EAMCETMedical Entrance Exam Solved Paper 2013 Physics 1. A wire of length L and linear density m is 5. Light of wavelenght λ from a point source stretched by a force T and the frequency is n1. Another wire of same material of length 2L and same linear density is stretched by a force 9T and its frequency is n2 . Then, the value of ( n2 / n1) is, falls on a small circular obstacle of diameter d. Dark and bright circular rings around a central bright spot are formed on a screen beyond the obstacle. The distance between the screen and obstacle is D. Then, the condition for the formation of rings is, (a) 4 : 1 (c) 3 : 2 (b) 1 : 3 (d) 1 : 2 (a) λ ≈ 2. A pipe closed at one end and open at the other end resonates with a sound of frequency 135 Hz and also with 165 Hz, but not at any other frequency intermediate between these two. Then, the frequency of the fundamental note of the pipe is (a) 15 Hz (c) 7.5 Hz (b) 60 Hz (d) 30 Hz 3. A small angled prism of refractive index 1.4 is combined with another small angled prism of refractive index 1.6 to produce dispersion without deviation. If the angle of first prism is 6°, the angle of the second prism is (a) 8° (c) 4° (b) 6° (d) 2° 4. The magnifying power of the astronomical telescope for normal adjustment is 50. The focal length of the eyepiece is 2 cm. The required length of the telescope for normal adjustment is (a) 102 cm (c) 98 cm (b) 100 cm (d) 25 cm (c) d ≈ d 4D λ2 D d2 4D D (d) λ ≈ 4 (b) λ ≈ 6. A bar magnet of moment of inertia I is vibrated in a magnetic field of induction 0 . 4 × 10−4 T. The time period of vibration is 12 s. The magnetic moment of the magnet is 120 Am 2 . The moment of inertia of the magnet is (in kgm 2 ) approximately (a) 172.8 × 10−4 (b) 2.1 π 2 10−2 (c) 1.57 × 10−2 (d) 1728 × 10−2 7. A short magnet of magnetic moment M, is placed on a straight line. The ratio of magnetic induction fields B1, B2 , B3 values on this line at points which are at distances 30 cm, 60 cm and 90 cm respectively from the centre of the magnet is (a) 27 : 3.37 : 1 (b) 37.3 : 1 : 27 (c) 27 : 8 : 3.37 (d) 1 : 2 : 3 2 | EAMCET (Medical) l Solved Paper 2013 8. Two capacitors having capacitances C1 and C 2 are charged with 120 V and 200 V batteries respectively. When they are connected in parallel now, it is found that the potential on each one of them is zero. Then, (a) 5C1 = 3C 2 (c) 9C1 = 5C 2 (b) 8C1 = 5C 2 (d) 3C1 = 5C 2 9. A parallel plate capacitor with a dielectric slab of dielectric constant 3 filling the space between the plates is charged to a potential V. The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced by another dielectric slab of dielectric constant 2. If the energies stored in the capacitor before and after the dielectric slab is changed are E1 and E2 , then E1 / E2 is 9 (a) 5 4 (b) 9 2 (c) 3 3 (d) 2 10. Copper and carbon wires are connected in series and the combined resistor is kept at 0° C. Assuming the combined resistance does not vary with temperature, the ratio of the resistances of carbon and copper wires at 0° C is (Temperature coefficients of resistivity of copper and carbon respectively are 4 × 10−3 / ° C and −0.5 × 10−3 / ° C) (a) 4 (c) 6 (b) 8 (d) 2 13. A body is thrown vertically upward from a point A 125 m above the ground. It goes up to a maximum height of 250 m above the ground and passes through A on its downward journey. The velocity of the body when it is at a height of 70 m above the ground is ( g = 10 ms −2 ) (a) 50 ms −1 (b) 60 ms −1 (c) 80 ms −1 (d) 20 ms −1 14. A body of mass 300 kg is moved through 10 m along a smooth inclined plane of angle 30°. The work done in moving in joules is ( g = 9.8 ms −2 ) (a) 4900 (c) 14,700 15. A balloon starting from rest ascends vertically with uniform acceleration to a height of 100 m in 10 s. The force on the bottom of the balloon by a mass of 50 kg is ( g = 10 ms −2 ) (a) 100 N (c) 600 N (b) 300 N (d) 400 N 16. Three particles, each of mass m, are placed at the vertices of a right angled triangle as shown in figure. The position vector of the centre of mass of the system is (O is the $ are unit vectors) origin and $i, $j, k 1 y , where p is the = pβ kB T pressure, y is the distance, kB is Boltzmann constant and T is the temperature. Dimensions of β are Y 11. In the equation (a) [M−1 L1 T2 ] (b) [M0 L2 T0 ] (c) [M1 L−1 T−2 ] (d) [M0 L0 T0 ] 12. A person reaches a point directly opposite on the other bank of a river. The velocity of the water in the river is 4 ms −1 and the velocity of the person in still water is 5 ms −1. If the width of the river is 84.6 m, time taken to cross the river in seconds is (a) 28.2 (c) 2 (b) 9.4 (d) 84.6 (b) 9800 (d) 2450 B m O m a 1 $ (a i − b$j ) 3 2 (b) (a $i − b$j ) 3 2 $ (c) (a i − b$j ) 3 1 (d) (a $i + b$j ) 3 (a) m A X 6 g/cc respectively. The moment of inertia of the combined system about an axis passing through ‘O’ and perpendicular to the rod is Mass m O x same material are k and 2k respectively.6 3 (a) (b) circle with uniform speed. then the ratio of the surface tension of water and mercury will be nearly (a) 13 : 2 (c) 16 : 5 (b) 5 : 16 (d) 2 : 13 25.5 20. Water rises in a capillary tube upto a height of 10 cm whereas mercury depresses in it by 3. If a horizontal force of 20 N is applied to 3 kg block. the ratio of pressure difference across the tubes is C a 2 1/ 2 r2 − r2 (d) 2 π 22 1 2 v1 − v 2 23. The height to which the combined mass rises (g = acceleration due to gravity) that the weight of a person at the equator becomes half the weight at the poles. A particle mass m is attached to a thin a uniform rod of length a at a distance of 4 from the mid-point C as shown in the figure. If the difference between the maximum and minimum tensions in the rope is 20 N. The mass of the rod is 4 m.EAMCET (Medical) l Solved Paper 2013 | 17. then the W ratio of work done in stretching A is WB 1 3 3 (c) 2 (a) 1 2 1 (d) 4 (b) 24. At displacements r1 and r2 from its mean position the velocities are v1 and v2 . Mass of the bob of the pendulum is also m. A stone tied to a rope is rotated in a vertical (a) 0.42 cm.3 4 14 (d) . During this collision the ball sticks with the bob of the pendulum. If earth were to rotate on its own axis such velocity v strikes the bob of a pendulum at rest. accelerations of the two blocks in ms −2 are ( g = 10 ms −2 ) 3 kg µ = 0. Then. Force constants of two wires A and B of the 19. A 3 kg block is placed over a 10 kg block and both are placed on a smooth horizontal surface. Assume the flow of the liquid through the tube is steady. A ball of mass m moving with a horizontal 21. 0. A particle is executing simple harmonic motion along a straight line.6 4 13 (c) . The time period of the particle is 1/ 2 r2 − r2 (a) 2 π 22 1 2 v 2 − v1 1/ 2 10 kg (c) 13 . If the angle of contact and density of mercury are 135° and 13.3 4 14 . If the two wires are stretched equally.0 (c) 1. mass of the stone in kg is ( g = 10 ms −2 ) (b) 1. The coefficient of friction between the blocks is 0.2 20 N (a) 2 π 2R g (b) 2 π R 2g (c) 2 π R 3g (d) 2 π R g 22. then its time period of rotation is (g = acceleration due to gravity near the poles and R is the radius of earth) (Ignore equatorial bulge) (a) v2 4g (b) v2 8g (c) v2 g (b) v2 2g 18.75 1 v12 + v 22 2 π r22 − r12 1/ 2 v 2 − v 22 (b) 2 π 12 2 r2 + r1 a 4 (a) 91 ma2 48 (b) 27 ma2 48 (c) 51 ma2 48 (d) 64 ma2 48 (a) (b) (c) (d) 1:8 1 :16 32 : 1 1:1 3 .2. 