
JEE Mains 2016 (CodeF)
Test Name: JEE Mains 2016 (CodeF)
Question 1:
A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of
hλ(where λ is wavelength associated with electron wave) is given by:
Option 1:  2 meV 
Option 2:  meV 
Option 3:  2meV 
Option 4:  meV 
Solutions:
The deBroglie wavelength is given by the expression
λ = hpHere, p refers to the momentum.⇒ p =hλ …..1
Also,
Kinetic energy, eV = 12p2m⇒ p = 2meV …..2From (1) and (2), we get
hλ= 2meV
Hence, the correct answer is option C.
Question 2:
2Chloro2methylpentane on reaction with sodium methoxide in methanol yields:
Option 1:  (a) and (c) 
Option 2:  (c) only 
Option 3:  (a) and (b) 
Option 4:  All of these 
Solutions:
In the reaction of 2chloro2methylpentane with sodium methoxide in methanol, the formation of all the given products is possible.
Hence, the correct answer is option D.
Question 3:
Which of the following compounds is metallic and ferromagnetic?
Option 1:  CrO_{2} 
Option 2:  VO_{2} 
Option 3:  MnO_{2} 
Option 4:  TiO_{2} 
Solutions:Among the given compounds, CrO_{2} is metallic and ferromagnetic.
Hence, the correct answer is option A.
Question 4:
Which of the following statements about low density polythene is FALSE?
Option 1:  It is a poor conductor of electricity. 
Option 2:  Its synthesis requires dioxygen or a peroxide initiator as a catalyst. 
Option 3:  It is used in the manufacture of buckets, dustbins, etc. 
Option 4:  Its synthesis requires high pressure. 
Solutions:Lowdensity polythene is inert, flexible and a poor conductor of electricity. It is not used in the manufacturing of buckets, dustbins, etc. These things are manufactured using highdensity polythene.
Hence, the correct answer is option C.
Question 5:
For a linear plot of log (x/m) versus log p in Freundlich adsorption isotherm, which of the following statements is correct? (k and n are constants.)
Option 1:  1/n appears as the intercept. 
Option 2:  Only 1/n appears as the slope. 
Option 3:  log (1/n) appears as the intercept. 
Option 4:  Both k and 1/n appear in the slope term. 
Solutions:
The Freundlich adsorption isotherm is represented by the expression
xm = kp1n⇒ log xm = log k + 1nlog p …..1
Upon comparing (1) with the equation of a straight line
y = mx + c,
it is clear that for a linear plot of log (x/m) versus log p, the slope of the line is given by the term 1/n.
Hence, the correct answer is option B.
Question 6:
The heats of combustion of carbon and carbon monoxide are –393.5 and –283.5 kJmol^{–1}, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :
Option 1:  676.5 
Option 2:  –676.5 
Option 3:  –110.5 
Option 4:  110.5 
Solutions:
Given,
Cs + O2g → CO2g; ∆H1 = 393.5 kJ mol1 …..1COg + 12O2g → CO2g; ∆H2 = 283.5 kJ mol1 …..2
Subtracting (2) from (1), we get
Cs + 12O2g → COg∆H = ∆H1 – ∆H2 = 393.5 – 283.5 = 110 kJ mol1
Hence, the correct answer is option C.
Question 7:
The hottest region of Bunsen flame shown in the figure below is :
Option 1:  region 2 
Option 2:  region 3 
Option 3:  region 4 
Option 4:  region 1 
Solutions:
In the given Bunsen flame, region 2 is the zone of primary combustion. It is the hottest region of Bunsen flame.
Hence, the correct answer is option A.
Question 8:
Which of the following is an anionic detergent?
Option 1:  Sodium lauryl sulphate 
Option 2:  Cetyltrimethyl ammonium bromide 
Option 3:  Glyceryl oleate 
Option 4:  Sodium stearate 
Solutions:
Among the given compounds, sodium lauryl sulphate is an anionic detergent.
CH3CH210CH2OSO3Na+
Hence, the correct answer is option A.
Question 9:
18 g glucose (C_{6}H_{12}O_{6}) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:
Option 1:  76.0 
Option 2:  752.4 
Option 3:  759.0 
Option 4:  7.6 
Solutions:
Number of moles of glucose, n_{1} =
18180 = 0.1Number of moles of water, n_{2} =
178.218 = 9.9For the given solution, the relative lowering of vapour pressure is given as
∆ppo = n1n1 + n2⇒∆ppo = 0.10.1 + 9.9 = 0.110⇒ ∆p = 0.01po = 0.01 × 760 = 7.6 torr∴ Vapour pressure of the solution = 760 – 7.6 = 752.4 torr
Hence, the correct answer is option B.
Question 10:
The distillation technique most suited for separating glycerol from spentlye in the soap industry is:
Option 1:  Fractional distillation 
Option 2:  Steam distillation 
Option 3:  Distillation under reduced pressure 
Option 4:  Simple distillation 
Solutions:
Distillation under reduced pressure is a separation technique that is used to separate liquids, which have very high boiling points and tend to decompose at or below their boiling points, from their mixtures.
Glycerol is a high boiling liquid. It can be separated from the spentlye in the soap industry by distillation under reduced pressure.
Hence, the correct answer is option C.
Question 11:
The species in which the N atom is in a state of sp hybridization is:
Option 1:  NO2 
Option 2:  NO3 
Option 3:  NO_{2} 
Option 4:  NO2+ 
Solutions:
Among the given species,
NO2+exists as a linear molecule.
O=N+=O
Thus, the N atom is present in a state of sp hybridisation.
Hence, the correct answer is option D.
Question 12:
Decomposition of H_{2}O_{2} follows a first order reaction. In fifty minutes the concentration of H_{2}O_{2} decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H_{2}O_{2} reaches 0.05 M, the rate of formation of O_{2} will be:
Option 1:  6.93 × 10^{–4 }mol min^{–1} 
Option 2:  2.66 L min^{–1} at STP 
Option 3:  1.34 × 10^{–2} mol min^{–1} 
Option 4:  6.93 × 10^{–2} mol min^{–1} 
Solutions:
The chemical equation for the decomposition of H_{2}O_{2} is
2 H2O2 → 2 H2O + O2
It is given that the concentration of H_{2}O_{2} reduces to onefourth of its initial value in 50 minutes.
⇒ 2t12 = 50 min⇒ t1/2 = 25 min
For a firstorder reaction,
Rate constant, k = 0.693t12⇒k = 0.69325 min1
Now, the rate law for the given reaction can be written as
dH2O2dt = kH2O2
When the concentration of H_{2}O_{2} reaches 0.05 M,
dH2O2dt = 0.69325 × 0.05 M min1
The rate of formation of O_{2} can be given as
dO2dt = 12dH2O2dt = 12 × 0.69325 × 0.05 M min1 = 6.93 × 104 M min1
Hence, the correct answer is option A.
