NSEJS – 2013-14
SOLUTIONS
1. One side of a glass slab is silvered as shown in the figure. A ray of light is incident on the other side at angle of incidence i = 45°. Refractive index of glass is given as 1.5. The deviation of the ray of light from its initial path when it comes out of the slab is (given sin–1 2 28
3 )
(a) 180° (b) 90°
(c) 120° (d) 45°
Ans. (b) Sol.
As shown in fig. we have angle of deviation of the light from its initial path when it comes out of the slab is,
= 90°.2. The number 5 41 lies between
(a) 32 and 33 (b) 30 and 31 (c) 31 and 32 (d) 29 and 30
Ans. (a)
Sol. 5 41 lies between 32 and 33.
3. Oxidation number and co-ordination number of Pt in cisplatin PtCl2(NH3)2 are respectively :
(a) +4 and 2 (b) 0 and 4 (c) +2 and 4 (d) +2 and 6
Ans. (c)
Sol. Oxidation no.
x – 2 – 0 = 0
x = + 2
coordination no.
4Passage for Q(4 – 6) : In a field one summer’s day a Grasshopper was hopping about, chirping and
singing to its heart’s content. An Ant passed by, bearing along with great toil an ear of pea he was taking to the nest. “Why not come and chat with me,” said the Grasshopper, “instead of toiling and moiling in that way ?” “I am helping to lay up food for the winter” said the Ant, “and recommend you to do the same.” Why bother about winter ?” Said the Grasshopper, “We have got plenty of food at present. “But the Ant went on its way and continued its toil. When the winter came the Grasshopper had no food and found itself dying of hunger – while it saw the ants distributing every day corn and grain from the stores they had collected in the summer. Then the Grasshopper knew : It is best to prepare for days of need.
4. In the passage given above there seems to be a factual error with respect to the ant carrying the food to the nest. The most probable reason for this mistake would be :
(a) Pea pod is too heavy for an ant to carry to its nest
(b) Ant couldn’t have passed b easily since it is the favorite food of grasshoppers. (c) Pea cannot be carried by an ant in the summer because it is a Rabi crop
Ans. (c)
Sol. Pea is a Rabi crop.
5. What could be the most plausible reason that all the ants that toiled and moiled in the summer were happy and content in the winter ? :
(a) Food that was procured was efficiently distributed and managed so that all the ants were fed equally (b) ants were probably happy since their food was not shared with Grasshopper
(c) Ants need not worry to work anymore since they had food stocked (d) Ants wre happy since they enjoyed working together in summer
Ans. (a)
Sol. Procured food is efficiently distributed and managed.
6. Grasshopper was at fault in this story mostly because : (a) Of its attitude towards ants who were working tirelessly (b) Of chirping and singing to its heart’s content in the summer
(c) Of not having a forethought to store food for the upcoming winter season
(d) It should have asked ants for the food and managed to surpass the winter somehow.
Ans. (c)
Sol. We have to prepare for future by savings.
7. An object of mass 1 kg is made to slide down a smooth inclined plane of length 20m. If the kinetic energy possessed by the body at the bottom of the plane is 100 J, then the inclination of the plane with the horizontal is (take g = 10 ms–2)
(a) 30° (b) 37° (c) 60° (d) 45°
Ans. (a) Sol.
As per question we have,
v = 2(gsin )
= 20 sin
Now, K.E = 1mv2 1 1 400 sin 100
2 2
sin
= 1 2
= 30°8. Two circles each of radius 3 touch each other externally in the plane. In how many way can a circle of radius 8 be placed in the plane touching each of these two circles ?
(a) 2 (b) 4 (c) 8 (d) 6
Ans. (b)
9. Which of these elements has the greatest electro negativity ?
(a) Br (b) N (c) S (d) O
Ans. (d)
Sol. On moving left to right in a period, electronegativity increases i.e. after F, O is most electronegative element.
