = −10 2ms(a) 0.75 (b) 1.0 (c) 1.5 (d) 0.5

20. A particle mass m is attached to a thinuniform rod of length a at a distance of a

4from the mid-point C as shown in the figure.The mass of the rod is 4 m. The moment ofinertia of the combined system about an axis passing through ‘O’ and perpendicular to the rod is

(a) 91

48

2ma (b) 27

48

2ma

(c) 51

48

2ma (d) 64

48

2ma

21. If earth were to rotate on its own axis suchthat the weight of a person at the equatorbecomes half the weight at the poles, thenits time period of rotation is (g = acceleration due to gravity near the poles and R is theradius of earth) (Ignore equatorial bulge)

(a) 22π R

g(b) 2

2π R

g

(c) 23

π R

g(d) 2 π R

g

22. A particle is executing simple harmonicmotion along a straight line. Atdisplacements r1 and r2 from its meanposition the velocities are v1 and v2 . The timeperiod of the particle is

(a) 2 22

12

22

12

1 2

π r r

v v

−−

/

(b) 2 12

22

22

12

1 2

π v v

r r

−+

/

(c) 1

212

22

22

12

1 2

πv v

r r

+−

/

(d) 2 22

12

12

22

1 2

π r r

v v

−−

/

23. Force constants of two wires A and B of thesame material are k and 2k respectively. Ifthe two wires are stretched equally, then the ratio of work done in stretching W

WA

B

is

(a) 1

3(b)

1

2

(c) 3

2(d)

1

4

24. Water rises in a capillary tube upto a heightof 10 cm whereas mercury depresses in it by3.42 cm. If the angle of contact and densityof mercury are 135° and 13.6 g/ccrespectively, then the ratio of the surfacetension of water and mercury will be nearly(a) 13 : 2 (b) 5 : 16

(c) 16 : 5 (d) 2 : 13

25. Two capillary tubes of lengths in the ratio2 : 1 and radii in the ratio 1 : 2 are connectedin series. Assume the flow of the liquidthrough the tube is steady. Then, the ratio of pressure difference across the tubes is(a) 1 : 8

(b) 1 :16

(c) 32 : 1

(d) 1 : 1

EAMCET (Medical) l Solved Paper 2013 | 3

OC

a2

a4

x

Mass m

µ = 0.2

10 kg

3 kg 20 N

26. When a liquid is heated in a glass vessel, itscoefficient of apparent expansion is 1.03 × °−10 3 / C. When the same liquid isheated in a copper vessel, its coefficient ofapparent expansion is 1.006 × °−10 3 / C. If the coefficient of linear expansion of copper is 17 10 6× °− / C, then the coefficient of linearexpansion of glass is(a) 8 5 10 4. /× °− C (b) 9 10 6× °− / C

(c) 27 10 6× °− / C (d) 10 10 4× °− / C

27. A horizontal uniform tube, open at both ends is containing a liquid of certain length atsome temperature. When the temperature is changed, the length of the liquid in the tubeis not changed. If α is the coefficient of linearexpansion of the material of the tube and γ is the coefficient of volume expansion of theliquid, then(a) γ α= 2 (b) γ α= 3

(c) γ α= 4 (d) γ α=

28. Match the following from table A, in the case of gases, with those in table B

Table A Table B

(a) Work done in isobaricprocess

(d)nRT

V

Velog 2

1

(b) Work done in isothermalprocess

(e) p V V( )2 1−

(c) Work done in adiabaticprocess

(f) nR T T( )1 2

1

−−γ

(g) zero

(a) a e, b d, c f→ → →

(b) a h, b g, c e→ → →(c) a e, b d, c g→ → →

(d) a g, b h, c e→ → →

29. Two cylinders A and B fitted with pistonscontain equal amounts of an ideal diatomicgas at 300 K. Piston A is free to move andpiston of B is fixed. Same amount of heat isgiven to the gases in the two cylinders.Temperature of the gas in cylinder Aincreases by 30 K. Then, increase intemperature of the gas in the cylinder B is ( .γ = 1 4 for diatomic gas)(a) 24 K (b) 36 K

(c) 54 K (d) 42 K

30. The temperature at the two ends A and B ofa rod of length 25 cm and circular cross-section are 100°C and 0°C respectively. Inthe steady state the temperature at a point10 cm from the end B is (Ignore loss of heatfrom curved surface of the rod)(a) 60°C (b) 80°C (c) 90°C (d) 40°C

31. The necessary condition in making of ajunction transistor (E-emitter, B-base and C-collector) (a) E and B are lightly doped and C is heavily doped

(b) E is heavily doped, B is thin and lightly dopedand C is moderately doped

(c) E and C are lightly doped and B is thick andheavily doped

(d) E and B are heavily doped and C is lightly doped

32. A p-n-p transistor is used in common emitter mode in an amplifier circuit. When basecurrent is changed by an amount ∆IB, thecollector current changes by 4 mA. If thecurrent amplification factor is 60, then thevalue of ∆IB is(a) 15 µA (b) 240 mA (c) 66 6. µA (d) 60 µA

33. A U235 reactor generates power at a rate of Pproducing 2 1018× fissions per second. Theenergy released per fission is 185 MeV. Thevalue of P is(a) 59.2 MW (b) 370 1018× MW

(c) 0.59 MW (d) 370 MW

34. The purpose of using heavy water in nuclear reactor is(a) to increase the energy released in nuclear fission

(b) to cool the reactor to room temperature

(c) to make the dynamo blades to work well

(d) to decrease the energy of fast neutrons to thethermal energy

35. The K α − X ray of Molybdenum has awavelength of 71 10 12× − m. If the energy of aMolybdenum atom with K-electron removedis 23.32 keV, then the energy ofMolybdenum atom when an L-electronremoved is (hc = 12.42 × −10 7 eV)(a) 17.5 keV (b) 40.82 keV

(c) 23.32 keV (d) 5.82 keV


36. In Moseley's law ν = −a z b( ), the values ofthe screening constant for K-series andL-series of X-rays are respectively(a) 1, 6.4 (b) 1, 4

(c) 4, 6 (d) 2, 4

37. An electron beam in a TV picture tube isaccelerated through a potential difference ofV volt. It passes through a region oftransverse magnetic induction field (B) andfollows a circular orbit of radius r. Theinduced magnetic field (B) is

(a) 2

2

mV

er(b)

2 mV

r(c)

2 mVr

e(d)

2 mV

er

38. A 0.01 H inductor and 3 π ohm resistanceare connected in series with a 220 V, 50 HzAC source. The phase difference betweenthe current and emf is(a)

π2

rad (b) π6

rad (c) π3

rad (d) π4

rad

39. When the temperature difference betweenthe junctions of a given thermocouple is120°C, the thermo emf is 30 mV. The

temperature of hot junction is decreased by20°C and cold junction’s temperature isincreased by 6°C. The percentage decreasein thermo emf is (assume thermo emf isdirectly proportional to the temperaturedifference)(a) 43 (b) 2.16 (c) 20.4 (d) 21.6

40. The emf of a cell E is 15 V as shown in thefigure with an internal resistance of 0.5 Ω.Then, the value of the current drawn fromthe cell is

(a) 3 A (b) 2 A (c) 5 A (d) 1 A

Chemistry1. Which one of the following statements is

correct?(a) Sucralose is an antiseptic(b) Lactic acid is an antimicrobial(c) Seconal is an antipyretic(d) Chloroxylenol is a tranquillizer

2. X Y← →HI HNOGlucose 3

What are X and Y? X Y

(a) n-hexane Gluconic acid

(b) Gluconic acid Saccharic acid

(c) n-hexanol Saccharic acid

(d) n-hexane Saccharic acid

3. Identify the pair of condensation polymersfrom the following(a) terylene and nylon-66(b) PVC and polystyrene(c) polyvinylether and polyisobutene(d) neoprene and PVP

4. The order of basic strength of C H NH6 5 2 (1), C H NH2 5 2 (2), (C H ) NH2 5 2 (3) and NH3 (4) is

(a) 1 4 2 3< < < (b) 1 3 2 4< < <(c) 4 2 3 1< < < (d) 3 2 4 1< < <

5. Which one of the following reactions is notcorrect?

(a) CH CHO +1

2O CH COOH3 2

23

Mn(OAc) air→

/

(b) CH CH OH + O3 2 2

Micoderma aceti→

CH COOH + H O3 2

(c) CH MgBr CH COOH + Mg(OH)Br32

3(ii) H O

(i) CO

3

→⊕

(d) CH OH + CO CH COOH3 3

Ca / , pressure→

∆

6. Which one of the following reactions is notcorrect?

(a) (HCOO) Ca + (H CCOO) Ca 2CH CHO2 3 2 3→∆

+ 2CaCO3

(b) CH CH(OH)CH CH COCH + H3 3 3 3 2

Cu/300 C→°


15 V

0.5 Ω

8 Ω 10 Ω

6 Ω 1 Ω

7 Ω2 Ω

(c) CH CH OH CH CHO + H3 2 3 2

Ag/300 C→°

(d) CH ==CH + PdCl + H O CH CHO2 2 2 22

3H

CaCl→

⊕

+ Pd + 2HCl

7. The pair of chemicals used as food for yeastin the fermentation of molasses is(a) (NH ) SO4 2 4, Na PO3 4

(b) (NH ) SO4 2 4, (NH ) PO4 3 4

(c) Na SO2 4, (NH ) PO4 3 4

(d) Na SO2 4, Na PO3 4

8. Which one of the following is chloropicrin?(a) CCl CHO3

(b) CCl —C(CH )