Question 13:
The pair having the same magnetic moment is : [At.No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
Option 1:  [Cr(H_{2}O)_{6}]^{2+} and [Fe(H_{2}O)_{6}]^{2+} 
Option 2:  [Mn(H_{2}O)_{6}]^{2+} and [Cr(H_{2}O)_{6}]^{2+} 
Option 3:  [CoCl_{4}]^{2–} and [Fe(H_{2}O)_{6}]^{2+} 
Option 4:  [Cr(H_{2}O)_{6}]^{2+} and [CoCl_{4}]^{2–} 
Solutions:
Complex  Metal Ion  dshell Electronic Configuration  Arrangement of Electrons of dorbitals in the Complex  Number of Unpaired Electrons 
[Cr(H_{2}O)_{6}]^{2+}  Cr^{2+}  d^{4}  4  
[Fe(H_{2}O)_{6}]^{2+}  Fe^{2+}  d^{6}  4  
[Mn(H_{2}O)_{6}]^{2+}  Mn^{2+}  d^{5}  5  
[CoCl_{4}]^{2–}  Co^{2+}  d^{7}  3 
The magnetic moment of a complex is dependent upon the number of unpaired electrons present in the dorbitals of the metal ion/atom in the complex.
Since, the complexes [Cr(H_{2}O)_{6}]^{2+} and [Fe(H_{2}O)_{6}]^{2+} contain the same number of unpaired electrons, they possess the same value of magnetic moment.
Hence, the correct answer is option A.
Question 14:
The absolute configuration of is
Option 1:  (2S, 3R) 
Option 2:  (2S, 3S) 
Option 3:  (2R, 3R) 
Option 4:  (2R, 3S) 
Solutions:The absolute configuration (R/S) of the given compound can be determined by using the CahnIngoldPrelog priority (CIP) rules.
The order of substituents at carbon 2 is OH > CH(Cl)(CH_{3}) > COOH > H. They are in the clockwise direction, as shown in the figure. But when hydrogen is kept away from the viewer, the order of priority becomes anticlockwise and thereby resulting in S configuration.
Similarly, the order of substituents at carbon 3 is Cl > CH(OH)(COOH) > CH_{3} > H. They are in the anticlockwise direction, as shown in the figure. But when hydrogen is kept away from the viewer, the order of priority becomes clockwise and thereby resulting in R configuration.
Hence, the correct answer is option A.
Question 15:
The equilibrium constant at 298 K for a reaction A + B
⇌ C + D is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L^{–1} ) will be:
Option 1:  0.818 
Option 2:  1.818 
Option 3:  1.182 
Option 4:  0.182 
Solutions:
A + B ⇌ C + DInitial conc. 1 1 1 1Eq. conc. 1x 1x 1+x 1+x
Keq = CDAB = 100⇒1+x1+x1x1x = 100⇒1+x21x2 = 100⇒1+x1x = 10⇒ x = 911
∴ Deq = 1 + 911 = 1.818 M
Hence, the correct answer is option B.
Question 16:
Which one of the following ores is best concentrated by froth floatation method?
Option 1:  Siderite 
Option 2:  Galena 
Option 3:  Malachite 
Option 4:  Magnetite 
Solutions:
Froth floatation method is carried out for the concentration of sulphide ores.
Among the given ores, only galena (PbS) is a sulphide ore.
Hence, the correct answer is option B.
Question 17:
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O_{2} by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:
Option 1:  C_{3}H_{8} 
Option 2:  C_{4}H_{8 } 
Option 3:  C_{4}H_{10} 
Option 4:  C_{3}H_{6} 
Solutions:
The formula of the hydrocarbon can be obtained by using the following equation and calculations.
CxHyg + x + y4 O2g → x CO2g + y2H2OlOxygen (O_{2}) used in the reaction = 20% of 375 mL = 75 mL Inert part of air in the reaction = 80% of 375 mL = 300 mL Total volume of gas left after combustion = 330 mL Volume of carbon dioxide (CO_{2}) gases after combustion = 330 mL − 300 mL = 30 mL
CxHyg15 mL + x + y4 O2g75 mL → x CO2g 30 mL+ y2H2Ol
x1 = 3015x = 2
x + y41 = 7515 ⇒ x + y4 = 5⇒y = 12⇒C2H12
Such compound is impossible and also, not provided in the given options.
Disclaimer: None of the options matches.
However, C_{3}H_{8} (option A) is closer to the answer.
Question 18:
The pair in which phosphorous atoms have a formal oxidation state of +3 is:
Option 1:  Pyrophosphorous and hypophosphoric acids 
Option 2:  Orthophosphorous and hypophosphoric acids 
Option 3:  Pyrophosphorous and pyrophosphoric acids 
Option 4:  Orthophosphorous and pyrophosphorous acids 
Solutions:Among the given pairs of oxoacids of phosphorus, phosphorus atoms have a formal oxidation state of +3 in orthophosphorous and pyrophosphorous acids.
Orthophosphorous Acid (H_{3}PO_{3})
Pyrophosphorous Acid (H_{4}P_{2}O_{5})
Hence, the correct answer is option D.
Question 19:
Which one of the following complexes shows optical isomerism?
Option 1:  cis [Co(en)_{2}Cl_{2}]Cl 
Option 2:  trans [Co(en)_{2}Cl_{2}]Cl 
Option 3:  [Co(NH_{3})_{4}Cl_{2}]Cl 
Option 4:  [Co(NH_{3})_{3}Cl_{3}] 
Solutions:
Complex [Co(en_{2})Cl_{2}]Cl exists in cis and trans formï¿½s, but only cis form shows optical isomerism. Complex [Co(NH_{3})_{4}Cl_{2}]Cl also exists in cis and trans formï¿½s and both are optically inactive. Complex [Co(NH_{3})_{3}Cl_{3}] exists in fac and mer forms and both are optically inactive.
Hence, the correct answer is option A.
Question 20:
The reaction of zinc with dilute and concentrated nitric acid, respectively, produces:
Option 1:  NO_{2} and NO 
Option 2:  NO and N_{2}O 
Option 3:  NO_{2} and N_{2}O 
Option 4:  N_{2}O and NO_{2} 
Solutions:
The chemical equations for the reactions of zinc with dilute and concentrated nitric acids are as follows:
Zn + 4 HNO3conc. → ZnNO32 + 2 H2O + 2 NO24 Zn + 10 HNO3dil. → 4 ZnNO32 + N2O + 5 H2O
Hence, the correct answer is option D.
Question 21:
Which one of the following statements about water is FALSE?
Option 1:  Water can act both as an acid and as a base. 
Option 2:  There is extensive intramolecular hydrogen bonding in the condensed phase. 
Option 3:  Ice formed by heavy water sinks in normal water. 
Option 4:  Water is oxidised to oxygen during photosynthesis. 
Solutions:
Among the given statements regarding water, the statement given in option B is false. Water shows extensive intermolecular hydrogen bonding in the condensed phase.
Hence, the correct answer is option B.
Question 22:
The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of:
Option 1:  Lead 
Option 2:  Nitrate 
Option 3:  Iron 
Option 4:  Fluoride 
Solutions:
The maximum permissible concentration of nitrate for a water sample suitable for drinking is about 50 ppm.
Since the given water sample was found to contain 100 ppm nitrate, it is not suitable for drinking.
Hence, the correct answer is option B.
Question 23:
The main oxides formed on combustion of Li, Na and K in excess of air are, respectively:
Option 1:  LiO_{2}, Na_{2}O_{2} and K_{2}O 
Option 2:  Li_{2}O_{2}, Na_{2}O_{2} and KO_{2} 
Option 3:  Li_{2}O, Na_{2}O_{2} and KO_{2} 
Option 4:  Li_{2}O, Na_{2}O and KO_{2} 
Solutions:
When made to undergo combustion in excess of air, Li mainly forms Li_{2}O, sodium forms Na_{2}O_{2} and potassium forms KO_{2}.
