NEET Previous Year Question Paper 2015 PDF

 

NATIONAL ELIGIBILITY CUM ENTRANCE TEST  
Phase-2 (Code:BB-QQ-XX)  
Answers & Solutions

PART A – CHEMISTRY.

NEET 2015 Chemistry Previous Year Question Paper

1. A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO₂ to give an unstable compound B. B, on treatment with phenol, forms a beautiful coloured compound C with the molecular formula C₁₂H₁₀N₂O. The structure of compound A is :  
(1)  
(2)  
(3)  
(4)  

Ans. (3)  

Sol.  
NO₂ → (Sn/HCl) → NH₂ → (HNO₂) → N₂⁺Cl⁻ → (Phenol) → N=N–OH (C₁₂H₁₀N₂O)

2. Consider the reaction :  
CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr  
This reaction will be the fastest in :  
(1) water  
(2) ethanol  
(3) methanol  
(4) N,N'-dimethylformamide (DMF)  

Ans. (4)  

Sol.  
H–C–N(CH₃)₂ (Polar aprotic solvent)  
This is a SN2 Reaction for which polar aprotic medium is suitable for faster rate of reaction.

3. The correct structure of the product A formed in the reaction:  
O (cyclohexenone) + H₂ (gas, 1 atmosphere) / Pd / carbon, ethanol → A  
is :  
(1)  
(2)  
(3)  
(4)  

Ans. (3)  

Sol.  
During hydrogenation of α,β unsaturated carbonyl compound by pd catalyst selective reduction is observed of double bond.

4. Which among the given molecules can exhibit tautomerism?  
I  
II  
III  
(1) Both II and III  
(2) III only  
(3) Both I and III  
(4) Both I and II  

Ans. (2)  

Sol.  
Only III  
α-H at bridge head carbon never show tautomerism.

5. The correct order of strengths of the carboxylic acids :  
I  
II  
III  
is :  
(1) II > I > III  
(2) I > II > III  
(3) II > III > I  
(4) III > II > I  

Ans. (3)  

Sol.  
(II > III > I)  
Acidic strength α –I, –M effect  
–I effect depend upon distance so II have stronger –I effect than III.

6. The compound that will react most readily with gaseous bromine has the formula :  
(1) C₂H₄  
(2) C₃H₆  
(3) C₂H₂  
(4) C₄H₁₀  

Ans. (4)  

Sol.  
Gaseous Bromine react fastest with alkane by free radical mechanism.  
C₄H₁₀ → (Br₂) → C₄H₉Br (free Radical substitution Reaction)

7. Which one of the following compounds shows the presence of intramolecular hydrogen bond ?  
(1) Concentrated acetic acid  
(2) H₂O₂  
(3) HCN  
(4) Cellulose  

Ans. (4)  

Sol.  
Cellulose is example of intramolecular H-bonding

8. The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 × 10⁻³ S cm⁻¹ at 298 K is  
(1) 28.8 S cm²/mol  
(2) 2.88 S cm²/mol  
(3) 11.52 S cm²/mol  
(4) 0.086 S cm²/mol  

Ans. (3)  

Sol.  
λₘ = (k × 1000)/M = (5.76 × 10⁻³ × 1000)/0.5 = 11.52 S cm² mol⁻¹.

9. The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the  
(1) Rate of decomposition is very slow  
(2) Rate is proportional to the surface coverage  
(3) Rate is inversely proportional to the surface coverage  
(4) Rate is independent of the surface coverage  

Ans. (4)  

Sol.  
PH₃ → (W) → P + (3/2)H₂  
Rate = k[PH₃].  
It is independent of the surface coverage because zero order reaction depend on surface area covered by reactant.

10. The coagulation values in millimoles per litre of the electrolytes used for the coagulation of As₂S₃ are given below :  
I. (NaCl) = 52, II. (BaCl₂) = 0.69, III. (MgSO₄) = 0.22  
The correct order of their coagulating power is  
(1) III > I > II  
(2) I > II > III  
(3) II > I > III  
(4) III > II > I  

Ans. (4)  

Sol.  
Coagulation power ∝ 1/Coagulation value  
Higher the coagulation power, lower is coagulation values in millimoles per litre.  
MgSO₄ > BaCl₂ > NaCl.

11. During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is  
(1) 330 minutes  
(2) 55 minutes  
(3) 110 minutes  
(4) 220 minutes  

Ans. (3)  

Sol.  
At anode : 2Cl⁻ → Cl₂ + 2e⁻  
0.1 × 2 = (3 × t(sec))/96500  
t = 6433 sec ≈ 107.2 min ≈ 110 min

12. How many electrons can fit in the orbital for which n = 3 and ℓ = 1 ?  
(1) 14  
(2) 2  
(3) 6  
(4) 10  

Ans. (2)  

Sol.  
n = 3, ℓ = 1  
3p orbital can have only 2 electron.

13. For a sample of perfect gas when its pressure is changed isothermally from pᵢ to pբ, the entropy change is given by  
(1) ΔS = RT ln(pᵢ/pբ)  
(2) ΔS = nR ln(pբ/pᵢ)  
(3) ΔS = nR ln(pᵢ/pբ)  
(4) ΔS = nRT ln(pբ/pᵢ)  

Ans. (3)  

Sol.  
ΔS_sys = nR ln(P₁/P₂) + nCp ln(T₂/T₁)  
In isothermal process T₁ = T₂  
ΔS_sys = nR ln(Pᵢ/Pբ)

14. The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is  
(1) 3  
(2) 0  
(3) 1  
(4) 2  

Ans. (1)  

Sol.  
Ba(OH)₂ → Ba²⁺ + 2OH⁻  
i = 3.

15. The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅N⁺H) in a 0.10 M aqueous pyridine solution (Kբ for C₅H₅N = 1.7 × 10⁻⁹) is  
(1) 1.6 %  
(2) 0.0060 %  
(3) 0.013 %  
(4) 0.77 %  

Ans. (3)  

Sol.  
C₅H₅N + H₂O ⇌ C₅H₅N⁺H + OH⁻  
α = √(Kբ/c) = √(1.7 × 10⁻⁹ / 0.1) = 1.3 × 10⁻⁴  
% = 1.3 × 10⁻⁴ × 100 = 0.013

16. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca²⁺) and fluoride ion (F⁻) are  
(1) 4 and 8  
(2) 4 and 2  
(3) 6 and 6  
(4) 8 and 4  

Ans. (4)  

Sol.  
Ca²⁺ is surrounded by 8F⁻  
F⁻ is surrounded by 4Ca²⁺

17. If the E°_Cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG° and K_eq ?  
(1) ΔG° < 0 ; K_eq < 1  
(2) ΔG° > 0; K_eq < 1  
(3) ΔG° > 0; K_eq > 1  
(4) ΔG° < 0; K_eq > 1  

Ans. (2)  

Sol.  
E°_cell < 0, so it is a non spontaneous process  
ΔG° = –nFE° = +ve, so ΔG° > 0  
ΔG° = –2.303RT log K  
So, K < 1

18. Which one of the following is incorrect for ideal solution ?  
(1) ΔG_mix = 0  
(2) ΔH_mix = 0  
(3) ΔU_mix = 0  
(4) ΔP = P_obs – P_calculated by Raoult's law = 0  

Ans. (1)  

Sol.  
For ideal solution inter molecular forces are identical so,  
ΔH_mix = 0, ΔV_mix = 0, ΔG_mix < 0  
So 1st option is incorrect.