10. An inclined plane of inclination v is placed in water as shown in figure given below. Consider a small area (
A) around point P at a depth h. If Density of water is
and acceleration due to gravity is g the force experienced by
A due to hydrostatic pressure is :(a) pgh (
A) sin
(b) pgh (
A) (c) pgh (d) pgh (
A) sec
Ans. (b)
Sol. Fluid pressure at the location of
A is, P =
ghNow, F = P ×
A =
gh
A11. If 3x + 3y – 1, 4x2 + y – 5, 4x + 2y are the sides of an equilateral triangle, its area is closed to the integer
(a) 84 (b) 86 (c) 85 (d) 87
Ans. (c)
Sol. 3x + 3y – 1 = 4x2 + y – 5 = 4x + 2y So, x = 2 & y = 3
So Area of triangle = 49 3 nearest to 85
12. The pH of a 0.025 M solution of KOH is :
(a) 12.40 (b) 3.69 (c) 10.31 (d) 1.0
Ans. (a)
Sol. Molarity of KOH = 0.025 M [OH–] = 25 × 10–3 M
pOH = 3 – log 25 [ pOH = – log10 [OH]]
POH 3 – 1.4 pOH 1.6
pH 14 – 1.6 = 12.4 [ pH + pOH = 14]
13. Consider the following two statements about light & sound. Choose the most appropriate option :
(i) When light and sound travel from air to water, light may bend towards normal while sound may bend away from normal
(ii) Sound is longitudinal wave while light is transverse wave (a) Statement (i) is correct while statement (ii) is incorrect (b) Statement (i) and statement (ii) are incorrect
(c) Statement (i) and statement (ii) are both correct and statement (ii) is not the reason for statement (i) (d) Statement (i) and statement (ii) are correct and statement (ii) is the correct reason for statement (i)
Ans. (c)
Sol. For Sound we have,
Air – denser medium Water – rarer medium
So, when sound moves from air to water it moves away from Normal. For light we have,
Air – rarer medium Water – denser medium
So, when light moves from Air to water it moves towards normal. And, light is transverse wave while sound is longitudinal wave in air.
14. If xy2 = a3, yz2 =b3 and zx2 = c3 then z3 equals (a) 4 2 bc a (b) 2 4 2 b c a (c) 4 2 b c a (d) 4 2 ab c Ans. (c) Sol. xy2 = a3 …(1) yz2 = b3 …(2) zx2 = c3 ….(3) (1) × (2) × (3) x3y3z3 = a3b3c3 …(4)
xyz = abcHence, solving these four equation,
z3 =4 2
b c a
15. How many H atoms are in 3.4 g of C12H22O11 ?
(a) 6.0 × 1023 (b) 3.8 × 1022 (c) 1.3 × 1023 (d) 6.0 × 1021
Ans. (c)
Sol. Molar weight of C12H22O11 = 12 × 12 + 22 × 1 + 11 × 16
=144 + 22 + 176 = 342 gm
i.e. no. of moles = 3.4 .01 342 no. of moles of H = .01 × 22 = .22 no. of atoms of H = .22 × NA = .22 × 6 × 1023 = 1.32 × 1023
Passage Q (16 -18) : Diatoms are the most common photosynthetic aquatic microorganisms group of
algae which are unicellular and can exist as colonies in the shape of filaments of ribbons, fans, zigzags or stars depending on the quality of the water. Diatom communities are a popular tool for studies of water quality and pollution management. Karthik from Bangalore recently went on a field trip from Bangalore to Mysore. On the way he stopped his car at a sewage canal, a lake and a mountain stream and collected water samples from all of these places for his lab work. After a careful analysis of his water samples, he observed that diatoms came with varying size/shape and the size/ shape increases ahs (have ?) Something to do with the water quality.
16. Below are the diatoms observed under a microscope by Karthik. help him to recognize the correct order of sample localities (Viz, Canal, Stream and Lake)
(a) A-Sewage Canal, B-Lake, C-Mountain Steam
(b) A-Mountain stream. B-Sewage Canal C-Lake (c) A-Lake, B-Mountain stream, C-Sewage Canal (d) A-Mountain stream, B-Lake, C-Sewage canal
Ans. (b)
Sol. A belongs to Mountain stream B belongs to Sewage canal and C belongs to Lake.
17. What is the take home message from the above experiment : (a) Diatoms come in different sizes and shapes
(b) Karthik enjoyed collecting samples from different locations
(c) The difference in size and shapes from different water samples is suggestive of the intensity of water pollution
(d) Nothing can be inferred from the above experiment.
Ans. (c)
Sol. Different type of diatoms are present in water having different level of pollution. They act as pollution indicator.
18. In the above experiment, difference in sizes and shapes of diatoms should be inferred as : (a) Different species of diatoms (b) Different genera of diatoms (c) Different families of diatoms (d) Different orders of diatoms
Ans. (b)
Sol. Different size and shapes of diatom show that they belong to different genus.