4 Li + O2 → 2 Li2O2 Na + O2 → Na2O2K + O2 → KO2
Hence, the correct answer is option C.
Question 24:
Thiol group is present in:
Option 1:  Cystine 
Option 2:  Cysteine 
Option 3:  Methionine 
Option 4:  Cytosine 
Solutions:Among the given biomolecules, the thiol group (−SH) is present in cysteine.
The structure of cysteine is
HSCH2CHNH2COOH
Hence, the correct answer is option B.
Question 25:
Galvanization is applying a coating of:
Option 1:  Cr 
Option 2:  Cu 
Option 3:  Zn 
Option 4:  Pb 
Solutions:
Galvanisation is applying a coating of zinc (Zn) to iron or steel to protect it from rusting.
Hence, the correct answer is option C.
Question 26:
Which of the following atoms has the highest first ionization energy?
Option 1:  Na 
Option 2:  K 
Option 3:  Sc 
Option 4:  Rb 
Solutions:
Among the given atoms, scandium (Sc), a dblock element, has the highest ionisation energy. It is due to the higher effective nuclear charge in the case of dblock elements than the sblock elements (Na, K and Rb).
Hence, the correct answer is option C.
Question 27:
In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br_{2} used per mole of amine produced are:
Option 1:  Four moles of NaOH and two moles of Br_{2} 
Option 2:  Two moles of NaOH and two moles of Br_{2} 
Option 3:  Four moles of NaOH and one mole of Br_{2} 
Option 4:  One mole of NaOH and one mole of Br_{2} 
Solutions:
The Hofmann bromamide degradation reaction is
RCONH2 + Br2 + 4 NaOH → RNH2 + 2 NaBr + Na2CO3 + 2 H2O
It is obvious that four moles of NaOH and one mole of Br_{2} are used in the given reaction.
Hence, the correct answer is option C.
Question 28:
Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p_{i }and temperature T_{1 }are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T_{2} . The final pressure p_{f} is:
Option 1:  2piT1T1+T2 
Option 2:  2piT2T1+T2 
Option 3:  2piT1T2T1+T2 
Option 4:  piT1T2T1+T2 
Solutions:The number of moles of a bulb can be calculated by using the ideal gas equation, PV = nRT.
⇒n = pVRTThere will be no change in the total number of moles of two bulbs before and after raising the temperature.
piVRT1 + piVRT1 = pfVRT2 + pfVRT2 piT1 + piT1 = pfT2 + pfT12piT1 = pf1T2 + 1T12piT1 = pfT1 + T2T1T2pf = 2piT2T1 + T2Hence, the correct answer is option B.
Question 29:
The reaction of propene with HOCl (Cl_{2 }+ H_{2}O) proceeds through the intermediate:
Option 1:  CH_{3} – CH^{+} – CH_{2} – Cl 
Option 2:  CH_{3} – CH(OH) – CH_{2}^{+} 
Option 3:  CH_{3} – CHCl – CH_{2}^{+} 
Option 4:  CH_{3} – CH^{+} – CH_{2} – OH 
Solutions:The reaction of propene with HOCl (Cl_{2}+H_{2}O) proceeds through the carbocation intermediate followed by an electrophilic addition reaction.
CH3CH=CH2 →HOCl CH3C+HCH2Cl Intermediate→OH CH3CHOHCH2Cl
Hence, the correct answer is option A.
Question 30:
The product of the reaction given below is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:The product of the reaction is in option A.
Hence, the correct answer is option A.
Question 31:
Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, – 2), then which one of the following is a vertex of this rhombus?
Option 1:  (–3, –8) 
Option 2:  13,83 
Option 3:  103,73 
Option 4:  (–3, –9) 
Solutions:
ABCD is a rhombus with two adjacent sides AB and AD along the lines x − y + 1 = 0 and 7x − y − 5 = 0, respectively.
xy+1=0 ⇒x=y1 …..17xy5=0 …..2Solving (1) and (2), we get
7y1y5=0⇒7y7y=5⇒6y=12⇒y=2∴ x = 1 So, the coordinates of A are (1, 2). We know that the diagonals of a rhombus bisect each other. Therefore, P(−1, −2) is the mid point of diagonal AC. Let the coordinates of C be
x1, y1. Using the midpoint formula, we get
1,2=1+x12,2+y12⇒x1=3, y1=6The coordinates of C are (−3, −6). Let the equations of the sides BC and CD be x − y + λ = 0 and 7x − y + μ = 0, respectively. The point (−3, −6) lies on the lines x − y + λ = 0 and 7x − y + μ = 0. So, λ = −3 and μ = 15. Therefore, the equations of the other sides of rhombus are x − y − 3 = 0 and 7x − y + 15 = 0. Solving x − y − 3 = 0 and 7x − y − 5 = 0, we get B =
13,83Solving x − y + 1 = 0 and 7x − y + 15 = 0, we get D =
73,43Hence, the correct answer is option B.
Question 32:
If the 2^{nd}, 5^{th} and 9^{th} terms of a nonconstant A.P. are in G.P., then the common ratio of this G.P. is :
Option 1:  43 
Option 2:  1 
Option 3:  74 
Option 4:  85 
Solutions:Let a and d be the first term and the common difference of AP, respectively. It is given that 2nd, 5th and 9th terms of the AP are in GP. Therefore, a_{2 }= a + d, a_{5 }= a + 4d and a_{9 }= a + 8d are in GP.
⇒a+4d2=a+d×a+8d⇒a2+16d2+8ad=a2+9ad+8d2⇒ad=8d2⇒a=8dTherefore, the required GP is 9d, 12d, 16d. Let r be the common ratio of the GP. Common ratio,
r=12d9d=43Hence, the correct answer is option A.
Question 33:
Let P be the point on the parabola, y^{2} = 8x which is at a minimum distance from the centre C of the circle, x^{2} + (y + 6)^{2} = 1. Then the equation of the circle, passing through C and having its centre at P is:
Option 1:  x^{2} + y^{2} – x + 4y –12 = 0 
Option 2:  x2+y2x4+2y24=0 
Option 3:  x^{2} + y^{2} – 4x + 9y + 18 = 0 
Option 4:  x^{2} + y^{2} – 4x + 8y + 12 = 0 
Solutions:The equation of the given parabola is y^{2} = 8x. Here, 4a = 8 ⇒ a = 2. The equation of the circle is x^{2} + (y + 6)^{2} = 1. Centre, C = (0, −6) Let P(2t^{2}, 4t) be the point on the parabola, which is at the shortest distance from the centre of the circle. It is known that the perpendicular distance between any two points is the shortest distance. So, the normal to the parabola at P passes through the centre of the circle.
The equation of the normal to the parabola at P(2t^{2}, 4t) is
y=tx++4t+2t3It passes through the centre C(0, −6).
∴6=4t+2t3⇒t3+2t+3=0Observe that, t = −1 satisfies the above equation. ∴ P = (2, −4) Radius of the required circle, CP
=202+4+62=22 [By the distance formula] Therefore, the equation of the required circle is
x22+y+42=222⇒x2+y24x+8y+12=0Hence, the correct answer is option D.
Question 34:
The system of linear equations
x + λy – z = 0
λx – y – z = 0
x + y – λz = 0
has a nontrivial solution for :
Option 1:  exactly one value of λ 
Option 2:  exactly two values of λ 
Option 3:  exactly three values of λ 
Option 4:  infinitely many values of λ 
Solutions:The given system of linear equations is
x+λyz=0λxyz=0x+yλz=0It is given that the system of linear equations has a nontrivial solution.
