AIEEE 2010 — 25 Apr Offline Set D (Previous Year Paper)

JEE Main · 2010 · with answers and explanations

questions
89
Total marks
424
Duration
180 minutes
Negative marking
−2

Subjects: Chemistry · Physics · Mathematics

Question 1

Chemistry · 8 marks · −2

The standard enthalpy of formation of NH3NH_3 is −46.0-46.0 kJ mol−1^{-1}. If the enthalpy of formation of H2H_2 from its atoms is −436-436 kJ mol−1^{-1} and that of N2N_2 is −712-712 kJ mol−1^{-1}, the average bond enthalpy of N-H bond in NH3NH_3 is

  1. A.−964-964 kJ mol−1^{-1}
  2. B.+352+352 kJ mol−1^{-1}
  3. C.+1056+1056 kJ mol−1^{-1}
  4. D.−1102-1102 kJ mol−1^{-1}

Correct answer: B

Explanation: 12(712)+32(436)−3x=−46⇒x=352\frac12(712)+\frac32(436)-3x=-46\Rightarrow x=352 kJ/mol.

Question 2

Chemistry · 8 marks · −2

The time for half life period of a certain reaction A→A \rightarrow products is 1 hour. When the initial concentration of the reactant 'A', is 2.0 mol L−1^{-1}, how much time does it take for its concentration to come from 0.50 to 0.25 mol L−1^{-1} if it is a zero order reaction ?

  1. A.4 h
  2. B.0.5 h
  3. C.0.25 h
  4. D.1 h

Correct answer: C

Explanation: t1/2=[A]02k⇒k=1t_{1/2}=\frac{[A]_0}{2k}\Rightarrow k=1 M/h; t=0.251=0.25t=\frac{0.25}{1}=0.25 h.

Question 3

Chemistry · 8 marks · −2

A solution containing 2.675 g of CoCl3⋅6NH3CoCl_3\cdot6NH_3 (molar mass = 267.5 g mol−1^{-1}) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3AgNO_3 to give 4.78 g of AgCl (molar mass = 143.5 g mol−1^{-1}). The formula of the complex is (At. Mass of Ag = 108 u)

  1. A.[Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3
  2. B.[CoCl2(NH3)4]Cl[CoCl_2(NH_3)_4]Cl
  3. C.[CoCl3(NH3)3][CoCl_3(NH_3)_3]
  4. D.[CoCl(NH3)5]Cl2[CoCl(NH_3)_5]Cl_2

Correct answer: A

Explanation: 0.01 mol complex gives 4.78143.5=0.0333\frac{4.78}{143.5}=0.0333 mol AgCl: 3 ionisable Cl−Cl^-.

Question 4

Chemistry · 4 marks · −1

Consider the reaction : Cl2(aq)+H2S(aq)→S(s)+2H+(aq)+2Cl−(aq)Cl_2(aq) + H_2S(aq) \rightarrow S(s) + 2H^+(aq) + 2Cl^-(aq). The rate equation for this reaction is rate = k[Cl2][H2S]k[Cl_2][H_2S]. Which of these mechanisms is/are consistent with this rate equation ? (A) Cl2+H2S→H++Cl−+Cl++HS−Cl_2 + H_2S \rightarrow H^+ + Cl^- + Cl^+ + HS^- (slow); Cl++HS−→H++Cl−+SCl^+ + HS^- \rightarrow H^+ + Cl^- + S (fast). (B) H2S⇌H++HS−H_2S \rightleftharpoons H^+ + HS^- (fast equilibrium); Cl2+HS−→2Cl−+H++SCl_2 + HS^- \rightarrow 2Cl^- + H^+ + S (slow)

  1. A.B only
  2. B.Both A and B
  3. C.Neither A nor B
  4. D.A only

Correct answer: D

Explanation: A's slow step gives k[Cl2][H2S]k[Cl_2][H_2S]; B gives k′[Cl2][H2S]/[H+]k'[Cl_2][H_2S]/[H^+].

Question 5

Chemistry · 4 marks · −1

If 10−410^{-4} dm3^3 of water is introduced into a 1.0 dm3^3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H2OH_2O at 300 K is 3170 Pa ; R=8.314R = 8.314 J K−1^{-1} mol−1^{-1})

  1. A.5.56×10−35.56\times10^{-3} mol
  2. B.1.53×10−21.53\times10^{-2} mol
  3. C.4.46×10−24.46\times10^{-2} mol
  4. D.1.27×10−31.27\times10^{-3} mol

Correct answer: D

Explanation: n=PVRT=3170⋅10−38.314⋅300=1.27×10−3n=\frac{PV}{RT}=\frac{3170\cdot10^{-3}}{8.314\cdot300}=1.27\times10^{-3} mol.

Question 6

Chemistry · 4 marks · −1

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is

  1. A.propene
  2. B.1-butene
  3. C.2-butene
  4. D.ethene

Correct answer: C

Explanation: 2-butene gives 2 CH3CHOCH_3CHO (44 u).

Question 7

Chemistry · 4 marks · −1

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (ΔTf)(\Delta T_f), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (KfK_f = 1.86 K kg mol−1^{-1})

  1. A.0.0372 K
  2. B.0.0558 K
  3. C.0.0744 K
  4. D.0.0186 K

Correct answer: B

Explanation: i=3i=3: 3⋅1.86⋅0.01=0.05583\cdot1.86\cdot0.01=0.0558 K.

Question 8

Chemistry · 4 marks · −1

From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2ZnCl_2, is

  1. A.2-Butanol
  2. B.2-Methylpropan-2-ol
  3. C.2-Methylpropanol
  4. D.1-Butanol

Correct answer: B

Explanation: The tertiary alcohol reacts fastest (Lucas test).

Question 9

Chemistry · 4 marks · −1

In the chemical reactions, aniline →NaNO2, HCl, 278 K\xrightarrow{NaNO_2,\ HCl,\ 278\ K} A →HBF4\xrightarrow{HBF_4} B, the compounds 'A' and 'B' respectively are

  1. A.nitrobenzene and fluorobenzene
  2. B.phenol and benzene
  3. C.benzene diazonium chloride and fluorobenzene
  4. D.nitrobenzene and chlorobenzene

Correct answer: C

Explanation: Diazotisation then Balz-Schiemann.

Question 10

Chemistry · 8 marks · −2

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is

  1. A.59.0
  2. B.47.4
  3. C.23.7
  4. D.29.5

Correct answer: C

Explanation: NH3=0.5NH_3=0.5 mmol, N =7=7 mg; 729.5=23.7%\frac{7}{29.5}=23.7\%.

