JEE Mains 2016 (Code-F)

Test Name: JEE Mains 2016 (Code-F)

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 de-Broglie 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:

2-Chloro-2-methylpentane 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 2-chloro-2-methylpentane 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: CrO2 Option 2: VO2 Option 3: MnO2 Option 4: TiO2

Solutions:Among the given compounds, CrO2 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, dust-bins, etc. Option 4: Its synthesis requires high pressure.

Solutions:Low-density 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 high-density 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 mol-1 …..1COg + 12O2g → CO2g; ∆H2 = -283.5 kJ mol-1 …..2

Subtracting (2) from (1), we get

Cs + 12O2g → COg∆H = ∆H1 – ∆H2       = -393.5 – -283.5 = -110 kJ mol-1

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.

CH3CH210CH2-OSO3-Na+

Hence, the correct answer is option A.

Question 9:

18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

 Option 1: 76 Option 2: 752.4 Option 3: 759 Option 4: 7.6

Solutions:

Number of moles of glucose, n1 =

18180 = 0.1Number of moles of water, n2 =

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 spent-lye 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 spent-lye 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: NO2 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 H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 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 H2O2 is

2 H2O2 → 2 H2O + O2

It is given that the concentration of H2O2 reduces to one-fourth of its initial value in 50 minutes.

⇒ 2t12 = 50 min⇒ t1/2 = 25 min

For a first-order reaction,

Rate constant, k = 0.693t12⇒k = 0.69325  min-1

Now, the rate law for the given reaction can be written as

-dH2O2dt = kH2O2

When the concentration of H2O2 reaches 0.05 M,

-dH2O2dt = 0.69325 × 0.05 M min-1

The rate of formation of O2 can be given as

dO2dt = -12dH2O2dt              = 12 × 0.69325 × 0.05 M min-1              = 6.93 × 10-4 M min-1

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(H2O)6]2+ and [Fe(H2O)6]2+ Option 2: [Mn(H2O)6]2+ and [Cr(H2O)6]2+ Option 3: [CoCl4]2– and [Fe(H2O)6]2+ Option 4: [Cr(H2O)6]2+ and [CoCl4]2–

Solutions:

 Complex Metal Ion d-shell Electronic Configuration Arrangement of Electrons of d-orbitals in the Complex Number of Unpaired Electrons [Cr(H2O)6]2+ Cr2+ d4 4 [Fe(H2O)6]2+ Fe2+ d6 4 [Mn(H2O)6]2+ Mn2+ d5 5 [CoCl4]2– Co2+ d7 3

The magnetic moment of a complex is dependent upon the number of unpaired electrons present in the d-orbitals of the metal ion/atom in the complex.

Since, the complexes [Cr(H2O)6]2+ and [Fe(H2O)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 Cahn-Ingold-Prelog priority (CIP) rules. The order of substituents at carbon 2 is OH > CH(Cl)(CH3) > 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 anti-clockwise and thereby resulting in S configuration.

Similarly, the order of substituents at carbon 3 is Cl > CH(OH)(COOH) > CH3 > H. They are in the anti-clockwise 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.      1-x   1-x    1+x   1+x

Keq = CDAB = 100⇒1+x1+x1-x1-x = 100⇒1+x21-x2 = 100⇒1+x1-x = 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% O2 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: C3H8 Option 2: C4H8 Option 3: C4H10 Option 4: C3H6

Solutions:

The formula of the hydrocarbon can be obtained by using the following equation and calculations.

CxHyg + x + y4 O2g → x CO2g + y2H2OlOxygen (O2) 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 (CO2) 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, C3H8 (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 (H3PO3) Pyrophosphorous Acid (H4P2O5)

Hence, the correct answer is option D.

Question 19:

Which one of the following complexes shows optical isomerism?

