JEE Main 2016 — 3 Apr Offline Set F (Previous Year Paper)

JEE Main · 2016 · with answers and explanations

questions
87
Total marks
348
Duration
180 minutes
Negative marking
−1

Subjects: Chemistry · Mathematics · Physics

Question 1

Chemistry · 4 marks · −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 h/λh/\lambda (where λ\lambda is wavelength associated with electron wave) is given by :

  1. A.2meV2meV
  2. B.meV\sqrt{meV}
  3. C.2meV\sqrt{2meV}
  4. D.meVmeV

Correct answer: C

Explanation: h/λ=p=2m KE=2meVh/\lambda=p=\sqrt{2m\,KE}=\sqrt{2meV}.

Question 2

Chemistry · 4 marks · −1

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields : (a) C2H5CH2C(CH3)2OCH3C_2H_5CH_2C(CH_3)_2OCH_3 (b) C2H5CH2C(CH3)=CH2C_2H_5CH_2C(CH_3)=CH_2 (c) C2H5CH=C(CH3)CH3C_2H_5CH=C(CH_3)CH_3

  1. A.(a) and (c)
  2. B.(c) only
  3. C.(a) and (b)
  4. D.All of these

Correct answer: D

Explanation: The tertiary halide gives the substitution ether via SN1S_N1 and both alkenes via elimination (E1/E2).

Question 3

Chemistry · 4 marks · −1

Which of the following compounds is metallic and ferromagnetic ?

  1. A.CrO2CrO_2
  2. B.VO2VO_2
  3. C.MnO2MnO_2
  4. D.TiO2TiO_2

Correct answer: A

Explanation: CrO2CrO_2 is a metallic, ferromagnetic oxide (used in magnetic tapes).

Question 4

Chemistry · 4 marks · −1

Which of the following statements about low density polythene is FALSE ?

  1. A.It is a poor conductor of electricity.
  2. B.Its synthesis requires dioxygen or a peroxide initiator as a catalyst.
  3. C.It is used in the manufacture of buckets, dust-bins etc.
  4. D.Its synthesis requires high pressure.

Correct answer: C

Explanation: Buckets and dustbins are made from high density polythene, not LDPE.

Question 5

Chemistry · 4 marks · −1

For a linear plot of log⁡(x/m)\log(x/m) versus log⁡p\log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constants)

  1. A.1/n1/n appears as the intercept.
  2. B.Only 1/n1/n appears as the slope
  3. C.log⁡(1/n)\log(1/n) appears as the intercept.
  4. D.Both k and 1/n1/n appear in the slope term.

Correct answer: B

Explanation: log⁡xm=log⁡k+1nlog⁡p\log\frac xm=\log k+\frac1n\log p: slope 1n\frac1n, intercept log⁡k\log k.

Question 6

Chemistry · 4 marks · −1

The heats of combustion of carbon and carbon monoxide are −393.5-393.5 and −283.5-283.5 kJ mol−1^{-1}, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :

  1. A.676.5
  2. B.−676.5-676.5
  3. C.−110.5-110.5
  4. D.110.5

Correct answer: C

Explanation: ΔfH(CO)=−393.5−(−283.5)=−110\Delta_fH(CO)=-393.5-(-283.5)=-110 kJ/mol (closest option −110.5-110.5).

Question 7

Chemistry · 4 marks · −1

The hottest region of Bunsen flame shown in the figure below is :

Figure for question 7
  1. A.region 2
  2. B.region 3
  3. C.region 4
  4. D.region 1

Correct answer: A

Explanation: Region 2 (just above the inner blue cone, the zone of complete combustion) is the hottest.

Question 8

Chemistry · 4 marks · −1

Which of the following is an anionic detergent ?

  1. A.Sodium lauryl sulphate
  2. B.Cetyltrimethyl ammonium bromide
  3. C.Glyceryl oleate
  4. D.Sodium stearate

Correct answer: A

Explanation: Sodium lauryl sulphate is an anionic synthetic detergent (sodium stearate is a soap).

Question 9

Chemistry · 4 marks · −1

18 g glucose (C6H12O6)(C_6H_{12}O_6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is :

  1. A.76.0
  2. B.752.4
  3. C.759.0
  4. D.7.6

Correct answer: B

Explanation: xsolute=0.10.1+9.9=0.01x_{solute}=\frac{0.1}{0.1+9.9}=0.01; taking p∘=760p^\circ=760 torr (water at its boiling point), p=760(0.99)=752.4p=760(0.99)=752.4 torr.

Question 10

Chemistry · 4 marks · −1

The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :

  1. A.Fractional distillation
  2. B.Steam distillation
  3. C.Distillation under reduced pressure
  4. D.Simple distillation

Correct answer: C

Explanation: Glycerol decomposes at its normal boiling point, so it is distilled under reduced pressure.

Question 11

Chemistry · 4 marks · −1

The species in which the N atom is in a state of sp hybridization is :

  1. A.NO2−NO_2^-
  2. B.NO3−NO_3^-
  3. C.NO2NO_2
  4. D.NO2+NO_2^+

Correct answer: D

Explanation: NO2+NO_2^+ is linear (O=N+=OO=N^+=O), sp hybridised.

Question 12

Chemistry · 4 marks · −1

Decomposition of H2O2H_2O_2 follows a first order reaction. In fifty minutes the concentration of H2O2H_2O_2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2H_2O_2 reaches 0.05 M, the rate of formation of O2O_2 will be :

  1. A.6.93×10−46.93\times10^{-4} mol min−1^{-1}
  2. B.2.66 L min−1^{-1} at STP
  3. C.1.34×10−21.34\times10^{-2} mol min−1^{-1}
  4. D.6.93×10−26.93\times10^{-2} mol min−1^{-1}

Correct answer: A

Explanation: Two half-lives in 50 min: t1/2=25t_{1/2}=25 min, k=0.0277k=0.0277 min−1^{-1}. Rate of H2O2H_2O_2 loss =1.39×10−3=1.39\times10^{-3}; O2O_2 forms at half this, 6.93×10−46.93\times10^{-4} (per litre).

Question 13

Chemistry · 4 marks · −1

The pair having the same magnetic moment is : [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]

  1. A.[Cr(H2O)6]2+[Cr(H_2O)_6]^{2+} and [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}
  2. B.[Mn(H2O)6]2+[Mn(H_2O)_6]^{2+} and [Cr(H2O)6]2+[Cr(H_2O)_6]^{2+}
  3. C.[CoCl4]2−[CoCl_4]^{2-} and [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}
  4. D.[Cr(H2O)6]2+[Cr(H_2O)_6]^{2+} and [CoCl4]2−[CoCl_4]^{2-}

Correct answer: A

Explanation: Cr2+Cr^{2+} (d4d^4) and Fe2+Fe^{2+} (d6d^6) with weak-field water both have 4 unpaired electrons.