OH

3 3 2

(c) CCl —NO3 2

(d) CHCl —CHCl2 2

9. The correct Fischer projection formula of ( , )2 3R R -2,3-dihydroxy butanoic acid is

(a) HH

CO H

CH

OHOH

2

3

(b) H

HO

CH

CO H

OHH

3

2

(c) HOHO

CH

CO H

HH

3

2

(d) H

HO

CO H

CH

OHH

2

3

10. Which one of the following causes cancer?(a) 1,2-benzpyrene (b) n-hexane

(c) 2-butene (d) Cyclohexane

(a) 2-methyl-2,3-diethyl pentane

(b) 4-ethyl-3,3-dimethyl hexane

(c) 3,4-diethyl-4-methyl pentane

(d) 3-ethyl-4,4-dimethyl hexane

12. Carbon and hydrogen in an organiccompound are detected as …… .(a) CaHCO ,3 CaCO3

(b) CaHCO3, CuSO 5H O4 2⋅(c) CaCO3, CuSO 5H O4 2⋅(d) CaCO , Cu(OH)3 2

13. Match the following.List I List II

(A) pH of unpollutedrain water

(I)H C==CH—CHO2

(B) Acid rain (II) 5.6(C) Acrolein (III) NO , CO2 2

(D) Freon (IV)CF Cl2 2

(V)CaOCl2

The correct answer is(A) (B) (C) (D)

(a) (II) (I) (III) (IV)

(b) (I) (II) (III) (IV)

(c) (III) (II) (IV) (I)

(d) (II) (III) (I) (IV)

14. During the manufacture of cast iron, theslag (CaSiO3) is formed in(a) zone of reduction only(b) zone of fusion only(c) zone of reduction and zone of fusion(d) zone of heat absorption

15. Observe the following statements,1. Lanthanides actively participate in

chemical reactions.2. The basic nature of hydroxides of

lanthanides increases from La(OH)3 to Lu(OH)3.

3. Lanthanides do not form coordinatecompounds as readily as d-block metals.

The correct statements are(a) 2 and 3 (b) 1, 2 and 3

(c) 1 and 3 (d) 1 and 2

16. Identify the correct set.

Molecule Hybridisation Shape

Number of lone pairs

ofelectrons

(a) XeO4 sp d3 2 pyramidal 1

(b) XeO3 sp3 pyramidal 1

(c) XeF4 sp d3 2 planar 3

(d) XeF2 sp d3 linear 2

17. Fluorine reacts with KHSO4 to form HF andX. Which one of the following is X?(a) SO3 (b) K SO2 4

(c) K S O2 2 8 (d) H S2


IUPAC name of11.

65

43

2

1

is

18. For the equilibrium, A g B g C g( ) ( ) ( );º + K p = 0.82 atm at 27°C. At thesame temperature, its K c in mol L−1 is ( )R = − −0.082 L atm mol K1 1

(a) 0.033 (b) 3.3

(c) 1.0 (d) 0.33

19. If the solubility product of MgF2 at a certaintemperature is 1.08 × −10 10 , its solubility in mol L−1 is

(a) 1.04 × −10 5 (b) 7.3 × −10 4

(c) 3.0 × −10 5 (d) 3.0 × −10 4

20. In the reaction,4NO ( ) O ( ) 2N O ( ),2 2 2 5g g g+ →

∆H = − 111 kJif N O2 5( )s is formed instead of N O2 5( ),g the ∆H value in kJ is(∆Hsublimation for N O = 54 kJ mol )2 5

–1

(a) – 165 (b) – 57

(c) + 219 (d) – 219

21. In which of the following reactions, H O2 2acts as oxidizing agent?(a) Cl + H O 2HCl + O2 2 2 2→(b) Ag O + H O 2Ag + H O + O2 2 2 2 2→(c) 2NaOH + H O Na O + 2H O2 2 2 2 2→(d) KNO + H O KNO + H O2 2 2 3 2→

22. Zinc reacts with hot and concentratedNaOH and forms(a) H2

(b) Zn(OH) + Na O2 2

(c) ZnO

(d) O2

23. Orange red coloured monoxide is(a) K O2 (b) Na O2

(c) Cs O2 (d) Li O2

24. Observe the following statements,1. H BO3 3 is used as antiseptic.2. In B H ,2 6 each boron is sp2 hybridized.3. Aqueous solution of borax is alkaline in

nature.The correct statements are(a) 2 and 3 (b) 1, 2 and 3

(c) 1 and 3 (d) 1 and 2

25. Assertion (A) GeF4 and SiCl4 act as Lewisbases.Reason (R) Ge and Si have d-orbitals toaccept electrons.The correct answer is(a) Both (A) and (R) are correct, (R) is not the correct

explanation of (A)

(b) (A) is correct but (R) is not correct

(c) (A) is not correct but (R) is correct

(d) Both (A) and (R) are correct, (R) is the correctexplanation of (A)

26. Observe the following statements,1. The solubility of group V hydrides in

water decreases from NH3 to BiH3.2. Phosphorous does not exhibit allotropy.3. The stability of group V hydrides

decreases from NH3 to BiH3.The correct statement(s) are(a) 1 and 3 (b) 2

(c) 1 and 2 (d) 1, 2 and 3

27. Sodium thiosulphate reacts with dil. HCl toform NaCl, H O,2 X and Y. X reacts with Na S2to form Na S O2 2 3 and Y. Which one of thefollowing is Y?(a) H S2 (b) SO2

(c) H SO2 4 (d) S

28. The energy required to overcome theattractive forces on the electrons, w, of somemetals is listed below. The number of metals showing photoelectric effect when light of300 nm wavelength falls on it is (1 10 19eV 1.6 J)= × −

Metal Li Na K Mg Cu Ag Fe Pt Ww (eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75(a) 6 (b) 8

(c) 5 (d) 4

29. If the uncertainty in velocity of a movingobject is 1.0 ms× − −10 6 1 and the uncertaintyin its position is 58 m, the mass of this objectis approximately equal to that of (h = × −6.626 Js)10 34

(a) helium (b) deuterium

(c) lithium (d) electron


30. The correct order of atomic radius of Li, Beand B is(a) B >Be >Li (b) B >Li >Be

(c) Li >B >Be (d) Li >Be >B

31. The number of σ and π bonds present inacetylene are respectively(a) 3, 2 (b) 3, 3 (c) 2, 2 (d) 2, 3

32. The total number of antibonding electronsin N2 and O2 molecules respectively is(a) 4, 6 (b) 6, 8 (c) 5, 8 (d) 4, 8

33. Consider the following statements for thereaction,KIO + 5KI + 6HCl 3I + 6KCl + 3H O3 2 2→

1. I− is reduced to I22. IO3

− is oxidized to I23. IO3

− is reduced to I24. Oxidation number of I increases from –1

(in KI) to zero (in I2)The correct statements are(a) 1, 2 and 4 (b) 3 and 4

(c) 1 and 4 (d) 1, 3 and 4

34. Two gases of molecular masses M A and MBare at temperatures TA and TB respectively.If T M T MA B B A= , which one of the followingproperties has the same value for both thegases?(a) Pressure (b) Kinetic energy

(c) RMS velocity (d) Density

35. The weight in gram, of the non-volatilesolute urea (NH CONH )2 2 to be dissolved in100 g of water in order to decrease its vapour pressure by 1.8% is(a) 6.0 (b) 0.3 (c) 3.0 (d) 0.6

36. The boiling point of a solution containing68.4 g of sucrose (molar mass = −342 1g mol )in 100 g of water is ( )K b for water = 0.512 K kg mol−1

(a) 98.98°C (b) 101.02°C

(c) 100.512°C (d) 100.02°C

37. A certain quantity of electricity is passedthrough aqueous Al (SO )2 4 3 and CuSO4solutions connected in series. 0.09 g of Al isdeposited on cathode during electrolysis.The amount of copper deposited on cathodein grams is (At. wt. of Al = 27; Cu 63.6)=(a) 0.318 (b) 31.8

(c) 0.636 (d) 3.18

38. Molar conductances of BaCl ,2 H SO2 4 andHCl at infinite dilution are X X1 2, and X3respectively. Molar conductance of BaSO4 atinfinite dilution is(a) ( )/X X X1 2 32 2+ − (b) X X X1 2 3+ −(c) X X X1 2 32+ − (d) ( )/X X X1 2 3 2+ −

39. Atoms of an element ‘A’ occupy 23

tetrahedral

voids in the hexagonal close packed (hcp)unit cell lattice formed by the element ‘B’.The formula of the compound formed by ‘A’and ‘B’ is(a) A B2 (b) AB2