19. The percentage change in acceleration due to gravity at an altitude equal to radius of earth compared to that on the surface of earth is given by :
(a) 25% increase (b) 35% decrease (c) 25% decrease (d) 75% decreases
Sol. As we know gh = 2 GM (Rh) and s 2 GM g R So, %decrease = s h s g – g 100 g
20. Let a, b be two positive real number such that a < b < 1 a and 1 1 x a – b . a b Then :
(a) x is always less than zero (b) x is always greater than zero
(c) x = 0 (d) No such definite conclusion can be drawn about x
Ans. (b) Sol. x = a – b 1– 1 a b X = (b – a)(1– ab) ab a & b are between 0 & 1
So x is always greater than zero.
21. Which of the following species has standard enthalpy of formation as 0 kj mol–1 ?
(a) H2O() (b) Na(g) (c) CO2(g) (d) Na(s)
Ans. (d)
Sol. Enthalpy of formation of pure natural species (element) is zero i.e. (Hf°) Na(s) = 0
22. A particle accelerates with a constant acceleration a0 and reaches a maximum velocity and then
decelerates with a0 and comes to rest. The position time graph describing the situation is best represented
by :
Ans. (b)
Sol. As we known slope of position-time graph gives velocity so graph (b) best represent the given situation.
23. Let m be the number distinct (non congruent) integer-sided triangles each with perimeter 15 and n be the number of distinct (non congruent) integer sided triangles each with perimeter 16. Then m-n equals :
(a) –2 (b) 0 (c) 4 (d) 2
24. What is the molality of a solution made by dissolving 100 g of bromothylmol blue (C27H28Br2O5S) in 1.00 L
of ethanol on a winter’s day at 10°C ? The density of ethanol at this temperature is 0.7979 kg L–1 : (a) 0.201 mol kg–1 (b) 0.128 mol kg–1 (c) 0.160 mol kg–1 (d) 0.100 mol kg–1
Ans. (a)
Sol. Morality = Mole of solute
wt. of solvent (kg) Molecular weight of C27H28Br2O5S = 27 × 12 + 28 × 1 + 2 × 80 + 5 × 16 + 1 × 32 = 624 gm Molality (m) 100 624 0.201 mol / kg 1 0.7979
25. Two bulbs of specifications 50W, 220V and a 100 W, 200V are connected first (i) in a parallel and then (ii) series across 220V power supply. Choose the correct statement:
(a) In (i) 50 W will glow brighter and in (ii) 100 W will glow brighter (b) In (i) 50 W will glow brighter and (ii) both will glow equality brighter. (c) In both the case the 50 W bulb will glow brighter.
(d) In (i) 100 W will glow brighter and (ii) 50 W will glow brighter
Ans. (d)
Sol. As per question we have,
B1 (50 W, 220 V)
R1= 2 1 1 V 968 P B2(100 W, 200 V)
R2= 2 2 2 V 400 P So, when both bulbs are connected m parallel B2 will glow brighter as
1 P
R And, when both bulbs are connected in series B1 will glow brighter as P
R.26. Let T be the number of 4-digit integers, each ending in 3 (in units place) and each divisible by 11. Then : (a) 70
T
79 (b) 90
T
99 (c) 80
T
89 (d) T
100Ans. (c)
Sol. First such no. is 1023 Last such no. is 9933
After 110 nos. this thing will be repeated, So, 80
T
8927. 1.000mL of 0.1000 mL–2 hydrochloric acid was diluted to 100.0 mL with deionised water. 10.00 mL of this solution was diluted to 100.0 mL again using deionised water. What is the pH of the final solution ?
(a) 2 (b) 3 (c) 5 (d) 4
Ans. (d)
Sol. M1V1 = M2V2 (Ist process)
0.1 × 1 = M2 × 100
M2 = 10–3 M2V2 = M3V3 (II nd process) 10–3 × 10 = M3 × 100
M3 = 10 –4 So pH = –log10 [H+] = –log10 [10–4] pH = 428. While playing football, Dimple fell down and was badly wounded on her left ankle. The Doctor prescribed her antibiotics for a week which should have healed her of the wound in a week. However, Dimples’ wound did not heal in the week. What among the following could have been the reason for inability of the wound to heal in the prescribed time frame given by the doctor ?
(a) Prescribed medicine’s date was expired
(b) Dimple wouldn’t have taken the full course of the antibiotics (c) Doctor’s inability to prescribe the correct medicine for the wound. (d) Both a & b could be the reason
Sol. Antibiotics are given for definite course of time to prevent wound from getting infected. Expired medicines do not show effective results.
