JEE Main 2013 — 7 Apr Offline Set Q (Previous Year Paper)

JEE Main · 2013 · with answers and explanations

questions
86
Total marks
344
Duration
180 minutes
Negative marking
−1

Subjects: Chemistry · Mathematics · Physics

Question 1

Chemistry · 4 marks · −1

An unknown alcohol is treated with the 'Lucas reagent' to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism:

  1. A.tertiary alcohol by SN1S_N1
  2. B.secondary alcohol by SN2S_N2
  3. C.tertiary alcohol by SN2S_N2
  4. D.secondary alcohol by SN1S_N1

Correct answer: A

Explanation: Tertiary alcohols react fastest via the stable tertiary carbocation (SN1S_N1).

Question 2

Chemistry · 4 marks · −1

The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+Na^+ will be:

  1. A.−5.1-5.1 eV
  2. B.−10.2-10.2 eV
  3. C.+2.55+2.55 eV
  4. D.−2.55-2.55 eV

Correct answer: A

Explanation: Na++e−→NaNa^++e^-\rightarrow Na is the reverse of ionisation, releasing 5.1 eV.

Question 3

Chemistry · 4 marks · −1

Stability of the species Li2Li_2, Li2−Li_2^- and Li2+Li_2^+ increases in the order of:

  1. A.Li2−<Li2+<Li2Li_2^- < Li_2^+ < Li_2
  2. B.Li2<Li2−<Li2+Li_2 < Li_2^- < Li_2^+
  3. C.Li2−<Li2<Li2+Li_2^- < Li_2 < Li_2^+
  4. D.Li2<Li2+<Li2−Li_2 < Li_2^+ < Li_2^-

Correct answer: A

Explanation: Bond orders: Li2Li_2 1, Li2+Li_2^+ 0.5, Li2−Li_2^- 0.5; Li2−Li_2^- has an antibonding electron so is less stable than Li2+Li_2^+.

Question 4

Chemistry · 4 marks · −1

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:

  1. A.1.00 M
  2. B.1.75 M
  3. C.0.975 M
  4. D.0.875 M

Correct answer: D

Explanation: 0.5⋅750+2⋅2501000=0.875\frac{0.5\cdot750+2\cdot250}{1000}=0.875 M.

Question 5

Chemistry · 4 marks · −1

Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest EM3+/M2+∘E^\circ_{M^{3+}/M^{2+}} value?

  1. A.Mn (Z = 25)
  2. B.Fe (Z = 26)
  3. C.Co (Z = 27)
  4. D.Cr (Z = 24)

Correct answer: C

Explanation: E∘E^\circ: Co 1.97 V > Mn 1.57 > Fe 0.77 > Cr −0.41-0.41 V.

Question 6

Chemistry · 4 marks · −1

A solution of (−)(-)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5SbCl_5, due to the formation of :

  1. A.carbene
  2. B.carbocation
  3. C.free radical
  4. D.carbanion

Correct answer: B

Explanation: SbCl5SbCl_5 abstracts Cl−Cl^- giving a planar benzylic carbocation, which recombines from either face.

Question 7

Chemistry · 4 marks · −1

The coagulating power of electrolytes having ions Na+Na^+, Al3+Al^{3+} and Ba2+Ba^{2+} for arsenic sulphide sol increases in the order:

  1. A.Na+<Ba2+<Al3+Na^+ < Ba^{2+} < Al^{3+}
  2. B.Ba2+<Na+<Al3+Ba^{2+} < Na^+ < Al^{3+}
  3. C.Al3+<Na+<Ba2+Al^{3+} < Na^+ < Ba^{2+}
  4. D.Al3+<Ba2+<Na+Al^{3+} < Ba^{2+} < Na^+

Correct answer: A

Explanation: As2S3As_2S_3 is a negative sol; by Hardy-Schulze, coagulating power rises with cation charge.

Question 8

Chemistry · 4 marks · −1

How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

  1. A.0.9 L
  2. B.2.0 L
  3. C.9.0 L
  4. D.0.1 L

Correct answer: C

Explanation: [H+][H^+] must fall tenfold: total volume 10 L, so add 9 L.

Question 9

Chemistry · 4 marks · −1

Experimentally it was found that a metal oxide has formula M0.98OM_{0.98}O. Metal M, is present as M2+M^{2+} and M3+M^{3+} in its oxide. Fraction of the metal which exists as M3+M^{3+} would be:

  1. A.4.08%
  2. B.6.05%
  3. C.5.08%
  4. D.7.01%

Correct answer: A

Explanation: 3x+2(0.98−x)=2⇒x=0.043x+2(0.98-x)=2\Rightarrow x=0.04; fraction =0.040.98=4.08%=\frac{0.04}{0.98}=4.08\%.

Question 10

Chemistry · 4 marks · −1

A compound with molecular mass 180 is acylated with CH3COClCH_3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is:

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

Correct answer: A

Explanation: Each acetylation adds 42; 390−18042=5\frac{390-180}{42}=5.

Question 11

Chemistry · 4 marks · −1

Given ECr3+/Cr∘=−0.74E^\circ_{Cr^{3+}/Cr} = -0.74 V; EMnO4−/Mn2+∘=1.51E^\circ_{MnO_4^-/Mn^{2+}} = 1.51 V; ECr2O72−/Cr3+∘=1.33E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 V; ECl/Cl−∘=1.36E^\circ_{Cl/Cl^-} = 1.36 V. Based on the data given above, strongest oxidising agent will be:

  1. A.Cr3+Cr^{3+}
  2. B.Mn2+Mn^{2+}
  3. C.MnO4−MnO_4^-
  4. D.Cl−Cl^-

Correct answer: C

Explanation: Highest reduction potential: MnO4−MnO_4^-.

Question 12

Chemistry · 4 marks · −1

Arrange the following compounds in order of decreasing acidity: (I) p-chlorophenol, (II) p-cresol (p-methylphenol), (III) p-nitrophenol, (IV) p-methoxyphenol

Figure for question 12
  1. A.I > II > III > IV
  2. B.III > I > II > IV
  3. C.IV > III > I > II
  4. D.II > IV > I > III

Correct answer: B

Explanation: Electron-withdrawing NO2NO_2 > Cl increase acidity; CH3CH_3 and especially OCH3OCH_3 (+R) decrease it.