 Option 1: cis [Co(en)2Cl2]Cl Option 2: trans [Co(en)2Cl2]Cl Option 3: [Co(NH3)4Cl2]Cl Option 4: [Co(NH3)3Cl3]

Solutions:

Complex [Co(en2)Cl2]Cl exists in cis and trans formï¿½s, but only cis form shows optical isomerism. Complex [Co(NH3)4Cl2]Cl also exists in cis and trans formï¿½s and both are optically inactive. Complex [Co(NH3)3Cl3] 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: NO2 and NO Option 2: NO and N2O Option 3: NO2 and N2O Option 4: N2O and NO2

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: LiO2, Na2O2 and K2O Option 2: Li2O2, Na2O2 and KO2 Option 3: Li2O, Na2O2 and KO2 Option 4: Li2O, Na2O and KO2

Solutions:

When made to undergo combustion in excess of air, Li mainly forms Li2O, sodium forms Na2O2 and potassium forms KO2.

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

HS-CH2-CHNH2-COOH

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 d-block element, has the highest ionisation energy. It is due to the higher effective nuclear charge in the case of d-block elements than the s-block 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 Br2 used per mole of amine produced are:

 Option 1: Four moles of NaOH and two moles of Br2 Option 2: Two moles of NaOH and two moles of Br2 Option 3: Four moles of NaOH and one mole of Br2 Option 4: One mole of NaOH and one mole of Br2

Solutions:

The Hofmann bromamide degradation reaction is

R-CO-NH2 + Br2 + 4 NaOH → R-NH2 + 2 NaBr + Na2CO3 + 2 H2O

It is obvious that four moles of NaOH and one mole of Br2 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 pi and temperature T1 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 T2 . The final pressure pf 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 (Cl2 + H2O) proceeds through the intermediate:

 Option 1: CH3 – CH+ – CH2 – Cl Option 2: CH3 – CH(OH) – CH2+ Option 3: CH3 – CHCl – CH2+ Option 4: CH3 – CH+ – CH2 – OH

Solutions:The reaction of propene with HOCl (Cl2+H2O) proceeds through the carbocation intermediate followed by an electrophilic addition reaction.

CH3-CH=CH2 →HOCl CH3-C+H-CH2-Cl Intermediate→OH- CH3-CHOH-CH2-Cl

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, xy + 1 = 0 and 7xy – 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 xy + 1 = 0 and 7x − y − 5 = 0, respectively.

x-y+1=0 ⇒x=y-1      …..17x-y-5=0                        …..2Solving (1) and (2), we get

7y-1-y-5=0⇒7y-7-y=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 mid-point 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 xy + λ = 0 and 7x − yμ = 0, respectively. The point (−3, −6) lies on the lines xy + λ = 0 and 7x − yμ = 0. So, λ = −3 and μ = 15. Therefore, the equations of the other sides of rhombus are xy − 3 = 0 and 7x − y + 15 = 0. Solving xy − 3 = 0 and 7x − − 5 = 0, we get B =

13,-83Solving xy + 1 = 0 and 7x − + 15 = 0, we get D =

-73,-43Hence, the correct answer is option B.

Question 32:

If the 2nd, 5th and 9th terms of a non-constant 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, a2 = a + d, a5 = a + 4d and a9 = + 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, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is:

 Option 1: x2 + y2 – x + 4y –12 = 0 Option 2: x2+y2-x4+2y-24=0 Option 3: x2 + y2 – 4x + 9y + 18 = 0 Option 4: x2 + y2 – 4x + 8y + 12 = 0

Solutions:The equation of the given parabola is y2 = 8x. Here, 4a = 8 ⇒ a = 2. The equation of the circle is x2 + (y + 6)2 = 1. Centre, C = (0, −6) Let P(2t2, 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(2t2, 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

=2-02+-4+62=22             [By the distance formula] Therefore, the equation of the required circle is

x-22+y+42=222⇒x2+y2-4x+8y+12=0Hence, the correct answer is option D.