0.5 (d) 0. Two capillary tubes of lengths in the ratio 2 : 1 and radii in the ratio 1 : 2 are connected in series. If α is the coefficient of linear expansion of the material of the tube and γ is the coefficient of volume expansion of the liquid. g. The temperature at the two ends A and B of coefficient of apparent expansion is 1. increase in temperature of the gas in the cylinder B is ( γ = 1. If the energy of a Molybdenum atom with K-electron removed is 23. A p-n-p transistor is used in common emitter mode in an amplifier circuit. In the steady state the temperature at a point 10 cm from the end B is (Ignore loss of heat from curved surface of the rod) (a) 8. B-base and C-collector) (a) E and B are lightly doped and C is heavily doped (b) E is heavily doped. Match the following from table A.59 MW (d) 370 MW 34. then the coefficient of linear expansion of glass is a rod of length 25 cm and circular crosssection are 100°C and 0°C respectively.32 keV (b) 40. b→ b→ b→ b→ d. then (a) γ = 2α (c) γ = 4α (b) γ = 3α (d) γ = α 28. The value of P is (a) 59. h. A U reactor generates power at a rate of P producing 2 × 1018 fissions per second.4 | EAMCET (Medical) l Solved Paper 2013 26. Temperature of the gas in cylinder A increases by 30 K. The energy released per fission is 185 MeV.2 MW (b) 370 × 1018 MW (c) 0. When the temperature is changed. with those in table B Table A Table B (a) Work done in isobaric process (d) (b) Work done in isothermal process (e) p( V2 − V1 ) (c) Work done in adiabatic process (f) V nRT loge 2 V1 nR(T1 − T2 ) γ −1 (g) zero (a) (b) (c) (d) a→ a→ a→ a→ e. open at both ends is containing a liquid of certain length at some temperature. Piston A is free to move and piston of B is fixed. g. its coefficient of apparent expansion is 1. The K α − X ray of Molybdenum has a wavelength of 71 × 10−12 m.82 keV .42 × 10−7 eV) (a) 17. h. Same amount of heat is given to the gases in the two cylinders. A horizontal uniform tube. If the coefficient of linear expansion of copper is 17 × 10−6 / ° C. the length of the liquid in the tube is not changed. its 30.5 × 10−4 / ° C (b) 9 × 10−6 / ° C (c) 27 × 10−6 / ° C (d) 10 × 10−4 / ° C 27.4 for diatomic gas) (a) 24 K (c) 54 K (b) 36 K (d) 42 K (a) 60°C (b) 80°C (c) 90°C (d) 40°C 31.006 × 10−3 / ° C.03 × 10−3 / ° C. When the same liquid is heated in a copper vessel. c→ c→ c→ c→ f e g e 29. the collector current changes by 4 mA. Then. then the energy of Molybdenum atom when an L-electron removed is (hc = 12. e. If the current amplification factor is 60. d.32 keV.6 µA (d) 60 µA 235 33. When a liquid is heated in a glass vessel. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. B is thin and lightly doped and C is moderately doped (c) E and C are lightly doped and B is thick and heavily doped (d) E and B are heavily doped and C is lightly doped 32. The purpose of using heavy water in nuclear reactor is (a) (b) (c) (d) to increase the energy released in nuclear fission to cool the reactor to room temperature to make the dynamo blades to work well to decrease the energy of fast neutrons to the thermal energy 35. When base current is changed by an amount ∆I B . in the case of gases.82 keV (d) 5.5 keV (c) 23. then the value of ∆I B is (a) 15 µA (b) 240 mA (c) 66. The necessary condition in making of a junction transistor (E-emitter. When the temperature difference between (d) 21.6 figure with an internal resistance of 0. The emf of a cell E is 15 V as shown in the 38. the value of the current drawn from the cell is 2 mVr 2 mV (d) e er are connected in series with a 220 V. An electron beam in a TV picture tube is accelerated through a potential difference of V volt. The induced magnetic field (B) is (a) 2 mV er 2 (b) 2 mV (c) r temperature of hot junction is decreased by 20°C and cold junction’s temperature is increased by 6°C. In Moseley's law ν = a( z − b).5 Ω. 6. The percentage decrease in thermo emf is (assume thermo emf is directly proportional to the temperature difference) (a) 43 π rad 2 (b) π rad 6 (c) π rad 3 (d) 7Ω 2Ω 15 V 1Ω 6Ω 0. A 0. Which one of the following reactions is not correct? ∆ (a) (HCOO)2 Ca + (H3CCOO)2 Ca → 2CH3CHO + 2CaCO 3 Cu/300°C (b) CH3CH(OH)CH3 → CH3COCH3 + H2 5 . 4 37. It passes through a region of transverse magnetic induction field (B) and follows a circular orbit of radius r.01 H inductor and 3 π ohm resistance (a) (b) 2. (C2H5 )2 NH (3) and NH 3 (4) is Sucralose is an antiseptic Lactic acid is an antimicrobial Seconal is an antipyretic Chloroxylenol is a tranquillizer HI 4.4 40. Which one of the following statements is correct? (a) (b) (c) (d) C2H5NH2 (2). 4 (c) 4. pressure (d) CH3OH + CO → CH3COOH 6. Identify the pair of condensation polymers from the following (a) (b) (c) (d) (b) 1 < 3 < 2 < 4 (d) 3 < 2 < 4 < 1 terylene and nylon-66 PVC and polystyrene polyvinylether and polyisobutene neoprene and PVP (b) CH3CH2OH + O 2 → CH3COOH + H2O (i) CO 2 (c) CH3MgBr → CH3COOH + Mg(OH)Br (ii) H 3O ⊕ Ca / ∆ . the thermo emf is 30 mV. The phase difference between the current and emf is (c) 20. Which one of the following reactions is not correct? (a) CH3CHO + What are X and Y? (a) (b) (c) (d) X n-hexane Gluconic acid n-hexanol n-hexane Mn(OAc)2 /air 1 O 2 → CH3COOH 2 Micoderma aceti Y Gluconic acid Saccharic acid Saccharic acid Saccharic acid 3.4 (b) 1. the values of the screening constant for K-series and L-series of X-rays are respectively (a) 1. The Chemistry 1. Then. 50 Hz AC source. X ← Glucose → Y (a) 1 < 4 < 2 < 3 (c) 4 < 2 < 3 < 1 5. 6 (d) 2.5 Ω π rad 4 39.16 8Ω (a) 3 A (b) 2 A 10 Ω (c) 5 A (d) 1 A the junctions of a given thermocouple is 120°C.EAMCET (Medical) l Solved Paper 2013 | 36. HNO 3 2. The order of basic strength of C6H5NH2 (1). Which one of the following is chloropicrin? (a) CCl 3CHO (b) CCl 3 —C(CH3 )2 OH (2R. IUPAC name of 2 zone of reduction only zone of fusion only zone of reduction and zone of fusion zone of heat absorption 15. List I CaCl 2 (d) CH2 ==CH2 + PdCl 2 + H2O → CH3CHO ⊕ H + Pd + 2HCl 7. Identify the correct set. The correct statements are (a) 2 and 3 (c) 1 and 3 1 3 (a) (b) (c) (d) is (b) 1. (NH4 )3 PO 4 (d) Na 2SO 4 . During the manufacture of cast iron. Shape Number of lone pairs of electrons pyramidal 1 5 6 (a) (b) (c) (d) 2-methyl-2. 1. Na 3PO 4 List II (A) pH of unpolluted (I) H2C==CH—CHO rain water (B) Acid rain (II) 5. 2 and 3 (d) 1 and 2 16.6 | EAMCET (Medical) l Solved Paper 2013 Ag/300°C (c) CH3CH2OH → CH3CHO + H2 13.6 (C) Acrolein (III) NO2. 2. Lanthanides do not form coordinate compounds as readily as d-block metals. (a) (b) (c) (d) CaHCO 3 . The correct Fischer projection formula of CO 2H OH OH CH3 (A) (II) (I) (III) (II) 14. CaCO 3 CaHCO 3 .3-dihydroxy butanoic acid is CH3 H (b) HO CH3 HO (c) HO H H CO 2H H (d) HO OH H CO 2H CO 2H OH H CH3 10. Fluorine reacts with KHSO4 to form HF and X.4-diethyl-4-methyl pentane 3-ethyl-4. the (c) CCl 3 —NO 2 (d) CHCl 2 —CHCl 2 H (a) H (a) (b) (c) (d) (b) n-hexane (d) Cyclohexane 4 11. Observe the following statements.4-dimethyl hexane 12. CuSO 4 ⋅ 5H2O CaCO 3 . The pair of chemicals used as food for yeast in the fermentation of molasses is (a) (NH4 )2 SO 4 . 3 R) -2.3-dimethyl hexane 3.3-diethyl pentane 4-ethyl-3. Match the following. CuSO 4 ⋅ 5H2O CaCO 3 . Carbon and hydrogen in an organic compound are detected as …… . Na 3PO 4 8.2-benzpyrene (c) 2-butene (B) (I) (II) (II) (III) (C) (III) (III) (IV) (I) (D) (IV) (IV) (I) (IV) slag (CaSiO3) is formed in 9. Which one of the following is X? (a) SO 3 (c) K 2S2O 8 (b) K 2SO 4 (d) H2S . The basic nature of hydroxides of lanthanides increases from La(OH) 3 to Lu(OH) 3. Cu(OH)2 Molecule Hybridisation (a) XeO 4 sp3d 2 3 (b) XeO 3 sp pyramidal 1 (c) XeF4 sp3d 2 planar 3 (d) XeF2 sp3d linear 2 17. Which one of the following causes cancer? (a) 1. CO2 (D) Freon (IV) CF2Cl2 (V) CaOCl2 (b) (NH4 )2 SO 4 . (NH4 )3 PO 4 The correct answer is (c) Na 2SO 4 . 3. Lanthanides actively participate in chemical reactions. In B2H 6 . The stability of group V hydrides decreases from NH 3 to BiH 3.082 L atm mol −1 K −1) (a) 0.82 atm same temperature. Observe the following statements. The correct statement(s) are (a) 1 and 3 (c) 1 and 2 (b) 2 (d) 1. The number of metals showing photoelectric effect when light of 300 nm wavelength falls on it is (1 eV = 1.75 (a) 6 (c) 5 (b) 8 (d) 4 29. 2.3 4.4 2. In the reaction. H2O. (R) is the correct explanation of (A) 26. 4NO2 ( g) + O2 ( g) → 2N2O5 ( g). Assertion (A) GeF4 and SiCl4 act as Lewis (b) 1.3 (d) 0.08 × 10−10 . The correct statements are (a) 2 and 3 (c) 1 and 3 bases.3 × 10−4 (c) 3. Which one of the following is Y? (a) H2S (c) H2SO 4 (b) SO 2 (d) S 28. of some metals is listed below. If the uncertainty in velocity of a moving object is 1. (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. its K c in mol L−1 is ( R = 0. 