Question 13

Chemistry · 4 marks · −1

The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be: (R=8.314R = 8.314 JK−1^{-1} mol−1^{-1} and log 2 = 0.301)

  1. A.48.6 kJ mol−1^{-1}
  2. B.58.5 kJ mol−1^{-1}
  3. C.60.5 kJ mol−1^{-1}
  4. D.53.6 kJ mol−1^{-1}

Correct answer: D

Explanation: 2.303log⁡2=EaR(1300−1310)⇒Ea≈53.62.303\log2=\frac{E_a}{R}\left(\frac1{300}-\frac1{310}\right)\Rightarrow E_a\approx53.6 kJ/mol.

Question 14

Chemistry · 4 marks · −1

Synthesis of each molecule of glucose in photosynthesis involves:

  1. A.10 molecules of ATP
  2. B.8 molecules of ATP
  3. C.6 molecules of ATP
  4. D.18 molecules of ATP

Correct answer: D

Explanation: The Calvin cycle consumes 18 ATP (and 12 NADPH) per glucose.

Question 15

Chemistry · 4 marks · −1

Which of the following complex species is not expected to exhibit optical isomerism?

  1. A.[Co(en)2Cl2]+[Co(en)_2Cl_2]^+
  2. B.[Co(NH3)3Cl3][Co(NH_3)_3Cl_3]
  3. C.[Co(en)(NH3)2Cl2]+[Co(en)(NH_3)_2Cl_2]^+
  4. D.[Co(en)3]3+[Co(en)_3]^{3+}

Correct answer: B

Explanation: Both fac and mer [Co(NH3)3Cl3][Co(NH_3)_3Cl_3] have mirror planes.

Question 16

Chemistry · 4 marks · −1

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0∘37.0^\circC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be: (R=8.314R = 8.314 J/mol K) (ln⁡7.5=2.01\ln7.5 = 2.01)

  1. A.q=−208q = -208 J, w=−208w = -208 J
  2. B.q=−208q = -208 J, w=+208w = +208 J
  3. C.q=+208q = +208 J, w=+208w = +208 J
  4. D.q=+208q = +208 J, w=−208w = -208 J

Correct answer: D

Explanation: Isothermal: ΔU=0\Delta U=0, heat absorbed q=+208q=+208 J, work done on gas w=−208w=-208 J.

Question 17

Chemistry · 4 marks · −1

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2CO_2. The empirical formula of the hydrocarbon is:

  1. A.C3H4C_3H_4
  2. B.C6H5C_6H_5
  3. C.C7H8C_7H_8
  4. D.C2H4C_2H_4

Correct answer: C

Explanation: C: 0.07 mol, H: 0.08 mol, ratio 7:8.

Question 18

Chemistry · 4 marks · −1

The order of stability of the following carbocations: (I) CH2=CH−CH2+CH_2=CH-CH_2^+ (allyl), (II) CH3CH2CH2+CH_3CH_2CH_2^+ (n-propyl), (III) C6H5CH2+C_6H_5CH_2^+ (benzyl) is:

Figure for question 18
  1. A.II > III > I
  2. B.I > II > III
  3. C.III > I > II
  4. D.III > II > I

Correct answer: C

Explanation: Benzyl (more delocalisation) > allyl > primary propyl.

Question 19

Chemistry · 4 marks · −1

Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar?

  1. A.S < Se < Ca < Ba < Ar
  2. B.Ba < Ca < Se < S < Ar
  3. C.Ca < Ba < S < Se < Ar
  4. D.Ca < S < Ba < Se < Ar

Correct answer: B

Explanation: IE increases up a group and across a period.

Question 20

Chemistry · 4 marks · −1

For gaseous state, if most probable speed is denoted by C∗C^*, average speed by Cˉ\bar{C} and mean square speed by C, then for a large number of molecules the ratios of these speeds are:

  1. A.C∗:Cˉ:C=1.128:1.225:1C^* : \bar{C} : C = 1.128 : 1.225 : 1
  2. B.C∗:Cˉ:C=1:1.128:1.225C^* : \bar{C} : C = 1 : 1.128 : 1.225
  3. C.C∗:Cˉ:C=1:1.125:1.128C^* : \bar{C} : C = 1 : 1.125 : 1.128
  4. D.C∗:Cˉ:C=1.225:1.128:1C^* : \bar{C} : C = 1.225 : 1.128 : 1

Correct answer: B

Explanation: 2:8/π:3=1:1.128:1.225\sqrt2:\sqrt{8/\pi}:\sqrt3=1:1.128:1.225 (C here is the root mean square speed).

Question 21

Chemistry · 4 marks · −1

The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:

  1. A.Methylamine
  2. B.Ammonia
  3. C.Phosgene
  4. D.Methylisocyanate

Correct answer: D

Explanation: Methyl isocyanate, CH3NCOCH_3NCO.

Question 22

Chemistry · 4 marks · −1

Consider the following reaction: xMnO4−+yC2O42−+zH+→xMn2++2yCO2+z2H2OxMnO_4^- + yC_2O_4^{2-} + zH^+ \rightarrow xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O. The values of x, y and z in the reaction are, respectively:

  1. A.2, 5 and 8
  2. B.2, 5 and 16
  3. C.5, 2 and 8
  4. D.5, 2 and 16

Correct answer: B

Explanation: 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2MnO_4^-+5C_2O_4^{2-}+16H^+\rightarrow2Mn^{2+}+10CO_2+8H_2O.

Question 23

Chemistry · 4 marks · −1

Which of the following exists as covalent crystals in the solid state?

  1. A.Silicon
  2. B.Sulphur
  3. C.Phosphorous
  4. D.Iodine

Correct answer: A

Explanation: Silicon is a covalent network solid; S8S_8, P4P_4, I2I_2 form molecular solids.