29. Which of the following situation is impossible?
(a) A body having constant velocity and variable acceleration (b) A body having zero velocity and non zero acceleration (c) A body having constant acceleration and variable velocity (d) A body having velocity and acceleration in opposite directions.
Ans. (a)
Sol. A body can’t have constant velocity with variable/uniform acceleration.
30. At what time (to the nearest second) immediately after 4 O’ clock will angle between the hands of the clock be the same as that at 4 O’ clock ?
(a) 4h 42m50s (b) 4h43m40s (c) 4h43m38s (d) 5h5m27s
Ans. (c)
Sol. Let the time be x minutes sin the angle must be corresponding to 20 minutes.
x 20 x 5 20 60
as minute hand is at x and hour hand at 20 + x 12 x = 43.6363 or x = 4h 43m 38s
31. 0.5755 g of a compound, containing sulfur and fluorine only, has a volume of 255.0 mL at 288.0 K and 50.01 kPa. What is the molecular formula of this compound ?
(a) S2F2 (b) SF2 (c) SF6 (d) SF4 Ans. (d) Sol. PV = nRT
PV = mRT M
M = mRT PV
M = –3 0.5755 0.0821 288 5 255 10 P = 50.01 kpa = 5 atm V = 255 ml = 255 × 10–3 lit T = 0.0821 lit atm mol k M = 0.57559 m
M = 106.7 and 4 SF M = 32 + 76 = 10832. Given here is a phylogenetic tree (family tree) of greater apes. Which of the following statement cannot be true from the tree ? (mya-million years ago)
(a) Humans are highly evolved among great apes (b) Humans did not evolve from chimpanzees
(c) Humans and chimpanzees are evolutionary cousins (d) Orangutans evolved much earlier than Humans
Ans. (d)
Sol. Orangutan evolved earlier than human but not much earlier.
33. A liquid (a) of density 1.6 gcm–3 and liquid (b) of unknown density is poured into a U-tubes as shown in the figure. The liquid are immiscible. If height of A is hA = 26.6 cm
and height of B is hB = 50 cm the density of B is :
(a) 3.01 g cm3 (b) 0.85 g cm3 (c) 0.33 g cm3 (d) 1.18 g cm3
Ans. (b)
Sol. As per Pascal’s law we have, PX =PY
agha =
bghb34. If a and b are two positive real numbers such that 2 2 a b 6 ab
, then a positive value of a
b lies between :
(a) 5 and 6 (b) 3 and 4 (c) 4 and 5 (d) 2 and 3
Ans. (a) Sol. 2 2 a b ab = 6 a b ba = 6 1 x x = 6 x2 + 1 – 6x = 0 x = 32 2 So, a
b lies between 5 & 6.
35. The isomerism which exists between CH3CHCl2 and CH2ClCH2Cl is :
(a) chain isomerism (b) functional group isomerism
(c) metamerism (d) positional isomerism
Ans. (d) Sol. CH3 CH Cl Cl & CH2 CH2 Cl Cl
are position isomers.
36. The term Biodiversity refers to :
(a) Species Diversity (b) Genetic diversity (c) Ecosystem diversity (d) All of these
Ans. (d)
Sol. Biodiversity refers to variety or diversity of life on earth which includes genetic, species, community and ecosystem diversity.
37. Diagram shown trajectory of a cricket ball. The set of arrows which show the direction of the acceleration of ball at points P and Q respectively is :
(a) (b)
(c) (d)
Ans. (c)
Sol. Direction of acceleration at point P and Q is shown in figure, so option (c) is correct.
38. Sucharitha purchases x pencils at Rs. x each, y pens at Rs. y each and z notebooks at Rs. z each. She purchases altogether 50 items and plays Rs. 1000/-. The cost y pencils, z pens and x notebook is :
(a) Rs. 600/- (b) Rs. 500/- (c) Rs. 750/- (d) Rs. 350/- Ans. (c) Sol. x + y + z = 50 x2 + y2 + z2 = 1000 (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
xy + yz + zx = 750.39. The metal that does not give H2 on treatment with dilute HCl is :
(a) Zn (b) Fe (c) Ca (d) Ag
Ans. (d)
40. On a field trip in North America, student noticed that when threatened, Horned lizards (Genus : Phrynosoma) squirt blood at the attackers, When the Professor asked what could have been the reason behind such behaviour of Horned lizards, one student said that certain sensory receptors had fired and triggered a neuronal reflex culminating in increasing the pressure in their sinus cavities unit the blood vessels in the corners of the eyes burst. Another student said that is was just an act to frighten off the predator. Thus it can be said that
(a) The first response is correct, while the second is incorrect (b) The first response is biological, while the second is philosophical (c) Both explanations are reasonable and can be scientifically tested
(d) The first explanation is testable as a scientific hypothesis, while the second in not
Ans. (c)
Sol. Horned lizard has such type of defence mechanism and proved scientifically.
