JEE Main 2015 — 4 Apr Offline Set B (Previous Year Paper)

JEE Main · 2015 · with answers and explanations

questions
88
Total marks
352
Duration
180 minutes
Negative marking
−1

Subjects: Chemistry · Mathematics · Physics

Question 1

Chemistry · 4 marks · −1

Which of the following is the energy of a possible excited state of hydrogen?

  1. A.−6.8-6.8 eV
  2. B.−3.4-3.4 eV
  3. C.+6.8+6.8 eV
  4. D.+13.6+13.6 eV

Correct answer: B

Explanation: En=−13.6n2E_n=-\frac{13.6}{n^2} eV; n=2n=2 gives −3.4-3.4 eV.

Question 2

Chemistry · 4 marks · −1

In the following sequence of reactions: Toluene →KMnO4\xrightarrow{KMnO_4} A →SOCl2\xrightarrow{SOCl_2} B →H2/Pd, BaSO4\xrightarrow{H_2/Pd,\ BaSO_4} C, the product C is:

  1. A.C6H5CH3C_6H_5CH_3
  2. B.C6H5CH2OHC_6H_5CH_2OH
  3. C.C6H5CHOC_6H_5CHO
  4. D.C6H5COOHC_6H_5COOH

Correct answer: C

Explanation: Benzoic acid, then benzoyl chloride, then Rosenmund reduction to benzaldehyde.

Question 3

Chemistry · 4 marks · −1

Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?

Figure for question 3
  1. A.1,3-dimethylcyclopentene (methyl on C1 of the double bond and on the ring carbon next to C2)
  2. B.1,4-dimethylcyclopentene
  3. C.1,5-dimethylcyclopentene (methyl on C1 and on the adjacent sp3 carbon)
  4. D.1,2-dimethylcyclopentene

Correct answer: A

Explanation: Ozonolysis of 1,3-dimethylcyclopentene cleaves C1=C2 giving CH3CO−CH2CH2−CH(CH3)−CHOCH_3CO-CH_2CH_2-CH(CH_3)-CHO, i.e. 5-oxo-2-methylhexanal.

Question 4

Chemistry · 4 marks · −1

The ionic radii (in Å) of N3−N^{3-}, O2−O^{2-} and F−F^- are respectively:

  1. A.1.36, 1.71 and 1.40
  2. B.1.71, 1.40 and 1.36
  3. C.1.71, 1.36 and 1.40
  4. D.1.36, 1.40 and 1.71

Correct answer: B

Explanation: Isoelectronic anions shrink as nuclear charge grows: N3−>O2−>F−N^{3-}>O^{2-}>F^-.

Question 5

Chemistry · 4 marks · −1

The color of KMnO4KMnO_4 is due to:

  1. A.d - d transition
  2. B.L →\rightarrow M charge transfer transition
  3. C.σ−σ∗\sigma - \sigma^* transition
  4. D.M →\rightarrow L charge transfer transition

Correct answer: B

Explanation: Mn(VII) is d0d^0; the intense colour arises from ligand (O) to metal charge transfer.

Question 6

Chemistry · 4 marks · −1

Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason : The reaction between nitrogen and oxygen requires high temperature.

  1. A.Both assertion and reason are correct, but the reason is not the correct explanation for the assertion
  2. B.The assertion is incorrect, but the reason is correct
  3. C.Both the assertion and reason are incorrect
  4. D.Both assertion and reason are correct, and the reason is the correct explanation for the assertion

Correct answer: D

Explanation: N2+O2→2NON_2+O_2\rightarrow2NO is endothermic and needs very high temperature (lightning, engines), which explains the assertion.

Question 7

Chemistry · 4 marks · −1

Which of the following compounds is not an antacid?

  1. A.Cimetidine
  2. B.Phenelzine
  3. C.Ranitidine
  4. D.Aluminium hydroxide

Correct answer: B

Explanation: Phenelzine is an antidepressant (MAO inhibitor); the others are antacids.

Question 8

Chemistry · 4 marks · −1

In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is false?

  1. A.Al2O3Al_2O_3 is mixed with CaF2CaF_2 which lowers the melting point of the mixture and brings conductivity
  2. B.Al3+Al^{3+} is reduced at the cathode to form Al
  3. C.Na3AlF6Na_3AlF_6 serves as the electrolyte
  4. D.CO and CO2CO_2 are produced in this process

Correct answer: C

Explanation: The electrolyte is alumina dissolved in molten cryolite; cryolite itself is not the species electrolysed.

Question 9

Chemistry · 4 marks · −1

Match the catalysts to the correct processes: Catalyst (A) TiCl3TiCl_3 (B) PdCl2PdCl_2 (C) CuCl2CuCl_2 (D) V2O5V_2O_5; Process (i) Wacker process (ii) Ziegler - Natta polymerization (iii) Contact process (iv) Deacon's process

  1. A.(A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)
  2. B.(A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)
  3. C.(A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)
  4. D.(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)

Correct answer: A

Explanation: TiCl3TiCl_3: Ziegler-Natta; PdCl2PdCl_2: Wacker; CuCl2CuCl_2: Deacon; V2O5V_2O_5: Contact process.

Question 10

Chemistry · 4 marks · −1

In the reaction: p-toluidine →NaNO2/HCl, 0−5∘C\xrightarrow{NaNO_2/HCl,\ 0-5^\circ C} D →CuCN/KCN, Δ\xrightarrow{CuCN/KCN,\ \Delta} E + N2N_2, the product E is

Figure for question 10
  1. A.4,4'-dimethylbiphenyl
  2. B.p-tolunitrile (4-methylbenzonitrile)
  3. C.Toluene
  4. D.p-toluic acid (4-methylbenzoic acid)

Correct answer: B

Explanation: Diazotisation then Sandmeyer cyanation gives p−CH3C6H4CNp-CH_3C_6H_4CN.

Question 11

Chemistry · 4 marks · −1

Which polymer is used in the manufacture of paints and lacquers?

  1. A.Glyptal
  2. B.Polypropene
  3. C.Poly vinyl chloride
  4. D.Bakelite

Correct answer: A

Explanation: Glyptal (glycerol-phthalic acid polyester) is used in paints and lacquers.