∴1λ1λ1111λ=0⇒1λ+1λλ2+11λ+1=0⇒λ+1+λ3λλ1=0⇒λ3λ=0
⇒λλ21=0⇒λ=0, 1, 1Thus, the given system of linear equations has a nontrivial solution for exactly three values of λ. Hence, the correct answer is option C.
Question 35:
If
fx+2f1x=3x, x≠0and
S=x∈R : fx=fx;then S :
Option 1:  contains exactly one element 
Option 2:  contains exactly two elements 
Option 3:  contains more than two elements 
Option 4:  is an empty set 
Solutions:Consider the set
S=x∈R : fx=fx. It is given that
fx+2f1x=3x, x≠0 …..(1) Now, replacing
x by 1x in
fx+2f1x=3x, we get
f1x+2fx=3x …..(2) Multiplying (2) by 2 and then subtracting it from (1), we get
fx+2f1x2f1x4fx=3x6x⇒3fx=3x6x⇒fx=2xxNow,
fx=fx
⇒2xx=2x+x⇒4x=2x⇒x2=2⇒x=±2
∴ S=x∈R : fx=fx=2, 2Thus, the set S contains exactly two elements. Hence, the correct answer is option B.
Question 36:
Let
p=limx→0+1+tan2x12xthen log p is equal to :
Option 1:  1 
Option 2:  12 
Option 3:  14 
Option 4:  2 
Solutions:
p=limx→0+1+tan2x12x =elimx→0+tan2x2x ∵limx→a1+fxgx=elimx→afxgx =elimx→0+tanx22x2 =e12 ∵limx→0tanxx=1Taking logarithm on both sides, we get
logp=loge12⇒logp=12Hence, the correct answer is option B.
Question 37:
A value of θ for which
2+3isinθ12isinθis purely imaginary, is :
Option 1:  π6 
Option 2:  sin1 34 
Option 3:  sin1 13 
Option 4:  π3 
Solutions:
2+3isinθ12isinθ=2+3isinθ12isinθ×1+2isinθ1+2isinθ=26sin2θ+7isinθ1+4sin2θ=26sin2θ1+4sin2θ+i7sinθ1+4sin2θFor
2+3isinθ12isinθto be purely imaginary,
26sin2θ1+4sin2θ=0⇒26sin2θ=0⇒sinθ=13⇒θ=sin113Hence, the correct answer is option C.
Question 38:
The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :
Option 1:  43 
Option 2:  23 
Option 3:  3 
Option 4:  43 
Solutions:Let the equation of the hyperbola be
x2a2y2b2=1. Suppose e be the eccentricity of the hyperbola. Length of the conjugate axis = 2b Distance between the foci = 2ae Given:
2b=122ae⇒2b=ae⇒4b2=a2e2⇒4a2e21=a2e2 ∵b2=a2e21⇒4a2e24a2=a2e2⇒3a2e2=4a2⇒e2=43⇒e=23Hence, the correct answer is option B.
Question 39:
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true :
Option 1:  3a^{2} – 32a + 84 = 0 
Option 2:  3a^{2} – 34a + 91 = 0 
Option 3:  3a^{2} – 23a + 44 = 0 
Option 4:  3a^{2} – 26a + 55 = 0 
Solutions:The given numbers are 2, 3, a and 11.
VarX=1n∑i=1nxi21n∑i=1nxi2 =22+32+a2+11242+3+a+1142 =4+9+a2+121416+a42 =3a232a+28016We know Variance = (Standard deviation)^{2}
∴ 3a232a+28016=3.52⇒3a232a+280=16×12.25⇒3a232a+84=0Hence, the correct answer is option A.
Question 40:
The integral
∫2×12+5x9x5+x3+13dxis equal to :
(Where C is an arbitrary constant)
Option 1:  x102x5+x3+12+C 
Option 2:  x52x5+x3+12+C 
Option 3:  x102x5+x3+12+C 
Option 4:  x5x5+x3+12+C 
Solutions:Let
I=∫2×12+5x9x5+x3+13dxDividing both numerator and denominator by x^{15}, we get
I=∫2×3+5×61+1×2+1x53dx
Let 1+1×2+1×5=t⇒2×35×6=dtdx⇒2×3+5x6dx=dt∴I=∫1t3dt=∫1t3dt =–12t2+C =121+1×2+1×52+C =x102x5+x3+12+CHence, the correct answer is option A.
Question 41:
If the line,
x32=y+21=z+43lies in the plane, lx + my – z = 9, then l^{2} + m^{2} is equal to:
Option 1:  18 
Option 2:  5 
Option 3:  2 
Option 4:  26 
Solutions:The equation of the given line is
x32=y+21=z+43. This line passes through the point P(3, –2, –4). Since the line lies on the plane, P also lies on the plane lx + my – z = 9.
∴ l×3+m×2–4=9⇒3l2m=5 …..(1)The line lies on the plane. So, the vector normal to the plane and the vector parallel to the line are perpendicular to each other.
∴2×l+m×1+3×1=0⇒2lm=3 …..(2)Solving (1) and (2), we get l = 1 and m = −1 ∴ l^{2} + m^{2} = 2
Hence, the correct answer is option C.
Question 42:
If 0 ≤ x < 2π , then the number of real values of x, which satisfy the equation cosx + cos2x + cos3x + cos4x = 0, is :
Option 1:  5 
Option 2:  7 
Option 3:  9 
Option 4:  3 
Solutions:The given equation is cosx + cos2x + cos3x + cos4x = 0.
cosx+cos4x+cos2x+cos3x=0⇒2cos5x2cos3x2+2cos5x2cosx2=0⇒2cos5x2cos3x2+cosx2=0⇒cos5x2=0 or cos3x2+cosx2=0Now,
cos5x2=0 ⇒5×2=2n+1π2, n∈Z⇒x=2n+1π5⇒x=π5, 3π5, π, 7π5, 9π5 0≤x<2πAlso,
cos3x2+cosx2=0⇒4cos3x23cosx2+cosx2=0⇒4cos3x22cosx2=0⇒2cosx22cos2x21=0
⇒2cosx2cosx=0⇒cosx2=0 or cosx=0cosx2=0⇒x2=2n+1π2
⇒x=2n+1π⇒x=π 0≤x<2πcosx=0⇒x=2n+1π2 ⇒x=π2,3π2 0≤x<2πThe solutions of the given trigonometric equation are
π5,3π5,π,7π5,9π5,π2,3π2.Thus, the number of real values of x satisfying the given equation is 7.
Hence, the correct answer is option B.
Question 43:
The area (in sq. units) of the region
x, y : y2≥2x and x2+y2≤4x, x≥0, y≥0is :
Option 1:  π83 
Option 2:  π423 
Option 3:  π2223 
Option 4:  π43 
Solutions:Consider the curves
y2=2x …..1×2+y2=4x …..2Solving (1) and (2), we get
x2+2x=4x⇒x22x=0⇒xx2=0⇒x=0 or x=2∴ y=0 or y=2 x, y≥0So, the points of the intersection of the curves are (0, 0) and (2, 2). The shaded region represents the region
x, y : y2≥2x and x2+y2≤4x, x≥0, y≥0.
∴ Required area=∫02ycircleyparaboladx=∫02ycircledx ∫02yparaboladx=πr24∫022x12dx=4π4232×3202
=π2232320=π83 square unitsHence, the correct answer is option A.