19. The solubility of AgCl(s) with solubility product 1.6 × 10⁻¹⁰ is in 0.1 M NaCl solution would be  
(1) Zero  
(2) 1.26 × 10⁻⁵ M  
(3) 1.6 × 10⁻⁹ M  
(4) 1.6 × 10⁻¹¹ M  

Ans. (3)  

Sol.  
AgCl ⇌ Ag⁺ + Cl⁻  
K_sp = [Ag⁺][Cl⁻]  
1.6 × 10⁻¹⁰ = S × 0.1  
S = 1.6 × 10⁻⁹ M

20. Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weights 10 g and 0.05 mole of X₃Y₂ weights 9 g, the atomic weights of X and Y are  
(1) 30, 20  
(2) 40, 30  
(3) 60, 40  
(4) 20, 30  

Ans. (2)  

Sol.  
For XY₂ : X + 2Y = 100  
For X₃Y₂ : 3X + 2Y = 180  
X = 40, Y = 30

21. The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 × 10⁻¹⁹ C)  
(1) 7.48 × 10²³  
(2) 6 × 10²³  
(3) 6 × 10²⁰  
(4) 3.75 × 10²⁰  

Ans. (4)  

Sol.  
W/E = (1 × 60)/96500 = 6/9650 = no. of mole e⁻  
no. of e⁻ = (6/9650) × 6.02 × 10²³ = 3.75 × 10²⁰

22. Boric acid is an acid because its molecule  
(1) Combines with proton from water molecule  
(2) Contains replaceable H⁺ ion  
(3) Gives up a proton  
(4) Accepts OH⁻ from water releasing proton  

Ans. (4)  

Sol.  
H₃BO₃ + H₂O ⇌ B(OH)₄⁻ + H⁺  
H₃BO₃ is Lewis acid and accept OH⁻ from H₂O and releases H⁺.

23. AlF₃ is soluble in HF only in presence of KF. It is due to the formation of  
(1) K[AlF₃H]  
(2) K₃[AlF₃H₃]  
(3) K₃[AlF₆]  
(4) AlH₃  

Ans. (3)  

Sol.  
AlF₃ + KF → (HF) → K₃[AlF₆]  
(maximum C.N. of Al³⁺ is six so it form AlF₆³⁻).

24. Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because  
(1) Zinc has higher negative electrode potential than iron  
(2) Zinc is lighter than iron  
(3) Zinc has lower melting point than iron  
(4) Zinc has lower negative electrode potential than iron  

Ans. (1)  

Sol.  
E°_Zn²⁺/Zn = –0.76 V  
E°_Fe²⁺/Fe = –0.44 V  
Zn has higher negative SRP so it work as anode and protect iron to make iron as cathode.

25. The suspension of slaked lime in water is known as  
(1) Aqueous solution of slaked lime  
(2) Limewater  
(3) Quicklime  
(4) Milk of lime  

Ans. (4)  

Sol.  
Suspension of slaked lime is called milk of lime.

26. The hybridizations of atomic orbitals of nitrogen in NO₂⁺, NO₃⁻ and NH₄⁺ respectively are  
(1) sp², sp and sp³  
(2) sp, sp³ and sp²  
(3) sp², sp³ and sp  
(4) sp, sp² and sp³  

Ans. (4)  

Sol.  
O=N=O (sp)  
NO₃⁻ (sp²)  
NH₄⁺ (sp³)

27. Which of the following fluoro-compounds is most likely to behave as a Lewis base ?  
(1) SiF₄  
(2) BF₃  
(3) PF₃  
(4) CF₄  

Ans. (3)  

Sol.  
Lewis base → ℓ.p donor  
PF₃

28. Which of the following pairs of ions is isoelectronic and isostructural ?  
(1) ClO₃⁻, SO₃²⁻  
(2) CO₃²⁻, NO₃⁻  
(3) ClO₃⁻, CO₃²⁻  
(4) SO₃²⁻, CO₃²⁻  

Ans. (2)  

Sol.  
Both trigonal planar and isoelectronic

29. In context with beryllium, which one of the following statements is incorrect ?  
(1) Its hydride is electron-deficient and polymeric.  
(2) It is rendered passive by nitric acid  
(3) It forms Be₂C  
(4) Its salts rarely hydrolyze.  

Ans. (2)  

Sol.  
Al, Cr are having passive nature with HNO₃ but Be dissolve in conc. HNO₃

30. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behaviour ?  
(1) CaF₂ + H₂SO₄ → CaSO₄ + 2HF  
(2) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O  
(3) 2S + 2H₂SO₄ → 2SO₂ + 2H₂O  
(4) C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O  

Ans. (1)  

Sol.  
1st reaction is not a redox reaction as the oxidation number of elements remains unchanged.

31. Which of the following pairs of d-orbitals will have electron density along the axis ?  
(1) d_xy , d_x²–y²  
(2) d_z² , d_xz  
(3) d_xz , d_yz  
(4) d_z² , d_x²–y²  

Ans. (4)  

Sol.  
d_z² and d_x²–y²

32. The correct geometry and hybridization for XeF₄ are -  
(1) square planar, sp³d²  
(2) Octahedral, sp³d²  
(3) trigonal bipyramidal, sp³d  
(4) planar triangle, sp³d³  

Ans. (2)  

Sol.  
Geometry → octahedral hybridization → sp³d²

33. Among the following, which one is a wrong statement?  
(1) I₃⁺ has bent geometry.  
(2) PH₅ and BiCl₅ do not exist.  
(3) pπ-dπ bonds are present in SO₂  
(4) SeF₄ and CH₄ have same shape.  

Ans. (4)  

Sol.  
SeF₄ (See saw) and CH₄ (Tetrahedral) – Not same shape

34. The correct increasing order of trans-effect of the following species is  
(1) CN⁻ > Br⁻ > C₆H₅⁻ > NH₃  
(2) NH₃ > CN⁻ > Br⁻ > C₆H₅⁻  
(3) CN⁻ > C₆H₅⁻ > Br⁻ > NH₃  
(4) Br⁻ > CN⁻ > NH₃ > C₆H₅⁻  

Ans. (3)  

Sol.  
Trans effect : F⁻ < NH₃ < Cl⁻ < Br⁻ < Ph⁻ < CH₃⁻ < CN⁻

35. Which one of the following statements related to lanthanons is incorrect ?  
(1) Ce (+4) solutions are widely used as oxidizing agent in volumetric analysis.  
(2) Europium shows +2 oxidation state.  
(3) The basicity decreases as the ionic radius decreases from Pr to Lu  
(4) All the lanthanons are much more reactive than aluminium  

Ans. (3)  

Sol.  
Ce⁴⁺ is strong oxidising agent and easily convert into Ce³⁺  
Eu²⁺ exist and behave as reducing agent lanthanons are much more reactive than aluminum.  
Lanthanoids are basic in nature and their acidity is three.

36. Jahn-Teller effect is not observed in high spin complexes of  
(1) d⁹  
(2) d⁷  
(3) d⁸  
(4) d⁴  

Ans. (3)  

Sol.  
Jahn teller effect : geometric distortion occur in unsymmetrical octahedral complexes  
d⁸ is symmetrical

37. Which of the following can be used as the halide component for Friedel–Crafts reaction?  
(1) Isopropyl chloride  
(2) Chlorobenzene  
(3) Bromobenzene  
(4) Chloroethene  

Ans. (1)  

Sol.  
CH₂=CH–Cl, Chlorobenzene, Bromobenzene not suitable. Isopropyl chloride is suitable.

38. In which of the following molecules, all atoms are coplanar ?  
(1)  
(2)  
(3)  
(4)  

Ans. (2)  

Sol.  
Biphenyl  
All carbon atom is sp² hybridised and its geometry is trigonal planar.

39. Which one of the following structures represent nylon 6,6 polymer ?  
(1)  
(2)  
(3)  
(4)  

Ans. (1)  

Sol.  
Nylon-66 → adipic acid + Hexamethylenediamine

40. In pyrrole the electron density is maximum on  
(1) 2 and 5  
(2) 2 and 3  
(3) 3 and 4  
(4) 2 and 4  

Ans. (4)  

Sol.  
Electron density is maximum on-2,4th carbon.

41. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction ?  
(1)  
(2)  
(3)  
(4)  

Ans. (4)  

Sol.  
CH₂=C=O + HBr → No reaction.