Question 12

Chemistry · 4 marks · −1

The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]+[Pt(Cl)(py)(NH_3)(NH_2OH)]^+ is (py = pyridine):

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

Correct answer: A

Explanation: Square planar [Mabcd][Mabcd] has 3 geometric isomers.

Question 13

Chemistry · 4 marks · −1

Higher order (>3) reactions are rare due to:

  1. A.increase in entropy and activation energy as more molecules are involved
  2. B.shifting of equilibrium towards reactants due to elastic collisions
  3. C.loss of active species on collision
  4. D.low probability of simultaneous collision of all the reacting species

Correct answer: D

Explanation: Simultaneous effective collision of more than three species is highly improbable.

Question 14

Chemistry · 4 marks · −1

Which among the following is the most reactive?

  1. A.Br2Br_2
  2. B.I2I_2
  3. C.ICl
  4. D.Cl2Cl_2

Correct answer: C

Explanation: The polar, weaker I-Cl bond makes the interhalogen ICl more reactive than the halogens.

Question 15

Chemistry · 4 marks · −1

Two Faraday of electricity is passed through a solution of CuSO4CuSO_4. The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu)

  1. A.63.5 g
  2. B.2 g
  3. C.127 g
  4. D.0 g

Correct answer: A

Explanation: Cu2++2e−→CuCu^{2+}+2e^-\rightarrow Cu: 2 F deposit 1 mol = 63.5 g.

Question 16

Chemistry · 4 marks · −1

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

  1. A.36 mg
  2. B.42 mg
  3. C.54 mg
  4. D.18 mg

Correct answer: D

Explanation: Adsorbed =50×(0.06−0.042)=0.9=50\times(0.06-0.042)=0.9 mmol =54=54 mg in total, i.e. 18 mg per gram.

Question 17

Chemistry · 4 marks · −1

The synthesis of alkyl fluorides is best accomplished by:

  1. A.Sandmeyer's reaction
  2. B.Finkelstein reaction
  3. C.Swarts reaction
  4. D.Free radical fluorination

Correct answer: C

Explanation: Swarts reaction: alkyl chloride/bromide with AgFAgF, Hg2F2Hg_2F_2, CoF2CoF_2 or SbF3SbF_3.

Question 18

Chemistry · 4 marks · −1

The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3NaC_8H_7SO_3Na (Mol. Wt. 206). What would be the maximum uptake of Ca2+Ca^{2+} ions by the resin when expressed in mole per gram resin?

  1. A.1206\frac{1}{206}
  2. B.2309\frac{2}{309}
  3. C.1412\frac{1}{412}
  4. D.1103\frac{1}{103}

Correct answer: C

Explanation: Two Na+Na^+ sites exchange one Ca2+Ca^{2+}: 12×206=1412\frac{1}{2\times206}=\frac1{412} mol per gram.

Question 19

Chemistry · 4 marks · −1

Which of the vitamins given below is water soluble?

  1. A.Vitamin D
  2. B.Vitamin E
  3. C.Vitamin K
  4. D.Vitamin C

Correct answer: D

Explanation: Vitamins B and C are water soluble; A, D, E, K are fat soluble.

Question 20

Chemistry · 4 marks · −1

The intermolecular interaction that is dependent on the inverse cube of distance between the molecule is:

  1. A.ion-dipole interaction
  2. B.London force
  3. C.hydrogen bond
  4. D.ion-ion interaction

Correct answer: C

Explanation: Interaction energies: ion-ion ∝1/r\propto1/r, ion-dipole ∝1/r2\propto1/r^2, dipole-dipole (fixed) ∝1/r3\propto1/r^3, London ∝1/r6\propto1/r^6. A hydrogen bond is a strong dipole-dipole interaction, hence the 1/r31/r^3 dependence.

Question 21

Chemistry · 4 marks · −1

The following reaction is performed at 298 K. 2NO(g)+O2(g)⇌2NO2(g)2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g). The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2NO_2(g) at 298 K? (KP=1.6×1012K_P = 1.6\times10^{12})

  1. A.86600+R(298)ln⁡(1.6×1012)86600 + R(298)\ln(1.6\times10^{12})
  2. B.86600−ln⁡(1.6×1012)R(298)86600 - \frac{\ln(1.6\times10^{12})}{R(298)}
  3. C.0.5[2×86,600−R(298)ln⁡(1.6×1012)]0.5[2\times86,600 - R(298)\ln(1.6\times10^{12})]
  4. D.R(298)ln⁡(1.6×1012)−86600R(298)\ln(1.6\times10^{12}) - 86600

Correct answer: C

Explanation: ΔGr∘=2Gf(NO2)−2Gf(NO)=−RTln⁡KP⇒Gf(NO2)=12[2(86600)−R(298)ln⁡KP]\Delta G^\circ_r=2G_f(NO_2)-2G_f(NO)=-RT\ln K_P\Rightarrow G_f(NO_2)=\frac12[2(86600)-R(298)\ln K_P].

Question 22

Chemistry · 4 marks · −1

Which of the following compounds is not colored yellow?

  1. A.K3[Co(NO2)6]K_3[Co(NO_2)_6]
  2. B.(NH4)3[As(Mo3O10)4](NH_4)_3[As(Mo_3O_{10})_4]
  3. C.BaCrO4BaCrO_4
  4. D.Zn2[Fe(CN)6]Zn_2[Fe(CN)_6]

Correct answer: D

Explanation: Zinc ferrocyanide is white; the other three are yellow precipitates.

Question 23

Chemistry · 4 marks · −1

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (at. Mass Ag = 108; Br = 80)

  1. A.36
  2. B.48
  3. C.60
  4. D.24

Correct answer: D

Explanation: %Br =80188⋅141250⋅100=24=\frac{80}{188}\cdot\frac{141}{250}\cdot100=24.

Question 24

Chemistry · 4 marks · −1

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately:

  1. A.3.22 Å
  2. B.5.72 Å
  3. C.0.93 Å
  4. D.1.86 Å

Correct answer: D

Explanation: bcc: r=3a4=1.86r=\frac{\sqrt3a}{4}=1.86 Å.