Question 24

Chemistry · 4 marks · −1

Compound (A), C8H9BrC_8H_9Br, gives a white precipitate when warmed with alcoholic AgNO3AgNO_3. Oxidation of (A) gives an acid (B), C8H6O4C_8H_6O_4. (B) easily forms anhydride on heating. Identify the compound (A).

Figure for question 24
  1. A.1-bromo-2-ethylbenzene
  2. B.4-methylbenzyl bromide
  3. C.2-methylbenzyl bromide
  4. D.3-methylbenzyl bromide

Correct answer: C

Explanation: Benzylic bromide reacts with AgNO3AgNO_3; oxidation gives phthalic acid (ortho), which forms phthalic anhydride, so A is o-methylbenzyl bromide.

Question 25

Chemistry · 4 marks · −1

Energy of an electron is given by E=−2.178×10−18E = -2.178\times10^{-18} J (Z2n2)\left(\frac{Z^2}{n^2}\right). Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h=6.62×10−34h = 6.62\times10^{-34} Js and c=3.0×108c = 3.0\times10^8 ms−1^{-1})

  1. A.2.816×10−72.816\times10^{-7} m
  2. B.6.500×10−76.500\times10^{-7} m
  3. C.8.500×10−78.500\times10^{-7} m
  4. D.1.214×10−71.214\times10^{-7} m

Correct answer: D

Explanation: ΔE=2.178×10−18⋅34\Delta E=2.178\times10^{-18}\cdot\frac34; λ=hcΔE≈1.214×10−7\lambda=\frac{hc}{\Delta E}\approx1.214\times10^{-7} m.

Question 26

Chemistry · 4 marks · −1

An organic compound A upon reacting with NH3NH_3 gives B. On heating B gives C. C in presence of KOH reacts with Br2Br_2 to give CH3CH2NH2CH_3CH_2NH_2. A is

  1. A.CH3CH2CH2COOHCH_3CH_2CH_2COOH
  2. B.CH3CH(CH3)COOHCH_3CH(CH_3)COOH
  3. C.CH3CH2COOHCH_3CH_2COOH
  4. D.CH3COOHCH_3COOH

Correct answer: C

Explanation: Propanoic acid gives ammonium salt, then propanamide, whose Hofmann degradation gives ethylamine.

Question 27

Chemistry · 4 marks · −1

In which of the following pairs of molecules/ions, both the species are not likely to exist?

  1. A.H2−H_2^-, He22−He_2^{2-}
  2. B.H22+H_2^{2+}, He2He_2
  3. C.H2−H_2^-, He22+He_2^{2+}
  4. D.H2+H_2^+, He22−He_2^{2-}

Correct answer: B

Explanation: Both H22+H_2^{2+} and He2He_2 have bond order zero.

Question 28

Mathematics · 4 marks · −1

The circle passing through (1,−2)(1, -2) and touching the axis of x at (3,0)(3, 0) also passes through the point

  1. A.(2,−5)(2, -5)
  2. B.(5,−2)(5, -2)
  3. C.(−2,5)(-2, 5)
  4. D.(−5,2)(-5, 2)

Correct answer: B

Explanation: (x−3)2+y2+λy=0(x-3)^2+y^2+\lambda y=0 through (1,−2)(1,-2): λ=4\lambda=4; (5,−2)(5,-2) satisfies it.

Question 29

Mathematics · 4 marks · −1

ABCD is a trapezium such that AB and CD are parallel and BC⊥CDBC \perp CD. If ∠ADB=θ\angle ADB = \theta, BC = p and CD = q, then AB is equal to

  1. A.p2+q2cos⁡θpcos⁡θ+qsin⁡θ\frac{p^2+q^2\cos\theta}{p\cos\theta+q\sin\theta}
  2. B.p2+q2p2cos⁡θ+q2sin⁡θ\frac{p^2+q^2}{p^2\cos\theta+q^2\sin\theta}
  3. C.(p2+q2)sin⁡θ(pcos⁡θ+qsin⁡θ)2\frac{(p^2+q^2)\sin\theta}{(p\cos\theta+q\sin\theta)^2}
  4. D.(p2+q2)sin⁡θpcos⁡θ+qsin⁡θ\frac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}

Correct answer: D

Explanation: Sine rule in △ABD\triangle ABD with BD=p2+q2BD=\sqrt{p^2+q^2} and ∠ABD=∠BDC=α\angle ABD=\angle BDC=\alpha (tan⁡α=p/q\tan\alpha=p/q) gives AB=(p2+q2)sin⁡θpcos⁡θ+qsin⁡θAB=\frac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}.

Question 30

Mathematics · 4 marks · −1

Given : A circle, 2x2+2y2=52x^2 + 2y^2 = 5 and a parabola, y2=45xy^2 = 4\sqrt{5}x. Statement-I : An equation of a common tangent to these curves is y=x+5y = x + \sqrt{5}. Statement-II : If the line, y=mx+5my = mx + \frac{\sqrt{5}}{m} (m≠0)(m \ne 0) is their common tangent, then m satisfies m4−3m2+2=0m^4 - 3m^2 + 2 = 0.

  1. A.Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I
  2. B.Statement-I is True; Statement-II is False.
  3. C.Statement-I is False; Statement-II is True
  4. D.Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Correct answer: A

Explanation: Tangency to the circle gives m4+m2−2=0m^4+m^2-2=0, i.e. m=±1m=\pm1, so y=x+5y=x+\sqrt5 is common (I true). m=±1m=\pm1 also satisfies m4−3m2+2=0m^4-3m^2+2=0 (II true) but that equation is not the governing condition.

Question 31

Mathematics · 4 marks · −1

A ray of light along x+3y=3x + \sqrt{3}y = \sqrt{3} gets reflected upon reaching x-axis, the equation of the reflected rays is

  1. A.3y=x−3\sqrt{3}y = x - \sqrt{3}
  2. B.y=3x−3y = \sqrt{3}x - \sqrt{3}
  3. C.3y=x−1\sqrt{3}y = x - 1
  4. D.y=x+3y = x + \sqrt{3}

Correct answer: A

Explanation: Incidence point (3,0)(\sqrt3,0); reflected slope 13\frac1{\sqrt3}: 3y=x−3\sqrt3y=x-\sqrt3.