Question 44:
Let , and be three unit vectors such that
a→×b→×c→=32b→+c→. If is not parallel to , then the angle between and is:
Option 1:  π2 
Option 2:  2π3 
Option 3:  5π6 
Option 4:  3π4 
Solutions:
a→×b→×c→=32b→+c→
⇒a→.c→b→a→.b→c→=32b→+32c→Equating the coefficients of and
c→, we get
a→.c→=32and
a→.b→=32Let θ be the angle between and .
∴a→.b→=32
⇒a→b→cosθ=32.
⇒cosθ=32 a→=b→=1
⇒θ=5π6Hence, the correct answer is option C.
Question 45:
A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:
Option 1:  (4 – π) x = πr 
Option 2:  x = 2r 
Option 3:  2x = r 
Option 4:  2x = (π + 4)r 
Solutions:Let the lengths of two parts be a and 2−a. According to the question,
a=4x and 2−a=2πr∴
x=a4 and r=2a2πArea of square =
a42=a216Area of circle =
π2a2π2=π4+a24a4π2=a24a+44π∴ The combined area of the square and circle
fa=a216+a24a+44π=a2π+4a216a+1616π∴ f’a=2aπ+8a1616πFor maximum or minimum, f ‘(a) = 0.
f’a=0 ⇒2aπ+8a16=0⇒2aπ+4=16⇒a=8π+4
∴ x=2π+4Also,
r=2a2π=28π+42π=1π+4So, x = 2r
Hence, the correct answer is option B.
Question 46:
The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :
Option 1:  103 
Option 2:  103 
Option 3:  203 
Option 4:  310 
Solutions:The equation of the line passing through P(1, −5, 9) and parallel to the vector having direction ratios 1, 1, 1 is
x11=y+51=z91=λAny point on this line can be taken as
Qλ+1, λ5, λ+9.If
Qλ+1, λ5, λ+9lies on the plane
xy+z=5, then
λ+1λ5+λ+9=5⇒λ=515=10∴ Q=9, 15, 1⇒PQ=1+92+5+152+9+12=300=103Hence, the correct answer is option A.
Question 47:
If a curve y = f (x) passes through the point (1, –1) and satisfies the differential equation, y (1 + xy) dx = xdy, then
f12is equal to :
Option 1:  45 
Option 2:  25 
Option 3:  45 
Option 4:  25 
Solutions:The given differential equation is
y1+xydx=xdy⇒dydx=yx+y2⇒dydxyx=y2⇒1y2dydx1xy=1Let 1y=v. Then,1y2dydx=dvdxSubstituting these values in the given differential equation, we get
dvdx1xv=1⇒dvdx+1xv=1 …..1This is a linear differential equation.
IF=e∫1xdx=elogx=xMultiplying both sides of (1) by IF and integrating w.r.t. x, we get
vx=∫xdx⇒vx=x22+C⇒xy=x22+CHere, x = 1, y = −1.
∴ 11=12+C⇒C=1+12=12
∴ xy=x2212When x =
12, we get
12y=14×212⇒12y=58⇒y=45Hence, the correct answer is option C.
Question 48:
If the number of terms in the expansion of
12x+4x2n, x≠0,is 28, then the sum of the coefficients of the terms in this expansion, is :
Option 1:  2187 
Option 2:  243 
Option 3:  729 
Option 4:  64 
Solutions:The number of terms in the given expansion =
C2n+2
∴ C2n+2=28⇒n+2n+12=28⇒n2+3n54=0⇒n+9n6=0⇒n=6 Because n cannot be negativeThe sum of the coefficients of all terms in the expansion can be obtained by putting x = 1. ∴ Sum of the coefficients of all terms in the expansion =
12+46=36=729Hence, the correct answer is option C.
Question 49:
Consider
fx=tan11+sinx1sinx, x∈0,π2.A normal to y = f (x) at
=π6also passes through the point :
Option 1:  0, 2π3 
Option 2:  π6, 0 
Option 3:  π4, 0 
Option 4:  (0, 0) 
Solutions:
fx=tan11+sinx1sinx; x∈0, π2 =tan1cos2x2+sin2x2+2cosx2sinx2cos2x2+sin2x22cosx2sinx2 =tan1cosx2+sinx2cosx2sinx2 =tan11+tanx21tanx2 =tan1tanπ4+x2∴ fx=π4+x2At x=π6,fx=π4+x2=π4+π12=π3Now, Slope of the tangent = f ‘(x) =
12∴ Slope of the normal =
112=2Thus, the equation of the normal at
π6, π3is
yπ3=2xπ6⇒yπ3=2x+π3⇒y+2x=2π3The point
0, 2π3satisfies the equation of the normal. So, the normal passes through the point
0, 2π3.
Hence, the correct answer is option A.
Question 50:
For x ∈R , f (x) = log 2 – sin x and g (x) = f(f (x)), then :
Option 1:  g’ (0) =cos(log 2) 
Option 2:  g’ (0) =– cos (log 2) 
Option 3:  g is differentiable at x = 0 and g‘ (0) = – sin(log 2) 
Option 4:  g is not differentiable at x = 0 
Solutions:Given:
fx=log2sinxg(x) = f (f (x))
∀ x∈R ∴ gx=log2sinlog2sinxAt x=0, gx=log2sinlog2sinx∴ g’x=coslog2sinxcosxAt x=0,g’0=coslog2Hence, the correct answer is option A.
Question 51:
Let two fair sixfaced dice A and B be thrown simultaneously. If E_{1} is the event that die A shows up four, E_{2} is the event that die B shows up two and E_{3} is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?
Option 1:  E_{2} and E_{3} are independent 
Option 2:  E_{1} and E_{3} are independent 
Option 3:  E_{1}, E_{2} and E_{3} are independent 
Option 4:  E_{1} and E_{2} are independent 
Solutions:Total number of elementary events = 36
E1=4, 1, 4, 2, …, 4, 6nE1=6E2=1, 2, 2, 2, …, 6, 2nE2=6E3=1, 2, 1, 4, …, 6, 5nE3=18E1∩E2 =4, 2 nE1∩E2=1E2∩E3=1, 2, 3, 2, 5, 2nE2∩E3=3E3∩E1=4, 1, 4, 3, 4, 5nE3∩E1=3E1∩E2∩E3=ϕnE1∩E2∩E3=0∴PE1=636=16=PE2PE3=1836=12PE1∩E2=136PE2∩E3=336=112PE3∩E1=336=112PE1∩E2∩E3=0Therefore, E_{1}, E_{2} and E_{3} are not independent events. Hence, the statement ‘E_{1}, E_{2} and E_{3} are independent’ is not TRUE.
Hence, the correct answer is option C.
Question 52:
If
A=5ab32and A adj A = AA^{T} , then 5a + b is equal to :
Option 1:  5 
Option 2:  4 
Option 3:  13 
Option 4:  –1 
Solutions:Consider the matrix:
A=5ab32Given: A adj A = AA^{T}
5ab322b35a=5ab325a3b2⇒10a+3b0010a+3b=25a2+b215a2b15a2b13Equating the above matrices,
10a+3b=25a2+b2, 10a+3b=13 and 15a2b=0Solving the above equations, we get
a=25 and b=3Therefore, the required value is
5a+b=5×25+3=5. Hence, the correct answer is option A.