Question 25

Chemistry · 4 marks · −1

Which of the following compounds will exhibit geometrical isomerism?

  1. A.3-Phenyl-1-butene
  2. B.2-Phenyl-1-butene
  3. C.1,1-Diphenyl-1-propane
  4. D.1-Phenyl-2-butene

Correct answer: D

Explanation: C6H5CH2CH=CHCH3C_6H_5CH_2CH=CHCH_3 has two different groups on each alkene carbon.

Question 26

Chemistry · 4 marks · −1

The vapour pressure of acetone at 20∘20^\circC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20∘20^\circC, its vapour pressure was 183 torr. The molar mass (g mol−1^{-1}) of the substance is:

  1. A.64
  2. B.128
  3. C.488
  4. D.32

Correct answer: A

Explanation: 2185≈1.2/M100/58⇒M≈64\frac{2}{185}\approx\frac{1.2/M}{100/58}\Rightarrow M\approx64 g/mol.

Question 27

Chemistry · 4 marks · −1

From the following statements regarding H2O2H_2O_2, choose the incorrect statement:

  1. A.It decomposes on exposure to light
  2. B.It has to be stored in plastic or wax lined glass bottles in dark
  3. C.It has to be kept away from dust
  4. D.It can act only as an oxidizing agent

Correct answer: D

Explanation: H2O2H_2O_2 acts as both oxidising and reducing agent.

Question 28

Chemistry · 4 marks · −1

Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?

  1. A.BeSO4BeSO_4
  2. B.BaSO4BaSO_4
  3. C.SrSO4SrSO_4
  4. D.CaSO4CaSO_4

Correct answer: A

Explanation: Small Be2+Be^{2+} is so strongly hydrated that BeSO4BeSO_4 is water soluble.

Question 29

Chemistry · 4 marks · −1

The standard Gibbs energy change at 300 K for the reaction 2A⇌B+C2A \rightleftharpoons B + C is 2494.2 J. At a given time, the composition of the reaction mixture is [A]=12[A] = \frac{1}{2}, [B]=2[B] = 2 and [C]=12[C] = \frac{1}{2}. The reaction proceeds in the: (R=8.314R = 8.314 J/K/mol, e=2.718e = 2.718)

  1. A.reverse direction because Q>KcQ > K_c
  2. B.forward direction because Q<KcQ < K_c
  3. C.reverse direction because Q<KcQ < K_c
  4. D.forward direction because Q>KcQ > K_c

Correct answer: A

Explanation: ln⁡K=−2494.28.314⋅300=−1\ln K=-\frac{2494.2}{8.314\cdot300}=-1, K=0.37K=0.37; Q=2⋅0.50.25=4>KQ=\frac{2\cdot0.5}{0.25}=4>K: reverse.

Question 30

Chemistry · 4 marks · −1

Which one has the highest boiling point?

  1. A.Ne
  2. B.Kr
  3. C.Xe
  4. D.He

Correct answer: C

Explanation: Largest, most polarisable noble gas Xe has the strongest dispersion forces.

Question 31

Mathematics · 4 marks · −1

The sum of coefficients of integral powers of x in the binomial expansion of (1−2x)50\left(1 - 2\sqrt{x}\right)^{50} is :

  1. A.12(350)\frac{1}{2}\left(3^{50}\right)
  2. B.12(350−1)\frac{1}{2}\left(3^{50}-1\right)
  3. C.12(250+1)\frac{1}{2}\left(2^{50}+1\right)
  4. D.12(350+1)\frac{1}{2}\left(3^{50}+1\right)

Correct answer: D

Explanation: Even-index terms: (1+2)50+(1−2)502=350+12\frac{(1+2)^{50}+(1-2)^{50}}{2}=\frac{3^{50}+1}{2}.

Question 32

Mathematics · 4 marks · −1

Let f(x) be a polynomial of degree four having extreme values at x=1x = 1 and x=2x = 2. If lim⁡x→0[1+f(x)x2]=3\lim_{x\to0}\left[1 + \frac{f(x)}{x^2}\right] = 3, then f(2) is equal to :

  1. A.−4-4
  2. B.0
  3. C.4
  4. D.−8-8

Correct answer: B

Explanation: f=ax4+bx3+2x2f=ax^4+bx^3+2x^2; f′(1)=f′(2)=0f'(1)=f'(2)=0 give a=12a=\frac12, b=−2b=-2; f(2)=8−16+8=0f(2)=8-16+8=0.

Question 33

Mathematics · 4 marks · −1

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :

  1. A.16.0
  2. B.15.8
  3. C.14.0
  4. D.16.8

Correct answer: C

Explanation: 256−16+1218=14\frac{256-16+12}{18}=14.

Question 34

Mathematics · 4 marks · −1

The sum of first 9 terms of the series 131+13+231+3+13+23+331+3+5+…\frac{1^3}{1} + \frac{1^3+2^3}{1+3} + \frac{1^3+2^3+3^3}{1+3+5} + \dots is :

  1. A.96
  2. B.142
  3. C.192
  4. D.71

Correct answer: A

Explanation: Tn=(n(n+1)/2)2n2=(n+1)24T_n=\frac{(n(n+1)/2)^2}{n^2}=\frac{(n+1)^2}{4}; ∑n=19=385−14=96\sum_{n=1}^9=\frac{385-1}{4}=96.

Question 35

Mathematics · 4 marks · −1

Let O be the vertex and Q be any point on the parabola, x2=8yx^2 = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then the locus of P is :

  1. A.y2=xy^2 = x
  2. B.y2=2xy^2 = 2x
  3. C.x2=2yx^2 = 2y
  4. D.x2=yx^2 = y

Correct answer: C

Explanation: Q=(4t,2t2)Q=(4t,2t^2), P=Q4=(t,t22)P=\frac Q4=(t,\frac{t^2}{2}): x2=2yx^2=2y.