Question 11

Chemistry · 8 marks · −2

The energy required to break one mole of Cl-Cl bonds in Cl2Cl_2 is 242 kJ mol−1^{-1}. The longest wavelength of light capable of breaking a single Cl - Cl bond is (c=3×108c = 3\times10^8 ms−1^{-1} and NA=6.02×1023N_A = 6.02\times10^{23} mol−1^{-1})

  1. A.594 nm
  2. B.640 nm
  3. C.700 nm
  4. D.494 nm

Correct answer: D

Explanation: λ=hcNAE=6.626×10−34⋅3×108⋅6.02×1023242000≈494\lambda=\frac{hcN_A}{E}=\frac{6.626\times10^{-34}\cdot3\times10^8\cdot6.02\times10^{23}}{242000}\approx494 nm.

Question 12

Chemistry · 8 marks · −2

Ionisation energy of He+He^+ is 19.6×10−1819.6\times10^{-18} J atom−1^{-1}. The energy of the first stationary state (n = 1) of Li2+Li^{2+} is

  1. A.4.41×10−164.41\times10^{-16} J atom−1^{-1}
  2. B.−4.41×10−17-4.41\times10^{-17} J atom−1^{-1}
  3. C.−2.2×10−15-2.2\times10^{-15} J atom−1^{-1}
  4. D.8.82×10−178.82\times10^{-17} J atom−1^{-1}

Correct answer: B

Explanation: E1∝Z2E_1\propto Z^2: −19.6×10−18⋅94=−4.41×10−17-19.6\times10^{-18}\cdot\frac94=-4.41\times10^{-17} J.

Question 13

Chemistry · 4 marks · −1

Consider the following bromides : (A) 1-bromobutane (CH3CH2CH2CH2BrCH_3CH_2CH_2CH_2Br), (B) 3-bromobut-1-ene (CH2=CHCH(Br)CH3CH_2=CHCH(Br)CH_3), (C) 2-bromobutane (CH3CH2CH(Br)CH3CH_3CH_2CH(Br)CH_3). The correct order of SN1S_N1 reactivity is

Figure for question 13
  1. A.B > C > A
  2. B.B > A > C
  3. C.C > B > A
  4. D.A > B > C

Correct answer: A

Explanation: Carbocation stability: allylic secondary > secondary > primary.

Question 14

Chemistry · 4 marks · −1

Which one of the following has an optical isomer ? (en = ethylenediamine)

  1. A.[Zn(en)(NH3)2]2+[Zn(en)(NH_3)_2]^{2+}
  2. B.[Co(en)3]3+[Co(en)_3]^{3+}
  3. C.[Co(H2O)4(en)]3+[Co(H_2O)_4(en)]^{3+}
  4. D.[Zn(en)2]2+[Zn(en)_2]^{2+}

Correct answer: B

Explanation: [Co(en)3]3+[Co(en)_3]^{3+} is chiral (tris-chelate).

Question 15

Chemistry · 4 marks · −1

On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol−1^{-1} and of octane = 114 g mol−1^{-1}).

  1. A.72.0 kPa
  2. B.36.1 kPa
  3. C.96.2 kPa
  4. D.144.5 kPa

Correct answer: A

Explanation: xhep=0.250.557=0.45x_{hep}=\frac{0.25}{0.557}=0.45; P=105(0.45)+45(0.55)=72P=105(0.45)+45(0.55)=72 kPa.

Question 16

Chemistry · 4 marks · −1

The main product of the following reaction is C6H5CH2CH(OH)CH(CH3)2→conc. H2SO4C_6H_5CH_2CH(OH)CH(CH_3)_2 \xrightarrow{conc.\ H_2SO_4} ?

Figure for question 16
  1. A.trans-C6H5CH=CHCH(CH3)2C_6H_5CH=CHCH(CH_3)_2
  2. B.C6H5CH2CH=C(CH3)2C_6H_5CH_2CH=C(CH_3)_2
  3. C.cis-C6H5CH=CHCH(CH3)2C_6H_5CH=CHCH(CH_3)_2
  4. D.C6H5CH2CH2C(CH3)=CH2C_6H_5CH_2CH_2C(CH_3)=CH_2

Correct answer: A

Explanation: E1 dehydration gives the alkene conjugated with the ring; the trans isomer is favoured.

Question 17

Chemistry · 4 marks · −1

Three reactions involving H2PO4−H_2PO_4^- are given below : (i) H3PO4+H2O→H3O++H2PO4−H_3PO_4 + H_2O \rightarrow H_3O^+ + H_2PO_4^- (ii) H2PO4−+H2O→HPO42−+H3O+H_2PO_4^- + H_2O \rightarrow HPO_4^{2-} + H_3O^+ (iii) H2PO4−+OH−→H3PO4+O2−H_2PO_4^- + OH^- \rightarrow H_3PO_4 + O^{2-}. In which of the above does H2PO4−H_2PO_4^- act as an acid ?

  1. A.(ii) only
  2. B.(i) and (ii)
  3. C.(iii) only
  4. D.(i) only

Correct answer: A

Explanation: Only in (ii) does H2PO4−H_2PO_4^- donate a proton.

Question 18

Chemistry · 4 marks · −1

In aqueous solution the ionization constants for carbonic acid are K1=4.2×10−7K_1 = 4.2\times10^{-7} and K2=4.8×10−11K_2 = 4.8\times10^{-11}. Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

  1. A.The concentration of CO32−CO_3^{2-} is 0.034 M.
  2. B.The concentration of CO32−CO_3^{2-} is greater than that of HCO3−HCO_3^-.
  3. C.The concentration of H+H^+ and HCO3−HCO_3^- are approximately equal.
  4. D.The concentration of H+H^+ is double that of CO32−CO_3^{2-}.

Correct answer: C

Explanation: K2≪K1K_2\ll K_1: first ionisation dominates, [H+]≈[HCO3−][H^+]\approx[HCO_3^-].

Question 19

Chemistry · 4 marks · −1

The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is

  1. A.288 pm
  2. B.398 pm
  3. C.618 pm
  4. D.144 pm

Correct answer: D

Explanation: 2(r++r−)=508⇒r−=1442(r^++r^-)=508\Rightarrow r^-=144 pm.

Question 20

Chemistry · 4 marks · −1

The correct order of increasing basicity of the given conjugate bases (R=CH3R = CH_3) is

  1. A.RCOO−<HC≡C−<R−<NH2−RCOO^- < HC\equiv C^- < R^- < NH_2^-
  2. B.R−<HC≡C−<RCOO−<NH2−R^- < HC\equiv C^- < RCOO^- < NH_2^-
  3. C.RCOO−<NH2−<HC≡C−<R−RCOO^- < NH_2^- < HC\equiv C^- < R^-
  4. D.RCOO−<HC≡C−<NH2−<R−RCOO^- < HC\equiv C^- < NH_2^- < R^-

Correct answer: D

Explanation: Conjugate acid strength: RCOOH>HC≡CH>NH3>CH4RCOOH>HC\equiv CH>NH_3>CH_4.