Question 32

Mathematics · 4 marks · −1

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ?

  1. A.median
  2. B.mode
  3. C.variance
  4. D.mean

Correct answer: C

Explanation: Variance is independent of a change of origin.

Question 33

Mathematics · 4 marks · −1

If x, y, z are in A.P. and tan⁡−1x\tan^{-1}x, tan⁡−1y\tan^{-1}y and tan⁡−1z\tan^{-1}z are also in A.P., then

  1. A.2x=3y=6z2x = 3y = 6z
  2. B.6x=3y=2z6x = 3y = 2z
  3. C.6x=4y=3z6x = 4y = 3z
  4. D.x=y=zx = y = z

Correct answer: D

Explanation: 2tan⁡−1y=tan⁡−1x+tan⁡−1z2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z with 2y=x+z2y=x+z forces y2=xzy^2=xz as well, so x=y=zx=y=z.

Question 34

Mathematics · 4 marks · −1

If ∫f(x) dx=Ψ(x)\int f(x)\,dx = \Psi(x), then ∫x5f(x3) dx\int x^5 f(x^3)\,dx is equal to

  1. A.13x3Ψ(x3)−3∫x3Ψ(x3) dx+C\frac{1}{3}x^3\Psi(x^3) - 3\int x^3\Psi(x^3)\,dx + C
  2. B.13x3Ψ(x3)−∫x2Ψ(x3) dx+C\frac{1}{3}x^3\Psi(x^3) - \int x^2\Psi(x^3)\,dx + C
  3. C.13[x3Ψ(x3)−∫x3Ψ(x3) dx]+C\frac{1}{3}\left[x^3\Psi(x^3) - \int x^3\Psi(x^3)\,dx\right] + C
  4. D.13[x3Ψ(x3)−∫x2Ψ(x3) dx]+C\frac{1}{3}\left[x^3\Psi(x^3) - \int x^2\Psi(x^3)\,dx\right] + C

Correct answer: B

Explanation: With t=x3t=x^3: 13∫tf(t)dt=13[tΨ(t)−∫Ψ(t)dt]=13x3Ψ(x3)−∫x2Ψ(x3)dx\frac13\int tf(t)dt=\frac13[t\Psi(t)-\int\Psi(t)dt]=\frac13x^3\Psi(x^3)-\int x^2\Psi(x^3)dx.

Question 35

Mathematics · 4 marks · −1

The equation of the circle passing through the foci of the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1, and having centre at (0,3)(0, 3) is

  1. A.x2+y2−6y+7=0x^2 + y^2 - 6y + 7 = 0
  2. B.x2+y2−6y−5=0x^2 + y^2 - 6y - 5 = 0
  3. C.x2+y2−6y+5=0x^2 + y^2 - 6y + 5 = 0
  4. D.x2+y2−6y−7=0x^2 + y^2 - 6y - 7 = 0

Correct answer: D

Explanation: Foci (±7,0)(\pm\sqrt7,0); r2=7+9=16r^2=7+9=16: x2+(y−3)2=16x^2+(y-3)^2=16.

Question 36

Mathematics · 4 marks · −1

The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0,1)(0, 1), (1,1)(1, 1) and (1,0)(1, 0) is

  1. A.2−22-\sqrt{2}
  2. B.1+21+\sqrt{2}
  3. C.1−21-\sqrt{2}
  4. D.2+22+\sqrt{2}

Correct answer: A

Explanation: Vertices (0,0),(2,0),(0,2)(0,0),(2,0),(0,2); inradius =2+2−222=2−2=\frac{2+2-2\sqrt2}{2}=2-\sqrt2, which is the x-coordinate.

Question 37

Mathematics · 4 marks · −1

The intercepts on x-axis made by tangents to the curve, y=∫0x∣t∣ dty = \int_0^x |t|\,dt, x∈Rx \in \mathbb{R}, which are parallel to the line y=2xy = 2x, are equal to

  1. A.±2\pm2
  2. B.±3\pm3
  3. C.±4\pm4
  4. D.±1\pm1

Correct answer: D

Explanation: y′=∣x∣=2⇒x=±2y'=|x|=2\Rightarrow x=\pm2, y=±2y=\pm2; tangents y=2x∓2y=2x\mp2 meet the x-axis at ±1\pm1.

Question 38

Mathematics · 4 marks · −1

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ....., is

  1. A.79(99−10−20)\frac{7}{9}(99 - 10^{-20})
  2. B.781(179+10−20)\frac{7}{81}(179 + 10^{-20})
  3. C.79(99+10−20)\frac{7}{9}(99 + 10^{-20})
  4. D.781(179−10−20)\frac{7}{81}(179 - 10^{-20})

Correct answer: B

Explanation: tr=79(1−10−r)t_r=\frac79(1-10^{-r}); S20=79[20−19(1−10−20)]=781(179+10−20)S_{20}=\frac79\left[20-\frac19(1-10^{-20})\right]=\frac{7}{81}(179+10^{-20}).

Question 39

Mathematics · 4 marks · −1

Consider : Statement-I : (p∧∼q)∧(∼p∧q)(p\wedge\sim q)\wedge(\sim p\wedge q) is a fallacy. Statement-II : (p→q)↔(∼q→∼p)(p \to q) \leftrightarrow (\sim q \to \sim p) is a tautology.

  1. A.Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I
  2. B.Statement-I is True; Statement-II is False.
  3. C.Statement-I is False; Statement-II is True
  4. D.Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Correct answer: A

Explanation: Both are verified by truth tables, but II (contrapositive) does not explain I.

Question 40

Mathematics · 4 marks · −1

The area (in square units) bounded by the curves y=xy = \sqrt{x}, 2y−x+3=02y - x + 3 = 0, x-axis, and lying in the first quadrant is

  1. A.36
  2. B.18
  3. C.274\frac{27}{4}
  4. D.9

Correct answer: D

Explanation: Curves meet at (9,3)(9,3); area =∫03[(2y+3)−y2]dy=9=\int_0^3[(2y+3)-y^2]dy=9.