41. If temperature of a certain mass of aluminum having specific heat capacity of 0.8 J/g°C is lowered by 6°C and heat lost is 96J, then mass of aluminum is :
(a) 20g (b) 48g (c) 60g (d) 16g Ans. (a) Sol. Q = m S
T
m = Q S T = 96 20 0.8 6 g42. The number of real values of a for which the cubic equation x3 – 3ax2 – a = 0 has all real roots one of which is a a itself is :
(a) 0 (b) 1 (c) 3 (d) 2
Ans. (d)
Sol. a3 – 3a(a)2 + 3a(a) – a = 0 A = 0, 1 & ½ a
1/2 as it does not give all real roots.43. The maximum number of isomers for an alkene with molecular formula C4H8 is :
(a) 5 (b) 2 (c) 4 (d) 3 Ans. (d) Sol. C4H8 CH3 CH CH CH3 CH3 C CH3 CH2 3 alkene isomers CH3–CH2–CH=CH2
44. People residing in coastal area usually do not face the problem of Thyroxin hormone deficiency because their food intake will be rich in one of the following minerals.
(a) Sodium (b) Chlorine (c) Phosphorus (d) Iodine
Ans. (d)
Sol. People residing in coastal area are dependent on sea food like crab, prawn, fishes and these are dependent on brown algae. Brown algae is rich in iodine.
45. In photoelectric effect, the maximum kinetic energy (Ek) of photoelectrons depends on frequency (f) of light
incident on a metal surface of work function (
). In an experiment f is varied end Ek is measured. Todetermine value for Planck’s constant (h).
(a) Plot Ek against
and find intercept of best fitted line(b) Plot Ek against
and find slope of line of best fit(c) Plot Ek against f and find slope of line of best fit
(d) Plot Ek against f and find intercept of best fitted line
Sol. By Einstein’s photoelectric equation, Ek = eVs = h – W0 Ek (in eV) = Vs = 0 h W (eV) e
The figure shows a plot of maximum determined kinetic energy (eV) of photoelectrons as a function of frequency of light (Hz). The slope of this line corresponds to the value of Planck’s constant h in eV s.
46. Around a lawn which is of semicircular shape a pavement of uniform width is laid. If the ratio of the area of the lawn to the area of the pavement is 25 : 24, then the ratio of the outer and inner perimeters of the pavement is : (a) 7 5 (b) 5 4 (c) 6 5 (d) 5 2 6 Ans. (a)
Sol. Ratio of outer and inner perimeter
= 49
25 = 7 5.
47. The method that cannot be used for removing permanent hardness of water is :
(a) boiling (b) distillation
(c) adding caustic soda (d) adding sodium carbonate
Ans. (a)
Sol. To remove permanent hardness, boiling cannot be used. It is used to remove temporary hardness.
48. In angiosperm plants, companion cell is associated with which one of the following elements? (a) Tracheids (b) Sieve tube (c) Vessels (d) Xylem fibre
Ans. (b)
Sol. In Angiosperms sieve tube is devoid of nucleus and it’s function is controlled by nucleus of companion cell.
49. Essential requirement for the operation of a step down transformer is : (a) Laminated iron core
(b) Electrical connection between primary and secondary coils (c) Non magnetic core
(d) Magnetic interaction between primary and secondary coils
Ans. (d)
Sol. Essential requirement for the operation of a step down transformer is Magnetic interaction between primary and secondary coils
50. Let ABC be a triangle in which AB = AC. Let D be a point on AC such that BD bisects angle B. Value of the ratio AB
BC is between :
(a) 1.0 and 1.5 (b) 2.0 and 2.5 (c) 1.5 and 2.0 (d) 2.5 and 3.0
Ans. (a,b,c,d) Sol. AB BC sinB sin 2B AB 1 BC 2 cosB 0 < B < 90º So AB 1 BC2 B/2 B/2 B A D C B
51. Consider the following reaction : 4 PCl3(g) P4(g) + 6Cl2(g). If the initial concentration of PCl3(g) is 1.0 M
and “x” is the equilibrium concentration of P4(g), what is the correct equilibrium relation ?