(c) A B4 3 (d) A B2 3

40. The temperature coefficient of a reaction is2.5. If its rate constant at T1 K is 2.5 s× − −10 3 1, the rate constant at T K2 in s−1

is ( )T T2 1>(a) 1.0 × −10 3 (b) 6.25 × −10 3

(c) 1.0 × −10 2 (d) 6.25 × −10 2

Zoology1. In amoeboid movement, according to

Goldacre and Lorsch cytoplasm solates due to(a) folding of protein molecules

(b) sliding of actin molecules

(c) sliding of myosin molecules

(d) unfolding of protein molecules

2. Which one of the branches of cranial nervesis not related to vagus?(a) Chorda tympani

(b) Cardiac depressor

(c) Recurrent laryngeal

(d) Pneumogastric


3. Match the following with reference toTaenia solium.

List I List II

(A) Mehlis glands (I) Tegument formation

(B) Vitelline gland (II) Osmoregulation andexcretion

(C) Mesenchymal cells (III) Lubricate passage of capsules into uterus

(D) Flame cells (IV) Secretion ofembryophore

(V) Capsule formationaround zygote

The correct match is(A) (B) (C) (D)

(a) (III) (V) (I) (II)

(b) (III) (V) (II) (IV)

(c) (V) (I) (II) (IV)

(d) (IV) (III) (I) (II)

4. In the life cycle of Wuchereria, 3rd and 4thmoults of microfilaria take place in(a) lymph vessels of man

(b) stomach of mosquito

(c) salivary glands of mosquito

(d) thoracic muscles of mosquito

5. Which one of the following became anendangered species as result of extinction ofRaphus cucullatus?(a) Sideroxylon grandiflorum

(b) Chrysanthemum

(c) Cinchona

(d) Bacillus thuringiensis

6. What is the resting membrane potential ofnerve fibre?(a) +75 mV (b) +45 mV

(c) − 70 mV (d) − 45 mV

7. Identify the correct combinations.(A) Housefly–Labellum–Pseudotrachea(B) Moth–labellum–Dutton’s membrane(C) Butterfly–1st Maxillae–Galea(D) Tse-tse fly–Rostrum–Haustellum(a) (A) and (B)

(b) (A) and (C)

(c) (B) and (C)

(d) (B) and (D)

8. Which of the following character is notrelated to Indian chain viper?(a) Large black rings occur in three rows on the

dorsal surface of the body

(b) Subcandals are present in two rows

(c) An arrow mark (↑) is present on the head

(d) Head is triangular covered by small scales

9. Arrange the ganglia of autonomous nervoussystem of cockroach in correct sequencefrom anterior to posterior end.(A) Frontal ganglion(B) Proventricular ganglion(C) Hypocerebral ganglion(D) Visceral ganglion(a) A–C–D–B (b) A–D–C–B

(c) B–C–D–A (d) B–D–A–C

10. Identify the typical dental formula of ametatherian.(a) I 5/4, C 1/1 PM 3/3, M 4/4

(b) I 3/3 C, 1/1 PM 4/4, M 3/3

(c) I 4/3 C, 1/1 PM 3/3, M 4/4

(d) I 4/5 C, 1/1 PM 3/3, M 4/4

11. Both hepatic and renal portal systems arefound in(a) Fishes, Amphibians, Reptiles

(b) Amphibians, Reptiles, Mammals

(c) Reptiles, Aves, Mammals

(d) Cyclostomes, Fishes, Amphibians

12. Unloading of oxygen from haemoglobin isenhanced under the following conditions.(a) Increase in pH, decrease in CO2, decrease in

temperature

(b) Increase in pH, increase in CO2, decrease intemperature

(c) Decrease in pH, increase in CO2, increase intemperature

(d) Decrease in pH, decrease in CO2, increase intemperature

13. Craspedote medusa is present in thefollowing pair.(a) Aurelia and Rhizostoma

(b) Pennatula and Aurelia

(c) Rhizostoma and Corallium

(d) Physalia and Halistemma


14. Identify the Nematodes, which have variousshapes of amphids but without phasmids.(a) Wuchereria and Ascaris

(b) Trichinella and Greeffiella

(c) Enterobius and Ancylostoma

(d) Ascaris and Ancylostoma

15. The part of adrenal cortex which secretescortisol is(a) Zona faciculata (b) Zona glomerulosa

(c) Zona reticularis (d) Zona pellucida

16. In the transverse section of Pheretima,identify the correct sequence of body walllayers from outer to inner side.(A) Circular muscles(B) Parietal peritonium(C) Epidermis(D) Cuticle(E) Longitudinal muscles(a) C B E A D→ → → →(b) D C E A B→ → → →(c) D C A E B→ → → →(d) D B A E C→ → → →

17. Statement (S) In gastropods, due totorsion mantle cavity is placed anteriorlybehind and above the head.Reason (R) During larval stage, visceralmass, shell and mantle cavity are twistedupto 180° counter clockwise with respect tothe head and foot.The correct answer is(a) Both S and R are true, and R is not a correct

explanation to S

(b) S is correct, but R is not correct

(c) S is not correct, but R is correct

(d) Both S and R are true, and R is a correctexplanation to S

18. Arrange the following in the ascendingorder of their number per cubic millimeter,present in the blood.(A) Basophils (B) Lymphocytes(C) Eosinophils (D) Neutrophils(E) Monocytes(a) A C B D E→ → → →(b) C A D E B→ → → →(c) C A D B E→ → → →(d) A C E B D→ → → →

19. In eukaryotes, which of the following genessynthesise heterogenous nuclear RNA(hnRNA)?(a) Holandric genes (b) Split genes

(c) A-gene (d) Z-genes

20. Match the correct pairs in list I and list IIwith reference to Pheretima.

List I List II

(A) PharyngealNephridia

(I) 14th segment

(B) Stomach (II) 18th segment

(C) Anterior loops (III) 9th and 14th segments

(D) Male genitalapertures

(IV) 4th, 5th and 6th segments

(V) 10th and 11th segments


(a) (IV) (I) (V) (III)

(b) (I) (III) (II) (V)

(c) (V) (IV) (III) (I)

(d) (IV) (III) (V) (II)

21. Match the following lists.

List I List II

(A) Simple squamousepithelium

(I) Ureters

(B) Simple cuboidalepithelium

(II) Epididymis

(C) Non-ciliated simplecolumnar epithelium

(III) Lining of alveoli oflungs

(D) Transitionalepithelium

(IV) Lining of thyroidvesicles

(E) Pseudostratifiednon-ciliatedcolumnar epithelium

(V) Mucosa of stomachand intestine

The correct match is(A) (B) (C) (D) (E)

(a) (III) (IV) (V) (I) (II)

(b) (I) (II) (IV) (III) (V)

(c) (V) (III) (II) (IV) (I)

(d) (II) (IV) (III) (V) (I)

22. What is the duration of one cardiac cycle inman when the heart beats for 75 times perminute?(a) 0.3 s (b) 0.4 s

(c) 0.8 s (d) 0.5 s


23. Identify the correct sequence of events withreference to conjugation of Vorticella.(A) Amphimixis(B) Disappearance of macronucleus(C) Attachment of the conjugants(D) Post conjugation fissions(E) Prezygotic nuclear divisions(F) Postzygotic nuclear divisions(a) C B A E D F→ → → → →(b) C B E A F D→ → → → →(c) F A D B C E→ → → → →(d) F D A E B C→ → → → →


List I List II

(A) Conjugate vaccine (I) Human papillomavirus

(B) Toxoid vaccine (II) Haemophilusinfluenzae

(C) Attenuated wholeagent vaccine

(III) Bubonic plague

(D) Inactivated wholeagent vaccine

(IV) Yellow fever

(V) Diphtheria


(a) (I) (IV) (V) (II)

(b) (II) (V) (IV) (III)

(c) (I) (II) (III) (V)

(d) (II) (III) (I) (IV)

25. In birds, foramen triosseum is present at the junction of(a) clavicle, scapula, corocoid

(b) scapula, carina, furcula

(c) scapula, corocoid, synsacrum

(d) clavicle, synsacrum, furcula

26. Identify the two small apertures present inthe auditory capsule of rabbit(a) foramen magnum and foramen ovale

(b) foramen ovale and fossa ovalis

(c) fenestra ovalis and obturator foramen

(d) fenestra ovalis and fenestra rotunda

27. Match the following

List I List II

(A) Founder effect (I) Long necked giraffes

(B) Bottle neck effect (II) Heterozygous forsickle-cell anaemia

(C) Genetic load (III) Pitcairn Islandhuman population

(D) Directional selection (IV) Polydactylic dwarfindividuals

(v) Sunflower population in California


(a) (IV) (II) (V) (III)

(b) (II) (I) (III) (IV)

(c) (V) (IV) (II) (III)

(d) (III) (IV) (II) (I)

28. Of the following statements about ‘antibodies’and ‘antigens’. Choose the correct set.(I) An antibody consists of four identical

light (L) chains, and two identical heavy (H) chains.

(II) The stem ‘Y’ of antibody is called ‘Fab ’fragment.

(III) The stem of antibody and lower part ofthe arms constitute the ‘constant’ (C)region.

(IV) The portion of the antigen to which anantibody binds is called epitope.