Question 53:
The Boolean Expression (p ∧ ~ q) ∨q∨ (~ p ∧ q) is equivalent to :
Option 1:  p ∧ q 
Option 2:  p ∨ q 
Option 3:  p∨ ~ q 
Option 4:  ~ p ∧ q 
Question 54:
The sum of all real values of x satisfying the equation
x25x+5×2+4x60=1is :
Option 1:  –4 
Option 2:  6 
Option 3:  5 
Option 4:  3 
Solutions:Consider the expression:
x25x+5×2+4x60=1This can be wrriten as
x25x+5×2+4x60=x25x+50Comparing by the property of exponents, we get
⇒x2+4x60=0⇒x=10,6Also,
x25x+5=1 or x25x+5=1If
x25x+5=1,then x=2, 3If
x25x+5=1, then x=4, 1Since â€‹3 does not satisfy the equation, the solutions are −10, 1, 2, 4 and 6. The sum of the solutions: −10 + 1 + 2 + 4 + 6 = 3 Hence, the correct answer is option D.
Question 55:
The centres of those circles which touch the circle, x^{2} + y^{2} – 8x – 8y – 4 = 0 , externally and also touch the xaxis, lie on :
Option 1:  an ellipse which is not a circle. 
Option 2:  a hyperbola 
Option 3:  a parabola 
Option 4:  a circle 
Solutions:Consider the given equation of a circle: x^{2} + y^{2} – 8x – 8y – 4 = 0 Compare the above equation of the circle with the general equation of a circle. Therefore, the ï¿½centre of the circle is (4, 4) and the radius is
42+42+4=6. Further, let us assume the centre of the circle to be (h, k). Therefore,
h42+k42=6+kh42+k42=6+k2h28h+16+k28k+16=36+k2+12kh28h20k4=0h28h+16=20k+20x42=20y+1This is an equation of a parabola. Hence, the correct answer is option C.
Question 56:
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :
Option 1:  59^{th} 
Option 2:  52^{nd} 
Option 3:  58^{th} 
Option 4:  46^{th} 
Solutions:There are 5 letters in the word SMALL, out of which 2 L’s are alike. Number of words starting with A =
4!2!=12Number of words starting with L = 4! = 24 Number of words starting with M =
4!2!=12Number of words starting with SA =
3!2!=3Number of words starting with SL = 3! = 6 The next number in the list would be SMALL. ∴ Position of the word SMALL = 12 + 24 + 12 + 3 + 6 + 1 = 58 Thus, the 58th word in the dictionary so formed is SMALL.
Hence, the correct answer is option C.
Question 57:
limn→∞ n+1 n+2……….3nn2n1/nis equal to :
Option 1:  27e2 
Option 2:  9e2 
Option 3:  3 log 3 – 2 
Option 4:  18e4 
Solutions:
Let L=limn→∞n+1n+2…3nn2n1nTaking logarithm on both sides, we get
logL=limn→∞1nlogn+1n+2…n+2nn.n.n…⏟2n terms⇒logL=limn→∞1nlog1+1n1+2n…1+rn… 2n terms⇒logL=limn→∞1nlog1+1n+log1+2n+…+log1+rn+… 2n terms⇒logL=1nlimn→∞∑r=12nlog1+rnPutting rn=x and 1n=dx, we get
logL=∫02log1+xdx⇒logL=xlog1+x02∫02×1+xdx⇒logL=2log30∫02111+xdx⇒logL=2log3x02+log1+x02⇒logL=2log32+log3⇒logL=3log32⇒L=e3log32=elog27e2=27e2Hence, the correct answer is option A.
Question 58:
If the sum of the first ten terms of the series
1352+2252+3152+42+4452+…., is 165m,then m is equal to :
Option 1:  101 
Option 2:  100 
Option 3:  99 
Option 4:  102 
Solutions:The given series is
1352+2252+3152+42+4452+…. This series can be rewritten as
852+1252+1652+2052+2452+…. nth term,
tn=4n+452=1625n+12
∴ S10=∑n=110tn⇒S10=1625∑n=110n+12=1625∑n=110n2+2n+1⇒S10=162510×10+1×2×10+16+2×10×10+12+10⇒S10=1625385+110+10=1625×505=165×101Given:
S10 =165m⇒165m=165×101⇒m=101Thus, the value of m is 101.
Hence, the correct answer is option A.
Question 59:
If one of the diameters of the circle, given by the equation, x^{2} + y^{2} – 4x + 6y – 12 = 0 is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is :
Option 1:  53 
Option 2:  5 
Option 3:  10 
Option 4:  52 
Solutions:The equation of the given circle is
x2+y24x+6y12=0.
The centre of the circle, A =
2, 3The radius of the circle, AB =
22+32–12=5Let R be the radius of the circle S.
OA=322+2+32=50=52In rt âˆ†OAB,
OB2=OA2+AB2⇒R2=50+25=75⇒R=53Thus, the radius of the circle S is
53units.
Hence, the correct answer is option A.
Question 60:
A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is :
Option 1:  10 
Option 2:  20 
Option 3:  5 
Option 4:  6 
Solutions:Let the uniform speed of the man be v m/min. Then, AB = 10v m
Suppose the height of the pillar, CD be h m and BC be x m. In âˆ†ADC,
tan30°=CDAC⇒13=hx+10v⇒x+10v=3h …..1In âˆ†BCD,
tan60°=CDBC⇒3=hx⇒h=3x …..2From (1) and (2), we get
x+10v=3×3x⇒2x=10v⇒xv=5Thus, the time taken by the man to reach the pillar from B is 5 min.
Hence, the correct answer is option C.
Question 61:
A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (Take g = 10 ms^{–2})
Option 1:  2 s 
Option 2:  22 s 
Option 3:  2 s 
Option 4:  2π2 s 
Solutions:
Let the mass per unit length of string be
λ.
⇒λ=MLTension at any point x is given as
T=MgxL=λgxVelocity of transverse wave on a stretched spring is given as
v=Tλ=λgxλ=gx ⇒v2=gxWe knowa= vdvdx⇒ a=gxdgxdx=g2⇒l=12g2t2⇒t=4lg=4×2010=22 sHence, the correct answer is option B.
Question 62:
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 ×10^{7} J of energy per kg which is converted to mechanical energy with a 20 % efficiency rate. Take g = 9.8 ms^{–2}.
Option 1:  6.45 ×10^{–3} kg 
Option 2:  9.89 ×10^{–3} kg 
Option 3:  12.89 × 10^{–3} kg 
Option 4:  2.45 × 10^{–3} kg 
Solutions:The potential energy in lifting mass of 10 kg = mgh
=10×9.8×1×1000 JLet M be the mass of fat burned. Then,
⇒M×(3.8×107) ×20100=10×9.8×1×1000⇒M=12.89×103 kgHence, the correct answer is option C.
Question 63:
A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction μ and the distance x (= QR) , are, respectively close to :
Option 1:  0.2 and 3.5 m 
Option 2:  0.29 and 3.5 m 
Option 3:  0.29 and 6.5 m 
Option 4:  0.2 and 6.5 m 
Solutions:Energy lost over part PQ =
μmgcosθ×2sinθ=2μmgtanθEnergy lost over part QR =
μmgxGiven:
⇒2μmgtanθ=μmgx⇒x=2tan30∘=23=3.5 mAlso, Loss in potential energy = mgh
∴4μmgtanθ=mgh=2mg⇒μ=2tan30∘4=tan30∘2=0.29Hence, the correct answer is option B.