Question 34:

The system of linear equations

x + λyz = 0

λxyz = 0

x + y – λz = 0

has a non-trivial 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+λy-z=0λx-y-z=0x+y-λz=0It is given that the system of linear equations has a non-trivial solution.

∴1λ-1λ-1-111-λ=0⇒1λ+1-λ-λ2+1-1λ+1=0⇒λ+1+λ3-λ-λ-1=0⇒λ3-λ=0

⇒λλ2-1=0⇒λ=0, 1, -1Thus, the given system of linear equations has a non-trivial 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=f-x;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=f-x. 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+2f1x-2f1x-4fx=3x-6x⇒-3fx=3x-6x⇒fx=2x-xNow,

fx=f-x

⇒2x-x=-2x+x⇒4x=2x⇒x2=2⇒x=±2

∴ S=x∈R : fx=f-x=-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+3isinθ1-2isinθis purely imaginary, is :

 Option 1: π6 Option 2: sin-1 34 Option 3: sin-1 13 Option 4: π3

Solutions:

2+3isinθ1-2isinθ=2+3isinθ1-2isinθ×1+2isinθ1+2isinθ=2-6sin2θ+7isinθ1+4sin2θ=2-6sin2θ1+4sin2θ+i7sinθ1+4sin2θFor

2+3isinθ1-2isinθto be purely imaginary,

2-6sin2θ1+4sin2θ=0⇒2-6sin2θ=0⇒sinθ=13⇒θ=sin-113Hence, 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

x2a2-y2b2=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⇒4a2e2-1=a2e2            ∵b2=a2e2-1⇒4a2e2-4a2=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: 3a2 – 32a + 84 = 0 Option 2: 3a2 – 34a + 91 = 0 Option 3: 3a2 – 23a + 44 = 0 Option 4: 3a2 – 26a + 55 = 0

Solutions:The given numbers are 2, 3, a and 11.

VarX=1n∑i=1nxi2-1n∑i=1nxi2             =22+32+a2+1124-2+3+a+1142             =4+9+a2+1214-16+a42             =3a2-32a+28016We know Variance = (Standard deviation)2

∴ 3a2-32a+28016=3.52⇒3a2-32a+280=16×12.25⇒3a2-32a+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 x15, we get

I=∫2×3+5×61+1×2+1x53dx

Let 1+1×2+1×5=t⇒-2×3-5×6=dtdx⇒2×3+5x6dx=-dt∴I=∫1t3-dt=-∫1t3dt                            =–12t2+C                            =121+1×2+1×52+C                            =x102x5+x3+12+CHence, the correct answer is option A.

Question 41:

If the line,

x-32=y+2-1=z+43lies in the plane, lx + myz = 9, then l2 + m2 is equal to:

 Option 1: 18 Option 2: 5 Option 3: 2 Option 4: 26

Solutions:The equation of the given line is

x-32=y+2-1=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⇒3l-2m=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⇒2l-m=3                …..(2)Solving (1) and (2), we get l = 1 and m = −1 ∴ l2 + m2 = 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⇒4cos3x2-3cosx2+cosx2=0⇒4cos3x2-2cosx2=0⇒2cosx22cos2x2-1=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: π-83 Option 2: π-423 Option 3: π2-223 Option 4: π-43

Solutions:Consider the curves

y2=2x                 …..1×2+y2=4x          …..2Solving (1) and (2), we get

x2+2x=4x⇒x2-2x=0⇒xx-2=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=∫02ycircle-yparaboladx=∫02ycircledx -∫02yparaboladx=πr24-∫022x12dx=4π4-232×3202

=π-223232-0=π-83 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=2-a2πArea of square =

a42=a216Area of circle =

π2-a2π2=π4+a2-4a4π2=a2-4a+44π∴ The combined area of the square and circle

fa=a216+a2-4a+44π=a2π+4a2-16a+1616π∴ f’a=2aπ+8a-1616πFor maximum or minimum, f ‘(a) = 0.