The correct match is(a) (II) and (IV) (b) (I) and (III)

(c) (III) and (IV) (d) (I) and (II)

29. Identify the tumour suppressor genes formthe following.(a) Oncogenes (b) P53 genes

(c) Pseudogenes (d) SRY genes

30. Choose the wrong statement with referenceto mutation theory.(a) Mutations are cumulated over generations

(b) There are no intermediate stages in the courseof evolution

(c) Mutants are markedly different from parents

(d) Mutations are subjected to natural selection


31. Which of the following hormones regulatesolute reabsorption during urine formationin rabbit?(a) Antidiuretic hormone and angiotensin I

(b) Angiotensin III and angiotensin I

(c) Nor-epinephrin and epinephrin

(d) Angiotensin II and aldosterone

32. In ECG a prolonged PR interval indicates(a) Hyperkalaemia and hypokalaemia

(b) Myocardial ischemia and hyperkalaemia

(c) Cardiac arrhythmia and hypokalaemia

(d) Coronary artery disease and rheumatic fever

33. Identify the organisms that belong to thebenthos of lake ecosystem.(a) Gerris and beetles

(b) Chironomid larvae and red annelids

(c) Daphnia and notonecta

(d) Ranatra and copepods

34. Statement (S) After implantation, theuterine myometrium undergoes changes tobecome ‘decidua’ in rabbit.Reason (R) The placenta of rabbit isdescribed as deciduous type.The correct answer is(a) Both S and R are true, and R is not a correct

explanation to S




35. Statement (S) Non-disjunction is thefailure of paired chromosomes to segregateduring the metaphase of meiotic divisions ofgametogenesis.Reason (R) Non-disjunction results inproduction of abnormal gametes.The correct answer is(a) Both S and R are true, and R is not a correct

explanation to S




36. Wax glands in honey bees are present onthese segments.(a) 1, 2, 3 and 4 abdominal segments of queens

(b) 2, 3 and 4 abdominal segments of queens

(c) 2, 3, 4 and 5 abdominal segments of workers

(d) 1, 2 and 3 abdominal segments of workers

37. Identify the correct pair.(A) Zinc – Essential for tissue repair(B) Cobalt – Essential for formation of

leucocytes(C) Iodine – Synthesis of thyroid hormones(D) Manganese – Synthesis of insulin(a) (B) and (C) (b) (A) and (D)

(c) (A) and (C) (d) (A) and (B)

38. Identify the correct sequence of stages in the ross cycle of Plasmodium.(A) Sporocyst (B) Ookinete (C) Sporozoites(D) Zygote (E) Oocyst(a) D B A E C→ → → → (b) C A E B D→ → → →(c) D E B A C→ → → → (d) D B E A C→ → → →

39. What is the phenotype of the offspring bornto a woman with normal vision(homozygous) and a colourblind man?(a) All the sons are with normal vision and the

daughters are colourblind

(b) All the sons and daughters are with normalvision

(c) All the sons and daughters are colourblind

(d) All the sons are colourblind and the daughtersare with normal vision


List I List II

(A) Georges Cuvier (I) Homeostasis

(B) Claude Bernard (II) Homology

(C) Louis de Buffon (III) Comparative Anatomy

(D) Richard Owen (IV) System ofNomenclature

(V) Natural History


(a) (I) (III) (II) (V)

(b) (III) (I) (V) (II)

(c) (II) (I) (III) (V)

(d) (III) (II) (I) (IV)


Botany1. Match the following lists.

List I List II

(A) G2 phase (I) Fusion of microtubules toform spindle apparatus

(B) Prometaphase (II) Production of energyrequired for spindleformation

(C) Anaphase (III) Recombination of genetic material

(D) Pachytene (IV) Contraction of tubulinproteins

(V) Reappearance ofplasmosome


(a) (V) (IV) (II) (III)

(b) (II) (IV) (I) (V)

(c) (V) (I) (IV) (II)

(d) (II) (I) (IV) (III)

2. In a DNA fragment, there are 8 turns, with40% of the bases are cytosine. What wouldbe the total number of hydrogen bondspresent in this DNA fragment?(a) 96 (b) 192

(c) 224 (d) 60

3. The cell organelle bounded by single unitmembrane, which was first reported byRhodin is involved in the following reaction.(a) conversion of serine to serine

(b) conversion of glycine to serine

(c) conversion of glycerate to PGA

(d) formation of glycolate from phosphoglycolate

4. Identify the correct pair of statements fromthe following.(I) Pericycle of dicot root parenchymatous

but sclerenchymatous in maturemonocot root.

(II) Pericycle cells of both dicot and monocotroots actively divide to produce lateralroots during secondary growth.

(III) All cells of endodermis are passage cellsin dicot root.

(IV) Xylem always produced in a centripetalmanner in the roots of fruit bearingplants.

(a) (II) and (III) (b) (III) and (IV)

(c) (I) and (II) (d) (I) and (IV)

5. Match the following lists.List I List II

(A) Tyloses (I) Coenocytic

(B) Periderm (II) Adaxial epidermis

(C) Motor cells (III) Complementary cells

(D) Laticifers (IV) Heart wood

(V) Conjunctive tissue


(a) (III) (II) (I) (V)

(b) (II) (V) (I) (III)

(c) (IV) (III) (II) (I)

(d) (IV) (I) (III) (V)

6. Collocytes which provide mechanicalstrength to the plants are present in(a) dicotyledonous root

(b) dicotyledonous stem

(c) monocotyledonous leaf

(d) monocotyledonous stem


List I List II

(A) A.G. Tansley (I) Classified plantcommunities

(B) Warming (II) Structure andfunction of nature

(C) Odum (III) Term–‘Biosphere’

(D) Reiter (IV) Term–‘Ecology’

(V) Term–‘Ecosystem’


(a) (V) (I) (II) (IV)

(b) (II) (V) (I) (III)

(c) (IV) (III) (V) (II)

(d) (IV) (III) (V) (I)


8. Entry of pollen tube into the ovule throughchalaza was first discovered by Treub in axerophyte. This plant is associated with apair of characters given below.(I) Unisexual flowers

(II) Sessile flowers(III) Centripetal arrangement of flowers(IV) Bisexual flowersThe correct pair is(a) (II) and (III) (b) (III) and (IV)

(c) (II) and (IV) (d) (I) and (II)

9. Assume that blue flower of a plant isdominant character over the white. When ablue flowered plant is crossed with whiteflowered plant, the progeny showed 50% ofplants with blue flowers and 50% of plantswith white flowers. The genotypes of blueand white parents respectively are(a) Bb, bb (b) bb, bb (c) BB, Bb (d) BB, bb


(A) Turill (I) Tissue culture

(B) Goethe (II) Pollen grains

(C) Salisbury (III) Nucleosome

(D) Oudet (IV) Flower-a modified stem

(V) Stomatal index


(a) (III) (I) (IV) (V)

(b) (V) (II) (III) (IV)

(c) (IV) (I) (II) (III)

(d) (II) (IV) (V) (III)


(A) Apices of undergroundbranches store food material

(I) Mentha

(B) Underground branches growobliquely upward from theaxillary buds of nodes of thestem below the soil

(II) Hydrocotyle

(C) Aerial branches grow obliquelydownwards and produceadventitious roots after touchingthe soil

(III) Agave

(D) Weak stemmed plants have acluster of leaves and roots atevery node

(IV) Stachys

(V) Jasminum


(a) (II) (V) (I) (III)

(b) (III) (IV) (II) (I)

(c) (IV) (V) (III) (II)

(d) (IV) (I) (V) (II)

12. Identify the wrong statement.(a) Stipules in Lathyrus are persistent.

(b) Phyllotaxy in Trillium is alternate.

(c) Venation in Caulophyllum is parallel.

(d) Cauline leaves are found in Cocos.

13. The type of inflorescence in a plant showingpolygamous condition and lateral style is(a) Compound corymb (b) Compound umbel

(c) Compound head (d) Compound raceme

14. Identify the correct combination ofcharacters found in Cucurbita.(a) Free central placentation, synandry, divergent

palmately reticulate venation

(b) Mesogamy, fusion of epicarp and thalamus toform rind of the fruit, seed germination epigeal

(c) Axile placentation, extrafloral nectaries, continuous,collenchymatous hypodermal ring in stem

(d) Bicarpellary gynoecium, compound sieve plate,unilocular ovary

15. There are 25 flowers in an inflorescence ofAllium. Each anther lobe of every stamencontains 60 pollen grains. What is the totalnumber of pollen grains produced in itsinflorescence?(a) 4500 (b) 9000 (c) 18000 (d) 1500


(A) Aggregate fruitlets of berriesappearing as a single fruit

(I) Rubus

(B) Achenes on the innermargins of receptacle

(II) Artabotrys

(C) Etaerio of drupes (III) Ficus

(D) Etaerio of follicles (IV) Magnolia

(V) Annona


(a) (V) (III) (I) (IV)

(b) (II) (I) (V) (IV)

(c) (II) (IV) (III) (V)

(d) (IV) (III) (I) (II)


17. Triple fusion was discovered in this plantpossessing naked bulb, fleshy leaf bases andloculicidal capsule.(a) Lilium (b) Trillium

(c) Aloe (d) Allium


List I List II

(A) Bicarpellary unilocular (I) Hyoscyamus

(B) Bicarpellary bilocular (II) Ulex

(C) Multicarpellary multilocular (III) Scilla

(D) Monocarpellary unilocular (IV) Citrus

(V) Capsicum


(a) (III) (V) (IV) (I)

(b) (IV) (I) (III) (V)

(c) (V) (I) (IV) (II)

(d) (V) (III) (I) (II)

19. Study the following and identify the correctpair of characters for Physalis.(I) Petiole adnate with the stem

(II) Thalamus cup-shaped(III) Seeds non-endospermic(IV) Flowers hypogynousThe correct answer is(a) (I) and (IV) (b) (II) and (IV)