Question 21

Chemistry · 4 marks · −1

The correct sequence which shows decreasing order of the ionic radii of the elements is

  1. A.Al3+>Mg2+>Na+>F−>O2−Al^{3+} > Mg^{2+} > Na^+ > F^- > O^{2-}
  2. B.Na+>Mg2+>Al3+>O2−>F−Na^+ > Mg^{2+} > Al^{3+} > O^{2-} > F^-
  3. C.Na+>F−>Mg2+>O2−>Al3+Na^+ > F^- > Mg^{2+} > O^{2-} > Al^{3+}
  4. D.O2−>F−>Na+>Mg2+>Al3+O^{2-} > F^- > Na^+ > Mg^{2+} > Al^{3+}

Correct answer: D

Explanation: Isoelectronic: radius falls as nuclear charge rises.

Question 22

Chemistry · 4 marks · −1

Solubility product of silver bromide is 5.0×10−135.0\times10^{-13}. The quantity of potassium bromide (molar mass taken as 120 g mol−1^{-1}) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

  1. A.1.2×10−101.2\times10^{-10} g
  2. B.1.2×10−91.2\times10^{-9} g
  3. C.6.2×10−56.2\times10^{-5} g
  4. D.5.0×10−85.0\times10^{-8} g

Correct answer: B

Explanation: [Br−]=5×10−130.05=10−11[Br^-]=\frac{5\times10^{-13}}{0.05}=10^{-11} M; mass 1.2×10−91.2\times10^{-9} g.

Question 23

Chemistry · 4 marks · −1

The Gibbs energy for the decomposition of Al2O3Al_2O_3 at 500∘500^\circC is as follows : 23Al2O3→43Al+O2\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2, ΔrG=+966\Delta_rG = +966 kJ mol−1^{-1}. The potential difference needed for electrolytic reduction of Al2O3Al_2O_3 at 500∘500^\circC is at least

  1. A.4.5 V
  2. B.3.0 V
  3. C.2.5 V
  4. D.5.0 V

Correct answer: C

Explanation: n=4n=4: E=9660004⋅96500≈2.5E=\frac{966000}{4\cdot96500}\approx2.5 V.

Question 24

Chemistry · 4 marks · −1

At 25∘25^\circC, the solubility product of Mg(OH)2Mg(OH)_2 is 1.0×10−111.0\times10^{-11}. At which pH, will Mg2+Mg^{2+} ions start precipitating in the form of Mg(OH)2Mg(OH)_2 from a solution of 0.001 M Mg2+Mg^{2+} ions ?

  1. A.9
  2. B.10
  3. C.11
  4. D.8

Correct answer: B

Explanation: [OH−]=10−11/10−3=10−4[OH^-]=\sqrt{10^{-11}/10^{-3}}=10^{-4}: pH 10.

Question 25

Chemistry · 4 marks · −1

Percentage of free space in cubic close packed structure and in body centred packed structure are respectively

  1. A.30% and 26%
  2. B.26% and 32%
  3. C.32% and 48%
  4. D.48% and 26%

Correct answer: B

Explanation: Packing fractions 0.74 and 0.68.

Question 26

Chemistry · 4 marks · −1

Out of the following, the alkene that exhibits optical isomerism is

  1. A.3-methyl-2-pentene
  2. B.4-methyl-1-pentene
  3. C.3-methyl-1-pentene
  4. D.2-methyl-2-pentene

Correct answer: C

Explanation: C3 of 3-methyl-1-pentene carries H, CH3CH_3, C2H5C_2H_5 and vinyl.

Question 27

Chemistry · 4 marks · −1

Biuret test is not given by

  1. A.carbohydrates
  2. B.polypeptides
  3. C.urea
  4. D.proteins

Correct answer: A

Explanation: Biuret test needs peptide/amide linkages, absent in carbohydrates.

Question 28

Chemistry · 4 marks · −1

The correct order of EM2+/M∘E^\circ_{M^{2+}/M} values with negative sign for the four successive elements Cr, Mn, Fe and Co is

  1. A.Mn > Cr > Fe > Co
  2. B.Cr > Fe > Mn > Co
  3. C.Fe > Mn > Cr > Co
  4. D.Cr > Mn > Fe > Co

Correct answer: A

Explanation: E∘E^\circ: Mn −1.18-1.18, Cr −0.91-0.91, Fe −0.44-0.44, Co −0.28-0.28 V.

Question 29

Chemistry · 4 marks · −1

The polymer containing strong intermolecular forces e.g. hydrogen bonding, is

  1. A.teflon
  2. B.nylon 6,6
  3. C.polystyrene
  4. D.natural rubber

Correct answer: B

Explanation: Polyamide N-H...O=C hydrogen bonds.

Question 30

Chemistry · 4 marks · −1

For a particular reversible reaction at temperature T, ΔH\Delta H and ΔS\Delta S were found to be both +ve. If TeT_e is the temperature at equilibrium, the reaction would be spontaneous when

  1. A.Te>TT_e > T
  2. B.T>TeT > T_e
  3. C.TeT_e is 5 times T
  4. D.T=TeT = T_e

Correct answer: B

Explanation: ΔG<0\Delta G<0 when T>ΔHΔS=TeT>\frac{\Delta H}{\Delta S}=T_e.

Question 31

Physics · 8 marks · −2

A rectangular loop has a sliding connector PQ of length ℓ\ell and resistance R Ω\Omega and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1I_1, I2I_2 and I are

Figure for question 31
  1. A.I1=−I2=BℓvRI_1 = -I_2 = \frac{B\ell v}{R}, I=2BℓvRI = \frac{2B\ell v}{R}
  2. B.I1=I2=Bℓv3RI_1 = I_2 = \frac{B\ell v}{3R}, I=2Bℓv3RI = \frac{2B\ell v}{3R}
  3. C.I1=I2=I=BℓvRI_1 = I_2 = I = \frac{B\ell v}{R}
  4. D.I1=I2=Bℓv6RI_1 = I_2 = \frac{B\ell v}{6R}, I=Bℓv3RI = \frac{B\ell v}{3R}

Correct answer: B

Explanation: PQ is an emf BℓvB\ell v with internal resistance R feeding two parallel R's: I=BℓvR+R/2=2Bℓv3RI=\frac{B\ell v}{R+R/2}=\frac{2B\ell v}{3R}, split equally.

Question 32

Physics · 8 marks · −2

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2t_1/t_2 will be

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

Correct answer: C

Explanation: t1=τ2ln⁡2t_1=\frac{\tau}{2}\ln2, t2=τln⁡4=2τln⁡2t_2=\tau\ln4=2\tau\ln2.

Question 33

Physics · 4 marks · −1

Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is false.

Correct answer: A

Explanation: Momentum is non-zero and conserved, so the combined body keeps some KE.

Question 34

Physics · 4 marks · −1

Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0V_0 and the maximum kinetic energy of the photoelectrons is KmaxK_{max}. When the ultraviolet light is replaced by X-rays, both V0V_0 and KmaxK_{max} increase. Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
  3. C.Statement-1 is false, Statement-2 is true.
  4. D.Statement-1 is true, Statement-2 is false.