Question 36

Mathematics · 4 marks · −1

Let α\alpha and β\beta be the roots of equation x2−6x−2=0x^2 - 6x - 2 = 0. If an=αn−βna_n = \alpha^n - \beta^n, for n≥1n \ge 1, then the value of a10−2a82a9\frac{a_{10} - 2a_8}{2a_9} is equal to :

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

Correct answer: B

Explanation: α2=6α+2⇒a10=6a9+2a8\alpha^2=6\alpha+2\Rightarrow a_{10}=6a_9+2a_8, so the ratio is 6a92a9=3\frac{6a_9}{2a_9}=3.

Question 37

Mathematics · 4 marks · −1

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is :

  1. A.55(23)1055\left(\frac{2}{3}\right)^{10}
  2. B.220(13)12220\left(\frac{1}{3}\right)^{12}
  3. C.22(13)1122\left(\frac{1}{3}\right)^{11}
  4. D.553(23)11\frac{55}{3}\left(\frac{2}{3}\right)^{11}

Correct answer: D

Explanation: Intended reading: balls placed independently, a specified box gets exactly 3: (123)(13)3(23)9=553(23)11\binom{12}{3}\left(\frac13\right)^3\left(\frac23\right)^9=\frac{55}{3}\left(\frac23\right)^{11}.

Question 38

Mathematics · 4 marks · −1

A complex number z is said to be unimodular if ∣z∣=1|z| = 1. Suppose z1z_1 and z2z_2 are complex numbers such that z1−2z22−z1zˉ2\frac{z_1 - 2z_2}{2 - z_1\bar{z}_2} is unimodular and z2z_2 is not unimodular. Then the point z1z_1 lies on a :

  1. A.straight line parallel to y-axis.
  2. B.circle of radius 2.
  3. C.circle of radius 2\sqrt{2}.
  4. D.straight line parallel to x-axis.

Correct answer: B

Explanation: ∣z1−2z2∣2=∣2−z1zˉ2∣2|z_1-2z_2|^2=|2-z_1\bar z_2|^2 simplifies to (∣z1∣2−4)(1−∣z2∣2)=0(|z_1|^2-4)(1-|z_2|^2)=0, so ∣z1∣=2|z_1|=2.

Question 39

Mathematics · 4 marks · −1

The integral ∫dxx2(x4+1)3/4\int \frac{dx}{x^2\left(x^4+1\right)^{3/4}} equals :

  1. A.(x4+1)1/4+c(x^4+1)^{1/4}+c
  2. B.−(x4+1)1/4+c-(x^4+1)^{1/4}+c
  3. C.−(x4+1x4)1/4+c-\left(\frac{x^4+1}{x^4}\right)^{1/4}+c
  4. D.(x4+1x4)1/4+c\left(\frac{x^4+1}{x^4}\right)^{1/4}+c

Correct answer: C

Explanation: =∫x−5dx(1+x−4)3/4=\int\frac{x^{-5}dx}{(1+x^{-4})^{3/4}}; with t=1+x−4t=1+x^{-4} it becomes −14∫t−3/4dt=−t1/4-\frac14\int t^{-3/4}dt=-t^{1/4}.

Question 40

Mathematics · 4 marks · −1

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0,0)(0, 0), (0,41)(0, 41) and (41,0)(41, 0), is :

  1. A.861
  2. B.820
  3. C.780
  4. D.901

Correct answer: C

Explanation: Need x,y≥1x,y\ge1, x+y≤40x+y\le40: ∑x=139(40−x)=780\sum_{x=1}^{39}(40-x)=780.

Question 41

Mathematics · 4 marks · −1

The distance of the point (1,0,2)(1, 0, 2) from the point of intersection of the line x−23=y+14=z−212\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} and the plane x−y+z=16x - y + z = 16, is :

  1. A.8
  2. B.3213\sqrt{21}
  3. C.13
  4. D.2142\sqrt{14}

Correct answer: C

Explanation: 5+11t=16⇒t=15+11t=16\Rightarrow t=1, point (5,3,14)(5,3,14); distance 16+9+144=13\sqrt{16+9+144}=13.

Question 42

Mathematics · 4 marks · −1

The equation of the plane containing the line 2x−5y+z=32x - 5y + z = 3; x+y+4z=5x + y + 4z = 5, and parallel to the plane, x+3y+6z=1x + 3y + 6z = 1, is :

  1. A.x+3y+6z=−7x + 3y + 6z = -7
  2. B.x+3y+6z=7x + 3y + 6z = 7
  3. C.2x+6y+12z=−132x + 6y + 12z = -13
  4. D.2x+6y+12z=132x + 6y + 12z = 13

Correct answer: B

Explanation: Plane x+3y+6z=kx+3y+6z=k through a point of the line, e.g. solving gives k=7k=7 (family method with λ=−112\lambda=-\frac{11}{2}).

Question 43

Mathematics · 4 marks · −1

The area (in sq. units) of the region described by {(x,y):y2≤2x and y≥4x−1}\{(x, y) : y^2 \le 2x \text{ and } y \ge 4x - 1\} is :

  1. A.564\frac{5}{64}
  2. B.1564\frac{15}{64}
  3. C.932\frac{9}{32}
  4. D.732\frac{7}{32}

Correct answer: C

Explanation: Curves meet at y=1,−12y=1,-\frac12. Area =∫−1/21(y+14−y22)dy=932=\int_{-1/2}^{1}\left(\frac{y+1}{4}-\frac{y^2}{2}\right)dy=\frac{9}{32}.

Question 44

Mathematics · 4 marks · −1

If m is the A.M. of two distinct real numbers ℓ\ell and n (ℓ,n>1)(\ell, n > 1) and G1G_1, G2G_2 and G3G_3 are three geometric means between ℓ\ell and n, then G14+2G24+G34G_1^4 + 2G_2^4 + G_3^4 equals :

  1. A.4ℓm2n4\ell m^2n
  2. B.4ℓmn24\ell mn^2
  3. C.4ℓ2m2n24\ell^2m^2n^2
  4. D.4ℓ2mn4\ell^2mn

Correct answer: A

Explanation: With r4=n/ℓr^4=n/\ell: sum =ℓ3n+2ℓ2n2+ℓn3=ℓn(ℓ+n)2=4ℓm2n=\ell^3n+2\ell^2n^2+\ell n^3=\ell n(\ell+n)^2=4\ell m^2n.