42. Which one of the following nitro-compounds does not react with nitrous acid  
(1)  
(2)  
(3)  
(4)  

Ans. (4)  

Sol.  
1° and 2° nitro compound react with HNO₂ but 3° nitro compound does not react with nitrous acid

43. the central dogma of molecular genetics states that the genetic information flows from  
(1) DNA → RNA → Carbohydrates  
(2) Amino acids → Proteins → DNA  
(3) DNA → Carbohydrates → Proteins  
(4) DNA → RNA → Proteins  

Ans. (4)  

Sol.  
DNA → RNA → Protein

44. The correct corresponding order of names of four aldoses with configuration given below  
respectively, is  
(1) D-erythrose, D-threose, L-erythrose, L-threose  
(2) L-erythrose, L-threose, L-erythrose, D-threose  
(3) D-threose, D-erythrose, L-threose, L-erythrose  
(4) L-erythrose, L-threose, D-erythrose, D-threose  

Ans. (1)  

Sol.  
D-erythrose, D-threose, L-erythrose, L-threose

45. In the given reaction  
+ → (HF, 0°C) → P  
the product P is  
(1)  
(2)  
(3)  
(4)  

Ans. (4)  

Sol.  
This is a Friedel – Craft reaction.

PART B – BIOLOGY

NEET 2015 Biology Previous Year Question Paper

46. A foreign DNA and plasmid cut by the same restriction endonuclease can be joined to form a recombinant plasmid using  
(1) ligase  
(2) Eco RI  
(3) Taq polymerase  
(4) polymerase II  

Ans. (1)  

Sol.  
Ligase are the enzymes used to join substrates. Here in case of DNA T₄ DNA ligase is used.

47. Which of the following is not a component of downstream processing?  
(1) Expression  
(2) Separation  
(3) Purification  
(4) Preservation  

Ans. (1)  

Sol.  
Expression of recombinant DNA is parts of upstream processing.

48. Which of the following restriction enzymes produces blunt ends?  
(1) Hind III  
(2) Sal I  
(3) Eco RV  
(4) Xho I  

Ans. (3)  

Sol.  
Eco RV has restriction sequence –  
5' – GAT ATC – 3'  
3' – CTA TAG – 5'

49. Which kind of therapy was given in 1990 to a four-year-old girl with adenosine deaminase (ADA) deficiency?  
(1) Radiation therapy  
(2) Gene therapy  
(3) Chemotherapy  
(4) Immunotherapy  

Ans. (2)

50. How many hot spots of biodiversity in the world have been identified till date by Norman Myers?  
(1) 43  
(2) 17  
(3) 25  
(4) 34  

Ans. (4)

51. The primary producers of the deep-sea hydrothermal vent ecosystem are  
(1) coral reefs  
(2) green algae  
(3) chemosynthetic bacteria  
(4) blue-green-algae  

Ans. (3)

52. Which of the following is correct for r-selected species?  
(1) Small number of progeny with large size  
(2) Large number of progeny with small size  
(3) Large number of progeny with large size  
(4) Small number of progeny with small size  

Ans. (2)

53. If '+' sign is assigned to beneficial interaction, '–' sign to detrimental and '0' sign to neutral interaction, then the population interaction represented by '+' '–' refers to  
(1) parasitism  
(2) mutualism  
(3) amensalism  
(4) commensalism  

Ans. (1)  

Sol.  
Parasitism +, –  
Mutualism +, +  
Amensalism 0, –  
Commensalism +, 0

54. Which of the following is correctly matched?  
(1) Stratification-Population  
(2) Aerenchyma-Opuntia  
(3) Age pyramid-Biome  
(4) Parthenium hysterophorus-Threat to biodiversity  

Ans. (4)  

Sol.  
Parthenium hysterophorus :- Exotic species do not that allow the growth of other plants near it.

55. Red List contains data or information on  
(1) marine vertebrates only  
(2) all economically important plants  
(3) plants whose products are in international trade  
(4) threatened species  

Sol.  
Red list of red data book IUCN (New name WCU) involve threatened species of plants & animals

56. Which one of the following is wrong for fungi?  
(1) They are both unicellular and multicellular.  
(2) They are eukaryotic.  
(3) All fungi possess a purely cellulosic cell wall.  
(4) They are heterotrophic  

Ans. (3)  

Sol.  
In fungi, cell wall is usually composed of Chitin. Cellulosic cell wall is found in oomycetes of phycomycetes in fungi.

57. Methanogens belong to  
(1) Slime moulds  
(2) Eubacteria  
(3) Archaebacteria  
(4) Dinoflagellates  

Ans. (3)  

Sol.  
Biogas producing, obligate anaerobe, methanogens are type of Archaebacteria.

58. Select the wrong statement.  
(1) Diatoms are microscopic and float passively in water.  
(2) The walls of diatoms are easily destructible.  
(3) 'Diatomaceous earth' is formed by the cell walls of diatoms.  
(4) Diatoms are chief producers in the oceans.  

Ans. (2)  

Sol.  
The wall of diatoms contain cellulose & Silica. It is called frustule. It is non-degradable.

59. The label of a herbarium sheet does not carry information on  
(1) height of the plant  
(2) date of collection  
(3) name of collector  
(4) local names  

Ans. (1)

60. Conifers are adapted to tolerate extreme environmental conditions because of  
(1) presence of vessels  
(2) broad hardy leaves  
(3) superficial stomata  
(4) thick cuticle  

Ans. (4)  

Sol.  
Presence of thick cuticle, presence of sunken stomata and needle like leaves are xerophytic adaptations of conifers.

61. Which one of the following statements is wrong?  
(1) Laminaria and Sargassum are used as food.  
(2) Algae increase the level of dissolved oxygen in the immediate environment.  
(3) Algin is obtained from red algae, and carrageenan from brown algae.  
(4) Agar-agar is obtained from Gelidium and Gracilaria.  

Ans. (3)  

Sol.  
Algin is obtained from Laminaria & Fucus –Brown algae while Carrageenan from chondrus crispus – Red algae

62. The term 'polyadelphous' is related to  
(1) calyx  
(2) gynoecium  
(3) androecium  
(4) corolla  

Ans. (3)  

Sol.  
If filaments of Androecium are joined to form more than two groups but their Anthers separate, it is called polyadelphous Eg :- Citrus.

63. How many plants among Indigofera, Sesbania, Salvia, Allium, Aloe, mustard, groundnut, radish, gram and turnip have stamens with different lengths in their flowers?  
(1) Six  
(2) Three  
(3) Four  
(4) Five  

Ans. (3)  

Sol.  
Salvia, Mustard, Radish and turnip have stamens with different lengths in their flowers

64. Radial symmetry is found in the flowers of  
(1) Cassia  
(2) Brassica  
(3) Trifolium  
(4) Pisum  

Ans. (2)  

Sol.  
Cassia, Trifolium & Pisum have zygomorphic flowers while Brassica has Actinomorphic flowers (Radial symmetry)

65. Free-central placentation is found in  
(1) Citrus  
(2) Dianthus  
(3) Argemone  
(4) Brassica  

Ans. (2)  

Sol.  
Free central placentation is found in Dianthus.

66. Cortex is the region found between  
(1) endodermis and vascular bundle  
(2) epidermis and stele  
(3) pericycle and endodermis  
(4) endodermis and pith  

Ans. (2)  

Sol.  
Sequence from outside to inside in T.S. of stem is epidermis, hypodermis, cortex, endodermis, stele (pericycle + vascular tissues + pith) hence cortex is present between epidermis & stele.

67. The balloon-shaped structures called tyloses  
(1) are linked to the ascent of sap through xylem vessels  
(2) originate in the lumen of vessels  
(3) characterize the sapwood  
(4) are extensions of xylem parenchyma cells into vessels  

Ans. (4)  

Sol.  
Ballon like parenchymatous ingrowth in vessels called tyloses which inhibits transportation of water & minerals in xylem

68. A non-proteinaceous enzyme is  
(1) deoxyribonuclease  
(2) lysozyme  
(3) Ribozyme  
(4) ligase  

Ans. (3)  

Sol.  
Ribozyme is non proteinaceous enzyme as it is 23S rRNA acts as a catalyst during protein synthesis.