Correct answer: D

Explanation: Higher frequency raises KmaxK_{max} and V0V_0; the spread of speeds is due to energy losses inside the metal, not a range of frequencies.

Question 35

Physics · 4 marks · −1

A ball is made of a material of density ρ\rho where ρoil<ρ<ρwater\rho_{oil} < \rho < \rho_{water} with ρoil\rho_{oil} and ρwater\rho_{water} representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position ?

Figure for question 35
  1. A.Picture (1)
  2. B.Picture (2)
  3. C.Picture (3)
  4. D.Picture (4)

Correct answer: B

Explanation: Oil floats on water; the ball sinks through oil but floats on water, so it rests at the interface with oil on top.

Question 36

Physics · 4 marks · −1

A particle is moving with velocity v⃗=K(yi^+xj^)\vec{v} = K(y\hat{i} + x\hat{j}), where K is a constant. The general equation for its path is

  1. A.y=x2+y = x^2 + constant
  2. B.y2=x+y^2 = x + constant
  3. C.xyxy = constant
  4. D.y2=x2+y^2 = x^2 + constant

Correct answer: D

Explanation: dydx=xy⇒y2=x2+c\frac{dy}{dx}=\frac xy\Rightarrow y^2=x^2+c.

Question 37

Physics · 4 marks · −1

Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line XX' is given by

Figure for question 37
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: A

Explanation: Fields cancel at the midpoint and reverse sign across it; outside the wires they add, diverging near each wire.

Question 38

Physics · 4 marks · −1

In the circuit shown below, the key K is closed at t = 0. The current through the battery is

Figure for question 38
  1. A.VR1R2R12+R22\frac{VR_1R_2}{\sqrt{R_1^2+R_2^2}} at t = 0 and VR2\frac{V}{R_2} at t=∞t = \infty
  2. B.VR2\frac{V}{R_2} at t = 0 and V(R1+R2)R1R2\frac{V(R_1+R_2)}{R_1R_2} at t=∞t = \infty
  3. C.VR2\frac{V}{R_2} at t = 0 and VR1R2R12+R22\frac{VR_1R_2}{\sqrt{R_1^2+R_2^2}} at t=∞t = \infty
  4. D.V(R1+R2)R1R2\frac{V(R_1+R_2)}{R_1R_2} at t = 0 and VR2\frac{V}{R_2} at t=∞t = \infty

Correct answer: B

Explanation: At t = 0 the inductor branch carries no current; at steady state it is a wire, R1 and R2 in parallel.

Question 39

Physics · 4 marks · −1

The figure shows the position - time (x - t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

Figure for question 39
  1. A.0.4 Ns
  2. B.0.8 Ns
  3. C.1.6 Ns
  4. D.0.2 Ns

Correct answer: B

Explanation: Velocity switches between +1+1 and −1-1 m/s: J=0.4⋅2=0.8J=0.4\cdot2=0.8 N s.

Question 40

Physics · 4 marks · −1

A nucleus of mass M+ΔmM + \Delta m is at rest and decays into two daughter nuclei of equal mass M2\frac{M}{2} each. Speed of light is c. The binding energy per nucleon for the parent nucleus is E1E_1 and that for the daughter nuclei is E2E_2. Then

  1. A.E2=2E1E_2 = 2E_1
  2. B.E1>E2E_1 > E_2
  3. C.E2>E1E_2 > E_1
  4. D.E1=2E2E_1 = 2E_2

Correct answer: C

Explanation: Energy is released, so the daughters are more tightly bound.

Question 41

Physics · 4 marks · −1

A nucleus of mass M+ΔmM + \Delta m is at rest and decays into two daughter nuclei of equal mass M2\frac{M}{2} each. Speed of light is c. The speed of daughter nuclei is

  1. A.cΔmM+Δmc\frac{\Delta m}{M+\Delta m}
  2. B.c2ΔmMc\sqrt{\frac{2\Delta m}{M}}
  3. C.cΔmMc\sqrt{\frac{\Delta m}{M}}
  4. D.cΔmM+Δmc\sqrt{\frac{\Delta m}{M+\Delta m}}

Correct answer: B

Explanation: Δmc2=2⋅12⋅M2v2⇒v=c2ΔmM\Delta mc^2=2\cdot\frac12\cdot\frac M2v^2\Rightarrow v=c\sqrt{\frac{2\Delta m}{M}}.

Question 42

Physics · 4 marks · −1

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α\alpha-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

  1. A.A−Z−8Z−4\frac{A-Z-8}{Z-4}
  2. B.A−Z−4Z−8\frac{A-Z-4}{Z-8}
  3. C.A−Z−12Z−4\frac{A-Z-12}{Z-4}
  4. D.A−Z−4Z−2\frac{A-Z-4}{Z-2}

Correct answer: B

Explanation: Protons: Z−6−2Z-6-2; neutrons: A−Z−6+2A-Z-6+2.

Question 43

Physics · 4 marks · −1

A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E⃗\vec{E} at the centre O is

Figure for question 43
  1. A.q4π2ϵ0r2j^\frac{q}{4\pi^2\epsilon_0r^2}\hat{j}
  2. B.−q4π2ϵ0r2j^-\frac{q}{4\pi^2\epsilon_0r^2}\hat{j}
  3. C.−q2π2ϵ0r2j^-\frac{q}{2\pi^2\epsilon_0r^2}\hat{j}
  4. D.q2π2ϵ0r2j^\frac{q}{2\pi^2\epsilon_0r^2}\hat{j}

Correct answer: C

Explanation: E=2kλr=q2π2ϵ0r2E=\frac{2k\lambda}{r}=\frac{q}{2\pi^2\epsilon_0r^2}, directed away from the ring (−j^-\hat j).

Question 44

Physics · 4 marks · −1

The combination of gates shown below yields

Figure for question 44
  1. A.OR gate
  2. B.NOT gate
  3. C.XOR gate
  4. D.NAND gate

Correct answer: A

Explanation: Each input passes through a NOT (NAND with tied inputs), then a NAND: AˉBˉ‾=A+B\overline{\bar A\bar B}=A+B.

Question 45

Physics · 4 marks · −1

A diatomic ideal gas is used in a Car engine as the working substance. If during the adiabatic expansion part of the cycle, volume of the gas increases from V to 32V the efficiency of the engine is

  1. A.0.5
  2. B.0.75
  3. C.0.99
  4. D.0.25

Correct answer: B

Explanation: T1T2=320.4=4\frac{T_1}{T_2}=32^{0.4}=4; η=1−14=0.75\eta=1-\frac14=0.75.