2 and 3 (d) 1 and 2 Metal Li Na K Mg Cu Ag Fe Pt W w (eV) 2. ∆H = − 111 kJ if N2O5 ( s) is formed instead of N2O5 ( g).7 4. HCl to form NaCl.033 (c) 1. at 27°C.0 (b) 3.2 3. If the solubility product of MgF2 at a certain temperature is 1.04 × 10−5 (b) 7.0 × 10−4 20. w.3 2. For the equilibrium.33 19. X reacts with Na2S to form Na2S2O3 and Y. Aqueous solution of borax is alkaline in nature.626 × 10−34 Js) (a) helium (c) lithium (b) deuterium (d) electron 7 . the ∆H value in kJ is (∆H sublimation for N2O5 = 54 kJ mol –1) (a) – 165 (c) + 219 (b) – 57 (d) – 219 21. Observe the following statements. H 3BO3 is used as antiseptic. 2. H2O2 acts as oxidizing agent? (a) (b) (c) (d) Cl 2 + H2O 2 → 2HCl + O 2 Ag 2O + H2O 2 → 2Ag + H2O + O 2 2NaOH + H2O 2 → Na 2O 2 + 2H2O KNO 2 + H2O 2 → KNO 3 + H2O 22. Phosphorous does not exhibit allotropy.8 4. The correct answer is (a) Both (A) and (R) are correct. its solubility in mol L−1 is (a) 1. 1.0 × 10−6 ms −1 and the uncertainty in its position is 58 m. 3.EAMCET (Medical) l Solved Paper 2013 | 18. X and Y. 3.3 4.6 × 10−19 J) H2 Zn(OH)2 + Na 2O ZnO O2 (a) K 2O (c) Cs 2O 25.7 6. Orange red coloured monoxide is (b) Na 2O (d) Li 2O 24. 2 and 3 27. 1. A( g) º B( g) + C( g). Reason (R) Ge and Si have d-orbitals to accept electrons. The energy required to overcome the attractive forces on the electrons. each boron is sp2 hybridized. Zinc reacts with hot and concentrated NaOH and forms (a) (b) (c) (d) 23. Sodium thiosulphate reacts with dil. the mass of this object is approximately equal to that of (h = 6. At the K p = 0. The solubility of group V hydrides in water decreases from NH 3 to BiH 3. In which of the following reactions.0 × 10−5 (d) 3. IO−3 is oxidized to I2 3. 2 (b) 3. X2 and X 3 respectively. 2 (d) 2.8 (d) 3. If TA M B = TB M A .5.512 K kg mol −1) (b) B >Li >Be (d) Li >Be >B 31.318 (c) 0. 8 (d) 4. of the non-volatile solute urea (NH2CONH2 ) to be dissolved in 100 g of water in order to decrease its vapour pressure by 1.0 × 10 Zoology 1. I− is reduced to I2 2. Molar conductances of BaCl2 . KIO3 + 5KI + 6HCl → 3I2 + 6KCl + 3H2O 1. IO−3 is reduced to I2 4.0 (b) 0. The boiling point of a solution containing (d) 0. of Al = 27.0 × 10−3 (b) 6. The temperature coefficient of a reaction is 2. H2SO4 and HCl at infinite dilution are X1. The total number of antibonding electrons in N2 and O2 molecules respectively is (a) 4. 3 32.25 × 10−3 −2 (d) 6. 3 and 4 are at temperatures TA and TB respectively.512°C (b) 101. The weight in gram.18 38.4 g of sucrose (molar mass = 342 g mol −1) in 100 g of water is ( K b for water = 0. Two gases of molecular masses M A and M B (a) Pressure (c) RMS velocity 36. 0.8 | EAMCET (Medical) l Solved Paper 2013 30. 8 (c) 5. according to Goldacre and Lorsch cytoplasm solates due to (a) (b) (c) (d) folding of protein molecules sliding of actin molecules sliding of myosin molecules unfolding of protein molecules 2.3 (c) 3. wt. 3 (c) 2. Oxidation number of I increases from –1 (in KI) to zero (in I2 ) The correct statements are (a) 1.02°C (d) 100. which one of the following properties has the same value for both the gases? (b) Kinetic energy (d) Density 35.98°C (c) 100. A certain quantity of electricity is passed through aqueous Al2 (SO4 ) 3 and CuSO4 solutions connected in series.5 × 10−3 s −1. The number of σ and π bonds present in acetylene are respectively (a) 3.636 (b) 31.6) (a) 0.0 (a) 98. Cu = 63.02°C 37. Atoms of an element ‘A’ occupy tetrahedral 34.25 × 10−2 (c) 1.6 (a) A2 B (c) A4 B3 (b) AB2 (d) A2 B3 40. Molar conductance of BaSO4 at infinite dilution is (a) ( X1 + X 2 − 2 X 3 )/2 (c) X1 + X 2 − 2 X 3 (b) X1 + X 2 − X 3 (d) ( X1 + X 2 − X 3 )/2 2 3 voids in the hexagonal close packed (hcp) unit cell lattice formed by the element ‘B’. 2 and 4 (c) 1 and 4 (b) 3 and 4 (d) 1.09 g of Al is deposited on cathode during electrolysis. 8 33. Be and B is (a) B >Be >Li (c) Li >B >Be 68. In amoeboid movement.8% is (a) 6. The formula of the compound formed by ‘A’ and ‘B’ is 39. Which one of the branches of cranial nerves is not related to vagus? (a) Chorda tympani (b) Cardiac depressor (c) Recurrent laryngeal (d) Pneumogastric . the rate constant at T2 K in s −1 is ( T2 > T1) (a) 1. 6 (b) 6. The amount of copper deposited on cathode in grams is (At. The correct order of atomic radius of Li. If its rate constant at T1 K is 2. Consider the following statements for the reaction. decrease in CO 2 . decrease in CO 2 . increase in CO 2 . Which one of the following became an endangered species as result of extinction of Raphus cucullatus? (a) (b) (c) (d) Sideroxylon grandiflorum Chrysanthemum Cinchona Bacillus thuringiensis 6. Reptiles. Mammals Reptiles. 3rd and 4th moults of microfilaria take place in (a) (b) (c) (d) lymph vessels of man stomach of mosquito salivary glands of mosquito thoracic muscles of mosquito 5. decrease temperature (c) Decrease in pH. related to Indian chain viper? List I List II (A) Mehlis glands (I) Tegument formation (B) Vitelline gland (II) Osmoregulation and excretion (C) Mesenchymal cells (III) Lubricate passage of capsules into uterus (D) Flame cells (IV) Secretion of embryophore (V) Capsule formation around zygote The correct match is (a) (b) (c) (d) (A) (III) (III) (V) (IV) (B) (V) (V) (I) (III) (C) (I) (II) (II) (I) (D) (II) (IV) (IV) (II) 4. (a) (b) (c) (d) I 5/4. Amphibians 12. M 3/3 I 4/3 C. decrease temperature (b) Increase in pH. (A) Frontal ganglion (B) Proventricular ganglion (C) Hypocerebral ganglion (D) Visceral ganglion (a) A–C–D–B (c) B–C–D–A (b) A–D–C–B (d) B–D–A–C 10. increase temperature (d) Decrease in pH. (a) (b) (c) (d) Aurelia and Rhizostoma Pennatula and Aurelia Rhizostoma and Corallium Physalia and Halistemma 9 . M 4/4 I 3/3 C. Identify the correct combinations. Craspedote medusa is present in the following pair. (a) Increase in pH. Both hepatic and renal portal systems are found in (a) (b) (c) (d) Fishes. Match the following with reference to Taenia solium. Reptiles Amphibians. In the life cycle of Wuchereria. Unloading of oxygen from haemoglobin is enhanced under the following conditions. 1/1 PM 4/4. Fishes. C 1/1 PM 3/3. M 4/4 11. Aves. 1/1 PM 3/3.EAMCET (Medical) l Solved Paper 2013 | 3. increase in CO 2 . 1/1 PM 3/3. increase temperature in in in in 13. Which of the following character is not (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. Identify the typical dental formula of a metatherian. Arrange the ganglia of autonomous nervous system of cockroach in correct sequence from anterior to posterior end. (A) (B) (C) (D) (a) (b) (c) (d) Housefly–Labellum–Pseudotrachea Moth–labellum–Dutton’s membrane Butterfly–1st Maxillae–Galea Tse-tse fly–Rostrum–Haustellum (A) and (B) (A) and (C) (B) and (C) (B) and (D) 8. Mammals Cyclostomes. What is the resting membrane potential of nerve fibre? (a) +75 mV (c) − 70 mV (b) +45 mV (d) − 45 mV 7. M 4/4 I 4/5 C. Amphibians. 8 s (b) 0. In eukaryotes. Match the correct pairs in list I and list II cortisol is (a) Zona faciculata (c) Zona reticularis heterogenous (a) (b) (c) (d) (A) (III) (I) (V) (II) (B) (IV) (II) (III) (IV) (C) (V) (IV) (II) (III) (D) (I) (III) (IV) (V) (E) (II) (V) (I) (I) 22. Match the following lists. which have various shapes of amphids but without phasmids. Arrange the following in the ascending order of their number per cubic millimeter. What is the duration of one cardiac cycle in man when the heart beats for 75 times per minute? (a) 0. shell and mantle cavity are twisted upto 180° counter clockwise with respect to the head and foot. but R is correct (d) Both S and R are true. The correct answer is (a) Both S and R are true. Identify the Nematodes. and R is a correct explanation to S (a) (b) (c) (d) A → C → B → D→ E C → A → D→ E→ B C → A → D→ B → E A → C → E→ B→ D RNA (b) Split genes (d) Z-genes with reference to Pheretima. The part of adrenal cortex which secretes 19. visceral mass. present in the blood. and R is not a correct explanation to S (b) S is correct.5 s . (a) (b) (c) (d) Wuchereria and Ascaris Trichinella and Greeffiella Enterobius and Ancylostoma Ascaris and Ancylostoma 15. identify the correct sequence of body wall layers from outer to inner side. (A) Basophils (B) Lymphocytes (C) Eosinophils (D) Neutrophils (E) Monocytes nuclear 20. 