69. Select the mismatch.  
(1) Methanogens-Prokaryotes  
(2) Gas vacuoles-Green bacteria  
(3) Large central vacuoles-Animal cells  
(4) Protists-Eukaryotes  

Ans. (3)  

Sol.  
Animal cells do not contain large central vacuole, their vacuole is poorly developed or absent.

70. Select the wrong statement.  
(1) Mycoplasma is a wall-less microorganism.  
(2) Bacterial cell wall is made up of peptidoglycan.  
(3) Pili and fimbriae are mainly involved in motility of bacterial cells.  
(4) Cyanobacteria lack flagellated cells.  

Ans. (3)  

Sol.  
Motility is performed by flagella only in bacterial cells while fimbriae provide attachment to base and pili forms conjugation tube during conjugation

71. A cell organelle containing hydrolytic enzymes is  
(1) mesosome  
(2) lysosome  
(3) microsome  
(4) ribosome  

Ans. (2)  

Sol.  
Hydrolytic enzyme containing vesicle is called lysosome.

72. During cell growth, DNA synthesis takes place in  
(1) M phase  
(2) S Phase  
(3) G₁ phase  
(4) G₂ phase  

Ans. (2)  

Sol.  
DNA Polymerase enzyme is synthesized in G₁ phase but activates in 'S' phase hence DNA replication takes place in 'S' phase.

73. Which of the following biomolecules is common to respiration-mediated breakdown of fats, carbohydrates and proteins?  
(1) Acetyl CoA  
(2) Glucose-6-phosphate  
(3) Fructose 1,6-bisphosphate  
(4) Pyruvic acid  

Ans. (1)  

Sol.  
Acetyl CoA is common intermediate of fats, carbohydrates & proteins during aerobic respiration.

74. A few drops of sap were collected by cutting across a plant stem by a suitable method. The sap was tested chemically. Which one of the following test results indicates that it is phloem sap?  
(1) Absence of sugar  
(2) Acidic  
(3) Alkaline  
(4) Low refractive index  

Ans. (3)  

Sol.  
Phloem sap is alkaline due to actively pumping of protons from companion cells to the outer cells.

75. You are given a tissue with its potential for differentiation in an artificial culture. Which of the following pairs of hormones would you add to the medium to secure shoots as well as roots?  
(1) Gibberellin and abscisic acid  
(2) IAA and gibberellin  
(3) Auxin and cytokinin  
(4) Auxin and abscisic acid  

Ans. (3)  

Sol.  
Auxin and cytokinin ratio regulates the growth of root & shoot as low concentration of Auxin and cytokinin promotes shoot growth while higher ratio promotes root growth

76. Phytochrome is a  
(1) chromoprotein  
(2) flavoprotein  
(3) glycoprotein  
(4) lipoprotein  

Ans. (1)  

Sol.  
Phytochromes are chromoproteins

77. Which is essential for the growth of root tip?  
(1) Mn  
(2) Zn  
(3) Fe  
(4) Ca  

Ans. (4)  

Sol.  
Ca promotes the growth of root tip.

78. The process which makes major difference between C₃ and C₄ plants is  
(1) respiration  
(2) glycolysis  
(3) Calvin cycle  
(4) photorespiration  

Ans. (4)  

Sol.  
Photorespiration differentiates C₃ plants from C₄ plants due to having high CO₂ concentration around RuBP in bundle sheath cells

79. Which one of the following statements is not correct?  
(1) Water hyacinth, growing in the standing water, drains oxygen from water that leads to the death of fishes.  
(2) Offspring produced by the asexual reproduction are called clone.  
(3) Microscopic, motile asexual reproductive structures are called zoospores.  
(4) In potato, banana and ginger, the plantlets arise from the internodes present in the modified stem.  

Ans. (4)  

Sol.  
Plantlet always arise from nodes of stem or modified stem

80. Which one of the following generates new genetic combinations leading to variation?  
(1) Nucellar polyembryony  
(2) Vegetative reproduction  
(3) Parthenogenesis  
(4) Sexual reproduction  

Ans. (4)  

Sol.  
New genetic combination develops after sexual reproduction due to following reasons  
(1) Crossing over during gamete formation  
(2) Chance combination of gametic fusion

81. Match Column-I with Column-II and select the correct option using the codes given below :  
Column-I  
a. Pistils fused together  
b. Formation of gametes  
c. Hyphae of higher Ascomycetes  
d. Unisexual female flower  
Column-II  
(i) Gametogenesis  
(ii) Pistillate  
(iii) Syncarpous  
(iv) Dikaryotic  

Codes:  
(1) (iii) (i) (iv) (ii)  
(2) (iv) (iii) (i) (ii)  
(3) (ii) (i) (iv) (iii)  
(4) (i) (ii) (iv) (iii)  

Ans. (1)

82. In majority of angiosperms  
(1) a small central cell is present in the embryo sac  
(2) egg has a filiform apparatus  
(3) there are numerous antipodal cells  
(4) reduction division occurs in the megaspore mother cells  

Ans. (4)

83. Pollination in water hyacinth and water lily is brought about by the agency of  
(1) bats  
(2) water  
(3) insects or wind  
(4) birds  

Ans. (3)

84. The ovule of an angiosperm is technically equivalent to  
(1) megaspore  
(2) megasporangium  
(3) megasporophyll  
(4) megaspore mother cell  

Ans. (2)

85. Taylor conducted the experiments to prove semiconservative mode of chromosome replication on  
(1) E. coli  
(2) Vinca rosea  
(3) Vicia faba  
(4) Drosophila melanogaster  

Ans. (3)

86. The mechanism that causes a gene to move from one linkage group to another is called  
(1) Crossing-over  
(2) inversion  
(3) duplication  
(4) translocation  

Ans. (4)

87. The equivalent of a structural gene is  
(1) recon  
(2) muton  
(3) cistron  
(4) operon  

Ans. (3)

88. A true breeding plant is  
(1) Always homozygous recessive in its genetic constitution  
(2) One that is able to breed on its own  
(3) Produced due to cross pollination among unrelated plants  
(4) near homozygous and produces offspring of its kind  

Ans. (4)

89. Which of the following rRNAs acts as structural RNA as well as ribozyme in bacterial  
(1) 5.8S rRNA  
(2) 5S rRNA  
(3) 18 S rRNA  
(4) 23S rRNA  

Ans. (4)

90. Stirred-tank bioreactors have been designed for  
(1) ensuring anaerobic conditions in the culture vessel  
(2) purification of product  
(3) addition of preservatives to the product  
(4) availability of oxygen throughout the process  

Ans. (4)

91. A molecule that can act as a genetic material must fulfill the traits given, except  
(1) it should provide the scope for slow changes that are required for evolution  
(2) it should be able to express itself in the form of 'Mendelian characters'  
(3) it should be able to generate its replica  
(4) it should be unstable structurally and chemically  

Ans. (4)  

Sol.  
Genetic material should be stable (chemically) otherwise its expression will change leading to loss in several metabolic functions or inconsistency in expression.

92. DNA-dependent RNA polymerase catalyzes transcription on the strand of the DNA which is called the  
(1) antistrand  
(2) template strand  
(3) coding strand  
(4) alpha strand  

Ans. (2)  

Sol.  
The strand of DNA on which RNA Polymerase binds to perform transcription is called template strand.

93. Interspecific hybridization is the mating of  
(1) more closely related individuals within same breed for 4-6 generations  
(2) animals within same breed without having common ancestors  
(3) two different related species  
(4) superior males and females of different breeds  

Ans. (3)

94. Which of the following is correct regarding AIDS causative HIV?  
(1) HIV does not escape but attacks the acquired immune response.  
(2) HIV is enveloped virus containing one molecule of single-stranded RNA and one molecule of reverse transcriptase.  
(3) HIV is enveloped virus that contains two identical molecules of single-stranded RNA and two molecules of reverse transcriptase.  
(4) HIV is unenveloped retrovirus.  