Question 14

Chemistry · 4 marks · −1

The absolute configuration of the Fischer projection with CO2HCO_2H at top, CH3CH_3 at bottom, C2 bearing H (left) and OH (right), C3 bearing H (left) and Cl (right) is :

Figure for question 14
  1. A.(2S, 3R)
  2. B.(2S, 3S)
  3. C.(2R, 3R)
  4. D.(2R, 3S)

Correct answer: A

Explanation: Assigning CIP priorities in the Fischer projection gives C2 = S and C3 = R.

Question 15

Chemistry · 4 marks · −1

The equilibrium constant at 298 K for a reaction A+B⇌C+DA + B \rightleftharpoons C + D is 100. If the initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol L−1^{-1}) will be:

  1. A.0.818
  2. B.1.818
  3. C.1.182
  4. D.0.182

Correct answer: B

Explanation: (1+x)2(1−x)2=100⇒x=911\frac{(1+x)^2}{(1-x)^2}=100\Rightarrow x=\frac9{11}; [D]=1.818[D]=1.818 M.

Question 16

Chemistry · 4 marks · −1

Which one of the following ores is best concentrated by froth floatation method?

  1. A.Siderite
  2. B.Galena
  3. C.Malachite
  4. D.Magnetite

Correct answer: B

Explanation: Froth flotation suits sulphide ores; galena is PbS.

Question 17

Chemistry · 4 marks · −1

The pair in which phosphorous atoms have a formal oxidation state of +3 is:

  1. A.Pyrophosphorous and hypophosphoric acids
  2. B.Orthophosphorous and hypophosphoric acids
  3. C.Pyrophosphorous and pyrophosphoric acids
  4. D.Orthophosphorous and pyrophosphorous acids

Correct answer: D

Explanation: H3PO3H_3PO_3 and H4P2O5H_4P_2O_5 both have P in +3; hypophosphoric is +4, pyrophosphoric +5.

Question 18

Chemistry · 4 marks · −1

Which one of the following complexes shows optical isomerism? (en = ethylenediamine)

  1. A.cis[Co(en)2Cl2]Cl[Co(en)_2Cl_2]Cl
  2. B.trans[Co(en)2Cl2]Cl[Co(en)_2Cl_2]Cl
  3. C.[Co(NH3)4Cl2]Cl[Co(NH_3)_4Cl_2]Cl
  4. D.[Co(NH3)3Cl3][Co(NH_3)_3Cl_3]

Correct answer: A

Explanation: cis-[Co(en)2Cl2]+[Co(en)_2Cl_2]^+ lacks a mirror plane and is chiral.

Question 19

Chemistry · 4 marks · −1

The reaction of zinc with dilute and concentrated nitric acid, respectively, produces:

  1. A.NO2NO_2 and NONO
  2. B.NONO and N2ON_2O
  3. C.NO2NO_2 and N2ON_2O
  4. D.N2ON_2O and NO2NO_2

Correct answer: D

Explanation: Dilute: 4Zn+10HNO3→4Zn(NO3)2+N2O+5H2O4Zn+10HNO_3\rightarrow4Zn(NO_3)_2+N_2O+5H_2O; concentrated gives NO2NO_2.

Question 20

Chemistry · 4 marks · −1

Which one of the following statements about water is FALSE?

  1. A.Water can act both as an acid and as a base.
  2. B.There is extensive intramolecular hydrogen bonding in the condensed phase.
  3. C.Ice formed by heavy water sinks in normal water
  4. D.Water is oxidized to oxygen during photosynthesis

Correct answer: B

Explanation: Hydrogen bonding in water/ice is intermolecular, not intramolecular.

Question 21

Chemistry · 4 marks · −1

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 :

  1. A.Lead
  2. B.Nitrate
  3. C.Iron
  4. D.Fluoride

Correct answer: B

Explanation: Nitrate limit is 50 ppm; 100 ppm exceeds it. Fluoride 1 ppm, lead 40 ppb... (lead limit 50 ppb) and iron 0.2 ppm are within limits.

Question 22

Chemistry · 4 marks · −1

The main oxides formed on combustion of Li, Na and K in excess of air are, respectively:

  1. A.LiO2LiO_2, Na2O2Na_2O_2 and K2OK_2O
  2. B.Li2O2Li_2O_2, Na2O2Na_2O_2 and KO2KO_2
  3. C.Li2OLi_2O, Na2O2Na_2O_2 and KO2KO_2
  4. D.Li2OLi_2O, Na2ONa_2O and KO2KO_2

Correct answer: C

Explanation: Li forms the monoxide, Na the peroxide, K the superoxide.

Question 23

Chemistry · 4 marks · −1

Thiol group is present in :

  1. A.Cystine
  2. B.Cysteine
  3. C.Methionine
  4. D.Cytosine

Correct answer: B

Explanation: Cysteine has a free −SH-SH group (cystine is its disulfide, methionine a thioether).

Question 24

Chemistry · 4 marks · −1

Galvanization is applying a coating of:

  1. A.Cr
  2. B.Cu
  3. C.Zn
  4. D.Pb

Correct answer: C

Explanation: Galvanisation is coating iron with zinc.

Question 25

Chemistry · 4 marks · −1

Which of the following atoms has the highest first ionization energy?

  1. A.Na
  2. B.K
  3. C.Sc
  4. D.Rb

Correct answer: C

Explanation: Sc (a d-block element with higher effective nuclear charge) has a higher IE than the alkali metals.

Question 26

Chemistry · 4 marks · −1

In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2Br_2 used per mole of amine produced are:

  1. A.Four moles of NaOH and two moles of Br2Br_2.
  2. B.Two moles of NaOH and two moles of Br2Br_2
  3. C.Four moles of NaOH and one mole of Br2Br_2
  4. D.One mole of NaOH and one mole of Br2Br_2

Correct answer: C

Explanation: RCONH2+Br2+4NaOH→RNH2+2NaBr+Na2CO3+2H2ORCONH_2+Br_2+4NaOH\rightarrow RNH_2+2NaBr+Na_2CO_3+2H_2O.