(c) (III) and (IV) (d) (II) and (III)

20. Identify the correct pairs from the following(I) Pusa moti of bajra – Mass selection

(II) IR-8 of rice – Clonal selection(III) TMV-3 of groundnut – Pureline

selection(IV) Aruna variety of castor– Polyploidy

breedingThe correct answer is(a) (III) and (IV) (b) (I) and (III)

(c) (I) and (IV) (d) (II) and (III)

21. Which one of the following SCP organismslack a membrane bound nucleus?(a) Spirulina (b) Paecilomyces

(c) Chlorella (d) Chaetomium

22. Arrange the following in sequential order oftheir usage in recombinant DNA technology.(I) Calcium chloride

(II) DNA ligase(III) Ethylene diamine tetra acetic acid(IV) Restriction endonucleaseThe correct answer is(a) (I), (IV), (III) and (II)

(b) (IV), (I), (II) and (III)

(c) (I), (IV), (II) and (III)

(d) (IV), (III), (II) and (I)

23. Assertion (A) Application of cytokininscauses the opening of stomata.Reason (R) Cytokinins induce the influx ofpotassium ions into guard cells.The correct answer is(a) Both A and B are true, but R is not the correct

explanation of A

(b) A is true, but R is false

(c) A is false, but R is true

(d) Both A and R are true, and R is the correctexplanation of A

24. What is the amino acid sequence in thepolypeptide segment translated from themRNA base sequence ofAGU-UUU-UCC-GGG-UCG?(a) Serine-Glycine-Serine-Phenylalanin-Serine

(b) Serine-Phenylalanine-Serine-Glycine-Serine

(c) Serine-Serine-Phenylalanine-Glycine-Serine

(d) Phenylalanin-Serine-Serine-Glycine-Serine

25. What is the ratio of ATP requirement for thefixation of 6 molecules of CO2 in sugarcaneand 5 molecules of N2 in bean?(a) 5 16: (b) 3 16:

(c) 5 8: (d) 3 8:

26. Identify the correct pair of statements.(I) Lutein is an oxygenated hydrocarbon.

(II) Inhibition of bacteroid respiration byoxygen is referred to as ‘Warburg effect’.

(III) Triplet code ‘UAA’ is called ‘Amber’.(IV) Delay of senescence by cytokinins is

known as ‘Richmond-Lang effect’.(a) (III) and (IV) (b) (II) and (III)

(c) (I) and (IV) (b) (I) and (III)


27. Arginosuccinase is an example to(a) Hydrolase

(b) Ligase

(c) Lyase

(d) Oxido-reductase

28. Identify the role of lectins in the formation of root nodules in legumes.(a) Formation of shepherd’s crook

(b) Recongnition of compatible Rhizobium by host

(c) Formation of peribacteroid membrane

(d) Formation of infection thread

29. Assertion (A) Azolla is used as biofertiliser in rice fields.Reason (R) Azolla contains nitrogen fixingcyanobacteria in its root nodules.The correct answer is(a) Both A and R are true, and R is not the correct

explanation of A

(b) (A) is true, but R is false

(c) (A) is false, but R is true


30. Identify the correct pair of statements fromthe following.(I) The attraction between two water

molecules in xylem vessels is called‘adhesion’.

(II) The number of molecules of O2 absorbedis more than the number of CO2molecules released when one moleculeof triolein is completely oxidised inrespiration.

(III) Bacillus mycoides is a nitrifyingbacterium.

(IV) Continuous system of cell walls andintercellular spaces in plant tissue iscalled ‘apoplast’.

The correct answer is(a) (II) and (III)

(b) (III) and (IV)

(c) (II) and (IV)

(d) (I) and (IV)

31. The number of stomata and epidermal cellsin 1 mm2 area of abaxial surface of theleaves of A, B, C and D plants are givenbelow.

PlantNumber ofStomata

Number ofepidermal cells

A 40 730

B 60 510

C 70 450

D 30 620

Identify the two plants having leaststomatal index.(a) B and C (b) A and D

(c) A and C (d) A and B

32. Study the following table showing thecomponents of water potential of four cells of an actively transpiring monocot plant.

CellOsmotic Potential

(MPa)Pressure potential

(MPa)

A – 0.9 0.5

B – 0.8 0.6

C – 0.6 0.1

D – 0.7 0.4

Identify the four cells as root hair, generalcortical cell and pericyclic cells of young rootrespectively (Assume symplast flow).(a) B, C, D, A (b) B, D, C, A

(c) B, D, A, C (d) A, D, B, C

33. What is the total number of amino acids,when the capsid of TMV contains 2130capsomers?(a) 336540 (b) 162622

(c) 948 (d) 387921


List I List II

(A) Syphilis (I) Acetobacter

(B) Pathogen of cattle (II) Agrobacterium

(C) Crown gall of apples (III) Corynebacterium

(D) Diphtheria (IV) Mycobacterium

(V) Treponema



(a) (III) (I) (IV) (II)

(b) (V) (IV) (II) (III)

(c) (V) (III) (II) (I)

(d) (II) (IV) (V) (III)

35. The microelement, which is an integral partof the electron carrier which transferselectrons from cyt-b-f complex to PS I is acomponent of(a) nitrate reductase

(b) cytochrome-‘C’-oxidase

(c) IAA oxidase

(d) dinitrogenase

36. Assertion (A) Dictyostele is present inrhizome of Pteris vittata.Reason (R) Meristeles are scattered in therhizome of Pteris vittata.The correct answer is(a) Both A and R are true, and R is not the correct

explanation of A

(b) (A) is true, but R is false

(c) (A) is false, but R is true


37. What is the ratio of mitotic divisions thattake place in the microspore of Cycas beforeand after liberation from themicrosporangium during the development of male gametophyte?(a) 1 2: (b) 2 1: (c) 3 2: (d) 1 1:

38. Identify the wrong statement in relation toFunaria.(a) Stomata are present in the epidermis of capsule

(b) Spores are viable for only one year

(c) Inner spore sac is one celled in thickness

(d) Trabeculae connect the innermost layer of thecapsule wall with the outer spore sac

39. Study the following lists

List I List II

(A) Anucleate (I) Spirogyra gametangium

(B) Uninucleate (II) Rhizopus chlamydospore

(C) Multinucleate (III) Pteris neck canal cell

(D) Binucleate (IV) Vitis matured sieveelement

(V) Akinite of Spirogyra


(a) (I) (IV) (V) (II)

(b) (IV) (I) (III) (II)

(c) (I) (III) (II) (IV)

(d) (IV) (I) (II) (III)

40. In F2-generation, 960 garden pea pods areproduced during dihybrid cross due to selfpollination of heterozygous parents. Howmany pods would be green and inflated?(a) 240

(b) 180

(c) 60

(d) 540


AnswersPhysics

Hints & Solutions

Physics1. We know that the frequency

nT= 1

2l µ

Here µ is a constant, so

n T∝l

Now, we can calculate the value of n /n2 1 as follow

n

n

T

T2

1

1

2

2

1

= ×l

l = ×1

2

9

1

= =9

2

3

2

Hence, the ratio of n

n 2

1

3

2is

2. The frequency of the fundamental note of the pipe is

n n v

2 1 24

− =

l

v n n

4 22 1

l= −

1. (c) 2. (a) 3. (c) 4. (c) 5. (b) 6. (a) 7. (*) 8. (d) 9. (c) 10. (b)

11. (b) 12. (a) 13. (a) 14. (d) 15. (c) 16. (d) 17. (b) 18. (d) 19. (b) 20. (a)

21. (a) 22. (a) 23. (b) 24. (d) 25. (c) 26. (c) 27. (a) 28. (a) 29. (d) 30. (d)

31. (b) 32. (c) 33. (a) 34. (d) 35. (d) 36. (a) 37. (a) 38. (b) 39. (d) 40. (d)

Chemistry1. (b) 2. (d) 3. (a) 4. (a) 5. (d) 6. (d) 7. (b) 8. (c) 9. (a) 10. (a)

11. (b) 12. (c) 13. (d) 14. (c) 15. (c) 16. (b) 17. (c) 18. (a) 19. (d) 20. (d)

21. (d) 22. (a) 23. (c) 24. (c) 25. (c) 26. (a) 27. (d) 28. (d) 29. (d) 30. (d)

31. (a) 32. (a) 33. (b) 34. (c) 35. (a) 36. (b) 37. (a) 38. (c) 39. (c) 40. (b)

Zoology1. (b) 2. (a) 3. (a) 4. (d) 5. (a) 6. (c) 7. (b) 8. (c) 9. (a) 10. (a)

11. (a) 12. (a) 13. (a) 14. (b) 15. (a) 16. (c) 17. (d) 18. (d) 19. (b) 20. (d)

21. (a) 22. (c) 23. (b) 24. (b) 25. (a) 26. (d) 27. (d) 28. (a) 29. (b) 30. (d)

31. (d) 32. (d) 33. (b) 34. (d) 35. (d) 36. (c) 37. (c) 38. (d) 39. (a) 40. (b)

Botany1. (d) 2. (c) 3. (d) 4. (d) 5. (c) 6. (b) 7. (a) 8. (d) 9. (a) 10. (d)

11. (d) 12. (b) 13. (a) 14. (b) 15. (c) 16. (a) 17. (a) 18. (a) 19. (b) 20. (b)

21. (a) 22. (d) 23. (d) 24. (b) 25. (b) 26. (c) 27. (c) 28. (b) 29. (b) 30. (c)

31. (b) 32. (c) 33. (a) 34. (b) 35. (b) 36. (d) 37. (d) 38. (b) 39. (d) 40. (d)

Given, n1 135= Hz and n2 165= Hz

Now, v

4

165 135

2

30

215

l= − = = Hz

3. We know about the thin prism,

δ µm A= −( )1

In given condition,

A A1 1 2 21 1( ) ( )µ µ− = −Given, µ µ1 2 11 4 6= = = °. , 1.6 and A

Now, 6 1 4 1 12( . ) ( )− = −A 1.6

A2 = ×6 0.4

0.6

A2

2 4

0 6= .