f’a=0 ⇒2aπ+8a-16=0⇒2aπ+4=16⇒a=8π+4

∴ x=2π+4Also,

r=2-a2π=2-8π+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 xy + 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

x-11=y+51=z-91=λAny point on this line can be taken as

Qλ+1, λ-5, λ+9.If

Qλ+1, λ-5, λ+9lies on the plane

x-y+z=5, then

λ+1-λ-5+λ+9=5⇒λ=5-15=-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

f-12is 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⇒dydx-yx=y2⇒1y2dydx-1xy=1Let 1y=v. Then,-1y2dydx=dvdxSubstituting these values in the given differential equation, we get

-dvdx-1xv=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.

∴ 1-1=-12+C⇒C=-1+12=-12

∴ xy=-x22-12When x =

-12, we get

-12y=-14×2-12⇒-12y=-58⇒y=45Hence, the correct answer is option C.

Question 48:

If the number of terms in the expansion of

1-2x+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+3n-54=0⇒n+9n-6=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 =

1-2+46=36=729Hence, the correct answer is option C.

Question 49:

Consider

fx=tan-11+sinx1-sinx, 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=tan-11+sinx1-sinx;  x∈0, π2       =tan-1cos2x2+sin2x2+2cosx2sinx2cos2x2+sin2x2-2cosx2sinx2       =tan-1cosx2+sinx2cosx2-sinx2       =tan-11+tanx21-tanx2       =tan-1tanπ4+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

y-π3=-2x-π6⇒y-π3=-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=log2-sinxg(x) = f (f (x))

∀ x∈R ∴ gx=log2-sinlog2-sinxAt x=0, gx=log2-sinlog2-sinx∴ g’x=coslog2-sinxcosxAt x=0,g’0=coslog2Hence, the correct answer is option A.

Question 51:

Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?

 Option 1: E2 and E3 are independent Option 2: E1 and E3 are independent Option 3: E1, E2 and E3 are independent Option 4: E1 and E2 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, E1, E2 and E3 are not independent events. Hence, the statement ‘E1, E2 and E3 are independent’ is not TRUE.

Hence, the correct answer is option C.

Question 52:

If

A=5a-b32and A adj A = AAT , then 5a + b is equal to :

 Option 1: 5 Option 2: 4 Option 3: 13 Option 4: –1

Solutions:Consider the matrix:

A=5a-b32Given: A adj A = AAT

5a-b322b-35a=5a-b325a3-b2⇒10a+3b0010a+3b=25a2+b215a-2b15a-2b13Equating the above matrices,

10a+3b=25a2+b2, 10a+3b=13 and 15a-2b=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∨ (~ pq) is equivalent to :

 Option 1: p ∧ q Option 2: p ∨ q Option 3: p∨ ~ q Option 4: ~ p ∧ q

Solutions:Consider the given expression:  (p ∧ ~ q) ∨q∨ (~ p ∧ q) By commutivity, the above expression is equivalent to [ (p ∧ ~ q) ∨q ] ∨ (~ p ∧ q) = (p ∨ q) ∧ ( ~ q ∨ ) ∨ (~ p ∧ q) = (p ∨ q) ∧ [ t ∨ (~ p ∧ q)] = (p ∨ q) ∧ t = p ∨ q Hence, the correct answer is option B.

Question 54:

The sum of all real values of x satisfying the equation

x2-5x+5×2+4x-60=1is :

 Option 1: –4 Option 2: 6 Option 3: 5 Option 4: 3

Solutions:Consider the expression:

x2-5x+5×2+4x-60=1This can be wrriten as

x2-5x+5×2+4x-60=x2-5x+50Comparing by the property of exponents, we get

⇒x2+4x-60=0⇒x=-10,6Also,

x2-5x+5=-1  or x2-5x+5=1If

x2-5x+5=-1,then x=2, 3If

x2-5x+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, x2 + y2 – 8x – 8y – 4 = 0 , externally and also touch the x-axis, 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: x2 + y2 – 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,

h-42+k-42=6+kh-42+k-42=6+k2h2-8h+16+k2-8k+16=36+k2+12kh2-8h-20k-4=0h2-8h+16=20k+20x-42=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: 59th Option 2: 52nd Option 3: 58th Option 4: 46th