Question 41

Mathematics · 4 marks · −1

The expression tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A} can be written as

  1. A.secA cosecA + 1
  2. B.tanA + cotA
  3. C.secA + cosecA
  4. D.sinA cosA + 1

Correct answer: A

Explanation: Simplifies to 1−cot⁡3Acot⁡A(1−cot⁡A)=1+sec⁡Acsc⁡A\frac{1-\cot^3A}{\cot A(1-\cot A)}=1+\sec A\csc A.

Question 42

Mathematics · 4 marks · −1

The real number k for which the equation, 2x3+3x+k=02x^3 + 3x + k = 0 has two distinct real roots in [0, 1]

  1. A.lies between 2 and 3
  2. B.lies between −1-1 and 0
  3. C.does not exist
  4. D.lies between 1 and 2

Correct answer: C

Explanation: f′(x)=6x2+3>0f'(x)=6x^2+3>0, so f is strictly increasing and cannot have two roots.

Question 43

Mathematics · 4 marks · −1

lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim_{x\to0}\frac{(1-\cos2x)(3+\cos x)}{x\tan4x} is equal to

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

Correct answer: C

Explanation: 2x2⋅44x2=2\frac{2x^2\cdot4}{4x^2}=2.

Question 44

Mathematics · 4 marks · −1

Let TnT_n be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1−Tn=10T_{n+1} - T_n = 10, then the value of n is

  1. A.5
  2. B.10
  3. C.8
  4. D.7

Correct answer: A

Explanation: (n+13)−(n3)=(n2)=10⇒n=5\binom{n+1}{3}-\binom n3=\binom n2=10\Rightarrow n=5.

Question 45

Mathematics · 4 marks · −1

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by dPdx=100−12x\frac{dP}{dx} = 100 - 12\sqrt{x}. If the firm employs 25 more workers, then the new level of production of items is

  1. A.3000
  2. B.3500
  3. C.4500
  4. D.2500

Correct answer: B

Explanation: P−2000=∫025(100−12x)dx=2500−1000=1500P-2000=\int_0^{25}(100-12\sqrt x)dx=2500-1000=1500.

Question 46

Mathematics · 4 marks · −1

Statement-I : The value of the integral ∫π/6π/3dx1+tan⁡x\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}} is equal to π6\frac{\pi}{6}. Statement-II : ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx.

  1. A.Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I
  2. B.Statement-I is True; Statement-II is False.
  3. C.Statement-I is False; Statement-II is True
  4. D.Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Correct answer: C

Explanation: Using II, 2I=π62I=\frac\pi6, I=π12I=\frac\pi{12}, so I is false and II true.

Question 47

Mathematics · 4 marks · −1

If P=[1α3133244]P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix} is the adjoint of a 3×33\times3 matrix A and ∣A∣=4|A| = 4, then α\alpha is equal to

  1. A.11
  2. B.5
  3. C.0
  4. D.4

Correct answer: A

Explanation: ∣P∣=∣A∣2=16|P|=|A|^2=16: ∣P∣=2α−6=16⇒α=11|P|=2\alpha-6=16\Rightarrow\alpha=11.

Question 48

Mathematics · 4 marks · −1

The number of values of k, for which the system of equations (k+1)x+8y=4k(k + 1)x + 8y = 4k, kx+(k+3)y=3k−1kx + (k + 3)y = 3k - 1 has no solution, is

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

Correct answer: A

Explanation: (k+1)(k+3)=8k⇒k=1,3(k+1)(k+3)=8k\Rightarrow k=1,3; k=1k=1 gives infinitely many solutions, k=3k=3 none: one value.

Question 49

Mathematics · 4 marks · −1

If y=sec⁡(tan⁡−1x)y = \sec(\tan^{-1}x), then dydx\frac{dy}{dx} at x=1x = 1 is equal to

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

Correct answer: D

Explanation: y=1+x2y=\sqrt{1+x^2}, y′=x1+x2=12y'=\frac{x}{\sqrt{1+x^2}}=\frac1{\sqrt2}.

Question 50

Mathematics · 4 marks · −1

If the lines x−21=y−31=z−4−k\frac{x-2}{1} = \frac{y-3}{1} = \frac{z-4}{-k} and x−1k=y−42=z−51\frac{x-1}{k} = \frac{y-4}{2} = \frac{z-5}{1} are coplanar, then k can have

  1. A.exactly one value
  2. B.exactly two values
  3. C.exactly three values
  4. D.any value

Correct answer: B

Explanation: Coplanarity determinant gives k2+3k=0k^2+3k=0: k=0,−3k=0,-3.

Question 51

Mathematics · 4 marks · −1

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A×BA \times B having 3 or more elements is

  1. A.220
  2. B.219
  3. C.211
  4. D.256

Correct answer: B

Explanation: 28−1−8−28=2192^8-1-8-28=219.

Question 52

Mathematics · 4 marks · −1

If the vectors AB⃗=3i^+4k^\vec{AB} = 3\hat{i} + 4\hat{k} and AC⃗=5i^−2j^+4k^\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k} are the sides of a triangle ABC, then the length of the median through A is

  1. A.72\sqrt{72}
  2. B.33\sqrt{33}
  3. C.45\sqrt{45}
  4. D.18\sqrt{18}

Correct answer: B

Explanation: AM⃗=AB⃗+AC⃗2=4i^−j^+4k^\vec{AM}=\frac{\vec{AB}+\vec{AC}}{2}=4\hat i-\hat j+4\hat k, length 33\sqrt{33}.

Question 53

Mathematics · 4 marks · −1

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

  1. A.1335\frac{13}{3^5}
  2. B.1135\frac{11}{3^5}
  3. C.1035\frac{10}{3^5}
  4. D.1735\frac{17}{3^5}

Correct answer: B

Explanation: (54)235+135=1135\binom54\frac{2}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}.

Question 54

Mathematics · 4 marks · −1

If z is a complex number of unit modulus and argument θ\theta, then arg⁡(1+z1+zˉ)\arg\left(\frac{1+z}{1+\bar{z}}\right) equals

  1. A.π2−θ\frac{\pi}{2}-\theta
  2. B.θ\theta
  3. C.π−θ\pi-\theta
  4. D.−θ-\theta

Correct answer: B

Explanation: zˉ=1/z\bar z=1/z, so 1+z1+1/z=z\frac{1+z}{1+1/z}=z.