.

A2 4= °Hence, the angle of the second prism is 4°.

4. We knows the magnifying power of theastronomical telescope is

M f

fo

e

= − and length of the telescope L f fo e= +

Given , M =50 and fe = 2 cm

Here, − =f

f o

e

50

− =f fo e50

So, the length of the telescope

L = – 50 f fe e+L = – 49 fe

L = – 49 × ( )2

L = – 98 cm

Hence, the required length of the telescope is98 cm.

5. Here the condition for the formation of rings is

λ ≈ d

D

2

4

6. Given, M = 120 Am2, T=12 s and B= 0.4 × −10 4 T

We know that,

TI

MB= 2π (squaring both sides)

So, I T MB=

2

24π

or I = × × ×× ×

−( ) ( ) ( )12 120 10

4

2 40.4

3.14 3.14

I = × −175.2 10 4 2kgm

In this question, the option (b) is more equivalentto 175 2 10 4. × − so that the answer is

I = × −1 10 4 272.8 kgm .

7. We know that the magnetic field induction at apoint

BM

r= µ

π0

34

According to question

B B B r

r

r

1 2 313

23

33

1 1 1: : : :=

Given r 1 30= =cm m0.3

r cm m2 60= = 0.6

and r3 90= =cm m0.9

Hence, B : B : B1 2 3 = 1

(0.3) :

1

(0.6) :

1

(0.9)3 3 3

B B B1 2 3: : = 1

0.027 :

1

0.236 :

1

0.729

B B B1 2 3: : = 37.03 : 4.629 : 1.37

No option is correct.

8. Given, V1 120= V and V2 200= V

Here given potential is zero for each capacitor,then

C V C V

C C1 1 2 2

1 2

0−+

=

C V C V1 1 2 2=120 2001 2V C=12 201 2V C=3 51 2C C=

9. Here the capacitance of dielectric capacitor

C1 3=and C2 2=We know that the energy stored in the capacitor

E q

C =

2

2

Hence,E

E

q

C

C

q 1

2

12

1

2

222

2= =

E

E

C

C 1

2

2

1

2

3= = ( )Qq q q1

2= −

10. Given, temperature coefficient of carbon

α 334 10= × °− / C

Temperature coefficient of copper

α 2310= − × °−0.5 / C


Hence, R R 1 1 2 2α α= −

R

R2

1

3

3

4 10

10= − ×

×

−

−0.5

R

R2

1

8

1=

11. Given equation,

1

p

y

k TBβ=

where, p = pressure, y = distance, kB = Boltzmann constant and T = temperature

Dimension of [ ]β

= [ ] [ ]

[ ] [

Dimension of Dimension of

Dimension of Di

s k s T

s pB

mension ofs y]

=−

− −[ ][ ]

[ ][ ]

ML T T

ML T L

2 3

1 2 = [ ]M L T0 2 0

12. Given,Velocity of water vw = −4 1ms

Velocity of person vp = −5 1ms

Width of the river = 84.6 m

We know that time taken

t s

v vw p

=−2 2

t =−

84.6

25 16

t = =84.6

3 28.6 s ≅ 28.25

13. The initial velocity of the body

u gh2 2=

u gh= = × × = = −2 2 10 125 2500 50 1ms

The final velocity from equation of motion

v u gh2 2 2= +

v = + × ×( )50 2 10 452

v = + = = −2500 900 3400 50 1ms

14. We know that

W = mg cos θ × 5

W = 300 9 8 30 10× × ° ×. cos

= 300 9 83

210× × ×.

= 2450 J

15. Given that, h = 100 m and t = 10 s

From equation of motion,

h at= 1

22

a h

t= × = −2 100

102

2 22=

2

( )ms

Now, F = 50 (10 + 2)

F = 600 N

16. Given, mass = m

and here rma mb m

mCM = + +$ $i j 0

3

r a bCM = +1

3( $ $)i j

17. From the conservation of momentum

mv m m m v+ × = + ′0 ( )

⇒ vmv

m m′ =

+( )

vv′ =2

and we knows

h u

g=

2

2

Here, u v= ′

So, h v

g

v

g= ′ ⇒( )2 2

2 8

18. We know that

F f ma− =Here for small body,

a1

20 3 10

3= − × ×0.2

a1214

3= −ms

Similarly, a126

10= = −0.6 ms

19. We know that the difference between themaximum and minimum tensions in the rope

T T mgmax min− = 2

Here, T Tmax min− = 20

Now, mass of the store

F ma=

mF

a= =

×=20

2 101kg


20. In given condition, the moment of inertia of thecombine system about an axis passing through Ois

I ma

ma= +4

3

22

I ma m= +

4

3

39

42

2

I m a= 94

482

21. Here, g g R 0 902= − ω

Given, gg

02

= and g g90 =

So,g

g R2

2= − ω

ω g

R=

2

We have, Tr

v=2π

T = 2πω

Q ω =

v

r

Hence, T R

g=2

2π

22. We know that, for SHM

v A r= −ω 2 2

In given condition,

v A r12 2

1= −ω

v A r22 2

2= −ω

Hence, v vT

A r A r22

12

2

22

12 2

224− = − − +π

[ ]

v vT

r r22

12

2

2 22

124− = −π

[ ]

Tr r

v v= −

−2 2

212

22

12

π ( )

( )

23. We know that the work done in a stretched wire

W kx= 1

22

Given, k kA = and k kB =2

So that W kxA = 1

22

and W k x kxB = =1

22 2 2( )

Hence, the ratio of work done in stretch wireW

Wkx kxA

B

= 1

22 2/

W

WA

B

= 1

2

24. Given, h 1 10= cm, h2 = 3.42 cm

θ2 135= and d2 = 13.6 g/cc

Here,T

T

h d

h d 1

2

1 1

2 2

2

1

= cos

cos

θθ

T

T1

2

10 1 135

45= ×

×°°3.43 13.6

cos

cos

T

T1

2 13= 2

25. We know that the pressure

pr∝

4

l

Given, l l1 2 2 1: := and r r1 2 1 2: :=Let l l1 2= , l l2 = and r r1 = , r r2 2=

Now,p

p r

l

l

r

1

2

14

1

2

24

= ×

p

p r

r

1

2

4

42 2= ×

l

l

( )

p

p 2

24

1

2 2=

× ( )

p

p 1

2

1

32=

26. Here, γ γ α γ αr ag g ac c= + = +3 3

1.03 1.006× + = × + × ×− − −10 3 10 3 17 103 3 6α g ( )

α g = × °−27 10 6 / C

27. The coefficient of linear expansion ( )α of the

material of the tube and the coefficient of volumeexpansion ( )γ of the liquid, then

γ α=2

28. We know that

(i) The work done in isobaric process is p V V( )2 1−(ii) The work done in isothermal process is

nRV

Vlog 2

1

(iii) The work done in adiabatic process is nR T T( )1 2

1

−−γ


29. We know that,

d nC dT nC Tp A V Bθ α= =nC T nC Tp A V Bα α=

C

C

T

T

p

V

B

A

= αα

γ = dT

dTB

A

Q γ =

C

C

p

V

αT Y TB A= ∝ = × =1 4 30 42. K

30. From the equation,θx

t= ∆l

Given, t t1 2100 0 25 0= ° = ° = =C C and 1 cm, , l π

θ10

100 0

25= −

θ10

4=

θ = °40 C

31. For making a junction transistor the emitter E isheavily doped, the base B is thin and lightlydoped and the collector C is moderately dopedsemiconductor materials and with together.

32. We know that, β = ∆∆

I

Ic

b

Given, β = 60 and ∆ Ic = = × −4 4 10 6mA A

Here, ∆ ∆I

Ib

c=β

= × −4 10

60

6

∆ I b = 66.6 µA

33. Pn E

t=

Given, n = ×2 10 18 fission per second

E =185 MeV

Here, P = × × ××

2 10 185 10

7 10

18 6

15

eV

P = 59.2 MeV

34. The purpose of using heavy water in nuclearreactor is to decrease the energy of fast neutronsto the thermal energy.

35. Given, Ek = = × × −23.32 k 23.32 1.6 eVeV 10 19

hc = × −12.42 eV10 7

λ αh = × −71 10 12 m

Now, E E hc

hL K

ka

− =

EL − × × = ××

−−

− 23.32 1.612.42

1010

71 10

17

12

EL = 5.82 keV

36. In Moseley's law v a z b= −( ), the values of the

screening constant for K-series and L-series ofX-rays are respectively 1 and 6.4.