(a) Kc = 6x 2 (b) Kc = 6x 7 /(1.0 – x)4 (c) Kc = x 7 / (1.0 – x)4 (d) Kc = (x)(6x) 6 /(1.0 – 4x)4 Ans. (d) Sol. 4PCl3(g)
P4(g) + 6Cl2(g) 1 0 0 1–4x x 6xKC = 6 4 2 4 3 [P ][Cl ] [PCl ] KC = 6 4 x(6x) (1 4x)
52. In pregnant women, foetus’s physiological functions like nourishment, respiration and excretion are taken up by :
(a) Stomach of mother (b) Umbilical cord (c) Placenta (d) Uterus
Ans. (c)
Sol. Placenta is the intimate connection between the foetus and uterine wall of the mother to exchange the materials.
53. On a rainy wet day, a thunder is heard 6 second after lightening. If speed of sound is 350ms–1 the altitude of the clouds is :
(a) 1.8 km (b) 1.9 km (c) 2.5 km (d) 2.1 km
Ans. (d)
Sol. Since, speed of light is much greater than speed of sound, Let altitude of clouds is x.
350 = x6
x = 2100 m = 2.1 km54. A certain principal becomes Rs. 96800/- in 2 years if compounded annually at certain rate of interest. The same principal becomes Rs. 97240/- in two years if compounded half yearly at the same rate of interest. The rate of interest is :
(a) 8% (b) 8 %1 3 (c) 1 12 % 2 (d) 10% Ans. (d) Sol. 96800 = P 2 R 1 100 97240 = P 4 R 1 200 on solving, w get R = 10%
55. Which properties of plastics make their disposal difficult : (I) PVC produces harmful combustion products; (II) polyalkenes are highly flammble; (III) polyalkenes are non-biodegradable :
(a) I and II only (b) II and III only (c) I and III only (d) I, II and III
Ans. (c)
Sol. PVC produces harmful combustion products and polyalkenes are non-biodegradble therefore plastics make their disposal difficult.
56. Suresh accidentally touched silencer of his two wheeler while parking and withdrew his leg immediately. Identify the correct order of the flow of message to the brain ?
(a) Sensory neutron
CNS
Motor neuron
Effectors
Receptors (b) Receptor
Sensory neuron
CNS
Motor neuron
Effectors (c) CNS
Motor neuron
Effectors
Receptors
Sensory neuron (d) Effectors
Receptors
Sensory neuron
Motor neuronAns. (b)
Sol. The path of reflex arc is as follows:
Receptor Sensory Neuron C.N.S. Motor Neuron Effector organ.
57. There are three bodies A, B and C. Body A when brought closer to B, attract. When body B is brought closer to C they repel. We can then conclude :
(a) Body A and B should have opposite charges while Body C should have the same charge s B (b) Body A and B should have same charge while Body C may have opposite charge
(c) Body B and C are definitely have same kind of charge while Body A may have opposite charge (d) Body C is neutral while Body A and B have opposite charge
Ans. (c)
Sol. A charged body attracts body which is oppositely charged and neutral body, but it repels body having same sign only.
58. A pen costs Rs. 13/- and a note book costs Rs. 35/-. Let m be the maximum number of items that can be bought for Rs. 1000/- and n be the minimum number of items that can be bought for the same amount. Then m + n is :
(a) 98 (b) 88 (c) 96 (d) 79
Ans. (a)
Sol. n < P + B < M 13 P + 35B = 1000 By hit & trial;
m = 60 at P = 50 & B = 10 n = 38 at P = 15 & B = 23
M + n = 60 + 38 = 9859. When the pH of the environment of a protein is changed, it is said to be denatured. This is due to : (a) breakage of peptide bonds (b) breakage of disulfide links
(c) breakdown of R groups (d) loss of tertiary structure
Ans. (d)
Sol. In denatured protein, tertiary structure of protein is distorted.
60. If Brain is controlling unit of an organism, then at cellular level which cell organelle can be comparable to Brain ?
(a) Chloroplast (b) Ribosome (c) Lysosome (d) Nucleus
Ans. (d)
Sol. Nucleus is control centre of the cell and it is also known as Brain of the cell.