Question 46

Physics · 4 marks · −1

If a source of power 4 kW produces 102010^{20} photons/second, the radiation belong to a part of the spectrum called

  1. A.X-rays
  2. B.ultraviolet rays
  3. C.microwaves
  4. D.γ\gamma-rays

Correct answer: A

Explanation: Photon energy 4×10−174\times10^{-17} J, f≈6×1016f\approx6\times10^{16} Hz: X-rays.

Question 47

Physics · 4 marks · −1

The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1×10−32.1\times10^{-3} are

  1. A.5, 1, 2
  2. B.5, 1, 5
  3. C.5, 5, 2
  4. D.4, 4, 2

Correct answer: A

Explanation: 5, 1, 2.

Question 48

Physics · 4 marks · −1

In a series LCR circuit R = 200 Ω\Omega and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30∘30^\circ. On taking out the inductor from the circuit the current leads the voltage by 30∘30^\circ. The power dissipated in the LCR circuit is

  1. A.305 W
  2. B.210 W
  3. C.Zero W
  4. D.242 W

Correct answer: D

Explanation: XL=XCX_L=X_C: resonance, P=V2R=242P=\frac{V^2}{R}=242 W.

Question 49

Physics · 4 marks · −1

Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0(54−rR)\rho(r) = \rho_0\left(\frac{5}{4} - \frac{r}{R}\right) upto r = R, and ρ(r)=0\rho(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r < R) from the origin is given by

  1. A.4πρ0r3ϵ0(53−rR)\frac{4\pi\rho_0r}{3\epsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)
  2. B.ρ0r4ϵ0(53−rR)\frac{\rho_0r}{4\epsilon_0}\left(\frac{5}{3}-\frac{r}{R}\right)
  3. C.4ρ0r3ϵ0(54−rR)\frac{4\rho_0r}{3\epsilon_0}\left(\frac{5}{4}-\frac{r}{R}\right)
  4. D.ρ0r3ϵ0(54−rR)\frac{\rho_0r}{3\epsilon_0}\left(\frac{5}{4}-\frac{r}{R}\right)

Correct answer: B

Explanation: q(r)=4πρ0(5r312−r44R)q(r)=4\pi\rho_0\left(\frac{5r^3}{12}-\frac{r^4}{4R}\right); E=q4πϵ0r2=ρ0r4ϵ0(53−rR)E=\frac{q}{4\pi\epsilon_0r^2}=\frac{\rho_0r}{4\epsilon_0}\left(\frac53-\frac rR\right).

Question 50

Physics · 8 marks · −2

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U(x)=ax12−bx6U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}, where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D=[U(x=∞)−Uat equilibrium]D = [U(x = \infty) - U_{at\ equilibrium}], D is

  1. A.b22a\frac{b^2}{2a}
  2. B.b212a\frac{b^2}{12a}
  3. C.b24a\frac{b^2}{4a}
  4. D.b26a\frac{b^2}{6a}

Correct answer: C

Explanation: Equilibrium x6=2abx^6=\frac{2a}{b}, U=−b24aU=-\frac{b^2}{4a}.

Question 51

Physics · 8 marks · −2

Two conductors have the same resistance at 0∘0^\circC but their temperature coefficients of resistance are α1\alpha_1 and α2\alpha_2. The respective temperature coefficients of their series and parallel combinations are nearly

  1. A.α1+α22\frac{\alpha_1+\alpha_2}{2}, α1+α2\alpha_1+\alpha_2
  2. B.α1+α2\alpha_1+\alpha_2, α1+α22\frac{\alpha_1+\alpha_2}{2}
  3. C.α1+α2\alpha_1+\alpha_2, α1α2α1+α2\frac{\alpha_1\alpha_2}{\alpha_1+\alpha_2}
  4. D.α1+α22\frac{\alpha_1+\alpha_2}{2}, α1+α22\frac{\alpha_1+\alpha_2}{2}

Correct answer: D

Explanation: For equal R0R_0 both combinations give α1+α22\frac{\alpha_1+\alpha_2}{2} to first order.

Question 52

Physics · 8 marks · −2

A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length s=t3+5s = t^3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly

Figure for question 52
  1. A.13 m/s2^2
  2. B.12 m/s2^2
  3. C.7.2 m/s2^2
  4. D.14 m/s2^2

Correct answer: D

Explanation: at=6t=12a_t=6t=12, v=12v=12, ac=14420=7.2a_c=\frac{144}{20}=7.2; a≈14a\approx14 m/s2^2.

Question 53

Physics · 4 marks · −1

Two fixed frictionless inclined plane making an angle 30∘30^\circ and 60∘60^\circ with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ?

Figure for question 53
  1. A.4.9 ms−2^{-2} in horizontal direction
  2. B.9.8 ms−2^{-2} in vertical direction
  3. C.zero
  4. D.4.9 ms−2^{-2} in vertical direction

Correct answer: D

Explanation: Vertical acceleration gsin⁡2θg\sin^2\theta (θ with horizontal): g(sin⁡260∘−sin⁡230∘)=g2=4.9g(\sin^260^\circ-\sin^230^\circ)=\frac g2=4.9 m/s2^2.

Question 54

Physics · 4 marks · −1

For a particle in uniform circular motion the acceleration a⃗\vec{a} at a point P(R,θ)P(R, \theta) on the circle of radius R is (here θ\theta is measured from the x-axis)

  1. A.−v2Rcos⁡θ i^+v2Rsin⁡θ j^-\frac{v^2}{R}\cos\theta\,\hat{i} + \frac{v^2}{R}\sin\theta\,\hat{j}
  2. B.−v2Rsin⁡θ i^+v2Rcos⁡θ j^-\frac{v^2}{R}\sin\theta\,\hat{i} + \frac{v^2}{R}\cos\theta\,\hat{j}
  3. C.−v2Rcos⁡θ i^−v2Rsin⁡θ j^-\frac{v^2}{R}\cos\theta\,\hat{i} - \frac{v^2}{R}\sin\theta\,\hat{j}
  4. D.v2Ri^+v2Rj^\frac{v^2}{R}\hat{i} + \frac{v^2}{R}\hat{j}

Correct answer: C

Explanation: Centripetal acceleration points to the centre: −v2R(cos⁡θi^+sin⁡θj^)-\frac{v^2}{R}(\cos\theta\hat i+\sin\theta\hat j).

Question 55

Physics · 4 marks · −1

An initially parallel cylindrical beam travels in a medium of refractive index μ(I)=μ0+μ2I\mu(I) = \mu_0 + \mu_2I, where μ0\mu_0 and μ2\mu_2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. As the beam enters the medium, it will

  1. A.diverge
  2. B.converge
  3. C.diverge near the axis and converge near the periphery
  4. D.travel as a cylindrical beam

Correct answer: B

Explanation: Refractive index is largest on the axis (self-focusing), so the beam converges.