Question 55

Mathematics · 4 marks · −1

If the equations x2+2x+3=0x^2 + 2x + 3 = 0 and ax2+bx+c=0ax^2 + bx + c = 0, a,b,c∈Ra, b, c \in \mathbb{R}, have a common root, then a:b:ca : b : c is

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

Correct answer: D

Explanation: The roots of the first are non-real, so both roots are common: a:b:c=1:2:3a:b:c=1:2:3.

Question 56

Mathematics · 4 marks · −1

Distance between two parallel planes 2x+y+2z=82x + y + 2z = 8 and 4x+2y+4z+5=04x + 2y + 4z + 5 = 0 is

  1. A.52\frac{5}{2}
  2. B.72\frac{7}{2}
  3. C.92\frac{9}{2}
  4. D.32\frac{3}{2}

Correct answer: B

Explanation: ∣16+5∣36=72\frac{|16+5|}{\sqrt{36}}=\frac72.

Question 57

Mathematics · 4 marks · −1

The term independent of x in expansion of (x+1x2/3−x1/3+1−x−1x−x1/2)10\left(\frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x - x^{1/2}}\right)^{10} is

  1. A.120
  2. B.210
  3. C.310
  4. D.4

Correct answer: B

Explanation: The base simplifies to x1/3−x−1/2x^{1/3}-x^{-1/2}; Tr+1∝x(20−5r)/6T_{r+1}\propto x^{(20-5r)/6}, r=4r=4: (104)=210\binom{10}{4}=210.

Question 58

Physics · 4 marks · −1

In an LCR circuit as shown below both switches are open initially. Now switch S1S_1 is closed, S2S_2 kept open. (q is charge on the capacitor and τ=RC\tau = RC is capacitive time constant). Which of the following statement is correct?

Figure for question 58
  1. A.At t=τt = \tau, q=CV/2q = CV/2
  2. B.At t=2τt = 2\tau, q=CV(1−e−2)q = CV(1 - e^{-2})
  3. C.At t=τ2t = \frac{\tau}{2}, q=CV(1−e−1)q = CV(1 - e^{-1})
  4. D.Work done by the battery is half of the energy dissipated in the resistor.

Correct answer: B

Explanation: With S1S_1 closed the RC branch charges: q=CV(1−e−t/τ)q=CV(1-e^{-t/\tau}).

Question 59

Physics · 4 marks · −1

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω6\,\Omega. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

  1. A.2.9 Volt
  2. B.13.3 Volt
  3. C.10.04 Volt
  4. D.zero volt

Correct answer: C

Explanation: Bulb 240 Ω240\,\Omega, heater 60 Ω60\,\Omega. Before: V=120⋅240246=117.1V=\frac{120\cdot240}{246}=117.1 V; after (parallel 48 Ω48\,\Omega): 120⋅4854=106.7\frac{120\cdot48}{54}=106.7 V; drop ≈10.04\approx10.04 V.

Question 60

Physics · 4 marks · −1

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ\sigma at equilibrium position. The extension x0x_0 of the spring when it is in equilibrium is: (Here k is spring constant)

  1. A.Mgk(1−LAσM)\frac{Mg}{k}\left(1-\frac{LA\sigma}{M}\right)
  2. B.Mgk(1−LAσ2M)\frac{Mg}{k}\left(1-\frac{LA\sigma}{2M}\right)
  3. C.Mgk(1+LAσM)\frac{Mg}{k}\left(1+\frac{LA\sigma}{M}\right)
  4. D.Mgk\frac{Mg}{k}

Correct answer: B

Explanation: kx0+AL2σg=Mgkx_0+\frac{AL}{2}\sigma g=Mg.

Question 61

Physics · 4 marks · −1

Two charges, each equal to q, are kept at x=−ax = -a and x=ax = a on the x-axis. A particle of mass m and charge q0=q2q_0 = \frac{q}{2} is placed at the origin. If charge q0q_0 is given a small displacement (y≪a)(y \ll a) along the y-axis, the net force acting on the particle is proportional to:

  1. A.−y-y
  2. B.1y\frac{1}{y}
  3. C.−1y-\frac{1}{y}
  4. D.yy

Correct answer: D

Explanation: F=2kq(q/2)a2+y2⋅ya2+y2≈kq2a3yF=2\frac{kq(q/2)}{a^2+y^2}\cdot\frac{y}{\sqrt{a^2+y^2}}\approx\frac{kq^2}{a^3}y (away from origin).

Question 62

Physics · 4 marks · −1

A beam of unpolarised light of intensity I0I_0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45∘45^\circ relative to that of A. The intensity of the emergent light is:

  1. A.I0/2I_0/2
  2. B.I0/4I_0/4
  3. C.I0/8I_0/8
  4. D.I0I_0

Correct answer: B

Explanation: I02cos⁡245∘=I04\frac{I_0}{2}\cos^245^\circ=\frac{I_0}{4}.

Question 63

Physics · 4 marks · −1

The anode voltage of a photocell is kept fixed. The wavelength λ\lambda of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows:

Figure for question 63
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: C

Explanation: As λ\lambda increases, photon energy falls; the current decreases and becomes zero at the threshold wavelength (graph 3).

Question 64

Physics · 4 marks · −1

Two coherent point sources S1S_1 and S2S_2 are separated by a small distance 'd' as shown. The fringes obtained on the screen will be:

Figure for question 64
  1. A.straight lines
  2. B.semi-circles
  3. C.concentric circles
  4. D.points

Correct answer: C

Explanation: The sources lie on a line perpendicular to the screen, so loci of constant path difference on the screen are circles about the foot of that line.