5th and 6th segments (V) 10th and 11th segments The correct match is (a) (b) (c) (d) (A) (IV) (I) (V) (IV) (B) (I) (III) (IV) (III) (C) (V) (II) (III) (V) (D) (III) (V) (I) (II) 21. In the transverse section of Pheretima.4 s (d) 0. List I List II (A) Pharyngeal Nephridia (I) 14th segment (B) Stomach (II) 18th segment (C) Anterior loops (III) 9th and 14th segments (D) Male genital apertures (IV) 4th. but R is not correct (c) S is not correct. List I List II (A) Simple squamous epithelium (I) Ureters (B) Simple cuboidal epithelium (II) Epididymis (C) Non-ciliated simple columnar epithelium (III) Lining of alveoli of lungs (D) Transitional epithelium (IV) Lining of thyroid vesicles (E) Pseudostratified non-ciliated columnar epithelium (V) Mucosa of stomach and intestine The correct match is 18.10 | EAMCET (Medical) l Solved Paper 2013 14. due to torsion mantle cavity is placed anteriorly behind and above the head. Statement (S) In gastropods.3 s (c) 0. (A) Circular muscles (B) Parietal peritonium (C) Epidermis (D) Cuticle (E) Longitudinal muscles (a) (b) (c) (d) C → B→ E→ A → D D→ C → E→ A → B D→ C → A → E→ B D→ B → A → E→ C 17. Reason (R) During larval stage. which of the following genes synthesise (hnRNA)? (a) Holandric genes (c) A-gene (b) Zona glomerulosa (d) Zona pellucida 16. The correct match is (a) (II) and (IV) (c) (III) and (IV) (b) (I) and (III) (d) (I) and (II) 29. Identify the two small apertures present in the auditory capsule of rabbit (a) (b) (c) (d) List I List II (A) Founder effect (I) Long necked giraffes (B) Bottle neck effect (II) Heterozygous for sickle-cell anaemia (C) Genetic load (III) Pitcairn Island human population (D) Directional selection (IV) Polydactylic dwarf individuals Sunflower population in California The correct match is (A) Conjugate vaccine (A) (I) (II) (I) (II) 27. (IV) The portion of the antigen to which an antibody binds is called epitope. Match the following lists. scapula. (a) Oncogenes (c) Pseudogenes (b) P53 genes (d) SRY genes 30. (III) The stem of antibody and lower part of the arms constitute the ‘constant’ (C) region. Match the following (v) 24. Of the following statements about ‘antibodies’ and ‘antigens’. (a) (b) (c) (d) 11 foramen magnum and foramen ovale foramen ovale and fossa ovalis fenestra ovalis and obturator foramen fenestra ovalis and fenestra rotunda (a) (b) (c) (d) (A) (IV) (II) (V) (III) (B) (II) (I) (IV) (IV) (C) (V) (III) (II) (II) (D) (III) (IV) (III) (I) 28. furcula scapula. Identify the correct sequence of events with reference to conjugation of Vorticella. corocoid scapula.EAMCET (Medical) l Solved Paper 2013 | 23. foramen triosseum is present at the junction of (a) (b) (c) (d) clavicle. Identify the tumour suppressor genes form the following. synsacrum. corocoid. In birds. (II) The stem ‘Y’ of antibody is called ‘Fab ’ fragment. (A) Amphimixis (B) Disappearance of macronucleus (C) Attachment of the conjugants (D) Post conjugation fissions (E) Prezygotic nuclear divisions (F) Postzygotic nuclear divisions (a) (b) (c) (d) C → B → A → E→ D→ F C → B→ E→ A → F→ D F→ A → D→ B → C → E F→ D→ A → E→ B → C List I List II (I) Human papilloma virus (B) Toxoid vaccine (II) Haemophilus influenzae (C) Attenuated whole agent vaccine (III) Bubonic plague (D) Inactivated whole agent vaccine (IV) Yellow fever (V) Diphtheria The correct answer is (B) (IV) (V) (II) (III) (C) (V) (IV) (III) (I) (D) (II) (III) (V) (IV) 25. furcula 26. carina. (I) An antibody consists of four identical light (L) chains. synsacrum clavicle. Choose the correct set. (a) Mutations are cumulated over generations (b) There are no intermediate stages in the course of evolution (c) Mutants are markedly different from parents (d) Mutations are subjected to natural selection . and two identical heavy (H) chains. Choose the wrong statement with reference to mutation theory. Reason (R) The placenta of rabbit is described as deciduous type. 2 and 3 abdominal segments of workers 37. (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) (c) (A) and (C) 38. and R is a correct explanation to S 36. Reason (R) Non-disjunction results in production of abnormal gametes. Identify the organisms that belong to the benthos of lake ecosystem. Statement (S) After implantation. and R is a correct explanation to S (b) (A) and (D) (d) (A) and (B) The correct match is (a) (b) (c) (d) (A) (I) (III) (II) (III) (B) (III) (I) (I) (II) (C) (II) (V) (III) (I) (D) (V) (II) (V) (IV) . but R is correct (d) Both S and R are true. 35. Identify the correct sequence of stages in the ross cycle of Plasmodium. Identify the correct pair. 2. The correct answer is (a) Both S and R are true. 4 and 5 abdominal segments of workers 1. Statement (S) Non-disjunction is the failure of paired chromosomes to segregate during the metaphase of meiotic divisions of gametogenesis. Wax glands in honey bees are present on these segments. Match the following lists. 3 and 4 abdominal segments of queens 2. (a) (b) (c) (d) 1. Which of the following hormones regulate solute reabsorption during urine formation in rabbit? (a) (b) (c) (d) Antidiuretic hormone and angiotensin I Angiotensin III and angiotensin I Nor-epinephrin and epinephrin Angiotensin II and aldosterone 32. 3 and 4 abdominal segments of queens 2. (a) (b) (c) (d) Gerris and beetles Chironomid larvae and red annelids Daphnia and notonecta Ranatra and copepods 34. What is the phenotype of the offspring born to 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 normal vision (c) All the sons and daughters are colourblind (d) All the sons are colourblind and the daughters are with normal vision 40. but R is not correct (c) S is not correct. List I List II (A) Georges Cuvier (B) Claude Bernard (II) Homology (C) Louis de Buffon (III) Comparative Anatomy (D) Richard Owen (IV) System of Nomenclature (I) Homeostasis (V) Natural History The correct answer is (a) Both S and R are true. and R is not a correct explanation to S (b) S is correct. 3. and R is not a correct explanation to S (b) S is correct. (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. but R is correct (d) Both S and R are true. but R is not correct (c) S is not correct. the uterine myometrium undergoes changes to become ‘decidua’ in rabbit. In ECG a prolonged PR interval indicates (a) (b) (c) (d) Hyperkalaemia and hypokalaemia Myocardial ischemia and hyperkalaemia Cardiac arrhythmia and hypokalaemia Coronary artery disease and rheumatic fever 33.12 | EAMCET (Medical) l Solved Paper 2013 31. EAMCET (Medical) l Solved Paper 2013 | 13 Botany 1. Match the following lists. List I List II (A) A. List I (A) (B) (C) (D) Tyloses Periderm Motor cells Laticifers List II (I) (II) (III) (IV) (V) Coenocytic Adaxial epidermis Complementary cells Heart wood Conjunctive tissue The correct match is (C) (II) (I) (IV) (IV) 2. What would be the total number of hydrogen bonds present in this DNA fragment? (b) 192 (d) 60 3.G. List I List II (A) G 2 phase (I) Fusion of microtubules to form spindle apparatus (B) Prometaphase (II) Production of energy required for spindle formation (C) Anaphase (III) Recombination of genetic material (D) Pachytene (IV) Contraction of tubulin proteins (V) Reappearance of plasmosome The correct match is (a) (b) (c) (d) (A) (V) (II) (V) (II) (B) (IV) (IV) (I) (I) (D) (III) (V) (II) (III) 40% of the bases are cytosine. which was first reported by Rhodin is involved in the following reaction. Tansley (I) Classified plant communities (B) Warming (II) Structure and function of nature 4. there are 8 turns. conversion of serine to serine conversion of glycine to serine conversion of glycerate to PGA formation of glycolate from phosphoglycolate (a) (b) (c) (d) (A) (III) (II) (IV) (IV) (B) (II) (V) (III) (I) (C) (I) (I) (II) (III) (D) (V) (III) (I) (V) 6. (D) Reiter (IV) Term–‘Ecology’ (V) Term–‘Ecosystem’ The correct match is (a) (b) (c) (d) (A) (V) (II) (IV) (IV) (B) (I) (V) (III) (III) (C) (II) (I) (V) (V) (D) (IV) (III) (II) (I) . In a DNA fragment. Collocytes which provide mechanical strength to the plants are present in (a) (b) (c) (d) dicotyledonous root dicotyledonous stem monocotyledonous leaf monocotyledonous stem 7. (II) Pericycle cells of both dicot and monocot roots actively divide to