37. We know that,

rmv

Bq= and v

qV

m= 2

Now, B mv

qr=

Bm

qr

qV

m= 2

B mV

qr= 2

2

BmV

er= 2

2 ( )∴ =q e

38. We know,

tan θ ω= L

R

Given, L= 0.01 H, R = 3 π

V = 220 V and f= 50 Hz

ω π= 2 f

ω π= ×50 2

tan θ ππ

= × ×50 2

3

0.01

tan θ = 1

3

θ π= / 6 rad

39. Here, E

E

t t

t tn c

n c

2

1

20 120 26

120

94

120= − − +

−= − =( ) ( )6

Percentage decrease in thermo emfE E

E2 1

1

10094 120

120100

− × = − × = 21.6

40. We know that

i E

r Ref

=+

= 15

0.5+ 14.5

= =15

151A


Chemistry1. (a) Sucralose is an artificial sweetening agent.

(b) Antimicrobials are commonly used in food

industry to ensure safe wholoesome food

products. Traditionally, lactic acid, which is a

weak organic acid and a natural food

preservative has been widely used to control

growth of pathogenic bacteria in food.

(c) Seconal is a tranquillizer.

(d) Chloroxylenol is an antiseptic.

2. Glucose on reduction with HI/red P at 373 K gives

n hexane and 2-iodo hexane, which suggests that

all the six carbon atoms in glucose are arranged

in a straight chain. Whereas, on oxidation with

HNO3 glucose gives saccharic acid (glycaric

acid), which indicates that glucose contains one

primary alcoholic group.

3. Condensation polymers are formed by repeated

condensaton reaction between two bifunctional

or trifunctional monomer units accompanied with

the elimination of small molecules like, water,

NH3, CO2, HCl, etc. For example, nylon 66 is

obtained by cndensation polymerisation of

hexamethylenediamine and adipic acid with the

loss of one water molecule. Thus,

n nH N — (CH ) — NH HOOC(CH COO2 2 6 2

hexamethylenediamine

+ 2 4) Hadipic acid

↓

—[NH—(CH ) —NH—C

O

—(CH ) —C

O

—]2 6 2 4nylon-66

n

+ −( )2 1n H O2

Similarly, terylene or dacron is a condensationpolymer of ethylene glycol and terephthalic acidwith elimination of water.

4. Aliphatic amines are stronger bases thanammonia. This is due to the reason that alkylgroups are electron-donating groups. As a resultthe electron density on the N-atom increases andthus they can donate the lone-pair of electronsmore easily than NH3.

H — N

H

— H

ammonia

• • C H N

H

— H2 5

1 amine

→−

• •

°

C H N

H

C H2 5 2 5

2 amine

→−

−←

°

But aniline is weaker base than NH3. The reasonbehind it is the resonance in aniline, due to whichthe lone pair of electrons on the nitrogen atomgets delocalised over the benzene ring and thusis less easily available for protonaton.

Thus, the basicity of amines increases in theorder.

C H NH < NH < C H NH < (C H ) NH.6 5 2 3 2 5 2 2 5 2


CH3

(CH )2 4

CH3n-hexane

HI/P

Reduction

CHO

(CHOH)4

CH OH2glucose

HNO3

Oxidation

COOH

(CHOH)4

COOHsaccharic

acid

n HO—CH —CH —OH2 2

+ n HO—C—

O

—C—OH

O

– (2n – 1) H O2

—OCH —CH —O—C—2 2

O

—C—

O

ethylene glycol

terephthalic acid

NH2 NH2 NH2

+

s

s

+

NH2

+

s

NH2

III III

IVV

5. Acetic acid is produced industrially by thecarbonylation of methanol in the presence of Rh/I2 catalyst.

CH OH + CO CH COOH32

3methanol

I –Rh catalyst

acetic ac→

id

6. Wacker process Alkenes can be easilyoxidised to their corresponding aldehydes andketones by treating them with a solution of PdCl2containing a catalytic amount of CuCl2 inpresence of O2.

CH ==CH H O + PdCl2 2 2 22

2

ethylene O

CuCl+ →

CH CHO + Pd + 2HCl3acetaldehyde

During this reaction, PdCl2 is reduced to metallicPd, therefore the reaction is carried out in thepresence of a cooxidant CuCl2. The function of CuCl2 is to reoxidise Pd to Pd(II) and itself isreoxidised to Cu(II) from Cu(I) by air so that O2 isthe only oxidising agent actually used up in thisprocess.

Pd + 2CuCl PdCl + 2CuCl2 2→4CuCl+ 4HCl+ O 4CuCl + 2H O2 2 2→

7. Ammonium salts, e.g., (NH ) SO ,4 2 4 (NH ) PO4 3 4

(but not NH Cl)4 act as food for yeast cells.

8. The hydrogen of the chloroform is replaced bynitro group, when it is treated with concentratednitric acid. The product formed is chloropicrin ortrichloronitromethane or nitro chloroform. It is aliquid, poisonous and used as an insecticide anda war gas.

CHCl + Conc HONO CNO Cl3 2 2 3chloroform chloropicrin

(tea

→

r gas)

+ H O2

9. Fischer projection It is three directionalrepresentation of optically active compounds and also said to be R-, S-system.

Here we hold the group of 3rd priority and rotatein triangular way be keeping —‘OH’ at place of—‘H’, —H at place of —‘COOH’ and —‘COOH’ at place of —‘OH’ to get structure as

as eye rotates, clockwise from (1) to (3) via (2),then it is R configuration at C2.

As we hold the group of 2nd priority and rotate intriangular manner to get the following structure

It is R configuration at C3.

10. 1,2-benzpyrene is a polynuclear aromatichydrocarbon and one of the most potentcarcinogenic compounds. It is the activecancer-producing agent in coal-tar and in the tarproduced during the incomplete combustion oforganic materials such as coal, petroleum,tobacoo, etc.


H—C—OH

COOH

CH3

H—C—OH

2

3

4

1

H—C—OH

COOH

CH(OH)CH3

2

4 1

3

The priority at C as2

HO—C—COOH

H

CH(OH)CH3

4

1 2

3

H—C—OH

CH(OH)COOH

CH3

2

4 1

3

The priority at C3

HO—C—CH3

CH(OH)COOH

CH4

2

1 3

1

21,2-benzypyrene

12. The detection of carbon and hydrogen in anorganic compound is done with the help of CuOtest.

C of organic compound

+ →2CuO CO + 2Cu2∆

H of organic compound + →CuO H O + Cu2∆

H O2 vapours turn anhydrous CuSO4 blue and CO2 evolved turn lime water milky as per following reactions.

CuSO + 5H O CuSO 5H O4 2 4 2hydrated copper sulphate

(blue)

→ ⋅

Ca(OH) + CO CaCO + H O2 2 3 2milky

→

13. (A) Pure wate has a pH of 7.0 (neutral), however,natural unpolluted rain water has a pH ofabout 5.6 (acidic), due to the naturalpresence of CO2, NO and SO2.

(B) Acid rain is the rain water containing H SO2 4

and HNO3 which are formed from the oxidesof S and N present in the air and has a pH of4-5. Although CO2 is present in a muchhigher concentration than NO and SO2, itdoes not form acid to the same extent as thetwo other gases.

(C) Acrolein enters the air from the burning offossil fuels and tobacco smoke. It is toxicand a strong irritant for the skin, eyes andnasal passages.

(D) Freons (CF Cl )2 2 are chlorofluoro carbons and has been completely banned because ofbeing destructive for ozone layer.

14. During the manufacture of cast iron, the slag isformed in slag formation zone which is alsoknown as zone of heat absorption, because thelime-stone present in charge needs a highertemperature to decompose into calcium oxideand CO2.

CaCO CaO + CO3 2lime-stone

1000 C→°

CaO act as a flux and combines with silicapresent as an impurity to form a fusible slag of CaSiO3.

CaO + SiO CaSiO2 3→15. The basic character of hydroxides of lanthanoids

decreases from La(OH)3 to Lu(OH)3. Due to thesmaller size of Lu, the Lu—OH bond aquiresmore covalent character.

16. The correct set are

Molecule Hybridisation Shape Number of lone pair of electrons

(1) XeO4 sp3 tetrahedral 0

(2) XeF4 sp d3 2 squareplanar

2

(3) XeF2 sp d3 linear 3

17. F2 oxidises potassium bisulphate to potassium

persulphate.

F + 2KHSO K S O + 2HF2 4 2 2 8→18. K K RTp c

ng= ( )∆

For the reaction,

A g B g C g( ) ( ) ( )º +

∆ng = gaseous products – gaseous reactants

= − =2 1 1

∴ KK

RTc

p

ng=

( )∆

=×

= = −0.82

(0.082 300)0.033 mol L

1

1

301


Xe

F3 2

XeF (sp d )4

[Square planar structure]

3XeF (sp d)2

[Linear structure]

Xe

O

O

O

O

Xe

OO

O3

XeO (sp )4

[Tetrahedral structure]

3XeO (sp )3

[Pyramidal orT-shaped structure]

F

Xe

FF

F F

43

2

1

65

4-ethyl-3,3-dimethyl hexane

11.