Question 56

Physics · 4 marks · −1

An initially parallel cylindrical beam travels in a medium of refractive index μ(I)=μ0+μ2I\mu(I) = \mu_0 + \mu_2I, where μ0\mu_0 and μ2\mu_2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. The initial shape of the wave front of the beam is

  1. A.convex
  2. B.concave
  3. C.convex near the axis and concave near the periphery
  4. D.planar

Correct answer: D

Explanation: A parallel beam has plane wavefronts.

Question 57

Physics · 4 marks · −1

An initially parallel cylindrical beam travels in a medium of refractive index μ(I)=μ0+μ2I\mu(I) = \mu_0 + \mu_2I, where μ0\mu_0 and μ2\mu_2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. The speed of light in the medium is

  1. A.minimum on the axis of the beam
  2. B.the same everywhere in the beam
  3. C.directly proportional to the intensity I
  4. D.maximum on the axis of the beam

Correct answer: A

Explanation: Largest μ\mu on the axis means lowest speed there.

Question 58

Physics · 4 marks · −1

A small particle of mass m is projected at an angle θ\theta with the x-axis with an initial velocity v0v_0 in the x-y plane as shown in the figure. At a time t<v0sin⁡θgt < \frac{v_0\sin\theta}{g}, the angular momentum of the particle is (where i^\hat{i}, j^\hat{j} and k^\hat{k} are unit vectors along x, y and z-axis respectively)

Figure for question 58
  1. A.−mgv0t2cos⁡θ j^-mgv_0t^2\cos\theta\,\hat{j}
  2. B.mgv0tcos⁡θ k^mgv_0t\cos\theta\,\hat{k}
  3. C.−12mgv0t2cos⁡θ k^-\frac{1}{2}mgv_0t^2\cos\theta\,\hat{k}
  4. D.12mgv0t2cos⁡θ i^\frac{1}{2}mgv_0t^2\cos\theta\,\hat{i}

Correct answer: C

Explanation: L⃗=mr⃗×v⃗=−12mgv0t2cos⁡θ k^\vec L=m\vec r\times\vec v=-\frac12mgv_0t^2\cos\theta\,\hat k about the launch point.

Question 59

Physics · 4 marks · −1

The equation of a wave on a string of linear mass density 0.04 kg m−1^{-1} is given by y=0.02 (m)sin⁡[2π(t0.04 (s)−x0.50 (m))]y = 0.02\,(m)\sin\left[2\pi\left(\frac{t}{0.04\,(s)} - \frac{x}{0.50\,(m)}\right)\right]. The tension in the string is

  1. A.4.0 N
  2. B.12.5 N
  3. C.0.5 N
  4. D.6.25 N

Correct answer: D

Explanation: v=0.50.04=12.5v=\frac{0.5}{0.04}=12.5 m/s; T=μv2=6.25T=\mu v^2=6.25 N.

Question 60

Mathematics · 4 marks · −1

Let cos⁡(α+β)=45\cos(\alpha + \beta) = \frac{4}{5} and let sin⁡(α−β)=513\sin(\alpha - \beta) = \frac{5}{13}, where 0≤α,β≤π40 \le \alpha, \beta \le \frac{\pi}{4}, then tan⁡2α\tan2\alpha =

  1. A.5633\frac{56}{33}
  2. B.1912\frac{19}{12}
  3. C.207\frac{20}{7}
  4. D.2516\frac{25}{16}

Correct answer: A

Explanation: tan⁡2α=3/4+5/121−5/16=5633\tan2\alpha=\frac{3/4+5/12}{1-5/16}=\frac{56}{33}.

Question 61

Mathematics · 4 marks · −1

Let S be a non-empty subset of R. Consider the following statement: P: There is a rational number x∈Sx \in S such that x>0x > 0. Which of the following statements is the negation of the statement P ?

  1. A.There is no rational number x∈Sx \in S such that x≤0x \le 0
  2. B.Every rational number x∈Sx \in S satisfies x≤0x \le 0
  3. C.x∈Sx \in S and x≤0⇒xx \le 0 \Rightarrow x is not rational
  4. D.There is a rational number x∈Sx \in S such that x≤0x \le 0

Correct answer: B

Explanation: Negate the existential: for all rational x in S, x≤0x\le0.

Question 62

Mathematics · 4 marks · −1

Let a⃗=j^−k^\vec{a} = \hat{j} - \hat{k} and c⃗=i^−j^−k^\vec{c} = \hat{i} - \hat{j} - \hat{k}. Then vector b⃗\vec{b} satisfying a⃗×b⃗+c⃗=0⃗\vec{a}\times\vec{b} + \vec{c} = \vec{0} and a⃗⋅b⃗=3\vec{a}\cdot\vec{b} = 3 is

  1. A.2i^−j^+2k^2\hat{i} - \hat{j} + 2\hat{k}
  2. B.i^−j^−2k^\hat{i} - \hat{j} - 2\hat{k}
  3. C.i^+j^−2k^\hat{i} + \hat{j} - 2\hat{k}
  4. D.−i^+j^−2k^-\hat{i} + \hat{j} - 2\hat{k}

Correct answer: D

Explanation: Check b⃗=(−1,1,−2)\vec b=(-1,1,-2): a⃗⋅b⃗=1+2=3\vec a\cdot\vec b=1+2=3, a⃗×b⃗=(−1,1,1)=−c⃗\vec a\times\vec b=(-1,1,1)=-\vec c.

Question 63

Mathematics · 4 marks · −1

The equation of the tangent to the curve y=x+4x2y = x + \frac{4}{x^2}, that is parallel to the x-axis, is

  1. A.y=1y = 1
  2. B.y=2y = 2
  3. C.y=3y = 3
  4. D.y=0y = 0

Correct answer: C

Explanation: 1−8x3=0⇒x=21-\frac8{x^3}=0\Rightarrow x=2, y=3y=3.

Question 64

Mathematics · 4 marks · −1

Solution of the differential equation cos⁡x dy=y(sin⁡x−y) dx\cos x\,dy = y(\sin x - y)\,dx, 0<x<π20 < x < \frac{\pi}{2} is

  1. A.ysec⁡x=tan⁡x+cy\sec x = \tan x + c
  2. B.ytan⁡x=sec⁡x+cy\tan x = \sec x + c
  3. C.tan⁡x=(sec⁡x+c)y\tan x = (\sec x + c)y
  4. D.sec⁡x=(tan⁡x+c)y\sec x = (\tan x + c)y

Correct answer: D

Explanation: Bernoulli: with t=1yt=\frac1y, t′+ttan⁡x=sec⁡xt'+t\tan x=\sec x; IF sec⁡x\sec x: sec⁡xy=tan⁡x+c\frac{\sec x}{y}=\tan x+c.