Question 27

Chemistry · 4 marks · −1

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pip_i and temperature T1T_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 T2T_2. The final pressure pfp_f is

Figure for question 27
  1. A.2pi(T1T1+T2)2p_i\left(\frac{T_1}{T_1+T_2}\right)
  2. B.2pi(T2T1+T2)2p_i\left(\frac{T_2}{T_1+T_2}\right)
  3. C.2pi(T1T2T1+T2)2p_i\left(\frac{T_1T_2}{T_1+T_2}\right)
  4. D.pi(T1T2T1+T2)p_i\left(\frac{T_1T_2}{T_1+T_2}\right)

Correct answer: B

Explanation: Moles conserved: 2piT1=pfT1+pfT2⇒pf=2piT2T1+T2\frac{2p_i}{T_1}=\frac{p_f}{T_1}+\frac{p_f}{T_2}\Rightarrow p_f=\frac{2p_iT_2}{T_1+T_2}.

Question 28

Chemistry · 4 marks · −1

The reaction of propene with HOCl (Cl2+H2O)(Cl_2 + H_2O) proceeds through the intermediate:

  1. A.CH3−CH+−CH2−ClCH_3-CH^+-CH_2-Cl
  2. B.CH3−CH(OH)−CH2+CH_3-CH(OH)-CH_2^+
  3. C.CH3−CHCl−CH2+CH_3-CHCl-CH_2^+
  4. D.CH3−CH+−CH2−OHCH_3-CH^+-CH_2-OH

Correct answer: A

Explanation: Cl+Cl^+ adds to the terminal carbon to give the more stable secondary carbocation (Markovnikov); water then attacks it.

Question 29

Chemistry · 4 marks · −1

The product of the reaction given below is: 4-tert-butylcyclohexene →1. NBS/hν; 2. H2O/K2CO3\xrightarrow{1.\ NBS/h\nu;\ 2.\ H_2O/K_2CO_3} X

Figure for question 29
  1. A.5-tert-butylcyclohex-2-en-1-ol
  2. B.5-tert-butylcyclohex-2-en-1-one
  3. C.5-tert-butylcyclohex-2-ene-1-carboxylic acid
  4. D.4-tert-butylcyclohexene (unchanged)

Correct answer: A

Explanation: NBS brominates the allylic position; aqueous K2CO3K_2CO_3 hydrolyses the allylic bromide to the allylic alcohol.

Question 30

Mathematics · 4 marks · −1

Two sides of a rhombus are along the lines, x−y+1=0x - y + 1 = 0 and 7x−y−5=07x - y - 5 = 0. If its diagonals intersect at (−1,−2)(-1, -2), then which one of the following is a vertex of this rhombus?

  1. A.(−3,−8)(-3, -8)
  2. B.(13,−83)\left(\frac{1}{3}, -\frac{8}{3}\right)
  3. C.(−103,−73)\left(-\frac{10}{3}, -\frac{7}{3}\right)
  4. D.(−3,−9)(-3, -9)

Correct answer: B

Explanation: The given sides meet at (1,2)(1,2); the opposite vertex is (−3,−6)(-3,-6). Opposite sides through it: x−y−3=0x-y-3=0, 7x−y+15=07x-y+15=0. Other vertices: (13,−83)(\frac13,-\frac83) and (−73,−43)(-\frac73,-\frac43).

Question 31

Mathematics · 4 marks · −1

If the 2nd2^{nd}, 5th5^{th} and 9th9^{th} terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

  1. A.43\frac{4}{3}
  2. B.1
  3. C.74\frac{7}{4}
  4. D.85\frac{8}{5}

Correct answer: A

Explanation: (a+4d)2=(a+d)(a+8d)⇒a=8d(a+4d)^2=(a+d)(a+8d)\Rightarrow a=8d; ratio a+4da+d=129=43\frac{a+4d}{a+d}=\frac{12}{9}=\frac43.

Question 32

Mathematics · 4 marks · −1

Let P be the point on the parabola, y2=8xy^2 = 8x which is at a minimum distance from the centre C of the circle, x2+(y+6)2=1x^2 + (y + 6)^2 = 1. Then the equation of the circle, passing through C and having its centre at P is:

  1. A.x2+y2−x+4y−12=0x^2 + y^2 - x + 4y - 12 = 0
  2. B.x2+y2−x4+2y−24=0x^2 + y^2 - \frac{x}{4} + 2y - 24 = 0
  3. C.x2+y2−4x+9y+18=0x^2 + y^2 - 4x + 9y + 18 = 0
  4. D.x2+y2−4x+8y+12=0x^2 + y^2 - 4x + 8y + 12 = 0

Correct answer: D

Explanation: Normal at (2t2,4t)(2t^2,4t): y=−tx+4t+2t3y=-tx+4t+2t^3 passes through (0,−6)(0,-6): t=−1t=-1, P =(2,−4)=(2,-4), radius =22=2\sqrt2: (x−2)2+(y+4)2=8(x-2)^2+(y+4)^2=8.

Question 33

Mathematics · 4 marks · −1

The system of linear equations x+λy−z=0x + \lambda y - z = 0, λx−y−z=0\lambda x - y - z = 0, x+y−λz=0x + y - \lambda z = 0 has a non-trivial solution for :

  1. A.exactly one value of λ\lambda.
  2. B.exactly two values of λ\lambda.
  3. C.exactly three values of λ\lambda.
  4. D.infinitely many values of λ\lambda.

Correct answer: C

Explanation: Determinant =λ(λ2−1)=0=\lambda(\lambda^2-1)=0: λ=0,±1\lambda=0,\pm1.

Question 34

Mathematics · 4 marks · −1

If f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x, x≠0x \ne 0, and S={x∈R:f(x)=f(−x)}S = \{x \in \mathbb{R} : f(x) = f(-x)\}; then S:

  1. A.contains exactly one element.
  2. B.contains exactly two elements.
  3. C.contains more than two elements.
  4. D.is an empty set.

Correct answer: B

Explanation: Solving, f(x)=2x−xf(x)=\frac2x-x; f(x)=f(−x)⇒4x=2x⇒x=±2f(x)=f(-x)\Rightarrow\frac4x=2x\Rightarrow x=\pm\sqrt2.

Question 35

Mathematics · 4 marks · −1

Let p=lim⁡x→0+(1+tan⁡2x)12xp = \lim_{x\to0^+}\left(1 + \tan^2\sqrt{x}\right)^{\frac{1}{2x}} then log⁡p\log p is equal to:

  1. A.1
  2. B.12\frac{1}{2}
  3. C.14\frac{1}{4}
  4. D.2

Correct answer: B

Explanation: log⁡p=lim⁡tan⁡2x2x=12\log p=\lim\frac{\tan^2\sqrt x}{2x}=\frac12.