61. Aldebarom, the brightest star in the constellation Taurus rises at local time 7:00 pm on 1st of October. On November 1st the star will rise at
(a) 6:00 pm (b) 5.00 pm (c) 9:00 pm (d) 8:34
Ans. (c)
62. In the xy-plane let A be the point (5, 0) and L be the line y x, 3
The number of point P on the line L such that triangle OAP is isosceles is (O being the origin)
(a) 1 (b) 3 (c) 2 (d) infinitely many
Ans. (b)
Sol. Let the point on the line P(
,
/3) case I OA = AP = 5 gives one value Case II OB = OP = 5 gives two value so total no. of triangle = 363. For the reaction, 2A + B
C which relationship is correct?(a) –
[A] = 2
[C] (b) –
[A] =
[C] (c) –
2
[A] =
[C] (d)
[A] =
[C]Ans. (a) Sol. 2A + B
C 1 [A] 2 t = [B] [C] t t [A] = 2 [C]64. In some societies, “woman were solely were solely held responsible for giving birth to female baby” assuming no role for men. But scientific advancement has proved men equally responsible for the birth of either sex. Armed with this information which of the following would be the most appropriate scenario for the birth of female child?
(a) Ovum with X chromosome and Sperm with Y chromosome is FEMALE (b) Ovum with Y chromosome and Sperm with Y chromosome is MALE (c) Ovum with X chromosome and Sperm without chromosome is FEMALE (d) Ovum with X chromosome and Sperm with X chromosome is FEMALE
Ans. (d)
Sol. X chromosome belongs to female characters so zygote formed by fusion of ovum and sperm having x chromosome will form female.
65. A block of mass 2 Kg placed on a floor experiences an external force in horizontal direction of 20N, frictional force of 6N and normal force of 20N. The body travels a distance of 10m under the combined effect of all these force. If initially body is at rest then what is the kinetic energy of the body at the end of 4m
(a) 260J (b) 140J (c) 60J (d) 460J
66. If x3 = a + 1 and x + (b/x) = a, then x equals (a) 2 a(b 1) a b (b) 2 ab 1 a b (c) 2 ab a 1 a b (d) 2 ab a 1 a b Ans. (d)
67. An electrochemical cell constructed for the reaction : Cu2+(aq) + M(s)
Cu(s)+ M 2+(aq) has an Eº= 0.75 V.
The standard reduction potential for Cu2+(aq) is 0.34 V. What is the standard reduction potential for M2+(aq)?
(a) 0.410 V (b) 1.09 V (c) –0.410 V (d) –1.09 V Ans. (c) Sol. º 2 Cu / Cu E 2 º M / M E = EºCell
0.34 – x = 0.75 x = 0.34 – 0.75 x = – 0.410 V68. Which one of the following is said to produce seeds exposed and they are called naked seed plant (a) Marsilea & Nostoc (b) Deodar & Pinus
(c) Maize & Garden pea plant (d) Spirogyra & Funaria
Ans. (b)
Sol. Deodar & Pinus are gymnosperms having naked ovules.
69. A body of mass 2 kg moving in the positive X-direction with a speed 4ms–1 collides head on with an another body of mass 3kg moving in the negative X-direction with a speed of 1ms–1. During collision a loud sound is heard and they both start moving together. The sound energy cannot be greater than
(a) 12J (b) 14J (b) 17.5J (d) 15J
Ans. (d)
Sol. By Momentum conservation we have, 2 × 4 + 3(–1) = 5v
v = 1 m/s So, loss in energy,
Ei – Ef = 1 2(2) (4) 2 + 1 2(3) (1) 2 – 1 2 (5) (1) 2 = 16 + 1.5 – 2.5 = 15 J
70. Let a, b, c be positive real numbers such that abc 1, (ab)2 = (bc)4 = (ca)x = abc. Then x equals
(a) 4 (b) 2 (c) 3 (d) 1
Ans. (b)
71. Which radiation is the easiest to shield?
(a) beta (b) alpha (c) gamma (d) neutron
Ans. (b)
Sol. Alpha particles is easiest to shield i.e. minimum penetration power.
72. Identify the correct order of sequence from exterior to interior. (a) Nucleus
Cell
Chromosome
DNA
Protein (b) Cell
Nucleus
DNA
Chromosome
Protein (c) Cell
Nucleus
DNA
Protein
Chromosome (d) Cell
Nucleus
Protein
DNA
ChromosomeAns. (b)
Sol. Cell has nucleus which has chromosomes. Chromosomes are made up of DNA which code for protein synthesis.