Question 65

Mathematics · 4 marks · −1

The area bounded by the curves y=cos⁡xy = \cos x and y=sin⁡xy = \sin x between the ordinates x=0x = 0 and x=3π2x = \frac{3\pi}{2} is

  1. A.42+24\sqrt{2} + 2
  2. B.42−14\sqrt{2} - 1
  3. C.42+14\sqrt{2} + 1
  4. D.42−24\sqrt{2} - 2

Correct answer: D

Explanation: ∫0π/4+∫π/45π/4+∫5π/43π/2∣cos⁡x−sin⁡x∣dx=(2−1)+22+(2−1)\int_0^{\pi/4}+\int_{\pi/4}^{5\pi/4}+\int_{5\pi/4}^{3\pi/2}|\cos x-\sin x|dx=(\sqrt2-1)+2\sqrt2+(\sqrt2-1).

Question 66

Mathematics · 4 marks · −1

If two tangents drawn from a point P to the parabola y2=4xy^2 = 4x are at right angles, then the locus of P is

  1. A.2x+1=02x + 1 = 0
  2. B.x=−1x = -1
  3. C.2x−1=02x - 1 = 0
  4. D.x=1x = 1

Correct answer: B

Explanation: Perpendicular tangents meet on the directrix x=−1x=-1.

Question 67

Mathematics · 4 marks · −1

If the vectors a⃗=i^−j^+2k^\vec{a} = \hat{i} - \hat{j} + 2\hat{k}, b⃗=2i^+4j^+k^\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k} and c⃗=λi^+j^+μk^\vec{c} = \lambda\hat{i} + \hat{j} + \mu\hat{k} are mutually orthogonal, then (λ,μ)(\lambda, \mu) =

  1. A.(2, -3)
  2. B.(-2, 3)
  3. C.(3, -2)
  4. D.(-3, 2)

Correct answer: D

Explanation: 2λ+4+μ=02\lambda+4+\mu=0, λ−1+2μ=0\lambda-1+2\mu=0: λ=−3\lambda=-3, μ=2\mu=2.

Question 68

Mathematics · 4 marks · −1

Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S = {(m/n, p/q) | m, n, p and q are integers such that n,q≠0n, q \ne 0 and qm = pn}. Then

  1. A.neither R nor S is an equivalence relation
  2. B.S is an equivalence relation but R is not an equivalence relation
  3. C.R and S both are equivalence relations
  4. D.R is an equivalence relation but S is not an equivalence relation

Correct answer: B

Explanation: R is not symmetric (0=0⋅y0=0\cdot y but not conversely); S is equality of fractions, an equivalence.

Question 69

Mathematics · 8 marks · −2

Let f:R→Rf : R \to R be defined by f(x)={k−2x,if x≤−12x+3,if x>−1f(x) = \begin{cases} k - 2x, & \text{if } x \le -1 \\ 2x + 3, & \text{if } x > -1 \end{cases}. If f has a local minimum at x=−1x = -1, then a possible value of k is

  1. A.0
  2. B.−12-\frac{1}{2}
  3. C.−1-1
  4. D.1

Correct answer: C

Explanation: Need f(−1)=k+2≤lim⁡x→−1+f=1f(-1)=k+2\le\lim_{x\to-1^+}f=1, i.e. k≤−1k\le-1.

Question 70

Mathematics · 8 marks · −2

The number of 3×33\times3 non-singular matrices, with four entries as 1 and all other entries as 0, is

  1. A.5
  2. B.6
  3. C.at least 7
  4. D.less than 4

Correct answer: C

Explanation: A permutation matrix plus one extra 1 in any of the 6 zero positions stays non-singular: 6⋅6=36≥76\cdot6=36\ge7.

Question 71

Mathematics · 8 marks · −2

Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ....., 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is 185\frac{1}{85}. Statement-2: If the four chosen numbers form an AP, then the set of all possible values of common difference is {±1,±2,±3,±4,±5}\{\pm1, \pm2, \pm3, \pm4, \pm5\}.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false
  3. C.Statement-1 is false, Statement-2 is true
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Correct answer: B

Explanation: Number of APs: 17+14+11+8+5+2=5717+14+11+8+5+2=57, 57(204)=185\frac{57}{\binom{20}{4}}=\frac1{85}; d can also be 6, so Statement-2 is false.

Question 72

Mathematics · 4 marks · −1

Statement-1: The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x−y+z=5x - y + z = 5. Statement-2: The plane x−y+z=5x - y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4).

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false
  3. C.Statement-1 is false, Statement-2 is true
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Correct answer: A

Explanation: Midpoint (2, 2, 5) lies on the plane and AB is along the normal; bisection alone does not explain the image.

Question 73

Mathematics · 4 marks · −1

Let S1=∑j=110j(j−1) 10CjS_1 = \sum_{j=1}^{10}j(j-1)\,{}^{10}C_j, S2=∑j=110j 10CjS_2 = \sum_{j=1}^{10}j\,{}^{10}C_j and S3=∑j=110j2 10CjS_3 = \sum_{j=1}^{10}j^2\,{}^{10}C_j. Statement-1: S3=55×29S_3 = 55\times2^9. Statement-2: S1=90×28S_1 = 90\times2^8 and S2=10×28S_2 = 10\times2^8.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false
  3. C.Statement-1 is false, Statement-2 is true
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Correct answer: B

Explanation: S1=90⋅28S_1=90\cdot2^8, S2=10⋅29S_2=10\cdot2^9 (not 282^8), S3=S1+S2=55⋅29S_3=S_1+S_2=55\cdot2^9.

Question 74

Mathematics · 4 marks · −1

Let A be a 2×22\times2 matrix with non-zero entries and let A2=IA^2 = I, where I is 2×22\times2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. Statement-1: Tr(A) = 0. Statement-2: |A| = 1.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false
  3. C.Statement-1 is false, Statement-2 is true
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Correct answer: B

Explanation: Off-diagonal equations force a+d=0a+d=0; then ∣A∣=−a2−bc=−1|A|=-a^2-bc=-1.

Question 75

Mathematics · 4 marks · −1

Let f:R→Rf : R \to R be a continuous function defined by f(x)=1ex+2e−xf(x) = \frac{1}{e^x + 2e^{-x}}. Statement-1: f(c)=13f(c) = \frac{1}{3}, for some c∈Rc \in R. Statement-2: 0<f(x)≤1220 < f(x) \le \frac{1}{2\sqrt{2}}, for all x∈Rx \in R.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is false
  3. C.Statement-1 is false, Statement-2 is true
  4. D.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Correct answer: D

Explanation: Max 122≈0.354>13\frac1{2\sqrt2}\approx0.354>\frac13 and f is continuous tending to 0, so by IVT 13\frac13 is attained.

Question 76

Mathematics · 4 marks · −1

For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is

  1. A.There is a regular polygon with rR=12\frac{r}{R} = \frac{1}{\sqrt{2}}
  2. B.There is a regular polygon with rR=23\frac{r}{R} = \frac{2}{3}
  3. C.There is a regular polygon with rR=32\frac{r}{R} = \frac{\sqrt{3}}{2}
  4. D.There is a regular polygon with rR=12\frac{r}{R} = \frac{1}{2}

Correct answer: B

Explanation: rR=cos⁡πn\frac rR=\cos\frac\pi n; 23\frac23 is not cos⁡πn\cos\frac\pi n for any n.