Ans. (1)  

Sol.  
HIV attacks helper T cells and not try to hide from them.

95. Among the following edible fishes, which one is a marine fish having rich source of omega-3 fatty acids?  
(1) Mackerel  
(2) Mystus  
(3) Mangur  
(4) Mrigala  

Ans. (1)

96. Match Column-I with Column-II and select the correct option using the codes given below:  
Column-I  
a. Citric acid  
b. Cyclosporin A  
c. Statins  
d. Butyric  
Column-II  
(i) Trichoderma  
(ii) Clostridium  
(iii) Aspergillus  
(iv) Monascus  

Codes:  
(1) (iii) (iv) (i) (ii)  
(2) (iii) (i) (ii) (iv)  
(3) (iii) (i) (iv) (ii)  
(4) (i) (iv) (ii) (iii)  

Ans. (3)

97. Biochemical Oxygen Demand (BOD) may not be good index for pollution for water bodies receiving effluents from  
(1) sugar industry  
(2) domestic sewage  
(3) dairy industry  
(4) petroleum industry  

Ans. (4)  

Sol.  
BOD is measure of Oxygen required by microbes to remove biodegradable chemicals

98. The principle of competitive exclusion was stated by  
(1) Verhulst and Pearl  
(2) C. Darwin  
(3) G. F. Gause  
(4) MacArthur  

Ans. (3)

99. Which of the following National Parks is home to the famous musk deer or hangul?  
(1) Dachigam National Park, Jammu & Kashmir  
(2) Keibul Lamjao National Park, Manipur  
(3) Bandhavgarh National Park, Madhya Pradesh  
(4) Eaglenest Wildlife Sanctuary, Arunachal Pradesh  

Ans. (1)

100. A lake which is rich in organic waste may result in  
(1) mortality of fish due to lack of oxygen  
(2) increased population of aquatic organisms due to minerals  
(3) drying of the lake due to algal bloom  
(4) increased population of fish due to lots of nutrients  

Ans. (1)  

Sol.  
Lake with large amount of organic waste will increase BOD of water since microbes will use more Dissolved Oxygen to degrade organic matter

101. The highest DDT concentration in aquatic food chain shall occur in  
(1) eel  
(2) phytoplankton  
(3) seagull  
(4) crab  

Ans. (3)  

Sol.  
Bioamplification of nondegradable chemicals is seen as we move upwards in trophic level and thus in Sea gull (bird) level of DDT will be maximum.

102. Which of the following sets of diseases is caused by bacteria?  
(1) Herpes and influenza  
(2) Cholera and tetanus  
(3) Typhoid and smallpox  
(4) Tetanus and mumps  

Ans. (2)

103. Match Column-I with Column-II for housefly classification and select the correct option using the codes given below :  
Column-I  
a. Family  
b. Order  
c. Class  
d. Phylum  
Column-II  
(i) Diptera  
(ii) Arthropoda  
(iii) Muscidae  
(iv) Insecta  

Codes:  
(1) (iv) (ii) (i) (iii)  
(2) (iii) (i) (iv) (ii)  
(3) (iii) (ii) (iv) (i)  
(4) (iv) (iii) (ii) (i)  

Ans. (2)

104. Choose the correct statement.  
(1) All Pisces have gills covered by an operculum.  
(2) All mammals are viviparous.  
(3) All cyclostomes do not possess jaws and paired fins.  
(4) All reptiles have a three-chambered heart.  

Ans. (3)

105. Study the four statements (A - D) given below and select the two correct ones out of them :  
A. Definition of biological species was given by Ernst Mayr.  
B. Photoperiod does not affect reproduction in plants.  
C. Binomial nomenclature system was given by R. H. Whittaker.  
D. In unicellular organisms, reproduction is synonymous with growth.  
The two correct statements are  
(1) A and B  
(2) B and C  
(3) C and D  
(4) A and D  

Ans. (4)

106. In male cockroaches, sperms are stored in which part of the reproductive system?  
(1) Vas deferens  
(2) Seminal vesicles  
(3) Mushroom glands  
(4) Testes  

Ans. (2)

107. Smooth muscles are  
(1) voluntary, spindle-shaped, uninucleate  
(2) involuntary, fusiform, non-striated  
(3) voluntary, multinucleate, cylindrical  
(4) involuntary, cylindrical, striated  

Ans. (2)

108. Oxidative phosphorylation is  
(1) formation of ATP energy released from electrons removed during substrate oxidation  
(2) formation of ATP by transfer of phosphate group from a substrate to ADP.  
(3) oxidation of phosphate group in ATP  
(4) addition of phosphate group to ATP  

Ans. (1)  

Sol.  
Oxidative phosphorylation occurs when NADH + H⁺ or FADH₂ are oxidized and their electron enters in ETC creating proton gradient which finally produce ATP in F₀-F₁ particle.

109. Which of the following is the least likely to be involved in stabilizing the three-dimensional folding of most proteins?  
(1) Ester bonds  
(2) Hydrogen bonds  
(3) Electrostatic interaction  
(4) Hydrophobic interaction  

Ans. (1)  

Sol.  
Ester bonds are formed in nucleic acids and lipids, but not proteins.

110. Which of the following describes the given graph correctly?  
(1) Exothermic reaction with energy A in absence of enzyme and B in presence of enzyme  
(2) Endothermic reaction with energy A in presence of enzyme and B in absence of enzyme  
(3) Exothermic reaction with energy A in presence of enzyme and B in absence of enzyme  
(4) Endothermic reaction with energy A in absence of enzyme and B in presence of enzyme.  

Ans. (3)

111. When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?  
(1) Both G₂/M and M  
(2) G₁/S  
(3) G₂/M  
(4) M  

Ans. (3)  

Sol.  
G₂/M check point ensures that DNA Replication is complete and no error is left.

112. Match the stages of meiosis in Column-I to their characteristic features in Column-II and select the correct option using the codes given below :  
Column-I  
a. Pachytene  
b. Metaphase I  
c. Diakinesis  
d. Zygotene  
Column-II  
(i) Pairing of homologous chromosomes  
(ii) Terminalization of chiasmata  
(iii) Crossing-over takes place  
(iv) Chromosomes align at equatorial plate  

Codes:  
(1) (iv) (iii) (ii) (i)  
(2) (iii) (iv) (ii) (i)  
(3) (i) (iv) (ii) (iii)  
(4) (ii) (iv) (iii) (i)  

Ans. (2)

113. Which hormones do stimulate the production of pancreatic juice and bicarbonate?  
(1) Insulin and glucagon  
(2) Angiotensin and epinephrine  
(3) Gastrin and insulin  
(4) Cholecystokinin and secretin  

Ans. (4)  

Sol.  
Pancreatic juice rich in enzymes is secreted under influence of cholecystokinin, while pancreatic juice rich in bicarbonates is secreted under influence of secretin.

114. The partial pressure of oxygen in the alveoli of the lungs is  
(1) less than that of carbon dioxide  
(2) equal to that in the blood  
(3) more than that in the blood  
(4) less than that in the blood  

Ans. (3)  

Sol.  
pO₂ in Alveoli is 104, while in oxygenated blood, it is 95.

115. Choose the correct statement.  
(1) Receptors do not produce graded potentials.  
(2) Nociceptors respond to changes in pressure.  
(3) Meissner's corpuscles are thermoreceptors.  
(4) Photoreceptors in the human eye are depolarized during darkness and become hyperpolarized in response to the light stimulus.  

Ans. (4)  

Sol.  
Photoreceptors in the human eye are depolarized during darkness and become hyperpolarized in response to the light stimulus.