73. In nuclear reactor, the electrons and protons are moving in opposite direction across a small hole in 2 second. If number of electron and protons are 2X1016 each, the current through the hole is given by
(a) 3.2 mA (b) 0mA (c) 6.4mA (d) 1.6mA
Ans. (a) Sol. I = Ie + Ip = qe qp t t = 16 19 2 10 1.6 10 2 + 2 × 1016 × 1.6 × 10–19 = 3.2 × 10–3 A = 3.2 mA
74. The sum 1 1 1 1 1 1 ... 1 1 equals
2 3 4 5 6 2012 20213 (a) 1 1 1 ... 1 100610071008 2013 (b) 1 1 1 ... 1 100610071008 2012 (c) 1 1 1 ... 1 100710081009 2013 (d) 1 1 1 ... 1 100710081009 2012 Ans. (c) Sol. x = 1 1 1 ... 1 1 1 1 ... 1 3 5 2013 2 4 6 2012 = 1 1 1 ... 1 1 1 1 1 ... 1 3 5 2013 2 2 3 1006 2x = 2 1 1 1 ... 1 1 1 ... 1 3 5 2013 2 3 1006 = 2 1 1 1 1 ... 1 1 1 ... 1 2 3 4 2013 1007 1008 2013 = x 1 1 ... 1 1007 1008 2013 x = x 1 1 ... 1 1007 1008 2013 .
75. Which species below has the same general shape as NH3?
(a) CO3 2– (b) SO3 2– (c) NO3 – (d) SO3 Ans. (b) Sol. NH3
sp 3hybridization with one lone pair. SO3
2–
sp3 hybridization with one lone pair.76. After hearing to an influential lecture on “how to conserve environment by avoiding usage of plastic”? Ghan Shyam resolved that he should also contribute towards protecting the environment from plastic menace. Can you suggest him the first step how should he go about doing this effectively
(a) He should urge his parents to stop using plastic materials at home.
(b) He should write a letter to the local civic body against selling plastic materials around his locality. (c) He should ask his teacher to advice people on his behalf to stop usage of plastics.
(d) He should practice minimizing plastic usage himself.
Ans. (d)
77. Two infinite wires are placed parallel to each other. They carry current l1 (12 = l1). The magnetic field is B1
and B2 respectively. Different situation are given in column 1. The comments on the direction and strength
of magnetic field are given in column II. Match the following
Column 1 Column 2
1. The point P is the mid point of the two conductors carrying current in same direction.
(p) B2>B1; Thus B2 + B1 getting into the plane of
the paper.
2. The point P is on right side of second conductor carrying current is same direction
(q) B2>B2; Thus B2 + B1 getting into the plane of
the paper.
3. The point P is at the mid point of the two conductors carrying current in opposite direction.
(r) B1 = B2; B2 – B1 = 0
4. The point P is at the right of the two conductors carrying current in opposite direction
(s) B1=B2; Thus B1+B2 = 2B getting into the plane
of paper (a) 1 P 2 Q 3 R 4 S (b) 1 S 2 R 3 P 4 Q (c) 1 R 2 P 3 S 4 Q (d) 1 S 2 R 3 Q 4 P Ans. (c)
Sol. Case – I B1
out side the plane of paperB2
In side the plane of paperSo, Net Magnetic field, B = 0
Case – II B1
out side the plane of paperB2
out side the plane of paperB2 > B1
so, Net magnetic field, B = B1 + B2
Inside the plane of paperCase – III B1
out side the plane of paperB2
out side the plane of paperSo, Net magnetic field, B = 2B1 = 2B2
inside the plane of paperCase – IV B1
out side the plane of paperB2
In side the plane of paperB2 > B1
So, Net magnetic field, B = B2 – B1
out side the plane of paper78. If the radius of the base of a cone is doubled then the slant area becomes 3 times the original slant area. Suppose when the radius of the base of the cone is quadrupled (that is increased to 4 times), the slantarea becomes k times the original slant area. Then the integer closest to k is :
(a) 11 (b) 8 (c) 10 (d) 6
79. The mass of 0.2 mole of Oxygen molecule is :
(a) 3.2 g (b) 6.4 g (c) 1.6 g (d) 2.75 g
Ans. (b)
Sol. Mole = Weight in grams GMM
Weight (gm) = 0.2 × 32 = 6.4 gm
80. Wuchereria is an example of :
(A) Nematoda (b) Annelida
(c) Arthropoda (d) Arthropoda
Ans. (a)
Sol. Wuchereria is a filarial worm which belongs to Aschelminthes (Nematoda).