Question 45

Mathematics · 4 marks · −1

Locus of the image of the point (2,3)(2, 3) in the line (2x−3y+4)+k(x−2y+3)=0(2x - 3y + 4) + k(x - 2y + 3) = 0, k∈Rk \in \mathbb{R}, is a :

  1. A.straight line parallel to y-axis.
  2. B.circle of radius 2\sqrt{2}.
  3. C.circle of radius 3\sqrt{3}.
  4. D.straight line parallel to x-axis.

Correct answer: B

Explanation: All lines pass through (1,2)(1,2); the image stays at distance 2\sqrt2 from (1,2)(1,2).

Question 46

Mathematics · 4 marks · −1

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse x29+y25=1\frac{x^2}{9} + \frac{y^2}{5} = 1, is :

  1. A.18
  2. B.272\frac{27}{2}
  3. C.27
  4. D.274\frac{27}{4}

Correct answer: C

Explanation: End (2,53)(2,\frac53): tangent 2x9+y3=1\frac{2x}{9}+\frac y3=1, intercepts 92\frac92 and 3. Area =4⋅12⋅92⋅3=27=4\cdot\frac12\cdot\frac92\cdot3=27.

Question 47

Mathematics · 4 marks · −1

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :

  1. A.192
  2. B.120
  3. C.72
  4. D.216

Correct answer: A

Explanation: 4-digit starting 6/7/8: 3⋅4⋅3⋅2=723\cdot4\cdot3\cdot2=72; all 5-digit: 120. Total 192.

Question 48

Mathematics · 4 marks · −1

Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A×BA \times B, each having at least three elements is :

  1. A.256
  2. B.275
  3. C.510
  4. D.219

Correct answer: D

Explanation: ∣A×B∣=8|A\times B|=8: 28−1−8−28=2192^8-1-8-28=219.

Question 49

Mathematics · 4 marks · −1

Let tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2)\tan^{-1}y = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right), where ∣x∣<13|x| < \frac{1}{\sqrt{3}}. Then a value of y is :

  1. A.3x+x31−3x2\frac{3x+x^3}{1-3x^2}
  2. B.3x−x31+3x2\frac{3x-x^3}{1+3x^2}
  3. C.3x+x31+3x2\frac{3x+x^3}{1+3x^2}
  4. D.3x−x31−3x2\frac{3x-x^3}{1-3x^2}

Correct answer: D

Explanation: RHS =3tan⁡−1x=3\tan^{-1}x for ∣x∣<13|x|<\frac1{\sqrt3}, so y=tan⁡3θ=3x−x31−3x2y=\tan3\theta=\frac{3x-x^3}{1-3x^2}.

Question 50

Mathematics · 4 marks · −1

The integral ∫24log⁡x2log⁡x2+log⁡(36−12x+x2) dx\int_2^4 \frac{\log x^2}{\log x^2 + \log(36 - 12x + x^2)}\,dx is equal to :

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

Correct answer: B

Explanation: 36−12x+x2=(6−x)236-12x+x^2=(6-x)^2; using x→6−xx\to6-x, 2I=∫24dx=22I=\int_2^4dx=2, I=1I=1.

Question 51

Mathematics · 4 marks · −1

The negation of ∼s∨(∼r∧s)\sim s \vee (\sim r \wedge s) is equivalent to :

  1. A.s∧(r∧∼s)s \wedge (r \wedge \sim s)
  2. B.s∨(r∨∼s)s \vee (r \vee \sim s)
  3. C.s∧rs \wedge r
  4. D.s∧∼rs \wedge \sim r

Correct answer: C

Explanation: ∼[∼s∨(∼r∧s)]=s∧(r∨∼s)=s∧r\sim[\sim s\vee(\sim r\wedge s)]=s\wedge(r\vee\sim s)=s\wedge r.

Question 52

Mathematics · 4 marks · −1

If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30∘30^\circ, 45∘45^\circ and 60∘60^\circ respectively, then the ratio, AB : BC, is :

  1. A.3:2\sqrt{3} : \sqrt{2}
  2. B.1:31 : \sqrt{3}
  3. C.2 : 3
  4. D.3:1\sqrt{3} : 1

Correct answer: D

Explanation: Distances h3h\sqrt3, hh, h3\frac{h}{\sqrt3}: AB=h(3−1)AB=h(\sqrt3-1), BC=h(3−1)3BC=\frac{h(\sqrt3-1)}{\sqrt3}.

Question 53

Mathematics · 4 marks · −1

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

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

Correct answer: B

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

Question 54

Mathematics · 4 marks · −1

Let a⃗\vec{a}, b⃗\vec{b} and c⃗\vec{c} be three non-zero vectors such that no two of them are collinear and (a⃗×b⃗)×c⃗=13∣b⃗∣∣c⃗∣a⃗(\vec{a}\times\vec{b})\times\vec{c} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}. If θ\theta is the angle between vectors b⃗\vec{b} and c⃗\vec{c}, then a value of sin⁡θ\sin\theta is :

  1. A.−23\frac{-\sqrt{2}}{3}
  2. B.23\frac{2}{3}
  3. C.−233\frac{-2\sqrt{3}}{3}
  4. D.223\frac{2\sqrt{2}}{3}

Correct answer: D

Explanation: (a⃗⋅c⃗)b⃗−(b⃗⋅c⃗)a⃗=13∣b⃗∣∣c⃗∣a⃗⇒cos⁡θ=−13(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a=\frac13|\vec b||\vec c|\vec a\Rightarrow\cos\theta=-\frac13, sin⁡θ=223\sin\theta=\frac{2\sqrt2}{3}.

Question 55

Mathematics · 4 marks · −1

If A=[12221−2a2b]A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} is a matrix satisfying the equation AAT=9IAA^T = 9I, where I is 3×33\times3 identity matrix, then ordered pair (a,b)(a, b) is equal to :

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

Correct answer: C

Explanation: Row orthogonality: a+4+2b=0a+4+2b=0 and 2a+2−2b=02a+2-2b=0 give a=−2a=-2, b=−1b=-1.