Question 36

Mathematics · 4 marks · −1

A value of θ\theta for which 2+3isin⁡θ1−2isin⁡θ\frac{2 + 3i\sin\theta}{1 - 2i\sin\theta} is purely imaginary, is:

  1. A.π6\frac{\pi}{6}
  2. B.sin⁡−1(34)\sin^{-1}\left(\frac{\sqrt{3}}{4}\right)
  3. C.sin⁡−1(13)\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)
  4. D.π3\frac{\pi}{3}

Correct answer: C

Explanation: Real part ∝2−6sin⁡2θ=0⇒sin⁡θ=13\propto 2-6\sin^2\theta=0\Rightarrow\sin\theta=\frac1{\sqrt3}.

Question 37

Mathematics · 4 marks · −1

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal half of the distance between its foci, is:

  1. A.43\frac{4}{\sqrt{3}}
  2. B.23\frac{2}{\sqrt{3}}
  3. C.3\sqrt{3}
  4. D.43\frac{4}{3}

Correct answer: B

Explanation: 2b=ae⇒4b2=a2e2=4a2(e2−1)2b=ae\Rightarrow4b^2=a^2e^2=4a^2(e^2-1), so e2=43e^2=\frac43.

Question 38

Mathematics · 4 marks · −1

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

  1. A.3a2−32a+84=03a^2 - 32a + 84 = 0
  2. B.3a2−34a+91=03a^2 - 34a + 91 = 0
  3. C.3a2−23a+44=03a^2 - 23a + 44 = 0
  4. D.3a2−26a+55=03a^2 - 26a + 55 = 0

Correct answer: A

Explanation: 134+a24−(16+a4)2=494\frac{134+a^2}{4}-\left(\frac{16+a}{4}\right)^2=\frac{49}{4} simplifies to 3a2−32a+84=03a^2-32a+84=0.

Question 39

Mathematics · 4 marks · −1

The integral ∫2x12+5x9(x5+x3+1)3 dx\int \frac{2x^{12} + 5x^9}{\left(x^5 + x^3 + 1\right)^3}\,dx is equal to: (where C is an arbitrary constant)

  1. A.x102(x5+x3+1)2+C\frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C
  2. B.x52(x5+x3+1)2+C\frac{x^5}{2\left(x^5+x^3+1\right)^2}+C
  3. C.−x102(x5+x3+1)2+C\frac{-x^{10}}{2\left(x^5+x^3+1\right)^2}+C
  4. D.−x5(x5+x3+1)2+C\frac{-x^5}{\left(x^5+x^3+1\right)^2}+C

Correct answer: A

Explanation: Divide by x15x^{15}: ∫2x−3+5x−6(1+x−2+x−5)3dx\int\frac{2x^{-3}+5x^{-6}}{(1+x^{-2}+x^{-5})^3}dx; with t=1+x−2+x−5t=1+x^{-2}+x^{-5} this is 12t2=x102(x5+x3+1)2\frac{1}{2t^2}=\frac{x^{10}}{2(x^5+x^3+1)^2}.

Question 40

Mathematics · 4 marks · −1

If the line, x−32=y+2−1=z+43\frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+4}{3} lies in the plane, lx+my−z=9lx + my - z = 9, then l2+m2l^2 + m^2 is equal to:

  1. A.18
  2. B.5
  3. C.2
  4. D.26

Correct answer: C

Explanation: 2l−m−3=02l-m-3=0 and 3l−2m+4=93l-2m+4=9 give l=1l=1, m=−1m=-1.

Question 41

Mathematics · 4 marks · −1

If 0≤x<2π0 \le x < 2\pi, then the number of real values of x, which satisfy the equation cos⁡x+cos⁡2x+cos⁡3x+cos⁡4x=0\cos x + \cos 2x + \cos 3x + \cos 4x = 0 is:

  1. A.5
  2. B.7
  3. C.9
  4. D.3

Correct answer: B

Explanation: 4cos⁡5x2cos⁡xcos⁡x2=04\cos\frac{5x}{2}\cos x\cos\frac x2=0: x=πx=\pi; π2,3π2\frac\pi2,\frac{3\pi}2; π5,3π5,7π5,9π5\frac\pi5,\frac{3\pi}5,\frac{7\pi}5,\frac{9\pi}5 (π\pi repeats): 7 values.

Question 42

Mathematics · 4 marks · −1

The area (in sq. units) of the region {(x,y):y2≥2x and x2+y2≤4x, x≥0, y≥0}\{(x, y) : y^2 \ge 2x \text{ and } x^2 + y^2 \le 4x,\ x \ge 0,\ y \ge 0\} is:

  1. A.π−83\pi - \frac{8}{3}
  2. B.π−423\pi - \frac{4\sqrt{2}}{3}
  3. C.π2−223\frac{\pi}{2} - \frac{2\sqrt{2}}{3}
  4. D.π−43\pi - \frac{4}{3}

Correct answer: A

Explanation: Curves meet at (2,2)(2,2). Area == quarter of the circle (π\pi) minus ∫022x dx=83\int_0^2\sqrt{2x}\,dx=\frac83.

Question 43

Mathematics · 4 marks · −1

Let a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} be three unit vectors such that a⃗×(b⃗×c⃗)=32(b⃗+c⃗)\vec{a}\times(\vec{b}\times\vec{c}) = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c}). If b⃗\vec{b} is not parallel to c⃗\vec{c}, then the angle between a⃗\vec{a} and b⃗\vec{b} is:

  1. A.π2\frac{\pi}{2}
  2. B.2π3\frac{2\pi}{3}
  3. C.5π6\frac{5\pi}{6}
  4. D.3π4\frac{3\pi}{4}

Correct answer: C

Explanation: (a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗=32(b⃗+c⃗)⇒a⃗⋅b⃗=−32(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c=\frac{\sqrt3}{2}(\vec b+\vec c)\Rightarrow\vec a\cdot\vec b=-\frac{\sqrt3}{2}, angle 5π6\frac{5\pi}{6}.

Question 44

Mathematics · 4 marks · −1

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:

  1. A.(4−π)x=πr(4 - \pi)x = \pi r
  2. B.x=2rx = 2r
  3. C.2x=r2x = r
  4. D.2x=(π+4)r2x = (\pi + 4)r

Correct answer: B

Explanation: 4x+2πr=24x+2\pi r=2; minimising x2+πr2x^2+\pi r^2 gives x=2π+4x=\frac{2}{\pi+4}, r=1π+4r=\frac{1}{\pi+4}, so x=2rx=2r.