Question 65

Physics · 4 marks · −1

A metallic rod of length ℓ\ell is tied to a string of length 2ℓ2\ell and made to rotate with angular speed ω\omega on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the e.m.f. induced across the ends of the rod is:

Figure for question 65
  1. A.3Bωℓ22\frac{3B\omega\ell^2}{2}
  2. B.4Bωℓ22\frac{4B\omega\ell^2}{2}
  3. C.5Bωℓ22\frac{5B\omega\ell^2}{2}
  4. D.2Bωℓ22\frac{2B\omega\ell^2}{2}

Correct answer: C

Explanation: e=∫2ℓ3ℓBωx dx=5Bωℓ22e=\int_{2\ell}^{3\ell}B\omega x\,dx=\frac{5B\omega\ell^2}{2}.

Question 66

Physics · 4 marks · −1

In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n - 1). If n≫1n \gg 1, the frequency of radiation emitted is proportional to

  1. A.1n2\frac{1}{n^2}
  2. B.1n3/2\frac{1}{n^{3/2}}
  3. C.1n3\frac{1}{n^3}
  4. D.1n\frac{1}{n}

Correct answer: C

Explanation: ν∝1(n−1)2−1n2=2n−1n2(n−1)2≈2n3\nu\propto\frac1{(n-1)^2}-\frac1{n^2}=\frac{2n-1}{n^2(n-1)^2}\approx\frac{2}{n^3}.

Question 67

Physics · 4 marks · −1

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ρ\rho and L is its latent heat of vaporization.

  1. A.T/ρL\sqrt{T/\rho L}
  2. B.T/ρLT/\rho L
  3. C.2T/ρL2T/\rho L
  4. D.ρL/T\rho L/T

Correct answer: C

Explanation: ρ4πR2dR L=T⋅8πR dR⇒R=2TρL\rho4\pi R^2dR\,L=T\cdot8\pi R\,dR\Rightarrow R=\frac{2T}{\rho L}.

Question 68

Physics · 4 marks · −1

The graph between angle of deviation (δ)(\delta) and angle of incidence (i) for a triangular prism is represented by:

Figure for question 68
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: B

Explanation: δ\delta first decreases to a minimum then increases smoothly (U-shaped curve).

Question 69

Physics · 4 marks · −1

Let [ϵ0][\epsilon_0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

  1. A.[ϵ0]=[M−1L−3T4A2][\epsilon_0] = [M^{-1}L^{-3}T^4A^2]
  2. B.[ϵ0]=[M−1L2T−1A−2][\epsilon_0] = [M^{-1}L^2T^{-1}A^{-2}]
  3. C.[ϵ0]=[M−1L2T−1A][\epsilon_0] = [M^{-1}L^2T^{-1}A]
  4. D.[ϵ0]=[M−1L−3T2A][\epsilon_0] = [M^{-1}L^{-3}T^2A]

Correct answer: A

Explanation: ϵ0=q24πFr2\epsilon_0=\frac{q^2}{4\pi Fr^2}: A2T2MLT−2L2=M−1L−3T4A2\frac{A^2T^2}{MLT^{-2}L^2}=M^{-1}L^{-3}T^4A^2.

Question 70

Physics · 4 marks · −1

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

Figure for question 70
  1. A.(132)p0v0\left(\frac{13}{2}\right)p_0v_0
  2. B.(112)p0v0\left(\frac{11}{2}\right)p_0v_0
  3. C.4p0v04p_0v_0
  4. D.p0v0p_0v_0

Correct answer: A

Explanation: Heat is absorbed on D to A (isochoric, 32p0v0\frac32p_0v_0) and A to B (isobaric, 52⋅2p0v0\frac52\cdot2p_0v_0): total 132p0v0\frac{13}{2}p_0v_0.

Question 71

Physics · 4 marks · −1

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %. What is the fundamental frequency of steel if density and elasticity of steel are 7.7×1037.7\times10^3 kg/m3^3 and 2.2×10112.2\times10^{11} N/m2^2 respectively?

  1. A.178.2 Hz
  2. B.200.5 Hz
  3. C.770 Hz
  4. D.188.5 Hz

Correct answer: A

Explanation: f=12ℓY⋅strainρ=132.2×1097.7×103≈178.2f=\frac1{2\ell}\sqrt{\frac{Y\cdot\text{strain}}{\rho}}=\frac{1}{3}\sqrt{\frac{2.2\times10^9}{7.7\times10^3}}\approx178.2 Hz.

Question 72

Physics · 4 marks · −1

This question has statement I and statement II. Statement-I: Higher the range, greater is the resistance of ammeter. Statement-II: To increase the range of ammeter, additional shunt needs to be used across it.

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

Correct answer: C

Explanation: Higher range needs a smaller shunt, so the ammeter resistance decreases (I false); an extra shunt in parallel increases the range (II true).

Question 73

Physics · 4 marks · −1

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

  1. A.2GmM3R\frac{2GmM}{3R}
  2. B.GmM2R\frac{GmM}{2R}
  3. C.GmM3R\frac{GmM}{3R}
  4. D.5GmM6R\frac{5GmM}{6R}

Correct answer: D

Explanation: Ef=−GMm6RE_f=-\frac{GMm}{6R}, Ei=−GMmRE_i=-\frac{GMm}{R}: difference 5GMm6R\frac{5GMm}{6R}.

Question 74

Physics · 4 marks · −1

A projectile is given an initial velocity of (i^+2j^)(\hat{i} + 2\hat{j}) m/s, where i^\hat{i} is along the ground and j^\hat{j} is along the vertical. If g=10g = 10 m/s2^2, the equation of its trajectory is:

  1. A.y=2x−5x2y = 2x - 5x^2
  2. B.4y=2x−5x24y = 2x - 5x^2
  3. C.4y=2x−25x24y = 2x - 25x^2
  4. D.y=x−5x2y = x - 5x^2

Correct answer: A

Explanation: x=tx=t, y=2t−5t2y=2t-5t^2.

Question 75

Physics · 4 marks · −1

Two capacitors C1C_1 and C2C_2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then :

  1. A.3C1=5C23C_1 = 5C_2
  2. B.3C1+5C2=03C_1 + 5C_2 = 0
  3. C.9C1=4C29C_1 = 4C_2
  4. D.5C1=3C25C_1 = 3C_2

Correct answer: A

Explanation: Connecting opposite plates, charges cancel: 120C1=200C2120C_1=200C_2.