116. Graves' disease is caused due to  
(1) hypersecretion of adrenal gland  
(2) hyposecretion of thyroid gland  
(3) hypersecretion of thyroid gland  
(4) hyposecretion of adrenal gland  

Ans. (3)  

Sol.  
Graves' disease is caused due to hypersecretion of thyroid gland

117. Name the ion responsible for unmasking of active sites for myosin for cross-bridge activity during muscle contraction.  
(1) Potassium  
(2) Calcium  
(3) Magnesium  
(4) Sodium  

Ans. (2)  

Sol.  
Calcium ion is responsible for unmasking of active sites for myosin for cross-bridge activity during muscle contraction

118. Name the blood cells, whose reduction in number can cause clotting disorder, leading to excessive loss of blood from the body.  
(1) Thrombocytes  
(2) Erythrocytes  
(3) Leucocytes  
(4) Neutrophils  

Ans. (1)  

Sol.  
Thrombocytes

119. Name a peptide hormone which acts mainly on hepatocytes, adipocytes and enhances cellular glucose uptake and utilization.  
(1) Gastrin  
(2) Insulin  
(3) Glucagon  
(4) Secretin  

Ans. (2)  

Sol.  
Insulin is a peptide hormone which acts mainly on hepatocytes, adipocytes and enhances cellular glucose uptake and utilization.

120. Osteoporosis, an age related disease of skeletal system, may occur due to  
(1) accumulation of uric acid leading to inflammation of joints  
(2) immune disorder affecting neuromuscular junction leading to fatigue  
(3) high concentration of Ca⁺⁺ and Na⁺  
(4) decreased level of estrogen  

Ans. (4)  

Sol.  
Osteoporosis, an age related disease of skeletal system, may occur due to decreased level of estrogen

121. Serum differs from blood in  
(1) lacking antibodies  
(2) lacking globulins  
(3) lacking albumins  
(4) lacking clotting factors  

Ans. (4)  

Sol.  
Blood Plasma – clotting factors = Serum

122. Lungs do not collapse between breaths and some air always remains in the lungs which can never be expelled because  
(1) pressure in the lungs is higher than the atmospheric pressure  
(2) there is a negative pressure in the lungs  
(3) there is a negative intrapleural pressure pulling at the lung walls  
(4) there is a positive intrapleural pressure  

Ans. (3)  

Sol.  
Lungs do not collapse between breaths and some air always remains in the lungs which can never be expelled because there is a negative intrapleural pressure pulling at the lung walls

123. The posterior pituitary gland is not a 'true' endocrine gland because  
(1) it secretes enzymes  
(2) it is provided with a duct  
(3) it only stores and releases hormones  
(4) it is under the regulation of hypothalamus  

Ans. (3)  

Sol.  
The posterior pituitary gland is not a 'true' endocrine gland because it only stores and releases hormones

124. The part of nephron involved in active reabsorption of sodium is  
(1) descending limb of Henle's loop  
(2) distal convoluted tubule  
(3) proximal convoluted tubule  
(4) Bowman's capsule  

Ans. (3)  

Sol.  
The part of nephron involved in active reabsorption of sodium is proximal convoluted tubule

125. Which of the following is hormone releasing IUD?  
(1) Cu7  
(2) LNG-20  
(3) Multiload 375  
(4) Lippes loop  

Ans. (2)  

Sol.  
LNG-20 is hormone releasing IUD

126. Which of the following is incorrect regarding vasectomy?  
(1) Irreversible sterility  
(2) No sperm occurs in seminal fluid  
(3) No sperm occurs in epididymis  
(4) Vasa deferentia is cut and tied  

Ans. (3)  

Sol.  
No sperm occurs in epididymis

127. Embryo with more than 16 blastomeres formed due to in vitro fertilization is transferred into  
(1) cervix  
(2) uterus  
(3) fallopian tube  
(4) fimbriae  

Ans. (2)  

Sol.  
Embryo with more than 16 blastomeres formed due to in vitro fertilization is transferred into uterus

128. Which of the following depicts the correct pathway of transport of sperms?  
(1) Efferent ductules → Rete testis → Vas deferens → Epididymis  
(2) Rete testis → Efferent ductules → Epididymis → Vas deferens  
(3) Rete testis → Epididymis → Efferent ductules → Vas deferens  
(4) Rete testis → Vas deferens → Efferent ductules → Epididymis  

Ans. (2)  

Sol.  
The correct pathway of transport of sperms is  
Rete testis → Efferent ductules → Epididymis → Vas deferens

129. Match Column-I with Column-II and select the correct option using the codes given below :  
Column-I  
a. Mons pubis  
b. Antrum  
c. Trophectoderm  
d. Nebenkern  
Column-II  
(i) Embryo formation  
(ii) Sperm  
(iii) Female external genitalia  
(iv) Graafian follicle  

Codes:  
(1) (i) (iv) (iii) (ii)  
(2) (iii) (iv) (ii) (i)  
(3) (iii) (iv) (i) (ii)  
(4) (iii) (i) (iv) (ii)  

Sol.  
a. Mons pubis – Female external genitalia  
b. Antrum – Graafian follicle  
c. Trophectoderm – Embryo formation  
d. Nebenkern – Sperm

130. Several hormones like hCG, hPL, estrogen, progesterone are produced by  
(1) pituitary  
(2) ovary  
(3) placenta  
(4) fallopian tube  

Ans. (3)

131. If a colour-blind man marries a woman who is homozygous for normal colour vision, the probability of their son being colour-blind is  
(1) 1  
(2) 0  
(3) 0.5  
(4) 0.75  

Ans. (2)  

Sol.  
Colour blind man XᶜY × Normal homozygous women XX  
Genotype of son – XY – normal vision  
Son receives Y chromosome from father and X chromosomes from mother

132. Genetic drift operates in  
(1) slow reproductive population  
(2) small isolated population  
(3) large isolated population  
(4) non-reproductive population  

Ans. (2)  

Sol.  
Genetic drift operates in small isolated population

133. In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by  
(1) q²  
(2) p²  
(3) 2pq  
(4) pq  

Ans. (3)  

Sol.  
In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by 2pq

134. The chronological order of human evolution from early to the recent is  
(1) Australopithecus → Homo habilis → Ramapithecus → Homo erectus  
(2) Australopithecus → Ramapithecus → Homo habilis → Homo erectus  
(3) Ramapithecus → Australopithecus → Homo habilis → Homo erectus  
(4) Ramapithecus → Homo habilis → Australopithecus → Homo erectus  

Ans. (3)  

Sol.  
The chronological order of human evolution from early to the recent is  
Ramapithecus → Australopithecus → Homo habilis → Homo erectus

135. Which of the following is the correct sequence of events in the origin of life?  
I. Formation of protobionts  
II. Synthesis of organic monomers  
III. Synthesis of organic polymers  
IV. Formation of DNA-based genetic systems  
(1) II, III, IV, I  
(2) II, III, IV  
(3) I, III, II, IV  
(4) II, III, I, IV  

Ans. (4)  

Sol.  
The correct sequence of events in the origin of life is  
Synthesis of organic monomers – Synthesis of organic polymers – Formation of protobionts – Formation of DNA-based genetic systems

PART C – PHYSICS

NEET 2015 Physics Previous Year Question Paper

136. A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens the person has to use, will be :  
(1) convex, +0.15 diopter  
(2) convex, +2.25 diopter  
(3) concave, –0.25 diopter  
(4) concave, –0.2 diopter  

Ans. (3)  

Sol.  
1/v – 1/u = 1/f  
1/∞ – 1/(–4) = 1/f  
f = –4 m  
power = 1/f = –0.25 D

137. A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10⁻⁵ cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :  
(1) 0.15 cm  
(2) 0.10 cm  
(3) 0.25 cm  
(4) 0.20 cm  

Ans. (1)  

Sol.  
Position of 1st minima  
y = (λD)/a = (5 × 10⁻⁵ × 0.6)/(0.02 × 10⁻²) = 0.15 cm

138. Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ₀) of the emitted X-ray is :  
(1) λ₀ = λ  
(2) λ₀ = (2mcλ²)/h  
(3) λ₀ = 2h/(mc)  
(4) λ₀ = (2m²c²λ³)/h²  

Ans. (2)  