Question 45

Mathematics · 4 marks · −1

The distance of the point (1,−5,9)(1, -5, 9) from the plane x−y+z=5x - y + z = 5 measured along the line x=y=zx = y = z is:

  1. A.10310\sqrt{3}
  2. B.103\frac{10}{\sqrt{3}}
  3. C.203\frac{20}{3}
  4. D.3103\sqrt{10}

Correct answer: A

Explanation: Point (1+t,−5+t,9+t)(1+t,-5+t,9+t) on the plane: 15+t=515+t=5, t=−10t=-10; distance 10310\sqrt3.

Question 46

Mathematics · 4 marks · −1

If a curve y=f(x)y = f(x) passes through the point (1,−1)(1, -1) and satisfies the differential equation, y(1+xy) dx=x dyy(1 + xy)\,dx = x\,dy, then f(−12)f\left(-\frac{1}{2}\right) is equal to:

  1. A.−45-\frac{4}{5}
  2. B.25\frac{2}{5}
  3. C.45\frac{4}{5}
  4. D.−25-\frac{2}{5}

Correct answer: C

Explanation: x dy−y dxy2=x dx⇒−xy=x22+c\frac{x\,dy-y\,dx}{y^2}=x\,dx\Rightarrow-\frac xy=\frac{x^2}{2}+c, c=12c=\frac12; y=−2xx2+1y=\frac{-2x}{x^2+1}, f(−12)=45f(-\frac12)=\frac45.

Question 47

Mathematics · 4 marks · −1

If the number of terms in the expansion of (1−2x+4x2)n\left(1 - \frac{2}{x} + \frac{4}{x^2}\right)^n, x≠0x \ne 0, is 28, then the sum of the coefficients of all the terms in this expansion, is:

  1. A.2187
  2. B.243
  3. C.729
  4. D.64

Correct answer: C

Explanation: Treating terms as distinct: (n+22)=28⇒n=6\binom{n+2}{2}=28\Rightarrow n=6; sum =(1−2+4)6=729=(1-2+4)^6=729.

Question 48

Mathematics · 4 marks · −1

Consider f(x)=tan⁡−1(1+sin⁡x1−sin⁡x)f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right). A normal to y=f(x)y = f(x) at x=π6x = \frac{\pi}{6} also passes through the point:

  1. A.(0,2π3)\left(0, \frac{2\pi}{3}\right)
  2. B.(π6,0)\left(\frac{\pi}{6}, 0\right)
  3. C.(π4,0)\left(\frac{\pi}{4}, 0\right)
  4. D.(0,0)(0, 0)

Correct answer: A

Explanation: f(x)=π4+x2f(x)=\frac\pi4+\frac x2; at π6\frac\pi6, f=π3f=\frac\pi3, slope 12\frac12. Normal y−π3=−2(x−π6)y-\frac\pi3=-2(x-\frac\pi6) passes (0,2π3)(0,\frac{2\pi}3).

Question 49

Mathematics · 4 marks · −1

For x∈Rx \in \mathbb{R}, f(x)=∣log⁡2−sin⁡x∣f(x) = |\log 2 - \sin x| and g(x)=f(f(x))g(x) = f(f(x)), then:

  1. A.g′(0)=cos⁡(log⁡2)g'(0) = \cos(\log 2)
  2. B.g′(0)=−cos⁡(log⁡2)g'(0) = -\cos(\log 2)
  3. C.g is differentiable at x=0x = 0 and g′(0)=−sin⁡(log⁡2)g'(0) = -\sin(\log 2)
  4. D.g is not differentiable at x=0x = 0

Correct answer: A

Explanation: Near 0, f(x)=log⁡2−sin⁡xf(x)=\log2-\sin x, f(0)=log⁡2f(0)=\log2, f′(0)=−1f'(0)=-1; near log⁡2\log2 also f=log⁡2−sin⁡xf=\log2-\sin x. g′(0)=f′(log⁡2)f′(0)=cos⁡(log⁡2)g'(0)=f'(\log2)f'(0)=\cos(\log2).

Question 50

Mathematics · 4 marks · −1

Let two fair six-faced dice A and B be thrown simultaneously. If E1E_1 is the event that die A shows up four, E2E_2 is the event that die B shows up two and E3E_3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

  1. A.E2E_2 and E3E_3 are independent.
  2. B.E1E_1 and E3E_3 are independent.
  3. C.E1E_1, E2E_2 and E3E_3 are independent.
  4. D.E1E_1 and E2E_2 are independent.

Correct answer: C

Explanation: Pairwise independent, but P(E1E2E3)=0≠172P(E_1E_2E_3)=0\ne\frac1{72}.

Question 51

Mathematics · 4 marks · −1

If A=[5a−b32]A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} and A adjA=AATA\,\text{adj}A = AA^T, then 5a+b5a + b is equal to:

  1. A.5
  2. B.4
  3. C.13
  4. D.−1-1

Correct answer: A

Explanation: ∣A∣I=AAT|A|I=AA^T: 15a−2b=015a-2b=0 and 10a+3b=1310a+3b=13, so a=25a=\frac25, b=3b=3, 5a+b=55a+b=5.

Question 52

Mathematics · 4 marks · −1

The Boolean Expression (p∧∼q)∨q∨(∼p∧q)(p\wedge\sim q)\vee q\vee(\sim p\wedge q) is equivalent to:

  1. A.p∧qp \wedge q
  2. B.p∨qp \vee q
  3. C.p∨∼qp \vee \sim q
  4. D.∼p∧q\sim p \wedge q

Correct answer: B

Explanation: (p∧∼q)∨q=p∨q(p\wedge\sim q)\vee q=p\vee q; adding (∼p∧q)⊆q(\sim p\wedge q)\subseteq q changes nothing.

Question 53

Mathematics · 4 marks · −1

The sum of all real values of x satisfying the equation (x2−5x+5)x2+4x−60=1\left(x^2 - 5x + 5\right)^{x^2 + 4x - 60} = 1 is:

  1. A.−4-4
  2. B.6
  3. C.5
  4. D.3

Correct answer: D

Explanation: Base 1: x=1,4x=1,4; exponent 0: x=−10,6x=-10,6; base −1-1 with even exponent: x=2x=2 (for x=3x=3 the exponent is odd). Sum =3=3.

Question 54

Mathematics · 4 marks · −1

The centres of those circles which touch the circle, x2+y2−8x−8y−4=0x^2 + y^2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:

  1. A.an ellipse which is not a circle.
  2. B.a hyperbola.
  3. C.a parabola.
  4. D.a circle.

Correct answer: C

Explanation: (h−4)2+(k−4)2=(∣k∣+6)2(h-4)^2+(k-4)^2=(|k|+6)^2 gives x2−8x−20y−4=0x^2-8x-20y-4=0 for y>0y>0: a parabola.