Question 56

Mathematics · 4 marks · −1

If the function g(x)={kx+1,0≤x≤3mx+2,3<x≤5g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 < x \le 5 \end{cases} is differentiable, then the value of k+mk + m is :

  1. A.165\frac{16}{5}
  2. B.103\frac{10}{3}
  3. C.4
  4. D.2

Correct answer: D

Explanation: Continuity 2k=3m+22k=3m+2; derivatives k4=m\frac k4=m. So m=25m=\frac25, k=85k=\frac85, k+m=2k+m=2.

Question 57

Mathematics · 4 marks · −1

The set of all values of λ\lambda for which the system of linear equations 2x1−2x2+x3=λx12x_1 - 2x_2 + x_3 = \lambda x_1, 2x1−3x2+2x3=λx22x_1 - 3x_2 + 2x_3 = \lambda x_2, −x1+2x2=λx3-x_1 + 2x_2 = \lambda x_3 has a non-trivial solution,

  1. A.is a singleton.
  2. B.contains two elements.
  3. C.contains more than two elements.
  4. D.is an empty set.

Correct answer: B

Explanation: Determinant gives λ3+λ2−5λ+3=(λ−1)2(λ+3)=0\lambda^3+\lambda^2-5\lambda+3=(\lambda-1)^2(\lambda+3)=0: λ=1,−3\lambda=1,-3.

Question 58

Mathematics · 4 marks · −1

The normal to the curve, x2+2xy−3y2=0x^2 + 2xy - 3y^2 = 0, at (1,1)(1, 1) :

  1. A.meets the curve again in the second quadrant.
  2. B.meets the curve again in the third quadrant.
  3. C.meets the curve again in the fourth quadrant.
  4. D.does not meet the curve again.

Correct answer: C

Explanation: Curve is the pair x=yx=y, x=−3yx=-3y. Normal at (1,1)(1,1): x+y=2x+y=2, meeting x=−3yx=-3y at (3,−1)(3,-1) in the fourth quadrant.

Question 59

Mathematics · 4 marks · −1

The number of common tangents to the circles x2+y2−4x−6y−12=0x^2 + y^2 - 4x - 6y - 12 = 0 and x2+y2+6x+18y+26=0x^2 + y^2 + 6x + 18y + 26 = 0, is :

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

Correct answer: B

Explanation: Centres (2,3),(−3,−9)(2,3),(-3,-9), radii 5 and 8; distance 13 =r1+r2=r_1+r_2: touching externally, 3 common tangents.

Question 60

Mathematics · 4 marks · −1

Let y(x) be the solution of the differential equation (xlog⁡x)dydx+y=2xlog⁡x(x\log x)\frac{dy}{dx} + y = 2x\log x, (x≥1)(x \ge 1). Then y(e) is equal to :

  1. A.0
  2. B.2
  3. C.2e
  4. D.e

Correct answer: B

Explanation: ddx(ylog⁡x)=2log⁡x⇒ylog⁡x=2xlog⁡x−2x+2\frac{d}{dx}(y\log x)=2\log x\Rightarrow y\log x=2x\log x-2x+2 (using x=1x=1). At x=ex=e: y=2y=2.

Question 61

Physics · 4 marks · −1

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion:

  1. A.kinetic energy, potential energy and total energy decrease
  2. B.kinetic energy decreases, potential energy increases but total energy remains same
  3. C.kinetic energy and total energy decrease but potential energy increases
  4. D.its kinetic energy increases but potential energy and total energy decrease

Correct answer: D

Explanation: Moving to lower n: KE=−EKE=-E increases while PE and total energy become more negative.

Question 62

Physics · 4 marks · −1

The period of oscillation of a simple pendulum is T=2πLgT = 2\pi\sqrt{\frac{L}{g}}. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

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

Correct answer: A

Explanation: Δgg=0.120+2⋅190≈2.7%≈3%\frac{\Delta g}{g}=\frac{0.1}{20}+2\cdot\frac{1}{90}\approx2.7\%\approx3\%.

Question 63

Physics · 4 marks · −1

A long cylindrical shell carries positive surface charge σ\sigma in the upper half and negative surface charge −σ-\sigma in the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale)

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

Correct answer: D

Explanation: Field lines start on the positive upper half and end on the negative lower half, curving around the cylinder like a dipole pattern (figure 4).

Question 64

Physics · 4 marks · −1

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:

  1. A.2005 kHz, and 1995 kHz
  2. B.2005 kHz, 2000 kHz and 1995 kHz
  3. C.2000 kHz and 1995 kHz
  4. D.2 MHz only

Correct answer: B

Explanation: AM wave contains the carrier and the two side bands fc±fmf_c\pm f_m.

Question 65

Physics · 4 marks · −1

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u=UV∝T4u = \frac{U}{V} \propto T^4 and pressure p=13(UV)p = \frac{1}{3}\left(\frac{U}{V}\right). If the shell now undergoes an adiabatic expansion the relation between T and R is:

  1. A.T∝e−3RT \propto e^{-3R}
  2. B.T∝1RT \propto \frac{1}{R}
  3. C.T∝1R3T \propto \frac{1}{R^3}
  4. D.T∝e−RT \propto e^{-R}

Correct answer: B

Explanation: Adiabatic: dU=−pdVdU=-pdV with U=aT4VU=aT^4V, p=aT43p=\frac{aT^4}{3} gives VT3=VT^3= const, so T∝1RT\propto\frac1R.

Question 66

Physics · 4 marks · −1

An inductor (L=0.03(L = 0.03 H)) and a resistor (R=0.15(R = 0.15 kΩ)\Omega) are connected in series to a battery of 15V EMF in a circuit shown. The key K1K_1 has been kept closed for a long time. Then at t=0t = 0, K1K_1 is opened and key K2K_2 is closed simultaneously. At t=1t = 1 ms, the current in the circuit will be: (e5≅150)(e^5 \cong 150)

Figure for question 66
  1. A.67 mA
  2. B.6.7 mA
  3. C.0.67 mA
  4. D.100 mA

Correct answer: C

Explanation: I0=15150=0.1I_0=\frac{15}{150}=0.1 A; decay I=I0e−Rt/L=0.1e−5≈0.67I=I_0e^{-Rt/L}=0.1e^{-5}\approx0.67 mA.