Question 77

Mathematics · 4 marks · −1

If α\alpha and β\beta are the roots of the equation x2−x+1=0x^2 - x + 1 = 0, then α2009+β2009\alpha^{2009} + \beta^{2009} =

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

Correct answer: B

Explanation: α,β=e±iπ/3\alpha,\beta=e^{\pm i\pi/3}; sum =2cos⁡2009π3=2cos⁡5π3=1=2\cos\frac{2009\pi}{3}=2\cos\frac{5\pi}3=1.

Question 78

Mathematics · 4 marks · −1

The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣|z - 1| = |z + 1| = |z - i| equals

  1. A.1
  2. B.2
  3. C.∞\infty
  4. D.0

Correct answer: A

Explanation: Only the circumcentre of 1, -1, i: z = 0.

Question 79

Mathematics · 4 marks · −1

A line AB in three-dimensional space makes angles 45∘45^\circ and 120∘120^\circ with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ\theta with the positive z-axis, then θ\theta equals

  1. A.45∘45^\circ
  2. B.60∘60^\circ
  3. C.75∘75^\circ
  4. D.30∘30^\circ

Correct answer: B

Explanation: 12+14+cos⁡2θ=1⇒θ=60∘\frac12+\frac14+\cos^2\theta=1\Rightarrow\theta=60^\circ.

Question 80

Mathematics · 4 marks · −1

The line L given by x5+yb=1\frac{x}{5} + \frac{y}{b} = 1 passes through the point (13, 32). The line K is parallel to L and has the equation xc+y3=1\frac{x}{c} + \frac{y}{3} = 1. Then the distance between L and K is

  1. A.17\sqrt{17}
  2. B.1715\frac{17}{\sqrt{15}}
  3. C.2317\frac{23}{\sqrt{17}}
  4. D.2315\frac{23}{\sqrt{15}}

Correct answer: C

Explanation: b=−20b=-20, L: 4x−y=204x-y=20; K: 4x−y=−34x-y=-3; distance 2317\frac{23}{\sqrt{17}}.

Question 81

Mathematics · 8 marks · −2

A person is to count 4500 currency notes. Let ana_n denote the number of notes he counts in the nthn^{th} minute. If a1=a2=⋯=a10=150a_1 = a_2 = \dots = a_{10} = 150 and a10,a11,…a_{10}, a_{11}, \dots are in A.P. with common difference −2-2, then the time taken by him to count all notes is

  1. A.34 minutes
  2. B.125 minutes
  3. C.135 minutes
  4. D.24 minutes

Correct answer: A

Explanation: After 10 min, 1500 counted; n2[2(148)−2(n−1)]=3000⇒n=24\frac n2[2(148)-2(n-1)]=3000\Rightarrow n=24; total 34 min.

Question 82

Mathematics · 8 marks · −2

Let f:R→Rf : R \to R be a positive increasing function with lim⁡x→∞f(3x)f(x)=1\lim_{x\to\infty}\frac{f(3x)}{f(x)} = 1. Then lim⁡x→∞f(2x)f(x)\lim_{x\to\infty}\frac{f(2x)}{f(x)} =

  1. A.23\frac{2}{3}
  2. B.32\frac{3}{2}
  3. C.3
  4. D.1

Correct answer: D

Explanation: 1≤f(2x)f(x)≤f(3x)f(x)→11\le\frac{f(2x)}{f(x)}\le\frac{f(3x)}{f(x)}\to1 (sandwich).

Question 83

Mathematics · 8 marks · −2

Let p(x) be a function defined on R such that p′(x)=p′(1−x)p'(x) = p'(1 - x), for all x∈[0,1]x \in [0, 1], p(0) = 1 and p(1) = 41. Then ∫01p(x) dx\int_0^1p(x)\,dx equals

  1. A.21
  2. B.41
  3. C.42
  4. D.41\sqrt{41}

Correct answer: A

Explanation: p(x)+p(1−x)=42p(x)+p(1-x)=42, so 2I=422I=42.

Question 84

Mathematics · 4 marks · −1

Let f:(−1,1)→Rf : (-1, 1) \to R be a differentiable function with f(0) = -1 and f'(0) = 1. Let g(x)=[f(2f(x)+2)]2g(x) = [f(2f(x) + 2)]^2. Then g'(0) =

  1. A.−4-4
  2. B.0
  3. C.−2-2
  4. D.4

Correct answer: A

Explanation: g′(0)=2f(0)⋅f′(0)⋅2f′(0)=−4g'(0)=2f(0)\cdot f'(0)\cdot2f'(0)=-4.

Question 85

Mathematics · 4 marks · −1

There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

  1. A.36
  2. B.66
  3. C.108
  4. D.3

Correct answer: C

Explanation: (32)(92)=108\binom32\binom92=108.

Question 86

Mathematics · 4 marks · −1

Consider the system of linear equations: x1+2x2+x3=3x_1 + 2x_2 + x_3 = 3, 2x1+3x2+x3=32x_1 + 3x_2 + x_3 = 3, 3x1+5x2+2x3=13x_1 + 5x_2 + 2x_3 = 1. The system has

  1. A.exactly 3 solutions
  2. B.a unique solution
  3. C.no solution
  4. D.infinite number of solutions

Correct answer: C

Explanation: Eq1 + Eq2 gives LHS of Eq3 but RHS 6 ≠ 1: inconsistent.

Question 87

Mathematics · 4 marks · −1

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colour is

  1. A.27\frac{2}{7}
  2. B.121\frac{1}{21}
  3. C.223\frac{2}{23}
  4. D.13\frac{1}{3}

Correct answer: A

Explanation: 3⋅4⋅2(93)=2484=27\frac{3\cdot4\cdot2}{\binom93}=\frac{24}{84}=\frac27.

Question 88

Mathematics · 4 marks · −1

For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is

  1. A.112\frac{11}{2}
  2. B.6
  3. C.132\frac{13}{2}
  4. D.52\frac{5}{2}

Correct answer: A

Explanation: ∑x2=40\sum x^2=40, ∑y2=105\sum y^2=105; combined variance 14510−9=112\frac{145}{10}-9=\frac{11}{2}.

Question 89

Mathematics · 4 marks · −1

The circle x2+y2=4x+8y+5x^2 + y^2 = 4x + 8y + 5 intersects the line 3x−4y=m3x - 4y = m at two distinct points if

  1. A.−35<m<15-35 < m < 15
  2. B.15<m<6515 < m < 65
  3. C.35<m<8535 < m < 85
  4. D.−85<m<−35-85 < m < -35

Correct answer: A

Explanation: Centre (2, 4), r = 5: ∣−10−m∣5<5⇒−35<m<15\frac{|-10-m|}{5}<5\Rightarrow-35<m<15.

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