Question 64:
Two identical wires A and B, each of length ‘l’ carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If B_{A} and B_{B} are the values of magnetic field at the centres of the circle and square respectively, then the ratio
BABBis:
Option 1:  π2162 
Option 2:  π216 
Option 3:  π282 
Option 4:  π28 
Solutions:It is given that the length of the wire is l.
Thus,
2πR=l⇒R=l2πAlso, 4a=l⇒a=l4â€‹
Magnetic field at the centre of the circle can be evaluated as
BA=μ0I2R⇒BA=μ0IπlMagnetic field at the centre of the square can be evaluated as
BB=4μ0I4πa2sin45∘+sin45∘⇒BB=16μ0I2lπ⇒BABB=μ0Iπl16μ0I2lπ=π282Hence, the correct answer is option C.
Question 65:
A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is :
Option 1:  2 Ω 
Option 2:  0.1 Ω 
Option 3:  3 Ω 
Option 4:  0.01 Ω 
Solutions:The value of shunt resistance required to convert it to ammeter can be evaluated as
S=igGiigHere, S = Shunt resistance i_{g} = Current through the galvanometer i = Current through the circuit G = Galvanometer resistance
⇒S=103×10210103≃102 ΩHence, the correct answer is option D.
Question 66:
An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears:
Option 1:  10 times nearer 
Option 2:  20 times taller 
Option 3:  20 times nearer 
Option 4:  10 times taller. 
Solutions:Angular magnification of the telescope is 20. Thus, the tree will appear 20 times taller to the observer.
Hence, the correct answer is option B.
Question 67:
The temperature dependence of resistances of Cu and undoped Si in the temperature range 300400K, is best described by:
Option 1:  Linear increase for Cu, exponential increase for Si 
Option 2:  Linear increase for Cu, exponential decrease for Si 
Option 3:  Linear decrease for Cu, linear decrease for Si 
Option 4:  Linear increase for Cu, linear increase for Si 
Solutions:The resistance increases linearly for the conductors and decreases exponentially for semiconductors in the given range of temperature. Thus, there will be a linear increase for Cu and exponential decrease for undoped Si.
Hence, the correct answer is option B.
Question 68:
Choose the correct statement:
Option 1:  In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal 
Option 2:  In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal 
Option 3:  In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal 
Option 4:  In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal 
Solutions:In frequency modulation, the frequency of the highfrequency carrier wave is made to vary in proportion to the amplitude of the audio signal. And in amplitude modulation, the amplitude of the highfrequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
Hence, the correct answer is option D.
Question 69:
Half – lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be :
Option 1:  4 : 1 
Option 2:  1 : 4 
Option 3:  5 : 4 
Option 4:  1 : 16 
Solutions:Given: Halflife of element A, T_{A} = 20 min Halflife of element B, T_{B} = 40 min Ratio of decayed nuclei for A and B will be â€‹
1NN0A1NN0B=112TTA112TTB=11280201128040=1116114=54Hence, the correct answer is option C.
Question 70:
‘n’ mole of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be:
Option 1:  3P0V02nR 
Option 2:  9P0V02nR 
Option 3:  9P0V0nR 
Option 4:  9P0V04nR 
Solutions:From graph,
PP0=P0V0V2V0⇒P=3P0P0V0V⇒nRTV=3P0P0V0V P=nRTVDifferentiating it wrt V, we get ⇒V=3V02Thus, ⇒P=3P0P0V0V⇒P=3P0P0V03V02=3P02Also, T=PVnR ⇒T=PVnR=9P0V04nRHence, the correct answer is option D.
Question 71:
An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :
Option 1:  0.08 H 
Option 2:  0.044 H 
Option 3:  0.065 H 
Option 4:  80 H 
Solutions:Here, resistance of the lamp =
8010=8 Ω
Irms=VrmsR2+XL2=10⇒XL=420Also, 2π×50×L=420⇒L≃0.065 HHence, the correct answer is option C.
Question 72:
A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :
Option 1:  3f4 
Option 2:  2f 
Option 3:  f 
Option 4:  f2 
Solutions:The fundamental frequency for an open organ pipe is
f=V2lWhen half of the pipe is immersed in water, this fundamental frequency can be evaluated as the fundamental frequency for a closed organ pipe as given below:
â€‹
f’=V4l=V4l2=V2l=fHence, the correct answer is option C.
Question 73:
The box of a pin hole camera, of length L, has a hole of radius a, It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_{min}) when :
Option 1:  a=λL and bmin=2λ2L 
Option 2:  a=λL and bmin=4λL 
Option 3:  a=λL2and bmin=22λL 
Option 4:  a=λ2Land bmin=2λ2L 
Solutions:It is given that the radius of the pinhole is a and distance of the screen is L. Then, the angular divergence of the incident beam is given as
d=λL2aThus, the spread of the spot will be
b=2a+λL2a …..(1) To determine the minimum value of b, we take
dbda=0⇒2λLa2=0⇒a=λL2Substituting in (1), the minimum value of b is found to be
bmin=2a+λLa=2λL2+λL2λL=2λL+2λL∴bmin=22λLHence, the correct answer is option C.
Disclaimer: In the original paper, none of the options were correct for this question. To make the question attemptable, we have changed option C.
Question 74:
A combination of capacitor is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal :
Option 1:  360 N/C 
Option 2:  420 N/C 
Option 3:  480 N/C 
Option 4:  240 N/C 
Solutions:Let charges on 4 μF, 9 μF and 3 μF capacitors are Q_{1}, Q_{2} and Q_{3}, respectively.
Then,
⇒Q1=8×4=24 μCTo evaluate Q2Q33=Q29 3 μF and 9 μF are in parallelAlso, Q3+Q2=24 μCThus, ⇒Q2=18 μCThus, the magnitude of the point charge Q will be
18 μC+24 μC=42 μCElectric field due to the point charge at a distance of 30 m will be
E=Q4πε0r2=9×109×42×106302=420 N/CHence, the correct answer is option B.
Question 75:
Arrange the following electromagnetic radiations per quantum in the order of increasing energy : A : Blue light B : Yellow light C : Xray D : Radiowave
Option 1:  A, B, D, C 
Option 2:  C, A B, D 
Option 3:  B, A, D, C 
Option 4:  D, B, A, C 
Solutions:The energy of the electromagnetic radiation increases with the increase in frequency. In the given case,
νD<νB<νA<νC∴ED<EB<EA<ECHence, the correct answer is option D.
Question 76:
Hysteresis loops for two magnetic materials A and B are given below:
These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:
Option 1:  A for electromagnets and B for electric generators. 
Option 2:  A for transformers and B for electric generators. 
Option 3:  B for electromagnets and transformers. 
Option 4:  A for electric generators and transformers. 
Question 77:
A pendulum clock loses 12 s a day if the temperature is 40 °C and gains 4 s a day if the temperature is 20 °C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively:
Option 1:  60°C; α = 1.85 × 10^{–4}/°C 
Option 2:  30°C; α = 1.85 × 10^{–3}/°C 
Option 3:  55°C; α = 1.85 × 10^{–2}/°C 
Option 4:  25°C; α = 1.85 × 10^{–5}/°C 
Solutions:Time loss per day is given as
∆t24×60×60=12α∆T=12α(T’T)Using the above, we can write
1286400=12α(T40) …..(1)486400=12α(T20) …..(2)Dividing (1) by (2), we get
3=T40T20⇒3T+60=T40⇒T=25 °CSubstituting this value in (1),
486400=12α×(2520)⇒α=486400×2×15=1.85×105 /°CHence, the correct answer is option D.