Question 55

Mathematics · 4 marks · −1

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:

  1. A.59th59^{th}
  2. B.52nd52^{nd}
  3. C.58th58^{th}
  4. D.46th46^{th}

Correct answer: C

Explanation: Starting with A, L, M: 12+24+12=4812+24+12=48; with SA, SL: 3+6=93+6=9; SMALL is next: 58th58^{th}.

Question 56

Mathematics · 4 marks · −1

lim⁡n→∞((n+1)(n+2)…3nn2n)1/n\lim_{n\to\infty}\left(\frac{(n+1)(n+2)\dots3n}{n^{2n}}\right)^{1/n} is equal to:

  1. A.27e2\frac{27}{e^2}
  2. B.9e2\frac{9}{e^2}
  3. C.3log⁡3−23\log3 - 2
  4. D.18e4\frac{18}{e^4}

Correct answer: A

Explanation: ln⁡y=∫02ln⁡(1+x)dx=3ln⁡3−2\ln y=\int_0^2\ln(1+x)dx=3\ln3-2, so y=27e2y=\frac{27}{e^2}.

Question 57

Mathematics · 4 marks · −1

If the sum of the first ten terms of the series (135)2+(225)2+(315)2+42+(445)2+…\left(1\frac{3}{5}\right)^2 + \left(2\frac{2}{5}\right)^2 + \left(3\frac{1}{5}\right)^2 + 4^2 + \left(4\frac{4}{5}\right)^2 + \dots, is 165m\frac{16}{5}m, then m is equal to:

  1. A.101
  2. B.100
  3. C.99
  4. D.102

Correct answer: A

Explanation: Terms are (4k+45)2\left(\frac{4k+4}{5}\right)^2: sum =1625(22+⋯+112)=1625⋅505=165⋅101=\frac{16}{25}(2^2+\dots+11^2)=\frac{16}{25}\cdot505=\frac{16}{5}\cdot101.

Question 58

Mathematics · 4 marks · −1

If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0, is a chord of a circle S, whose centre is at (−3,2)(-3, 2), then the radius of S is:

  1. A.535\sqrt{3}
  2. B.5
  3. C.10
  4. D.525\sqrt{2}

Correct answer: A

Explanation: Given circle: centre (2,−3)(2,-3), radius 5. Distance between centres 525\sqrt2; r2=25+50=75r^2=25+50=75.

Question 59

Mathematics · 4 marks · −1

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∘30^\circ. 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∘60^\circ. Then the time taken (in minutes) by him, from B to reach the pillar, is:

  1. A.10
  2. B.20
  3. C.5
  4. D.6

Correct answer: C

Explanation: h=3xh=\sqrt3x and h=x+y3⇒y=2xh=\frac{x+y}{\sqrt3}\Rightarrow y=2x. AB takes 10 min, so B to pillar takes 5 min.

Question 60

Physics · 4 marks · −1

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=10g = 10 ms−2^{-2})

  1. A.2s
  2. B.222\sqrt{2} s
  3. C.2\sqrt{2} s
  4. D.2π22\pi\sqrt{2} s

Correct answer: B

Explanation: v=gxv=\sqrt{gx} at height x from the bottom; t=∫0Ldxgx=2L/g=22t=\int_0^L\frac{dx}{\sqrt{gx}}=2\sqrt{L/g}=2\sqrt2 s.

Question 61

Physics · 4 marks · −1

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 3.8×1073.8\times10^7 J of energy per kg which is converted to mechanical energy with a 20 % efficiency rate. Take g=9.8g = 9.8 ms−2^{-2} :

  1. A.6.45×10−36.45\times10^{-3} kg
  2. B.9.89×10−39.89\times10^{-3} kg
  3. C.12.89×10−312.89\times10^{-3} kg
  4. D.2.45×10−32.45\times10^{-3} kg

Correct answer: C

Explanation: Work =10⋅9.8⋅1000=9.8×104=10\cdot9.8\cdot1000=9.8\times10^4 J =x⋅3.8×107⋅0.2=x\cdot3.8\times10^7\cdot0.2, so x≈12.89×10−3x\approx12.89\times10^{-3} kg.

Question 62

Physics · 4 marks · −1

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\mu. 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\mu and the distance x(=QR)x(=QR), are, respectively close to :

Figure for question 62
  1. A.0.2 and 3.5 m
  2. B.0.29 and 3.5 m
  3. C.0.29 and 6.5 m
  4. D.0.2 and 6.5 m

Correct answer: B

Explanation: Incline PQ has height 2 m at 30∘30^\circ, length 4 m. Equal losses: μmgcos⁡30∘⋅4=μmgx⇒x=23≈3.5\mu mg\cos30^\circ\cdot4=\mu mgx\Rightarrow x=2\sqrt3\approx3.5 m. Energy: mg⋅2=2μmg⋅23⇒μ=123≈0.29mg\cdot2=2\mu mg\cdot2\sqrt3\Rightarrow\mu=\frac1{2\sqrt3}\approx0.29.

Question 63

Physics · 4 marks · −1

Two identical wires A and B, each of length ℓ\ell, 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 BAB_A and BBB_B are the values of magnetic field at the centres of the circle and square respectively, then the ratio BABB\frac{B_A}{B_B} is :

  1. A.π2162\frac{\pi^2}{16\sqrt{2}}
  2. B.π216\frac{\pi^2}{16}
  3. C.π282\frac{\pi^2}{8\sqrt{2}}
  4. D.π28\frac{\pi^2}{8}

Correct answer: C

Explanation: BA=μ0IπℓB_A=\frac{\mu_0I\pi}{\ell}; BB=4⋅μ0I4π(ℓ/8)⋅2=82μ0IπℓB_B=4\cdot\frac{\mu_0I}{4\pi(\ell/8)}\cdot\sqrt2=\frac{8\sqrt2\mu_0I}{\pi\ell}; ratio π282\frac{\pi^2}{8\sqrt2}.

Question 64

Physics · 4 marks · −1

A galvanometer having a coil resistance of 100 Ω100\,\Omega 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 :

  1. A.2 Ω2\,\Omega
  2. B.0.1 Ω0.1\,\Omega
  3. C.3 Ω3\,\Omega
  4. D.0.01 Ω0.01\,\Omega

Correct answer: D

Explanation: S=igGi−ig=10−3⋅10010≈0.01 ΩS=\frac{i_gG}{i-i_g}=\frac{10^{-3}\cdot100}{10}\approx0.01\,\Omega.

Question 65

Physics · 4 marks · −1

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 :

  1. A.10 times nearer.
  2. B.20 times taller.
  3. C.20 times nearer.
  4. D.10 times taller.

Correct answer: C

Explanation: A telescope magnifies the angle subtended by 20, which makes the object appear 20 times nearer.