Question 76

Physics · 4 marks · −1

A hoop of radius r and mass m rotating with an angular velocity ω0\omega_0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

  1. A.rω03\frac{r\omega_0}{3}
  2. B.rω02\frac{r\omega_0}{2}
  3. C.rω0r\omega_0
  4. D.rω04\frac{r\omega_0}{4}

Correct answer: B

Explanation: Angular momentum about the contact point: mr2ω0=2mr2ω⇒v=rω02mr^2\omega_0=2mr^2\omega\Rightarrow v=\frac{r\omega_0}{2}.

Question 77

Physics · 4 marks · −1

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0V_0 and its pressure is P0P_0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency:

  1. A.12πV0MP0A2γ\frac{1}{2\pi}\frac{V_0MP_0}{A^2\gamma}
  2. B.12πA2γP0MV0\frac{1}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}
  3. C.12πMV0AγP0\frac{1}{2\pi}\sqrt{\frac{MV_0}{A\gamma P_0}}
  4. D.12πAγP0V0M\frac{1}{2\pi}\frac{A\gamma P_0}{V_0M}

Correct answer: B

Explanation: Adiabatic: dP=−γP0V0AxdP=-\frac{\gamma P_0}{V_0}Ax, restoring force γP0A2V0x\frac{\gamma P_0A^2}{V_0}x.

Question 78

Physics · 4 marks · −1

A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is:

Figure for question 78
  1. A.3Q4πϵ0L\frac{3Q}{4\pi\epsilon_0L}
  2. B.Q4πϵ0Lln⁡2\frac{Q}{4\pi\epsilon_0L\ln2}
  3. C.Qln⁡24πϵ0L\frac{Q\ln2}{4\pi\epsilon_0L}
  4. D.Q8πϵ0L\frac{Q}{8\pi\epsilon_0L}

Correct answer: C

Explanation: V=∫L2LkQLxdx=Qln⁡24πϵ0LV=\int_L^{2L}\frac{kQ}{Lx}dx=\frac{Q\ln2}{4\pi\epsilon_0L}.

Question 79

Physics · 4 marks · −1

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

  1. A.6×10−116\times10^{-11} weber
  2. B.3.3×10−113.3\times10^{-11} weber
  3. C.6.6×10−96.6\times10^{-9} weber
  4. D.9.1×10−119.1\times10^{-11} weber

Correct answer: D

Explanation: M=μ0R2πr22(d2+R2)3/2M=\frac{\mu_0R^2\pi r^2}{2(d^2+R^2)^{3/2}}; ϕ=Mi≈9.1×10−11\phi=Mi\approx9.1\times10^{-11} Wb.

Question 80

Physics · 4 marks · −1

If a piece of metal is heated to temperature θ\theta and then allowed to cool in a room which is at temperature θ0\theta_0 the graph between the temperature T of the metal and time t will be closest to :

Figure for question 80
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: B

Explanation: Newton's cooling: exponential decay towards θ0\theta_0 (graph 2).

Question 81

Physics · 4 marks · −1

The I - V characteristic of an LED is

Figure for question 81
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: D

Explanation: In forward bias, current rises steeply beyond a threshold that increases from red to blue (graph 4).

Question 82

Physics · 4 marks · −1

This question has Statement I and Statement II. Statement-I : A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as f(12mv2)f\left(\frac{1}{2}mv^2\right) then f=(mM+m)f = \left(\frac{m}{M+m}\right). Statement-II : Maximum energy loss occurs when the particles get stuck together as a result of the collision.

  1. A.Statement-I is true, Statement-II is true, Statement-II is not a correct explanation of Statement-I.
  2. B.Statement-I is true, Statement-II is false.
  3. C.Statement-I is false, Statement-II is true
  4. D.Statement-I is true, Statement-II is true, Statement-II is a correct explanation of Statement-I.

Correct answer: C

Explanation: Max loss (perfectly inelastic) =12mMM+mv2=\frac12\frac{mM}{M+m}v^2, so f=MM+mf=\frac{M}{M+m}: I false, II true.

Question 83

Physics · 4 marks · −1

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5s. In another 10s it will decrease to α\alpha times its original magnitude, where α\alpha equals.

  1. A.0.81
  2. B.0.729
  3. C.0.6
  4. D.0.7

Correct answer: B

Explanation: Every 5 s multiplies by 0.9; after 15 s total: 0.93=0.7290.9^3=0.729.

Question 84

Physics · 4 marks · −1

Diameter of plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2×1082\times10^8 m/s, the focal length of the lens is :

  1. A.20 cm
  2. B.30 cm
  3. C.10 cm
  4. D.15 cm

Correct answer: B

Explanation: R=d22t=90.6=15R=\frac{d^2}{2t}=\frac{9}{0.6}=15 cm; μ=1.5\mu=1.5; f=Rμ−1=30f=\frac{R}{\mu-1}=30 cm.

Question 85

Physics · 4 marks · −1

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

  1. A.6 V/m
  2. B.9 V/m
  3. C.12 V/m
  4. D.3 V/m

Correct answer: A

Explanation: E0=cB0=6E_0=cB_0=6 V/m.

Question 86

Physics · 4 marks · −1

Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2^2 and 1.00 Am2^2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of earth's magnetic induction is 3.6×10−53.6\times10^{-5} Wb/m2^2)

  1. A.2.56×10−42.56\times10^{-4} Wb/m2^2
  2. B.3.50×10−43.50\times10^{-4} Wb/m2^2
  3. C.5.80×10−45.80\times10^{-4} Wb/m2^2
  4. D.3.6×10−53.6\times10^{-5} Wb/m2^2

Correct answer: A

Explanation: At O (equatorial, 10 cm from each): B=μ04πM1+M2x3+BH=2.2×10−4+0.36×10−4=2.56×10−4B=\frac{\mu_0}{4\pi}\frac{M_1+M_2}{x^3}+B_H=2.2\times10^{-4}+0.36\times10^{-4}=2.56\times10^{-4} T.

View all papers — JEE Main

Practice this paper with a timer

Attempt it online, get scored instantly, and see which subjects are costing you marks.

Start practising free