produce lateral roots during secondary growth. Identify the correct pair of statements from (C) Odum (III) Term–‘Biosphere’ the following. (a) (b) (c) (d) (a) (II) and (III) (c) (I) and (II) (b) (III) and (IV) (d) (I) and (IV) 5. (III) All cells of endodermis are passage cells in dicot root. Match the following lists. Match the following lists. with (a) 96 (c) 224 (IV) Xylem always produced in a centripetal manner in the roots of fruit bearing plants. (I) Pericycle of dicot root parenchymatous but sclerenchymatous in mature monocot root. The cell organelle bounded by single unit membrane. synandry. List I 11. Assume that blue flower of a plant is dominant character over the white. extrafloral nectaries. What is the total number of pollen grains produced in its inflorescence? The correct match is (a) (b) (c) (d) The correct match is List II (A) Apices of underground (I) Mentha branches store food material (B) Underground branches grow (II) Hydrocotyle obliquely upward from the axillary buds of nodes of the stem below the soil (C) Aerial branches grow obliquely (III) Agave downwards and produce adventitious roots after touching the soil (D) Weak stemmed plants have a (IV) Stachys cluster of leaves and roots at every node (V) Jasminum List II (A) Aggregate fruitlets of berries (I) Rubus appearing as a single fruit (B) Achenes on the inner (II) Artabotrys margins of receptacle (C) Etaerio of drupes (III) Ficus (D) Etaerio of follicles (IV) Magnolia (V) Annona The correct match is (a) (b) (c) (d) (A) (V) (II) (II) (IV) (B) (III) (I) (IV) (III) (C) (I) (V) (III) (I) (D) (IV) (IV) (V) (II) . the progeny showed 50% of plants with blue flowers and 50% of plants with white flowers. Match the following lists. continuous. Identify the wrong statement. Match the following lists. unilocular ovary 15. (I) Unisexual flowers (II) Sessile flowers (III) Centripetal arrangement of flowers (IV) Bisexual flowers The correct pair is (a) (II) and (III) (c) (II) and (IV) (b) (III) and (IV) (d) (I) and (II) 9. This plant is associated with a pair of characters given below. The genotypes of blue and white parents respectively are (a) Bb. The type of inflorescence in a plant showing polygamous condition and lateral style is (a) Compound corymb (c) Compound head (b) Compound umbel (d) Compound raceme 14. of (a) Free central placentation. Venation in Caulophyllum is parallel. Identify the correct combination characters found in Cucurbita. bb (b) bb. List I (A) (B) (C) (D) List II Turill Goethe Salisbury Oudet (I) (II) (III) (IV) (V) Tissue culture Pollen grains Nucleosome Flower-a modified stem Stomatal index (A) (III) (V) (IV) (II) (B) (I) (II) (I) (IV) (C) (IV) (III) (II) (V) (a) (b) (c) (d) (A) (II) (III) (IV) (IV) (C) (I) (II) (III) (V) (D) (III) (I) (II) (II) 12. Match the following lists. bb 10. Phyllotaxy in Trillium is alternate. bb (c) BB. fusion of epicarp and thalamus to form rind of the fruit. (a) (b) (c) (d) Stipules in Lathyrus are persistent. List I (B) (V) (IV) (V) (I) Allium. divergent palmately reticulate venation (b) Mesogamy. When a blue flowered plant is crossed with white flowered plant. 13. Cauline leaves are found in Cocos. There are 25 flowers in an inflorescence of (D) (V) (IV) (III) (III) (a) 4500 (b) 9000 (c) 18000 (d) 1500 16. Bb (d) BB. seed germination epigeal (c) Axile placentation. compound sieve plate. collenchymatous hypodermal ring in stem (d) Bicarpellary gynoecium. Entry of pollen tube into the ovule through chalaza was first discovered by Treub in a xerophyte. Each anther lobe of every stamen contains 60 pollen grains.14 | EAMCET (Medical) l Solved Paper 2013 8. (a) (III) and (IV) (c) (I) and (IV) (b) (II) and (III) (b) (I) and (III) . The correct answer is (a) Both A and B are true. (IV). (I) Petiole adnate with the stem (II) Thalamus cup-shaped (III) Seeds non-endospermic (IV) Flowers hypogynous The correct answer is (a) (I) and (IV) (c) (III) and (IV) (b) (II) and (IV) (d) (II) and (III) 20. Study the following and identify the correct pair of characters for Physalis. (I) Calcium chloride (II) DNA ligase (III) Ethylene diamine tetra acetic acid (IV) Restriction endonuclease The correct answer is (a) Lilium (c) Aloe (b) Trillium (d) Allium 18. (I). (III) Triplet code ‘UAA’ is called ‘Amber’. (IV) Delay of senescence by cytokinins is known as ‘Richmond-Lang effect’. (IV). Match the following lists.EAMCET (Medical) l Solved Paper 2013 | 15 17. Assertion (A) Application of cytokinins causes the opening of stomata. What is the ratio of ATP requirement for the fixation of 6 molecules of CO2 in sugarcane and 5 molecules of N2 in bean? (a) 5 : 16 (c) 5 : 8 (b) 3 : 16 (d) 3 : 8 26. but R is true (d) Both A and R are true. but R is false (c) A is false. (II) Inhibition of bacteroid respiration by oxygen is referred to as ‘Warburg effect’. fleshy leaf bases and loculicidal capsule. their usage in recombinant DNA technology. (II) and (III) (I). 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 breeding The correct answer is (a) (III) and (IV) (c) (I) and (IV) (b) (I) and (III) (d) (II) and (III) 21. (II) and (III) (IV). (I) Lutein is an oxygenated hydrocarbon. and R is the correct explanation of A 24. (III). Identify the correct pair of statements. Reason (R) Cytokinins induce the influx of potassium ions into guard cells. Arrange the following in sequential order of possessing naked bulb. Triple fusion was discovered in this plant 22. 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 The correct match is (a) (b) (c) (d) (A) (III) (IV) (V) (V) (B) (V) (I) (I) (III) (C) (IV) (III) (IV) (I) (D) (I) (V) (II) (II) 19. What is the amino acid sequence in the polypeptide segment translated from the mRNA base sequence of AGU-UUU-UCC-GGG-UCG? (a) (b) (c) (d) Serine-Glycine-Serine-Phenylalanin-Serine Serine-Phenylalanine-Serine-Glycine-Serine Serine-Serine-Phenylalanine-Glycine-Serine Phenylalanin-Serine-Serine-Glycine-Serine 25. (III) and (II) (IV). (II) and (I) 23. Which one of the following SCP organisms lack a membrane bound nucleus? (a) Spirulina (c) Chlorella (b) Paecilomyces (d) Chaetomium (a) (b) (c) (d) (I). but R is not the correct explanation of A (b) A is true. 7 0. Assertion (A) Azolla is used as biofertiliser in rice fields. general cortical cell and pericyclic cells of young root respectively (Assume symplast flow). is a nitrifying (IV) Continuous system of cell walls and intercellular spaces in plant tissue is called ‘apoplast’. C and D plants are given below.6 C – 0. C (b) B. Arginosuccinase is an example to (a) (b) (c) (d) 31. but R is false (c) (A) is false. Identify the role of lectins in the formation of Plant Number of Stomata Number of epidermal cells A 40 730 root nodules in legumes. (a) (b) (c) (d) Formation of shepherd’s crook Recongnition of compatible Rhizobium by host Formation of peribacteroid membrane Formation of infection thread 29. A (c) B. (a) B. (II) The number of molecules of O2 absorbed is more than the number of CO2 molecules released when one molecule of triolein is completely oxidised in respiration.1 D – 0. A.9 0. The number of stomata and epidermal cells in 1 mm 2 area of abaxial surface of the leaves of A. C. C 33. Study the following table showing the components of water potential of four cells of an actively transpiring monocot plant.8 0. D. A (d) A. B C Osmotic Potential Pressure potential (MPa) (MPa) A – 0. and R is the correct explanation of A (III) Bacillus mycoides bacterium. The correct answer is (a) Both A and R are true.16 | EAMCET (Medical) l Solved Paper 2013 27. Reason (R) Azolla contains nitrogen fixing cyanobacteria in its root nodules. Hydrolase Ligase Lyase Oxido-reductase 28. Identify the correct pair of statements from (I) The attraction between two water molecules in xylem vessels is called ‘adhesion’. B. and R is not the correct explanation of A (b) (A) is true. but R is true (d) Both A and R are true. C. Match the following lists.6 0. The correct answer is (a) (b) (c) (d) (II) and (III) (III) and (IV) (II) and (IV) (I) and (IV) 510 70 450 D 30 620 plants (a) B and C (c) A and C having least (b) A and D (d) A and B 32. D. 60 Identify the two stomatal index. D. What is the total number of amino acids.5 B – 0.4 Identify the four cells as root hair. Cell the following. B. 