Question 67

Physics · 4 marks · −1

A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TMT_M. If the Young's modulus of the material of the wire is Y then 1Y\frac{1}{Y} is equal to: (g = gravitational acceleration)

  1. A.[(TMT)2−1]MgA\left[\left(\frac{T_M}{T}\right)^2 - 1\right]\frac{Mg}{A}
  2. B.[1−(TMT)2]AMg\left[1 - \left(\frac{T_M}{T}\right)^2\right]\frac{A}{Mg}
  3. C.[1−(TTM)2]AMg\left[1 - \left(\frac{T}{T_M}\right)^2\right]\frac{A}{Mg}
  4. D.[(TMT)2−1]AMg\left[\left(\frac{T_M}{T}\right)^2 - 1\right]\frac{A}{Mg}

Correct answer: D

Explanation: TM2T2=L+ΔLL=1+MgAY\frac{T_M^2}{T^2}=\frac{L+\Delta L}{L}=1+\frac{Mg}{AY}, so 1Y=[(TMT)2−1]AMg\frac1Y=\left[\left(\frac{T_M}{T}\right)^2-1\right]\frac{A}{Mg}.

Question 68

Physics · 4 marks · −1

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

  1. A.2.45 V/m
  2. B.5.48 V/m
  3. C.7.75 V/m
  4. D.1.73 V/m

Correct answer: A

Explanation: I=0.14πI=\frac{0.1}{4\pi}; E0=2Icϵ0≈2.45E_0=\sqrt{\frac{2I}{c\epsilon_0}}\approx2.45 V/m.

Question 69

Physics · 4 marks · −1

Two coaxial solenoids of different radii carry current I in the same direction. Let F⃗1\vec{F}_1 be the magnetic force on the inner solenoid due to the outer one and F⃗2\vec{F}_2 be the magnetic force on the outer solenoid due to the inner one. Then:

  1. A.F⃗1\vec{F}_1 is radially inwards and F⃗2\vec{F}_2 is radially outwards
  2. B.F⃗1\vec{F}_1 is radially inwards and F⃗2=0\vec{F}_2 = 0
  3. C.F⃗1\vec{F}_1 is radially outwards and F⃗2=0\vec{F}_2 = 0
  4. D.F⃗1=F⃗2=0\vec{F}_1 = \vec{F}_2 = 0

Correct answer: C

Explanation: The inner solenoid's azimuthal currents sit in the outer solenoid's uniform axial field; Idl⃗×B⃗Id\vec l\times\vec B on each element points radially outward (parallel currents attract the windings outward). The inner solenoid produces no field outside itself, so F⃗2=0\vec F_2=0.

Question 70

Physics · 4 marks · −1

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion the average time of collision between molecules increases as VqV^q, where V is the volume of the gas. The value of q is : (γ=CpCv)\left(\gamma = \frac{C_p}{C_v}\right)

  1. A.3γ−56\frac{3\gamma-5}{6}
  2. B.γ+12\frac{\gamma+1}{2}
  3. C.γ−12\frac{\gamma-1}{2}
  4. D.3γ+56\frac{3\gamma+5}{6}

Correct answer: B

Explanation: τ∝λvrms∝VT\tau\propto\frac{\lambda}{v_{rms}}\propto\frac{V}{\sqrt T} and T∝V1−γT\propto V^{1-\gamma}: τ∝V(γ+1)/2\tau\propto V^{(\gamma+1)/2}.

Question 71

Physics · 4 marks · −1

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0Q_0 and then connected to the L and R as shown. If a student plots graphs of the square of maximum charge (QMax2)(Q^2_{Max}) on the capacitor with time (t) for two different values L1L_1 and L2L_2 (L1>L2)(L_1 > L_2) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)

Figure for question 71
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: D

Explanation: Qmax2∝e−Rt/LQ_{max}^2\propto e^{-Rt/L}: exponential decay, slower for the larger inductance L1L_1 (graph 4).

Question 72

Physics · 4 marks · −1

From a solid sphere of mass M and radius R, a spherical portion of radius R/2R/2 is removed, as shown in the figure. Taking gravitational potential V=0V = 0 at r=∞r = \infty, the potential at the centre of the cavity thus formed is (G = gravitational constant)

Figure for question 72
  1. A.−GMR-\frac{GM}{R}
  2. B.−2GM3R-\frac{2GM}{3R}
  3. C.−2GMR-\frac{2GM}{R}
  4. D.−GM2R\frac{-GM}{2R}

Correct answer: A

Explanation: Full sphere at r=R2r=\frac R2: −11GM8R-\frac{11GM}{8R}; removed sphere (mass M8\frac M8) at its own centre: −3GM8R-\frac{3GM}{8R}. Difference −GMR-\frac{GM}{R}.

Question 73

Physics · 4 marks · −1

A train is moving on a straight track with speed 20 ms−1^{-1}. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms−1^{-1}) close to :

  1. A.12 %
  2. B.18 %
  3. C.24 %
  4. D.6 %

Correct answer: A

Explanation: Approaching: 1000⋅320300=1066.71000\cdot\frac{320}{300}=1066.7 Hz; receding: 1000⋅320340=941.21000\cdot\frac{320}{340}=941.2 Hz. Change ≈12%\approx12\%.

Question 74

Physics · 4 marks · −1

Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

Figure for question 74
  1. A.80 N
  2. B.120 N
  3. C.150 N
  4. D.100 N

Correct answer: B

Explanation: The system is in equilibrium; the wall's friction must support the total weight 20+100=12020+100=120 N.