Sol.  
K.E. of electrons = h²/(2mλ²)  
So maximum energy of photon will also be this much.  
hc/λ₀ = h²/(2mλ²) ⇒ λ₀ = (2mcλ²)/h

139. Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is :  
(1) –3 V  
(2) +3 V  
(3) +4 V  
(4) –1 V  

Ans. (1)  

Sol.  
k_max = hν – ϕ  
2 eV = 5 eV – ϕ ⇒ ϕ = 3 eV  
So V_st = 3 volt  
V_anode – V_cathode = –3 volt

140. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be :  
(1) (20/13)λ  
(2) (16/25)λ  
(3) (9/16)λ  
(4) (20/7)λ  

Ans. (4)  

Sol.  
1/λ = R[(1/2²) – (1/3²)]  
1/λ' = R[(1/3²) – (1/4²)]  
dividing λ' = (20/7)λ

141. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is :  
(1) 60  
(2) 15  
(3) 30  
(4) 45  

Ans. (1)  

Sol.  
N₁ = 0.6 N₀  
N₂ = 0.15 N₀  
N₂/N₁ = (1/2)² so two half life period has passed  
so time taken = 2t₁/₂ = 2 × 30 = 60 minutes

142. For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is :  
(1) 15 mV  
(2) 10 mV  
(3) 20 mV  
(4) 30 mV  

Ans. (3)  

Sol.  
A_V = β (R_out / R_in) = 100 × (2kΩ / 1kΩ) = 200  
(V_out)_AC / (V_in)_AC = 200 ⇒ (V_in)_AC = 4/200 = 20 mV

143. The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance R₁ will be :  
(1) 3.13 A  
(2) 2.5 A  
(3) 10.0 A  
(4) 1.43 A  

Ans. (2)  

Sol.  
The diode D₁ will be in reverse bias, so it will block the current and diode D₂ will be in forward bias, so it will pass the current  
i = 10/(2+2) = 2.5 A

144. What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1 ?  
(1) 1, 1  
(2) 0, 1  
(3) 0, 0  
(4) 1, 0  

Ans. (4)

145. Planck's constant (h), speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations of these has dimension of length?  
(1) √(Gc / h^{3/2})  
(2) √(hG / c^{3/2})  
(3) √(hG / c^{5/2})  
(4) √(hc / G)  

Ans. (2)  

Sol.  
L = (h)^a (c)^b (G)^c  
m⁰L¹T⁰ = (m¹L²T⁻¹)^a (L¹T⁻¹)^b (m⁻¹L³T⁻²)^c  
a – c = 0, 2a + b + 3c = 1, –a – b – 2c = 0  
solving b = –3/2, a = 1/2, c = 1/2  
L = √(hG)/c^{3/2}

146. Two cars P and Q start from a point at the same time in a straight line and their positions are represented by x_P(t) = at + bt² and x_Q(t) = ft – t². At what time do the cars have the same velocity  
(1) (f – a)/[2(1 + b)]  
(2) (a – f)/(1 + b)  
(3) (a + f)/[2(b – 1)]  
(4) (a + f)/[2(1 + b)]  

Ans. (1)  

Sol.  
V_P = V_Q  
a + 2bt = f – 2t ⇒ t = (f – a)/[2(b + 1)]

147. In the given figure, a = 15 m/s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is  
(1) 6.2 m/s  
(2) 4.5 m/s  
(3) 5.0 m/s  
(4) 5.7 m/s  

Ans. (4)  

Sol.  
a_c = V²/r  
15 cos 30° = V²/2.5  
V² = 32.73  
V = 5.7 m/sec

148. A rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be  
(1) mV/3  
(2) mV  
(3) 2mV  
(4) mV/2  

Ans. (2)  

Sol.  
J = 2mV cos 60° = mV

149. A bullet of mass 10 g moving horizontally with a velocity of 400 ms⁻¹ strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be  
(1) 160 ms⁻¹  
(2) 100 ms⁻¹  
(3) 80 ms⁻¹  
(4) 120 ms⁻¹  

Ans. (4)  

Sol.  
During the collision, apply momentum conservation  
(0.01)(400) + 0 = (2)V + (0.01)V'  
where V = √(2gh) = √(2 × 10 × 0.1) = √2  
solving V' = 120 m/sec.

150. Two identical balls A and B having velocities of 0.5 m/s and –0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be  
(1) 0.3 m/s and 0.5 m/s  
(2) –0.5 m/s and 0.3 m/s  
(3) 0.5 m/s and –0.3 m/s  
(4) –0.3 m/s and 0.5 m/s  

Ans. (4)  

Sol.  
Mass of balls are same and the collision is perfectly elastic, so their velocity will be interchanged.  
So, V_A = –0.3 m/s, V_B = 0.5 m/s

151. A particle moves from a point (–2î + 5ĵ) to (4ĵ + 3k̂) when a force of (4î + 3ĵ)N is applied. How much work has been done by the force?  
(1) 2 J  
(2) 8 J  
(3) 11 J  
(4) 5 J  

Ans. (4)  

Sol.  
S⃗ = r⃗_f – r⃗_i = (4ĵ + 3k̂) – (–2î + 5ĵ) = 2î – ĵ + 3k̂  
F⃗ = 4î + 3ĵ  
W = F⃗ · S⃗ = 8 – 3 = 5 J

152. Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_B > I_A) have equal kinetic energy of rotation. If L_A and L_B be their angular momenta respectively, then  
(1) L_A > L_B  
(2) L_A = L_B / 2  
(3) L_A = 2 L_B  
(4) L_B > L_A  

Ans. (4)  

Sol.  
KE_A = KE_B  
(1/2) I_A ω_A² = (1/2) I_B ω_B²  
since I_B > I_A so ω_B < ω_A  
(1/2) L_A ω_A = (1/2) L_B ω_B ⇒ L_B > L_A

153. A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (E_sphere / E_cylinder) will be :  
(1) 3 : 1  
(2) 2 : 3  
(3) 1 : 5  
(4) 1 : 4  

Ans. (3)  

Sol.  
KE of sphere = (1/2)(2/5 mR²)ω² = (1/5)mR²ω²  
KE of cylinder = (1/2)(mR²/2)(2ω)² = mR²ω²  
So, KE_sphere / KE_cylinder = 1/5

154. A light rod of length ℓ has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is :  
(1) √(m₁m₂) ℓ²  
(2) (m₁m₂)/(m₁ + m₂) ℓ²  
(3) (m₁ + m₂) ℓ²  
(4) (m₁ + m₂) ℓ²  

Ans. (2)  

Sol.  
I = m₁r₁² + m₂r₂² = [m₁m₂/(m₁ + m₂)] ℓ²

155. Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by  
(1)  
(2)  
(3)  
(4)  

Ans. (3)  

Sol.  
g_in = g₀ (r/R)  
g₀ is 'g' at surface  
g_out = g₀ (R²/r²)

156. A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g₀, the value of acceleration due to gravity at the earth's surface, is  
(1) – (mg₀ R²)/[2(R + h)]  
(2) (mg₀ R²)/[2(R + h)]  
(3) – (mg₀ R²)/[2(R + h)]  
(4) (R mg₀)/[2(R + h)]  

Ans. (3)  

Sol.  
TE = – (GMm)/[2(R + h)] = – (g₀ m R²)/[2(R + h)]

157. A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3 × 10⁻⁴ J, the value of the surface tension of the liquid is  
(1) 8.0 Nm⁻¹  
(2) 0.250 N m⁻¹  
(3) 0.125 Nm⁻¹  
(4) 0.2 Nm⁻¹  

Ans. (3)  

Sol.  
Increase in surface area = (20 – 8) × 2 = 24 cm²  
So work done = T · ΔS = T × 24 × 10⁻⁴ = 3 × 10⁻⁴  
T = 0.125 N/m