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 . 30. D. when the capsid of TMV contains 2130 capsomers? (a) 336540 (c) 948 (b) 162622 (d) 387921 34. The correct match is The correct answer is (a) Both A and R are true. but R is true (d) Both A and R are true. The microelement. and R is the correct explanation of A 37. Assertion (A) Dictyostele is present in (II) Rhizopus chlamydospore (C) Multinucleate (III) Pteris neck canal cell (D) Binucleate (IV) Vitis matured sieve element (V) Akinite of Spirogyra Reason (R) Meristeles are scattered in the rhizome of Pteris vittata.EAMCET (Medical) l Solved Paper 2013 | 38. In F2 -generation. 960 garden pea pods are produced during dihybrid cross due to self pollination of heterozygous parents. Identify the wrong statement in relation to The correct answer is (a) (b) (c) (d) (A) (III) (V) (V) (II) (B) (I) (IV) (III) (IV) (C) (IV) (II) (II) (V) Funaria. How many pods would be green and inflated? (a) (b) (c) (d) 240 180 60 540 . (b) 2 : 1 List II (A) Anucleate (a) nitrate reductase (b) cytochrome-‘C’-oxidase (c) IAA oxidase (d) dinitrogenase (a) 1 : 2 17 (d) 1 : 1 (a) (b) (c) (d) (A) (I) (IV) (I) (IV) (B) (IV) (I) (III) (I) (C) (V) (III) (II) (II) (D) (II) (II) (IV) (III) 40. but R is false (c) (A) is false. and R is not the correct explanation of A (b) (A) is true. What is the ratio of mitotic divisions that take place in the microspore of Cycas before and after liberation from the microsporangium during the development of male gametophyte? (c) 3 : 2 (I) Spirogyra gametangium (B) Uninucleate rhizome of Pteris vittata. Study the following lists List I 36. (D) (II) (III) (I) (III) (a) (b) (c) (d) 35. which is an integral part of the electron carrier which transfers electrons from cyt-b-f complex to PS I is a component of Stomata are present in the epidermis of capsule Spores are viable for only one year Inner spore sac is one celled in thickness Trabeculae connect the innermost layer of the capsule wall with the outer spore sac 39. (c) 35. the ratio of 2 is n1 2 = 2. (d) 40. (a) 19. (a) 29. (d) 4. (b) 11. (d) 21. (b) 24. (a) 4. (d) 5. We know that the frequency n= 1 T 2l µ Here µ is a constant. (c) 27. (d) 11. (a) 18. (a) 10. (a) 14. (c) 26. (a) 23. (a) 4. (a) 3. (b) 34. (a) 26. (c) 17. (c) 6. (c) 38. (c) 16. (b) 7. (a) 34. (c) 2. (a) 11. (a) 28. (d) 6. (a) 22. (a) 8. (d) 20. (a) 22. (b) 2. (a) 33. (b) 8. The frequency of the fundamental note of the pipe is v n2 − n1 = 2 4l v n2 − n1 = 4l 2 . (c) 16. (b) 20. (b) 35. (b) 15. (a) 40. (b) 30. (c) 9. (c) 24. (a) 6. (d) 2. (d) 31. (b) 12. (a) 10. (a) 22. so n ∝ T l Now. (*) 8. (a) 16. (c) 33. we can calculate the value of n2 /n1 as follow n2 l1 T 1 9 = × 2 = × n1 l2 T1 2 1 9 3 = 2 2 n 3 Hence. (d) 23. (a) 34. (b) 25. (d) 35. (d) 32. (c) 4. (b) 12. (b) 32. (b) 39. (d) 36. (b) 39. (b) 29. (b) 37. (a) 29. (b) 13. (d) 19. (d) 14. (d) 9. (a) 37.Answers Physics 1. (a) 12. (d) 28. (b) 21. (b) 20. (d) 24. (b) 32. (a) 7. (d) 15. (b) 34. (c) 15. (c) 39. (d) 29. (a) 19. (d) 7. (d) 9. (a) 38. (d) 17. (d) 39. (d) 25. (b) 15. (d) 18. (d) 28. (c) 31. (d) 30. (a) 14. (b) 2. (d) 37. (c) 27. (b) 24. (b) 18. (c) 28. (a) 11. (b) 26. (d) 19. (a) 38. (d) 21. (a) 5. (d) 27. (b) 1. (c) 25. (d) 30. (a) 13. (d) 35. (b) 36. (d) 36. (c) 7. (d) 3. (b) Zoology Botany 1. (c) 10. (c) 13. (a) 3. (b) 8. (c) 3. (a) 21. (a) 23. (d) 33. (c) 23. (a) 27. (a) 10. (d) 12. (d) Chemistry 1. (c) 37. (c) 26. (b) 20. (d) 40. (a) 13. (d) 5. (b) 17. (a) 36. (d) 38. (d) 31. (c) 33. (a) 32. (b) 30. (d) 22. (a) 14. (c) 18. (b) 25. (d) 31. (a) 17. (c) 5. (b) 6. (c) 9. (d) Hints & Solutions Physics 1. (c) 40. (c) 16. µ1 = 1.6 and A1 = 6° Now.14 I = 175.6 2. = = = 15 Hz 4l 2 2 3.4 A2 = 0. We knows the magnifying power of the astronomical telescope is f M = − o and length of the telescope L = fo + fe fe Given .9 m 1 1 1 Hence.729 B1 : B2 : B3 = 37.4. 8. the angle of the second prism is 4°. We know about the thin prism.8 × 10 −4 kgm 2 . Given.9)3 1 1 1 B1 : B2 : B3 = : : 0. V1 = 120 V and V2 = 200 V Here given potential is zero for each capacitor.37 No option is correct. 6 (1.3)3 (0. B1 : B2 : B3 = : : (0. Here the capacitance of dielectric capacitor C1 = 3 and C2 = 2 We know that the energy stored in the capacitor q2 E = 2C E1 q12 2C2 Hence. Given. n1 = 135 Hz and n2 = 165 Hz v 165 − 135 30 Now.236 0.4 A2 = 0.027 0. T=12 s and B= 0. A1 (µ1 − 1) = A2 (µ 2 − 1) Given. M = 120 Am 2 . δ m = (µ − 1) A In given condition. 2 × 10 −4 so that the answer is I = 172. 4. then C1V1 − C2V2 =0 C1 + C2 L = – 50 fe + fe C1V1 = C2V2 L = – 49 fe 120 V1 = 200 C2 L = – 49 × ( 2 ) 12 V1 = 20 C2 L = – 98 cm Hence. 5. 7.6 m and r3 = 90 cm = 0. I = or (12 )2 × (120 ) × ( 0. We know that the magnetic field induction at a point µ0 M 4π r 3 According to question B= B1 : B2 : B3 = Given 1 1 1 : : r13 r23 r33 r1 = 30 cm = 0. Here the condition for the formation of rings is d2 λ≈ 4D 6.5 × 10 −3 / ° C . µ 2 = 1.2 × 10 −4 kgm 2 19 3 C1 = 5 C2 9. M =50 and fe = 2 cm − fo Here.4 × 10 −4 ) I= 4 × 3. the length of the telescope In this question. 4 − 1) = A2 (1. the required length of the telescope is 98 cm.3 m r2 = 60 cm = 0.14 × 3.4 × 10 −4 T We know that. the option (b) is more equivalent to 175 . T = 2π I (squaring both sides) MB T 2 MB 4π 2 So.6 A2 = 4° Hence.03 : 4. = 50 fe − fo = 50 fe So. Given.629 : 1.EAMCET (Medical) l Solved Paper 2013 | Given.6)3 (0.6 − 1) 6 × 0. temperature coefficient of carbon α 3 = 4 × 10 −3 /°C Temperature coefficient of copper α 2 = − 0. = = E2 2C1 q 22 E1 C 2 = 2 = (Qq1 = q 2 − q ) E2 C1 3 10. p = pressure. From the conservation of momentum mv + m × 0 = (m + m)v′ mv ⇒ v′ = ( m + m) v v′ = 2 and we knows u2 h = 2g Here. 8 × cos 30 ° × 10 3 = 300 × 9 . R1α 1 = − R2 α 2 R2 −4 × 10 −3 = R1 0.6 ms −2 10 19.2 × 3 × 10 a1 = 3 14 a1 = ms −2 3 6 Similarly. Given that. 1 2 h = at 2 2 h 2 × 100 a = 2 = = 2 ms −2 t (10 )2 Now.5 × 10 −3 R2 8 = R1 1 11. We know that W = mg cos θ × 5 W = 300 × 9 .6 = 28. Velocity of water vw = 4 ms −1 Velocity of person vp = 5 ms −1 Width of the river = 84.20 | EAMCET (Medical) l Solved Paper 2013 Hence. We know that the difference between the maximum and minimum tensions in the rope Tmax − Tmin = 2 mg Here.6 s ≅ 28. h = 100 m and t = 10 s From equation of motion. kB = Boltzmann constant and T = temperature 15. 8 × × 10 2 = 2450 J F = 50 (10 + 2) F = 600 N rCM = ma $i + mb$j + m0 3m rCM = 1 $ (a i + b$j ) 3 17. y = distance. u = v′ (v′ )2 h = 2g ⇒ v2 8g 18. a1 = = 0. 20 − 0. So. 16. Given equation. mass of the store F = ma F 20 m= = = 1kg a 2 × 10 .6 t= 25 − 16 t= 84. We know that F − f = ma Here for small body. 1 y = pβ kBT where.25 3 13.6 m We know that time taken s t = vw2 − vp2 84. Given. The initial velocity of the body u 2 = 2 gh u = 2 gh = 2 × 10 × 125 = 2500 = 50 ms −1 The final velocity from equation of motion v 2 = u 2 + 2 gh v = (50 )2 + 2 × 10 × 45 v = 2500 + 900 = 3400 = 50 ms −1 14. Given. mass = m and here Dimension of [β ] [ Dimensions of kB ] [ Dimensions of T ] = [ Dimensions of p] [ Dimensions of y] = [ ML2 T −3 ][T ] = [ M0L2 T 0 ] [ ML−1 T −2 ][ L ] 12. Tmax − Tmin = 20 Now. EAMCET (Medical) l Solved Paper 2013 | 20.43 × 13. p1 r4 l = × p2 2l (2 r )4 22. v1 = ω A2 − r1 2 v 2 = ω A2 − r 2 2 Hence. the moment of inertia of the combine system about an axis passing through O is 4 ma2 + ma2 3 I = 4 39 ma2 + m 4 3 94 I= m a2 48 2 21.42 cm θ 2 = 135 and d 2 = 13. kA = k and kB = 2 k 1 So that WA = kx 2 2 1 and WB = (2 k ) x 2 = kx 2 2 21 p1 1 = p2 32 26. In given condition.006 × 10 −3 + 3 × (17 × 10 −6 ) α g = 27 × 10 −6 / ° C 27. l1 : l2 = 2 : 1 and r1 : r2 = 1 : 2 Let l1 = 2 l. h2 = 3. Hence. Here. I = p1 r4 l = 1 × 24 p2 l1 r2 Now. h1 = 10 cm. 1.03 × 10 γ r = γ ag + 3α g = γ ac + 3α c −3 + 3α g = 1. 