Question 66

Physics · 4 marks · −1

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300 - 400 K, is best described by :

  1. A.Linear increase for Cu, exponential increase for Si.
  2. B.Linear increase for Cu, exponential decrease for Si.
  3. C.Linear decrease for Cu, linear decrease for Si.
  4. D.Linear increase for Cu, linear increase for Si.

Correct answer: B

Explanation: Metal resistance rises roughly linearly; an intrinsic semiconductor's resistance falls exponentially with temperature.

Question 67

Physics · 4 marks · −1

Choose the correct statement :

  1. A.In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
  2. B.In frequency modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
  3. C.In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
  4. D.In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

Correct answer: D

Explanation: AM varies the carrier's amplitude in proportion to the signal amplitude. (Statement 2 also describes FM correctly; the key is 4.)

Question 68

Physics · 4 marks · −1

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 :

  1. A.4 : 1
  2. B.1 : 4
  3. C.5 : 4
  4. D.1 : 16

Correct answer: C

Explanation: Decayed: A =1516N0=\frac{15}{16}N_0, B =34N0=\frac34N_0; ratio 5:45:4.

Question 69

Physics · 4 marks · −1

'n' moles of an ideal gas undergoes a process A→BA \to B as shown in the figure. The maximum temperature of the gas during the process will be :

Figure for question 69
  1. A.3P0V02nR\frac{3P_0V_0}{2nR}
  2. B.9P0V02nR\frac{9P_0V_0}{2nR}
  3. C.9P0V0nR\frac{9P_0V_0}{nR}
  4. D.9P0V04nR\frac{9P_0V_0}{4nR}

Correct answer: D

Explanation: Line: P=3P0−P0V0VP=3P_0-\frac{P_0}{V_0}V; PVPV is maximum at V=32V0V=\frac32V_0, P=32P0P=\frac32P_0: Tmax=9P0V04nRT_{max}=\frac{9P_0V_0}{4nR}.

Question 70

Physics · 4 marks · −1

An arc lamp requires a direct current of 10 A and 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 :

  1. A.0.08 H
  2. B.0.044 H
  3. C.0.065 H
  4. D.80 H

Correct answer: C

Explanation: R=8 ΩR=8\,\Omega; need Z=22 ΩZ=22\,\Omega: ωL=484−64=420\omega L=\sqrt{484-64}=\sqrt{420}, L≈0.065L\approx0.065 H.

Question 71

Physics · 4 marks · −1

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 :

  1. A.3f4\frac{3f}{4}
  2. B.2f
  3. C.f
  4. D.f2\frac{f}{2}

Correct answer: C

Explanation: Open pipe: f=v2ℓf=\frac{v}{2\ell}. Closed pipe of length ℓ2\frac\ell2: v4(ℓ/2)=f\frac{v}{4(\ell/2)}=f.

Question 72

Physics · 4 marks · −1

The box of 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\lambda 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 bminb_{min}) when :

  1. A.a=λLa = \sqrt{\lambda L} and bmin=(2λ2L)b_{min} = \left(\frac{2\lambda^2}{L}\right)
  2. B.a=λLa = \sqrt{\lambda L} and bmin=4λLb_{min} = \sqrt{4\lambda L}
  3. C.a=λ2La = \frac{\lambda^2}{L} and bmin=4λLb_{min} = \sqrt{4\lambda L}
  4. D.a=λ2La = \frac{\lambda^2}{L} and bmin=(2λ2L)b_{min} = \left(\frac{2\lambda^2}{L}\right)

Correct answer: B

Explanation: b=a+λLab=a+\frac{\lambda L}{a} is minimum at a=λLa=\sqrt{\lambda L}, giving bmin=2λL=4λLb_{min}=2\sqrt{\lambda L}=\sqrt{4\lambda L}.

Question 73

Physics · 4 marks · −1

A combination of capacitors 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 charges on the 4 μ\muF and 9 μ\muF capacitors), at a point distant 30 m from it, would equal :

Figure for question 73
  1. A.360 N/C
  2. B.420 N/C
  3. C.480 N/C
  4. D.240 N/C

Correct answer: B

Explanation: 4 μ4\,\muF in series with (3+9) μ(3+9)\,\muF across 8 V: charge 24 μ24\,\muC; the parallel pair has 2 V, so 9 μ9\,\muF holds 18 μ18\,\muC. Q=42 μQ=42\,\muC, E=9×109⋅42×10−6900=420E=\frac{9\times10^9\cdot42\times10^{-6}}{900}=420 N/C.

Question 74

Physics · 4 marks · −1

Arrange the following electromagnetic radiations per quantum in the order of increasing energy : A : Blue light, B : Yellow light, C : X-ray, D : Radiowave

  1. A.A, B, D, C
  2. B.C, A, B, D
  3. C.B, A, D, C
  4. D.D, B, A, C

Correct answer: D

Explanation: Energy rises with frequency: radio < yellow < blue < X-ray.

Question 75

Physics · 4 marks · −1

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 :

Figure for question 75
  1. A.A for electromagnets and B for electric generators
  2. B.A for transformers and B for electric generators
  3. C.B for electromagnets and transformers
  4. D.A for electric generators and transformers

Correct answer: C

Explanation: B has the narrow loop (low coercivity, low hysteresis loss), suitable for electromagnets and transformer cores.

Question 76

Physics · 4 marks · −1

A pendulum clock loses 12 s a day if the temperature is 40∘40^\circC and gains 4 s a day if the temperature is 20∘20^\circC. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α)(\alpha) of the metal of the pendulum shaft are respectively :

  1. A.60∘60^\circC; α=1.85×10−4\alpha = 1.85\times10^{-4}/∘^\circC
  2. B.30∘30^\circC; α=1.85×10−3\alpha = 1.85\times10^{-3}/∘^\circC
  3. C.55∘55^\circC; α=1.85×10−2\alpha = 1.85\times10^{-2}/∘^\circC
  4. D.25∘25^\circC; α=1.85×10−5\alpha = 1.85\times10^{-5}/∘^\circC

Correct answer: D

Explanation: Loss/gain ∝12αΔθ⋅86400\propto\frac12\alpha\Delta\theta\cdot86400: 124=40−θθ−20⇒θ=25∘\frac{12}{4}=\frac{40-\theta}{\theta-20}\Rightarrow\theta=25^\circC; 12=12α⋅15⋅86400⇒α≈1.85×10−512=\frac12\alpha\cdot15\cdot86400\Rightarrow\alpha\approx1.85\times10^{-5}/∘^\circC.