158. Three liquids of densities ρ₁, ρ₂ and ρ₃ (with ρ₁ > ρ₂ > ρ₃), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ₁, θ₂ and θ₃ obey  
(1) π > θ₁ > θ₂ > θ₃ > π/2  
(2) π/2 > θ₁ > θ₂ > θ₃ ≥ 0  
(3) 0 ≤ θ₁ < θ₂ < θ₃ < π/2  
(4) π/2 < θ₁ < θ₂ < θ₃ < π  

Ans. (3)  

Sol.  
h = 2T cosθ /(ρ g r)  
cosθ₁ / ρ₁ = cosθ₂ / ρ₂ = cosθ₃ / ρ₃  
cosθ₁ > cosθ₂ > cosθ₃ as ρ₁ > ρ₂ > ρ₃  
0 ≤ θ₁ < θ₂ < θ₃ < π/2

159. Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100°C, while the other one is at 0°C. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is  
(1) 0°C  
(2) 50°C  
(3) more than 50°C  
(4) less than 50°C but greater than 0°C  

Ans. (3)  

Sol.  
Body at 100°C temperature has greater heat capacity than body at 0°C so final temperature will be closer to 100°C. So T_c > 50°C

160. A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be  
(1) T  
(2) (7/4)T  
(3) (3/2)T  
(4) (4/3)T  

Ans. (3)  

Sol.  
ΔT = ΔT₀ e^(–λt)  
T = 2T e^(–λ(10 min))  
ΔT' = 2T e^(–λ(20 min)) = T/2  
So T_f = T + T/2 = (3/2)T

161. One mole of an ideal monatomic gas undergoes a process described by the equation PV³ = constant. The heat capacity of the gas during this process is :  
(1) R  
(2) (3/2)R  
(3) (5/2)R  
(4) 2R  

Ans. (1)  

Sol.  
PV³ = constant  
for a polytropic process. PV^α = constant  
C = C_v + R/(1 – α) = (3/2)R + R/(1 – 3) = R

162. The temperature inside a refrigerator is t₂ °C and the room temperature is t₁ °C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be  
(1) (t₁ + t₂)/(t₁ – 273)  
(2) t₁/(t₁ – t₂)  
(3) (t₁ + 273)/(t₁ – t₂)  
(4) (t₂ + 273)/(t₁ – t₂)  

Ans. (3)  

Sol.  
Q_more / W = T_more / (T_more – T_less) = (t₁ + 273)/(t₁ – t₂)

163. A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas ?  
(1) mKT  
(2) P/(kT)  
(3) Pm/(kT)  
(4) P/(kTV)  

Ans. (3)  

Sol.  
density = mP / (kT)

164. A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is :  
(1) 9/16  
(2) 3/4  
(3) 4/3  
(4) 16/9  

Ans. (1)  

Sol.  
T = 2π √(m/k) = 3 sec  
T' = 2π √[(m+1)/k] = 5 sec  
(m/(m+1)) = (3/5)² = 9/25  
25m = 9m + 9 ⇒ m = 9/16 kg

165. The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L meter long. The length of the open pipe will be  
(1) 4L  
(2) L  
(3) 2L  
(4) L/2  

Ans. (3)  

Sol.  
ℓ₀ = 2L

166. Three sound waves of equal amplitudes have frequencies (n – 1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be  
(1) 2  
(2) 1  
(3) 4  
(4) 3  

Ans. (2)  

Sol.  
no. of beats = 1 (HCF of beat frequencies)

167. An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10⁵ N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is  
(1) 7 μC  
(2) 8 mC  
(3) 2 mC  
(4) 5 mC  

Ans. (3)  

Sol.  
τ = PE sinθ  
4 = P × 2 × 10⁵ × (1/2)  
P = 4 × 10⁻⁵ = q × 2 × 10⁻²  
q = 2 × 10⁻³ coulomb = 2 mC

168. A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k₁, k₂, k₃ and k₄ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by  
(1) 1/k = 1/k₁ + 1/k₂ + 1/k₃ + 3/(2k₄)  
(2) k = k₁ + k₂ + k₃ + 3k₄  
(3) k = (2/3)(k₁ + k₂ + k₃) + 2k₄  
(4) 2/k = 3/(k₁ + k₂ + k₃) + 1/k₄  

Ans. (4 or bonus)

169. The potential difference (V_A – V_B) between the points A and B in the given figure is :  
(1) +9 V  
(2) –3 V  
(3) +3 V  
(4) +6 V  

Ans. (1)  

Sol.  
V_A – V_B = 9 volt

170. A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is :  
(1) 13 Ω  
(2) 230 Ω  
(3) 46 Ω  
(4) 26 Ω  

Ans. (4)  

Sol.  
i = 500/100 = 5 A  
so 130 = 5R  
R = 26 Ω

171. A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be :  
(1) 2n²B  
(2) nB  
(3) n²B  
(4) 2nB  

Ans. (3)  

Sol.  
B' = n²B

172. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is :  
(1) (2W)/√3  
(2) W/√3  
(3) √3 W  
(4) (√3 W)/2  

Ans. (3)  

Sol.  
W_ext = U_f – U_i = MB/2 = W  
τ = MB sin 60° = √3 W

173. An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10⁻² T. If the value of e/m is 1.76 × 10¹¹ C/kg, the frequency of revolution of the electron is :  
(1) 6.82 MHz  
(2) 1 GHz  
(3) 100 MHz  
(4) 62.8 MHz  

Ans. (2)  

Sol.  
f = (1/(2π)) (qB/m) = 10⁹ Hz = 1 GHz

174. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?  
(1) R = 25 Ω, L = 1.5 H, C = 45 μF  
(2) R = 20 Ω, L = 1.5 H, C = 35 μF  
(3) R = 25 Ω, L = 2.5 H, C = 45 μF  
(4) R = 15 Ω, L = 3.5 H, C = 30 μF  

Ans. (4)  

Sol.  
Option with highest quality factor should be chosen as most appropriate answer.  
Q = (1/R) √(L/C)

175. A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate dB/dt. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is :  
(1) –(dB/dt) π r² in loop 1 and zero in loop 2  
(2) zero in loop 1 and zero in loop 2  
(3) –(dB/dt) π r² in loop 1 and –(dB/dt) π r² in loop 2  
(4) –(dB/dt) π R² in loop 1 and zero in loop 2  

Ans. (1)  

Sol.  
e = –dϕ/dt = –π r² (dB/dt) in loop 1 & zero in loop 2.

176. The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is  
(1) 1.0  
(2) 0.4  
(3) 0.5  
(4) 0.8  

Ans. (4)  

Sol.  
Power factor = R/Z = 80 / √(80² + 60²) = 0.8

177. A 100 Ω resistance and a capacitor of 100 Ω reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is  
(1) 11√2 A  
(2) 2.2 A  
(3) 11 A  
(4) 4.4 A  

Ans. (2)  

Sol.  
Z = √(R² + X_C²) = 100√2  
i_max = V_max / Z = (220√2)/(100√2) = 2.2 A

178. Two identical glass (μ_g = 3/2) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (μ_w = 4/3). The focal length of the combination is  
(1) 3f/4  
(2) f/3  
(3) f  
(4) 4f/3  

Ans. (1)  

Sol.  
1/f = (3/2 – 1)(2/R) = 1/R  
1/f' = (4/3 – 1)(–2/R) = –2/(3R)  
1/f_eq = 2/f + 1/f' = 4/(3f)  
f_eq = 3f/4

179. An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is  
(1) 16  
(2) 8  
(3) 10  
(4) 12  

Ans. (4)  

Sol.  
x/μ + (ℓ – x)/μ = 5 + 3  
ℓ/μ = 8  
ℓ = 8 × 3/2 = 12 cm

180. The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio (I_max – I_min)/(I_max + I_min) will be  
(1) √n / (n + 1)²  
(2) √n / (n + 1)  
(3) 2√n / (n + 1)  
(4) √n / (n + 1)²  

Ans. (3)  

Sol.  
I_max = (√n + 1)² I  
I_min = (√n – 1)² I  
(I_max – I_min)/(I_max + I_min) = 2√n / (n + 1)

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