2 g and g 90 = g 2 g = g − Rω 2 2 g ω = 2R r T = 2π v 2π T = ω So. the ratio of work done in stretch wire WA 1 2 = kx / kx 2 WB 2 WA 1 = WB 2 24. Here. We know that the work done in a stretched wire W= 1 2 kx 2 Given. g 0 = g 90 − Rω Given. We know that the pressure r4 l Given. l2 = l and r1 = r . v2 v2 2 2 − v12 − v12 4π 2 = 2 [ A2 − r1 2 − A2 + r2 2 ] T 4π 2 = 2 [r22 − r12 ] T 2 T =2 π (r2 − r12 ) (v2 2 − v12 ) 23.6 g/cc T1 h1 d1 cos θ 2 = T2 h2d 2 cos θ1 T1 10 × 1 cos 135° = T2 3. We know that (i) The work done in isobaric process is p(V2 − V1 ) (ii) The work done in isothermal process is V nR log 2 V1 (iii) The work done in adiabatic process is nR (T1 − T2 ) γ −1 . Given. then γ = 2α 28. g0 = We have. We know that. The coefficient of linear expansion (α ) of the material of the tube and the coefficient of volume expansion (γ ) of the liquid. for SHM v = ω A2 − r 2 p2 1 = p2 2 × (2 )4 In given condition. Here. r2 = 2 r p∝ v Q ω = r 2R T =2 π g Hence.6 cos 45° T1 2 = T2 13 25. 2 × 1018 × 185 × 10 6 eV P= 7 × 1015 P = 59. β = c ∆ Ib Given.32 × 1. 35. From the equation. The purpose of using heavy water in nuclear reactor is to decrease the energy of fast neutrons to the thermal energy. 4 × 30 = 42 K 30. R = 3 π tan θ = Given. mv 2qV and v = r= Bq m mv Now.32 × 1.6 µA nE P= t 33. B = qr 31. Given. β = 60 and ∆ Ic = 4 mA = 4 × 10 −6 A ∆I 4 × 10 −6 ∆ Ib = c = β 60 ∆ I b = 66.5 15 = = 1A 15 .32 keV = 23. Given. = = = E1 t n − tc 120 120 Percentage decrease in thermo emf E2 − E1 94 − 120 × 100 = × 100 = 21. Now. Here. For making a junction transistor the emitter E is Here. Ek = 23. 37.01 H.6 E1 120 40.6 × 10 αTB = Y ∝ TA = 1.4. ∆I 32. l = 25 and π = 10 cm θ 100 − 0 = 10 25 θ =4 10 θ = 40 ° C heavily doped.5+ 14. dθ = nCpdTA = nCVαTB nCpαTA = nCVαTB Cp αTB = CV αTA γ= EL − 23. We know.82 keV Cp Q γ = CV dTB dTA EL − EK = B= m 2qV qr m B = 2 mV qr 2 B= 2 mV er 2 (∴q = e ) 38. the base B is thin and lightly doped and the collector C is moderately doped semiconductor materials and with together. −1 hc hka EL = 5. t 2 = 0 ° C.01 tan θ = 3π 1 tan θ = 3 θ = π / 6 rad E2 ( t n − 20 ) − ( tc + 6) 120 − 26 94 39. the values of the screening constant for K-series and L-series of X-rays are respectively 1 and 6.2 MeV 34.22 | EAMCET (Medical) l Solved Paper 2013 29. n = 2 × 10 18 fission per second E = 185 MeV Here. ωL R L= 0.42 × 10 −7 71 × 10 −12 36. We know that. In Moseley's law v = a ( z − b). We know that. θ ∆t = x l Given.6 × 10 −19 eV −7 hc = 12. V = 220 V and f= 50 Hz ω = 2 πf ω = 50 × 2 π 50 × 2 π × 0.42 × 10 eV λ hα = 71 × 10 −12 m = 12. t1 = 100 ° C. We know that. We know that i = E 15 = r + Ref 0. Glucose on reduction with HI/red P at 373 K gives n hexane and 2-iodo hexane. which indicates that glucose contains one primary alcoholic group. 2. . n HO—CH2—CH2—OH ethylene glycol O O + n HO—C— —C—OH terephthalic acid – (2n – 1) H2O O O —OCH2—CH2—O—C— —C— 4. (b) Antimicrobials are commonly used in food industry to ensure safe wholoesome food products.EAMCET (Medical) l Solved Paper 2013 | 23 Chemistry 1. NH 3 . on oxidation with HNO 3 glucose gives saccharic acid (glycaric acid). CH3 (CH2)4 CHO HI/P Reduction (CHOH)4 COOH HNO3 Oxidation (CHOH)4 CH3 CH2OH COOH n-hexane glucose saccharic acid 3. As a result the electron density on the N-atom increases and thus they can donate the lone-pair of electrons more easily than NH 3 . (d) Chloroxylenol is an antiseptic. The reason behind it is the resonance in aniline. This is due to the reason that alkyl groups are electron-donating groups. Traditionally. nylon 66 is obtained by cndensation polymerisation of hexamethylenediamine and adipic acid with the loss of one water molecule. C 6H 5NH 2 < NH 3 < C 2H 5NH 2 < (C 2H 5 )2 NH. etc. For example. which suggests that all the six carbon atoms in glucose are arranged in a straight chain. water. HCl. Thus. which is a weak organic acid and a natural food preservative has been widely used to control growth of pathogenic bacteria in food. Aliphatic amines are stronger bases than ammonia. the basicity of amines increases in the order. due to which the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus is less easily available for protonaton. Whereas. •• •• H — N — H C 2H 5 →− N — H C 2H 5 →− N −←C 2H 5 H H H ammonia 2 ° amine 1° amine But aniline is weaker base than NH 3 . (c) Seconal is a tranquillizer. lactic acid. V IV Thus. CO 2 . NH2 + + NH2 NH2 s I s II III NH2 NH2 nH 2N — (CH 2 )6 — NH 2 + nHOOC(CH 2 )4 COOH hexamethylenediamine adipic acid ↓ + s O O —[NH—(CH 2 )6 —NH—C —(CH 2 )4 —C —]n nylon-66 + (2 n − 1)H 2O Similarly. (a) Sucralose is an artificial sweetening agent. Condensation polymers are formed by repeated condensaton reaction between two bifunctional or trifunctional monomer units accompanied with the elimination of small molecules like. terylene or dacron is a condensation polymer of ethylene glycol and terephthalic acid with elimination of water. The function of CuCl 2 is to reoxidise Pd to Pd(II) and itself is reoxidised to Cu(II) from Cu(I) by air so that O 2 is the only oxidising agent actually used up in this process. The product formed is chloropicrin or trichloronitromethane or nitro chloroform. It is the active cancer-producing agent in coal-tar and in the tar produced during the incomplete combustion of organic materials such as coal. S-system. Ammonium salts. The hydrogen of the chloroform is replaced by chloroform 4 HO—C—COOH O2 ethylene CH(OH)CH3 Here we hold the group of 3rd priority and rotate in triangular way be keeping —‘OH’ at place of —‘H’. 2 H—C—OH 3 H—C—OH 4 1 CH3 2 1. Fischer projection It is three directional representation of optically active compounds and also said to be R-.g. 1 The priority at C2 as H—C—OH I 2 –Rh catalyst 3 CH 3OH + CO → CH 3COOH acetic acid methanol 6. therefore the reaction is carried out in the presence of a cooxidant CuCl 2 . etc. CHCl 3 + Conc HONO 2 → CNO 2Cl 3 chloropicrin (tear gas) + H 2O 9.24 | EAMCET (Medical) l Solved Paper 2013 2 5. poisonous and used as an insecticide and a war gas. (NH 4 )2 SO 4 . nitro group. It is a liquid. when it is treated with concentrated nitric acid.. 10. Acetic acid is produced industrially by the carbonylation of methanol in the presence of Rh/I2 catalyst. CuCl H 2 3 CH 3CHO + Pd + 2HCl acetaldehyde During this reaction. —H at place of —‘COOH’ and —‘COOH’ at place of —‘OH’ to get structure as 1 2 CH 2 ==CH 2 + H 2O + PdCl 2 → COOH 4 2 CH(OH)COOH 3 1 HO—C—CH3 CH4 It is R configuration at C 3 . PdCl 2 is reduced to metallic Pd. Pd + 2CuCl 2 → PdCl 2 + 2CuCl 4CuCl + 4HCl + O 2 → 4CuCl 2 + 2H 2O 7. (NH 4 )3 PO 4 (but not NH 4Cl) act as food for yeast cells. 1 COOH CH(OH)CH3 as eye rotates. tobacoo.2-benzpyrene is a polynuclear aromatic hydrocarbon and one of the most potent carcinogenic compounds.2-benzypyrene . then it is R configuration at C 2 . clockwise from (1) to (3) via (2). 1. 2 The priority at C3 4 CH(OH)COOH 1 H—C—OH 3 CH3 As we hold the group of 2nd priority and rotate in triangular manner to get the following structure 8. petroleum. e. Wacker process Alkenes can be easily oxidised to their corresponding aldehydes and ketones by treating them with a solution of PdCl 2 containing a catalytic amount of CuCl 2 in presence of O 2 . CuSO 4 + 5H 2O → CuSO 4 ⋅ 5H 2O (2) XeF4 sp3d 2 (3) XeF2 sp3d Ca(OH)2 + CO 2 → CaCO 3 + H 2O milky 14. Although CO 2 is present in a much higher concentration than NO and SO 2 . Due to the smaller size of Lu. due to the natural presence of CO 2 . however. NO and SO 2 . (A) Pure wate has a pH of 7. F2 oxidises potassium bisulphate to potassium persulphate. K p = Kc (RT )∆ng For the reaction. 6 5 4-ethyl-3. It is toxic and a strong irritant for the skin. lime -stone 1000°C → CaO + CO 2 square planar linear Xe 2 3 Xe O O O 13.082 × 300) . The detection of carbon and hydrogen in an organic compound is done with the help of CuO test. During the manufacture of cast iron. (C) Acrolein enters the air from the burning of fossil fuels and tobacco smoke. it does not form acid to the same extent as the two other gases.0 (neutral).033 mol L−1 1 30 (0. natural unpolluted rain water has a pH of about 5.3-dimethyl hexane 15. (D) Freons (CF2Cl 2 ) are chlorofluoro carbons and has been completely banned because of being destructive for ozone layer. Number of lone pair of electrons tetrahedral 0 O hydrated copper sulphate (blue) CaCO 3 25 3 XeO4 (sp ) [Tetrahedral structure] F F Xe F O O O 3 XeO3 (sp ) [Pyramidal or T-shaped structure] F Xe F 3 2 XeF4 (sp d ) [Square planar structure] F 3 XeF2 (sp d) [Linear structure] 17. CaO + SiO 2 → CaSiO 3 1 4 2 11. 