Question 78:
The region between two concentric spheres of radii ‘a‘ and ‘b‘, respectively (see figure), has volume charge density
ρ=Ar, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is:
Option 1:  Q2πb2a2 
Option 2:  2Qπb2a2 
Option 3:  2Qπa2 
Option 4:  Q2πa2 
Solutions:Imagine a Gaussian sphere of radius r such that a < r < b.
Applying Gauss’ law, we can write
E.4πr2=1ε0Q+∫arAr4πr2dr=Qε0+4πA∫arrdr=Qε0+4πAr2a2⇒E=Q4πε0r2+A2ε01a2r2Since E is constant,
dEdr=0⇒Q2πε0r3+0+Aa2ε0r3=0⇒Q2π=Aa2⇒A=Q2πa2Hence, the correct answer is option D.
Question 79:
In an experiment for determination of refractive index of glass of a prism by i – δ, plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index?
Option 1:  1.6 
Option 2:  1.7 
Option 3:  1.8 
Option 4:  1.5 
Solutions:When i = 35° and e = 79°, δ = 40°
So, from
δ=i+eAA=74°the refractive index can be evaluated as
μ=sinδmin +A2sin A2⇒μ=sin40+742sin742=sin57∘sin37∘=1.4Since δmin <40∘, the nearest value of n will be 1.5.Hence, the correct answer is option D.
Question 80:
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :
Option 1:  92 ± 5.0 s 
Option 2:  92 ± 1.8 s 
Option 3:  92 ± 3 s 
Option 4:  92 ± 2 s 
Solutions:The mean time can be calculated as
tmean=90+91+95+924=92 sAbsolute error = Mean time − Measured reading
Thus, absolute errors in the readings will be 2, 1, 3 and 0.
Mean error =
2+1+3+04=1.5 sAs the least count is 1 s, the mean error would be 2 s.
Thus, the reported mean time will be
t=92±2 sHence, the correct answer is option D.
Question 81:
Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d):
Option 1:  Zener diode, Simple diode, Light dependent resistance, Solar cell 
Option 2:  Solar cell, Light dependent resistance, Zener diode, Simple diode 
Option 3:  Zener diode, solar cell, Simple diode, Light dependent resistance 
Option 4:  Simple diode, Zener diode, Solar cell, Light dependent resistance 
Solutions:The characteristics shown in the figures (a), (b), (c) and (d) are for simple diode, zener diode, solar cell and lightdependent resistance, respectively.
Hence, the correct answer is the option D.
Question 82:
Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to
3λ4, the speed of the fastest emitted electron will be:
Option 1:  <υ4312 
Option 2:  =υ4312 
Option 3:  =υ3412 
Option 4:  >υ3412 
Solutions:We know hcλ=w+mv22 …..(1) Here, w=Work function of photocellv=Speed of the fastest electronNow,hc3λ4=w+mv’22 …..(2) Multiplying (1) by 43 and subtracting (2), we getw3=46mv2mv’22⇒mv’22=46mv2+w3⇒mv’22>46mv2⇒v’>43vHence, the correct answer is option D.
Question 83:
A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance
2A3from equilibrium position. The new amplitude of the motion is:
Option 1:  3A 
Option 2:  A3 
Option 3:  7A3 
Option 4:  A341 
Solutions:
We know
v=ωA2X2(in SHM)
Given:
v=ωA22A32⇒v=5Aω3Also, vnew =ωAnew2x2=3v⇒vnew =ωAnew22A32=35Aω3⇒Anew=7A3Hence, the correct answer is option C.
Question 84:
A particle of mass m is moving along the side of square of side ‘a‘, with a uniform speed v in the xy plane as shown in the figure:
Which of the following statements is false for the angular momentum
L→about the origin?
Option 1:  L→=mυR2a k^when the particle is moving from C to D. 
Option 2:  L→=mυR2+a k^when the particle is moving from B to C. 
Option 3:  L→=mυ2Rk^when the particle is moving from D to A. 
Option 4:  L→=mυRRk^when the particle is moving from A to B. 
Solutions:The angular momentum of a particle moving in straight line is evaluated as
L→=mr→×v⇀Thus, the angular momentum of the particle moving from C to D will be
L→=mvR2+ak^The angular momentum of the particle moving from B to C will be
L→=mvR2+ak^The angular momentum of the particle moving from D to A will be
L→=mvR2k^The angular momentum of the particle moving from A to B will be
L→=mvR2k^Hence, the correct answer is option A.
Disclaimer: To make this question attemptable, we have changed the option (C) because in the original paper, both options (A) and (C) are correct.
Question 85:
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PV^{ n} = constant, then n is given by (Here C_{P }and C_{V} are molar specific heat at constant pressure and constant volume, respectively):
Option 1:  n=CCPCCV 
Option 2:  n=CPCCCV 
Option 3:  n=CCVCCP 
Option 4:  n=CVCP 
Solutions:We know
C=CV+R1n⇒CCV=CPCV1n⇒1n=CPCVCCV⇒n=1CPCVCCV=CCPCCVHence, the correct answer is option A.
Question 86:
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45^{th} division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25^{th} division coincides with the main scale line?
Option 1:  0.80 mm 
Option 2:  0.70 mm 
Option 3:  0.50 mm 
Option 4:  0.75 mm 
Solutions:Least count of screw gauge =
PitchNo. of circular scale divisions=0.550=0.01 mmZero error of the screw gauge = Main scale reading + Circular reading × Least count
⇒0.5 +45×0.01 = 0.05 mmMeasured reading (with zero error) =
0.5 + 25 × 0.01=0.75 mmActual reading or thickness of sheet = Measured reading − Zero error =
0.75 –0.05=0.80 mmHence, the correct answer is option A.
Question 87:
A roller is made by joining, together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:
Option 1:  turn right. 
Option 2:  go straight. 
Option 3:  turn left and right alternately. 
Option 4:  turn left. 
Solutions:When roller moves further, the radius of the cone on rail AB decreases and the distance covered by left cone is smaller for the same number of rotations. Thus, it turns towards left.
Hence, the correct answer is option D.
Question 88:
If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is:
Option 1:  AND 
Option 2:  OR 
Option 3:  NAND 
Option 4:  NOT 
Solutions:When all the inputs are 0, the output is 0. When any of the inputs is 1, the output is 1. Thus, the given gate is the OR gate.
Hence, the correct answer is option B.
Question 89:
For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β
Option 1:  α=β1β 
Option 2:  α=β1+β 
Option 3:  α=β21+β2+2β 
Option 4:  1α=1β+1 
Solutions:
β=α1α⇒1α=1β+1Hence, the correct answer is option A.
Disclaimer: In the original paper, there were two correct options for this question. To make this question attemptable (single correct), we have changed option C.
Question 90:
A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius or earth R; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere)
Option 1:  gR 
Option 2:  gR/2 
Option 3:  gR21 
Option 4:  2gR 
Solutions:The orbital velocity of the satellite,
v0=GMR+hThe escape velocity from the surface of the Earth,
ve=2GMR+hTo escape from the orbit, the velocity required by the satellite is
vev0=2GMR+hGMR+h⇒vev0=GMR+h21=21gRHence, the correct answer is option C.