19. MgF Mg 2F2s s sº 2

2

+ −+

K s s ssp = =( ) ( )2 42 3

∴ sK3

10

4

10

4= = × −

sp 1.08

= × −27 10 12

s = × −27 10 123

= × − −3 10 4 1mol L

20. The reaction given :

4NO O 2N O2 52 2( ) ( ) ( );g g g+ → ∆H = −111kJ

…(i)

N O N O2 5 2 5( ) ( );s g→ ∆Hsub kJ= 54 …(ii)

Eq. (ii) is multiplied by 2 and then divided by

Eq. (i) gives

4NO O 2N O2 52 2( ) ( ) ( )g g s+ →

∴ ∆H = − − ×111 54 2( )

= − 219 kJ

21. (a) H O2 2 reduces chlorine to HCl.

H O + Cl 2HCl+ O2 2 2 2→(b) H O2 2 reduces silver oxide to silver in the

alkaline medium.

Ag O + H O 2Ag + H O + O2 2 2 2 2→(c) H O2 2 is a weak acid and show neutralization

reactions with hydroxides.

2NaOH+ H O Na O + H O2 2 2 2 2sodium peroxide

→

(d) H O2 2 oxidises nitrites to nitrates in alkaline

medium.

KNO + H O KNO2 2 2 3potassium nitrite potassium nitrate

→

+ H O2

22. With hot and concentrated NaOH, Zn forms

sodium zincate and H2 gas is liberated.

Zn+ 2NaOH Na ZnO + H2 2 2sodium zincate

→ ↑

23. Li O2 and Na O2 are white solids as expected but

surprisingly K O2 is pale yellow, Rb O2 is bright

yellow and Cs O2 is orange. This is due to the

gradually increased ionic character from Li O2 to

Cs O.2

24. (a) Orthoboric acid (H BO )3 3 is used as an

antiseptic and eye wash under the name‘boric lotion’.

(b) In B H ,2 6 both B atoms undergo sp3-hybridisation.

(c) The aqueous solution of borax is weaklyalkaline in nature due to hydrolysis of B O4 7

2–

ions.

Na B O + 7H O 2NaOH + 4H BO2 4 7 2 3 3strong alkali weak acid

º25. The tetrahalides of Si, Ge, Sn and Pb form

hexahalocomplexes like [SiF62] ,− [GeF6

2] ,−

[SiCl62] ,− [SnCl6

2] ,− [PbCl62] − with the

corresponding halide ions by increasing theircoordination number from 4 to 6, due to theavailability of the vacant d-orbitals. The propertyof accepting electron pair from donor makes thetetrahalides of group 14 elements (except carbonhalides) to act as strong Lewis acids.

26. (a) The solubility of group V hydrides in waterdecreases from NH3 to BiH3. NH3 is highlysoluble in water due to H-bonding but otherhydrides are less soluble.

(b) All the elements of group V except Bi showallotropy. Phosphorus exists in a number ofallotropic forms such as white, red, scarlet,α-black, β-black and violet phosphorus.

(c) Thermal stability of hydrides of group Vdecreases gradually from NH3 to BiH3 due to decrease in M—H bond strength down thegroup on account of increase in size ofcentral atom.

27. Sodium thiosulphate decomposes on treatmentwith dilute HCl with the evolution of SO2 andprecipitation of sulphur.

Na S O + 2HCl 2NaCl+ SO2 2 3 2sodium thiosulphate [ ]

→ ↑X

+ S + H O[ ]

2↓Y


B atom(Ground state)

B atom(Excited state)

3sp -hybridisation

2s 2p

SO2, when reacts with Na S,2 form sodium

thiosulphate and sulphur.

3SO + 2Na S 2Na S O + S2 2 2 2 3[ ] sulphur

→ ↓Y

28. Energy of a photon of radiation of wavelength

300 nm will be

E h hc= =νλ

= × ××

− −

−( ( )

(

6.626 Js) 3.0 ms

m)

10 10

300 10

34 8 1

9

= × −6.626 J10 19

= ××

−

−6.626

1.6eV

10

10

19

19

[As 1 10 19eV 1.6 J]= × −

= 4.14 eV

Thus, the electron will be detached from the metal

atom only if the energy of the photon (4.14 eV) is

sufficient to overcome the force of attraction of

the electron by the nucleus and it is evident from

the given table that the photon of 300 nm has

greater energy than the values of work function

( )w for 4 metal atoms (Li, Na, K, Mg). Thus, these 4

metal atoms will show photoelectric effect when

light of 300 nm wavelength falls on them.

29. Here, we are given,

∆x = 58 m

h = × −6.626 Js10 34

∆v = × − −1.0 ms10 6 1

According to Heisenberg’s uncertainty principle

∆ ∆x m vh⋅ =( )

4π

i.e., mh

x v=

⋅ ⋅4π ∆ ∆

= ×× × × ×

−

−6.626

3.14 1.0

10

4 58 10

34

6

= × −9.09 kg10 31

Thus, the mass of moving object is approximately

equal to that of electron.

30. As we move from left to right in a period, nuclear

charge increases by one unit in each succeeding

element while the number of the shells remains

the same. Due to this enhanced nuclear charge

the electrons of all the shells are pulled little closer

to the nucleus thereby making each individual

shell smaller and smaller. This results in a

decrease of the atomic radius as we move from

left to right in a period, i.e.,

Li> Be > B

31. Structure of acetylene

H—C C—H2σ

σ

π σ≡≡

32. N2 molecule The electronic configuration of N is

1 2 22 2 3s s p, . The MO formed during the

combination of atomic orbitals of two N atoms are

σ1s, σ σ σ π π* , , * , ,1 2 2 2 2s s s p py z=

σ π π π2 2 2 2p p p px x y z, * , * *=MO configuration of N2 σ σ σ1 1 22 2 2s s s, * , ,

σ * ,2 2s π σ2 2 22 2 2p p py z x= ,

∴ The total number of antibonding electrons = 4.

O2 molecule The electronic configuration of O is

1 2 22 2 4s s p, . The MO formed during the

combination of atomic orbitals of two O-atoms

are σ σ σ σ σ1 1 2 2 2s s s s px, * , , * , , σ σ2 2p py z= ,

π π* * ,2 2p py z= π * 2px

MO configuration of O2 σ σ σ1 1 22 2 2s s s, * , ,

σ σ π π π* , , , *2 2 2 2 22 2 2 2 1s p p p px y z y= = π * .2 1pz

∴ The total number of antibonding electrons = 6.

Oxidation

33. 5I I O 6H 3I––1 –

0

( ) ( ) ( ) ( )aq aq aq aq+ + →+

− +

53 2

0

Reduction +3H O2 ( )l

34. T M T MA B B A=

∴ T

M

T

MA

A

B

A

=

RMS velocity = =( )vRT

MRMS

3

As T

M

T

MA

A

B

B

= is equal for both the gases, thus

both have equal RMS velocity.

35. According to Raoult’s law;p p

p

n

n ns° −

°=

+2

1 2

Let p° = vapour pressure of pure water = 100 mm


∴ p ps° − = Lowering in vapour pressure on dissolving solute = 1.8 mm

n2 = moles of solute = =w

M

w2

2

2

60mol

n1 = moles of solvent = =w

M1

1

100

18mol

Hence, 1.8

10060

100

18 60

2

2

=+

w

w

or1.8

100

100

18 60 602 2+

=w w

or1

10

3

10000 602 2+ =w w

or1

10 60

3

100002 2= =w w

This, gives w2 = −6.1 6.0 g.~

36. Here, we have

w2 = 68.4 g; M21342= −g mol

w1 100= g; Tb° = °100 C

Kb = −0.512 K kg mol 1

∴ ∆TK w

w Mb

b= × ××

1000 2

1 2

= × ××

1000

100 342

0.512 68.4

= °1.024 C

Hence, boiling point of solution

( )T T Tb b b= ° + ∆ = + = °100 1.024 101.024 C

37. For deposition of Al, reaction is

Al Al3 3+ −+ →e

Thus, 1 mol = 27 g. Al is deposited by

3F C= ×3 96500

∴ 0.09 g of Al will be deposited by

= × ×3 96500

270.09 = 965 C

For deposition of Cu, the reaction is,

Cu Cu2 2+ −+ →e

Thus, 2F C= ×2 96500 deposit Cu = 1mol

= 63.6 g

∴ 965 C will deposit Cu63.6= ×

×965

2 96500

= 0.318 g

38. Λ0 for BaCl2 1= X

Λ0 for H SO2 4 = X2

Λ0 for HCl = X3

By Kohlrausch’s law

Λ0 for BaSOBa SO2+

42–4

0 0= +λ λ

= + −Λ Λ Λ02

0 02(BaCl (H SO ) (HCl)2 4)

= + −X X X1 2 32

39. Let the number of atoms B in hcp lattice = N

As number of tetrahedral voids is double the

number of atoms in the close packing, thus

number of tetrahedral voids = 2N

Since, atoms A occupy 2/3 rd of the tetrahedral

voids, the number of atoms A in the lattice

= × =2

32

4

3N

N.

∴ Ratio of A BN

N: : : := = =4

3

4

31 4 3

Hence, the formula of the compound = A B4 3.

40. The temperature coefficient of a chemical

reaction is defined as the ratio of the specific

reaction rates of a reaction at two temperatures

differing by 10°C.

Temperature coefficient,

µ = °°

rate constant at + 10 C

rate constant at C

T

T

or µ = >T

TT T2

12 1( )

or T T2 1= ×µ

= × × − −2.5 2.5 s10 3 1

= × − −6.25 s10 3 1


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