16. The correct set are Molecule Hybridisation Shape ∆ + 2CuO → CO 2 + 2Cu ∆ H of organic compound + CuO → H 2O + Cu (1) XeO 4 sp3 H 2O vapours turn anhydrous CuSO 4 blue and CO 2 evolved turn lime water milky as per following reactions. C of organic compound decreases from La(OH)3 to Lu(OH)3 . eyes and nasal passages.EAMCET (Medical) l Solved Paper 2013 | 3 CaO act as a flux and combines with silica present as an impurity to form a fusible slag of CaSiO 3 . A(g ) º B(g ) + C(g ) ∆ng = gaseous products – gaseous reactants = 2 − 1= 1 Kp ∴ Kc = ∆n (RT ) g 0. the Lu—OH bond aquires more covalent character. because the lime-stone present in charge needs a higher temperature to decompose into calcium oxide and CO 2 .82 1 = = = 0. (B) Acid rain is the rain water containing H 2SO 4 and HNO 3 which are formed from the oxides of S and N present in the air and has a pH of 4-5. The basic character of hydroxides of lanthanoids 12. F2 + 2KHSO 4 → K 2S 2O 8 + 2HF 18.6 (acidic). the slag is formed in slag formation zone which is also known as zone of heat absorption. (a) Orthoboric acid (H 3BO 3 ) is used as an + 2F− 2s s antiseptic and eye wash under the name ‘boric lotion’. With hot and concentrated NaOH. (b) All the elements of group V except Bi show allotropy. with the [SiCl 6 ]2 − . Zn + 2NaOH → Na 2 ZnO 2 + H 2 ↑ sodium zincate 23. Ag 2O + H 2O 2 → 2Ag + H 2O + O 2 (c) H 2O 2 is a weak acid and show neutralization reactions with hydroxides. (c) Thermal stability of hydrides of group V decreases gradually from NH 3 to BiH 3 due to decrease in M—H bond strength down the group on account of increase in size of central atom. Zn forms sodium zincate and H 2 gas is liberated. MgF2 s º Mg 2+ 24. due to the availability of the vacant d-orbitals. Rb 2O is bright yellow and Cs 2O is orange. β-black and violet phosphorus.26 | EAMCET (Medical) l Solved Paper 2013 19. [GeF6 ]2 − . ∆Hsub = 54 kJ …(ii) Eq. Li 2O and Na 2O are white solids as expected but surprisingly K 2O is pale yellow. This is due to the gradually increased ionic character from Li 2O to Cs 2O. KNO 2 + H 2O 2 → KNO 3 potassium nitrite 2p B atom (Ground state) potassium nitrate + H 2O 22. Na 2B4O 7 + 7H 2O º 2NaOH + 4H 3BO 3 strong alkali weak acid 25. (b) In B2H 6 . (a) The solubility of group V hydrides in water decreases from NH 3 to BiH 3 . Sodium thiosulphate decomposes on treatment with dilute HCl with the evolution of SO 2 and precipitation of sulphur. 27. H 2O 2 + Cl 2 → 2HCl + O 2 (b) H 2O 2 reduces silver oxide to silver in the alkaline medium. 26. The property of accepting electron pair from donor makes the tetrahalides of group 14 elements (except carbon halides) to act as strong Lewis acids. [PbCl 6 ]2 − corresponding halide ions by increasing their coordination number from 4 to 6. both B atoms undergo sp3 -hybridisation. (ii) is multiplied by 2 and then divided by Eq. The tetrahalides of Si. The reaction given : 4NO 2 (g ) + O 2 (g ) → 2N 2O 5 (g ). NH 3 is highly soluble in water due to H-bonding but other hydrides are less soluble. Ksp = (2 s)2 (s) = 4s3 ∴ s3 = Ksp 4 = 1. α-black. 2NaOH + H 2O 2 → Na 2O 2 + H 2O sodium peroxide (d) H 2O 2 oxidises nitrites to nitrates in alkaline medium. ∆H = − 111kJ …(i) N 2O 5 (s) → N 2O 5 (g ).08 × 10 −10 4 2s = 27 × 10 −12 s = 3 27 × 10 −12 = 3 × 10 −4 mol L−1 B atom (Excited state) 20. Na 2S 2O 3 + 2HCl → 2NaCl + SO 2 ↑ sodium thiosulphate [X] + S ↓ + H 2O [Y ] . (i) gives 4NO 2 (g ) + O 2 (g ) → 2N 2O 5 (s) ∴ sp3-hybridisation (c) The aqueous solution of borax is weakly alkaline in nature due to hydrolysis of B4O 2– 7 ions. Ge. (a) H 2O 2 reduces chlorine to HCl. red. scarlet. [SnCl 6 ]2 − . Phosphorus exists in a number of allotropic forms such as white. Sn and Pb form ∆H = − 111 − (54 × 2 ) = − 219 kJ 21. hexahalocomplexes like [SiF6 ]2 − . O2 molecule The electronic configuration of O is 1s2 . π 2 py2 = 2 pz2 . π * 2 py = π * 2 pz MO configuration of N2 σ1s2 .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. when reacts with Na 2S. i. we are given. σ2 s. σ2 p x .14 × 58 × 1. σ2 p2x . σ * 2 s2 . σ * 1s2 . N2 molecule The electronic configuration of N is 1s2 .0 × 10 −6 ms −1 According to Heisenberg’s uncertainty principle h ∆x ⋅ (m∆v ) = 4π h i. 29. 30.626 × 10 −34 4 × 3. π * 2 p x MO configuration of O2 σ1s2 . 2 s2 2 p3 . the electron will be detached from the metal atom only if the energy of the photon (4.626 × 10 eV 1. σ2 s2 . σ * 1s. m= 4π ⋅ ∆x ⋅ ∆v = the electrons of all the shells are pulled little closer to the nucleus thereby making each individual shell smaller and smaller.e. σ2 p x . ∴ The total number of antibonding electrons = 6. π 2 py2 = π 2 pz2 . form sodium thiosulphate and sulphur. As 35. σ * 2 s. Na.. 3SO 2 + 2Na 2S → 2Na 2S 2O 3 + S↓ [ Y ] sulphur 28. 5I (aq ) + I O −3 (aq ) + 6H + (aq ) → 3I2 (aq ) +5 0 Reduction +3H 2O(l ) 34. TA MB = TB MA ∴ TA T = B MA MA RMS velocity = (vRMS ) = 3RT M TA T = B is equal for both the gases. Li > Be > B 31. −19 27 The MO formed during the combination of atomic orbitals of two N atoms are σ1s. the mass of moving object is approximately equal to that of electron.626 × 10 −34 Js)(3. This results in a decrease of the atomic radius as we move from left to right in a period.0 × 10 −6 = 9. π 2 py = π 2 pz .626 × 10 −34 Js ∆v = 1.. Energy of a photon of radiation of wavelength 300 nm will be c λ (6. σ * 1s.e. As we move from left to right in a period. p° − p s n2 = p° n1 + n2 Let p° = vapour pressure of pure water = 100 mm .626 × 10 −19 J = 6. σ2 s. Oxidation –1 – – 0 33. Due to this enhanced nuclear charge 2π σ H—C ≡≡C—H σ 32. σ * 2 s. Here. Thus.6 × 10 −19 J] = 4. K. π * 2 p1y = π * 2 p1z .0 × 10 8 ms −1 ) = (300 × 10 −9 m) E = hν = h = 6. π * 2 py = π * 2 pz .14 eV Thus. these 4 metal atoms will show photoelectric effect when light of 300 nm wavelength falls on them. nuclear charge increases by one unit in each succeeding element while the number of the shells remains the same. σ * 1s2 . 2 s2 2 p4 . σ2 py = σ2 pz . Structure of acetylene σ 6. σ2 p2x ∴ The total number of antibonding electrons = 4. ∆x = 58 m h = 6.6 × 10 −19 [As 1 eV = 1. σ2 s2 . σ * 2 s2 . Mg). According to Raoult’s law. π * 2 p x .09 × 10 −31 kg Thus.EAMCET (Medical) l Solved Paper 2013 | SO 2 . thus MA MB both have equal RMS velocity. The MO formed during the combination of atomic orbitals of two O-atoms are σ1s. 1 ~ − 6.024 = 101.024° C Hence.28 | EAMCET (Medical) l Solved Paper 2013 ∴ p° − p s = Lowering in vapour pressure on dissolving solute = 1.5 × 10 −3 s −1 = 6. Temperature coefficient. K b = 0. boiling point of solution (Tb ) = Tb ° + ∆Tb = 100 + 1. the formula of the compound = A4 B3 .0 g. Λ0 for BaCl 2 = X1 1. 3 3 4N 4 ∴ Ratio of A : B = : N = :1= 4: 3 3 3 Hence. As number of tetrahedral voids is double the number of atoms in the close packing. the number of atoms A in the lattice 2 4N = × 2N = . 1 mol = 27 g. 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. atoms A occupy 2/3 rd of the tetrahedral voids. rate constant at T + 10 ° C µ= rate constant at T ° C T2 (T2 > T1 ) T1 or µ= or T2 = µ × T1 = 2. gives w2 = 6. the reaction is. or w1 100 = mol M1 18 Λ0 for HCl = X3 = Λ (BaCl 2 ) + Λ0 (H 2SO 4 ) − 2 Λ0 (HCl) = X1 + X2 − 2 X3 39.09 = 965 C 27 For deposition of Cu.8 mm w w n2 = moles of solute = 2 = 2 mol M2 60 n1 = moles of solvent = w2 1. Here. Cu 2 + + 2e − → Cu 1000 × 0. we have ∴ = 0. 40. Al is deposited by 3F = 3 × 96500 C ∴ 0.512 K kg mol Thus. Let the number of atoms B in hcp lattice = N Tb ° = 100 ° C −1 1000 × K b × w2 w1 × M2 = Λ0 for BaSO 4 = λ0Ba 2+ + λ0SO 2– 4 w2 = 68. 2F = 2 × 96500 C deposit Cu = 1mol = 63.8 100 w2 w2 + = 100 18 60 60 w1 = 100 g.024° C 37. reaction is Al 3 + + 3 e − → Al Thus.8 60 = 100 100 + w2 18 60 Hence.4 g.6 × 965 ∴ 965 C will deposit Cu = 2 × 96500 38.5 × 2.25 × 10 −3 s −1 .09 g of Al will be deposited by 3 × 96500 = × 0.6 g 63.4 100 × 342 = 1. thus number of tetrahedral voids = 2N Since.512 × 68. M2 = 342 g mol −1 ∆Tb = By Kohlrausch’s law 0 36.318 g Λ0 for H 2SO 4 = X2 1 3w2 w or + = 2 10 10000 60 1 w2 3w2 or = = 10 60 10000 This. For deposition of Al.