Question 77

Physics · 4 marks · −1

The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density ρ=Ar\rho = \frac{A}{r}, 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 :

Figure for question 77
  1. A.Q2π(b2−a2)\frac{Q}{2\pi(b^2-a^2)}
  2. B.2Qπ(a2−b2)\frac{2Q}{\pi(a^2-b^2)}
  3. C.2Qπa2\frac{2Q}{\pi a^2}
  4. D.Q2πa2\frac{Q}{2\pi a^2}

Correct answer: D

Explanation: Enclosed charge Q+2πA(r2−a2)Q+2\pi A(r^2-a^2); EE is constant when Q−2πAa2=0Q-2\pi Aa^2=0, A=Q2πa2A=\frac{Q}{2\pi a^2}.

Question 78

Physics · 4 marks · −1

In an experiment for determination of refractive index of glass of a prism by i−δi - \delta, plot, it was found that a ray incident at angle 35∘35^\circ, suffers a deviation of 40∘40^\circ and that it emerges at angle 79∘79^\circ. In that case which of the following is closest to the maximum possible value of the refractive index ?

  1. A.1.6
  2. B.1.7
  3. C.1.8
  4. D.1.5

Correct answer: D

Explanation: A=i+e−δ=74∘A=i+e-\delta=74^\circ. μ=sin⁡A+δm2sin⁡37∘\mu=\frac{\sin\frac{A+\delta_m}{2}}{\sin37^\circ} with δm<40∘\delta_m<40^\circ: μ<53sin⁡57∘≈1.4\mu<\frac53\sin57^\circ\approx1.4; nearest option 1.5.

Question 79

Physics · 4 marks · −1

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95s and 92s. If the minimum division in the measuring clock is 1s, then the reported mean time should be :

  1. A.92±5.092 \pm 5.0s
  2. B.92±1.892 \pm 1.8s
  3. C.92±392 \pm 3s
  4. D.92±292 \pm 2s

Correct answer: D

Explanation: Mean 92 s; mean absolute deviation 1.5 s, which is reported to the clock's 1 s resolution as ±2\pm2 s.

Question 80

Physics · 4 marks · −1

Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d) :

Figure for question 80
  1. A.Zener diode, Simple diode, Light dependent resistance, Solar cell
  2. B.Solar cell, Light dependent resistance, Zener diode, Simple diode
  3. C.Zener diode, Solar cell, Simple diode, Light dependent resistance
  4. D.Simple diode, Zener diode, Solar cell, Light dependent resistance

Correct answer: D

Explanation: (a) forward exponential only: simple diode; (b) sharp reverse breakdown: Zener; (c) curve shifts into the fourth quadrant when illuminated: solar cell; (d) resistance falling with intensity: LDR.

Question 81

Physics · 4 marks · −1

Radiation of wavelength λ\lambda, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3λ4\frac{3\lambda}{4}, the speed of the fastest emitted electron will be :

  1. A.<v(43)1/2< v\left(\frac{4}{3}\right)^{1/2}
  2. B.=v(43)1/2= v\left(\frac{4}{3}\right)^{1/2}
  3. C.=v(34)1/2= v\left(\frac{3}{4}\right)^{1/2}
  4. D.>v(43)1/2> v\left(\frac{4}{3}\right)^{1/2}

Correct answer: D

Explanation: 12mv12=43⋅hcλ−ϕ=43⋅12mv2+ϕ3>43⋅12mv2\frac12mv_1^2=\frac43\cdot\frac{hc}{\lambda}-\phi=\frac43\cdot\frac12mv^2+\frac\phi3>\frac43\cdot\frac12mv^2.

Question 82

Physics · 4 marks · −1

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A3\frac{2A}{3} from equilibrium position. The new amplitude of the motion is :

  1. A.3A
  2. B.A3A\sqrt{3}
  3. C.7A3\frac{7A}{3}
  4. D.A341\frac{A}{3}\sqrt{41}

Correct answer: C

Explanation: 3ωA2−4A29=ωA12−4A29⇒A1=7A33\omega\sqrt{A^2-\frac{4A^2}{9}}=\omega\sqrt{A_1^2-\frac{4A^2}{9}}\Rightarrow A_1=\frac{7A}{3}.

Question 83

Physics · 4 marks · −1

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 PVnPV^n = constant, then n is given by (Here CPC_P and CVC_V are molar specific heat at constant pressure and constant volume, respectively):

  1. A.n=C−CPC−CVn = \frac{C - C_P}{C - C_V}
  2. B.n=CP−CC−CVn = \frac{C_P - C}{C - C_V}
  3. C.n=C−CVC−CPn = \frac{C - C_V}{C - C_P}
  4. D.n=CPCVn = \frac{C_P}{C_V}

Correct answer: A

Explanation: For a polytropic process C=CV−Rn−1C=C_V-\frac{R}{n-1}; with R=CP−CVR=C_P-C_V: n=C−CPC−CVn=\frac{C-C_P}{C-C_V}.

Question 84

Physics · 4 marks · −1

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 45th45^{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 25th25^{th} division coincides with the main scale line ?

  1. A.0.80 mm
  2. B.0.70 mm
  3. C.0.50 mm
  4. D.0.75 mm

Correct answer: A

Explanation: LC =0.01=0.01 mm; zero error =−(50−45)×0.01=−0.05=-(50-45)\times0.01=-0.05 mm. Reading =0.5+0.25=0.75=0.5+0.25=0.75 mm; corrected 0.800.80 mm.

Question 85

Physics · 4 marks · −1

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 :

Figure for question 85
  1. A.turn right
  2. B.go straight
  3. C.turn left and right alternately
  4. D.turn left

Correct answer: D

Explanation: As it rolls, the rail AB (inclined) makes the left cone contact at a larger radius than the right, so the left side covers more distance per turn and the roller turns left.

Question 86

Physics · 4 marks · −1

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 :

Figure for question 86
  1. A.AND
  2. B.OR
  3. C.NAND
  4. D.NOT

Correct answer: B

Explanation: x is high whenever any input is high (it goes high with the first high input and stays high): OR gate.

Question 87

Physics · 4 marks · −1

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h≪Rh \ll 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.)

  1. A.gR\sqrt{gR}
  2. B.gR/2\sqrt{gR/2}
  3. C.gR(2−1)\sqrt{gR}\left(\sqrt{2}-1\right)
  4. D.2gR\sqrt{2gR}

Correct answer: C

Explanation: ve−vo=2gR−gR=gR(2−1)v_e-v_o=\sqrt{2gR}-\sqrt{gR}=\sqrt{gR}(\sqrt2-1) for h≪Rh\ll R.

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