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=2log3-0-∫021-11+xdx⇒logL=2log3-x02+log1+x02⇒logL=2log3-2+log3⇒logL=3log3-2⇒L=e3log3-2=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 re-written 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, x2 + y2 – 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+y2-4x+6y-12=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=-3-22+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 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 ×107 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 be the mass of fat burned. Then,

⇒M×(3.8×107) ×20100=10×9.8×1×1000⇒M=12.89×10-3 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 BA and BB 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=igGi-igHere, S = Shunt resistance ig = Current through the galvanometer i = Current through the circuit G = Galvanometer resistance

⇒S=10-3×10210-10-3≃10-2 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 300-400K, 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 high-frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. And in amplitude modulation, the amplitude of the high-frequency 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: Half-life of element A, TA = 20 min Half-life of element B, TB = 40 min Ratio of decayed nuclei for A and B will be â€‹

1-NN0A1-NN0B=1-12TTA1-12TTB=1-1280201-128040=1-1161-14=54Hence, the correct answer is option C.

Question 70:

n’ mole of an ideal gas undergoes a process AB 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,

P-P0=-P0V0V-2V0⇒P=3P0-P0V0V⇒nRTV=3P0-P0V0V       P=nRTVDifferentiating it wrt V, we get ⇒V=3V02Thus,  ⇒P=3P0-P0V0V⇒P=3P0-P0V03V02=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:

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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 bmin) 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⇒2-λLa2=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 Q1Q2 and Q3, 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 will be

18 μC+24 μC=42 μCElectric field due to the point charge at a distance of 30 m will be

E=Q4πε0r2=9×109×42×10-6302=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 : X-ray 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.

Solutions:Electromagnets and transformers require a magnetic material, which has low energy losses. Hysteresis loop for such a magnetic material is represented by the figure B. Hence, the correct answer is option C.

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 co-efficient 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(T-40)     …..(1)486400=12α(T-20)     …..(2)Dividing (1) by (2), we get

-3=T-40T-20⇒-3T+60=T-40⇒T=25 °CSubstituting this value in (1),

486400=12α×(25-20)⇒α=486400×2×15=1.85×10-5 /°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: Q2πb2-a2 Option 2: 2Qπb2-a2 Option 3: 2Qπa2 Option 4: Q2πa2

Solutions:Imagine a Gaussian sphere of radius r  such that arb.

Applying Gauss’ law, we can write

E.4πr2=1ε0Q+∫arAr4πr2dr=Qε0+4πA∫arrdr=Qε0+4πAr2-a2⇒E=Q4πε0r2+A2ε01-a2r2Since 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+e-AA=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 light-dependent 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 get-w3=46mv2-mv’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=ωA2-X2(in SHM)

Given:

v=ωA2-2A32⇒v=5Aω3Also, vnew =ωAnew2-x2=3v⇒vnew =ωAnew2-2A32=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 x-y plane as shown in the figure: Which of the following statements is false for the angular momentum

 Option 1: L→=mυR2-a 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 CP and CV are molar specific heat at constant pressure and constant volume, respectively):

 Option 1: n=C-CPC-CV Option 2: n=CP-CC-CV Option 3: n=C-CVC-CP Option 4: n=CVCP

Solutions:We know

C=CV+R1-n⇒C-CV=CP-CV1-n⇒1-n=CP-CVC-CV⇒n=1-CP-CVC-CV=C-CPC-CVHence, 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 45th 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 25th 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=β1-β Option 2: α=β1+β Option 3: α=β21+β2+2β Option 4: 1α=1β+1

Solutions:

β=α1-α⇒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: gR2-1 Option 4: 2gR