Question 75

Physics · 4 marks · −1

Distance of the centre of mass of a solid uniform cone from its vertex is z0z_0. If the radius of its base is R and its height is h then z0z_0 is equal to :

  1. A.3h4\frac{3h}{4}
  2. B.5h8\frac{5h}{8}
  3. C.3h28R\frac{3h^2}{8R}
  4. D.h24R\frac{h^2}{4R}

Correct answer: A

Explanation: Centre of mass of a solid cone lies at h4\frac h4 from the base, i.e. 3h4\frac{3h}{4} from the vertex.

Question 76

Physics · 4 marks · −1

A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below. If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

Figure for question 76
  1. A.(a) and (c), respectively
  2. B.(b) and (d), respectively
  3. C.(b) and (c), respectively
  4. D.(a) and (b), respectively

Correct answer: B

Explanation: In (b) the loop lies in the x-y plane with magnetic moment along +z+z (parallel to B): stable. In (d) the current sense is reversed, moment along −z-z: unstable. In (a) and (c) the moment is perpendicular to B (torque acts).

Question 77

Physics · 4 marks · −1

In the circuit shown, the current in the 1 Ω1\,\Omega resistor is :

Figure for question 77
  1. A.0 A
  2. B.0.13 A, from Q to P
  3. C.0.13 A, from P to Q
  4. D.1.3 A, from P to Q

Correct answer: B

Explanation: Applying Kirchhoff's laws to the two meshes (6 V with 3 ohm on the left, 9 V with 2 ohm and 3 ohm on the right, 1 ohm shared) gives about 0.13 A in the 1 ohm resistor, flowing from Q to P.

Question 78

Physics · 4 marks · −1

In the given circuit, charge Q2Q_2 on the 2 μ2\,\muF capacitor changes as C is varied from 1 μ1\,\muF to 3 μ3\,\muF. Q2Q_2 as a function of 'C' is given properly by : (figures are drawn schematically and are not to scale)

Figure for question 78
  1. A.Graph (1): increasing, concave down
  2. B.Graph (2): increasing, concave up
  3. C.Graph (3): decreasing
  4. D.Graph (4): increasing linearly

Correct answer: A

Explanation: C in series with (1+2) μ(1+2)\,\muF: Q=3CEC+3Q=\frac{3CE}{C+3}, Q2=23Q=2CEC+3Q_2=\frac23Q=\frac{2CE}{C+3}, increasing and concave down.

Question 79

Physics · 4 marks · −1

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

  1. A.50%
  2. B.56%
  3. C.62%
  4. D.44%

Correct answer: B

Explanation: Initial KE =3mv2=3mv^2; final velocity 223v\frac{2\sqrt2}{3}v with mass 3m: KE =43mv2=\frac43mv^2. Loss =59≈56%=\frac59\approx56\%.

Question 80

Physics · 4 marks · −1

Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is μ\mu, a ray, incident at an angle θ\theta, on the face AB would get transmitted through the face AC of the prism provided:

  1. A.θ<sin⁡−1[μsin⁡(A−sin⁡−1(1μ))]\theta < \sin^{-1}\left[\mu\sin\left(A - \sin^{-1}\left(\frac{1}{\mu}\right)\right)\right]
  2. B.θ>cos⁡−1[μsin⁡(A+sin⁡−1(1μ))]\theta > \cos^{-1}\left[\mu\sin\left(A + \sin^{-1}\left(\frac{1}{\mu}\right)\right)\right]
  3. C.θ<cos⁡−1[μsin⁡(A+sin⁡−1(1μ))]\theta < \cos^{-1}\left[\mu\sin\left(A + \sin^{-1}\left(\frac{1}{\mu}\right)\right)\right]
  4. D.θ>sin⁡−1[μsin⁡(A−sin⁡−1(1μ))]\theta > \sin^{-1}\left[\mu\sin\left(A - \sin^{-1}\left(\frac{1}{\mu}\right)\right)\right]

Correct answer: D

Explanation: Need r2=A−r1<θcr_2=A-r_1<\theta_c, i.e. r1>A−sin⁡−11μr_1>A-\sin^{-1}\frac1\mu, so sin⁡θ>μsin⁡(A−sin⁡−11μ)\sin\theta>\mu\sin\left(A-\sin^{-1}\frac1\mu\right).

Question 81

Physics · 4 marks · −1

From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is :

  1. A.MR2162π\frac{MR^2}{16\sqrt{2}\pi}
  2. B.4MR293π\frac{4MR^2}{9\sqrt{3}\pi}
  3. C.4MR233π\frac{4MR^2}{3\sqrt{3}\pi}
  4. D.MR2322π\frac{MR^2}{32\sqrt{2}\pi}

Correct answer: B

Explanation: Side a=2R3a=\frac{2R}{\sqrt3}, mass m=2M3πm=\frac{2M}{\sqrt3\pi}; I=ma26=4MR293πI=\frac{ma^2}{6}=\frac{4MR^2}{9\sqrt3\pi}.

Question 82

Physics · 4 marks · −1

Match List - I (Fundamental Experiment) with List - II (its conclusion): (A) Franck-Hertz Experiment, (B) Photo-electric experiment, (C) Davison - Germer Experiment; (i) Particle nature of light, (ii) Discrete energy levels of atom, (iii) Wave nature of electron, (iv) Structure of atom

  1. A.(A)-(ii), (B)-(iv), (C)-(iii)
  2. B.(A)-(ii), (B)-(i), (C)-(iii)
  3. C.(A)-(iv), (B)-(iii), (C)-(ii)
  4. D.(A)-(i), (B)-(iv), (C)-(iii)

Correct answer: B

Explanation: Franck-Hertz: discrete levels; photoelectric: particle nature of light; Davisson-Germer: electron waves.

Question 83

Physics · 4 marks · −1

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5×10−42.5\times10^{-4} ms−1^{-1}. If the electron density in the wire is 8×10288\times10^{28} m−3^{-3}, the resistivity of the material is close to:

  1. A.1.6×10−7 Ω1.6\times10^{-7}\,\Omegam
  2. B.1.6×10−6 Ω1.6\times10^{-6}\,\Omegam
  3. C.1.6×10−5 Ω1.6\times10^{-5}\,\Omegam
  4. D.1.6×10−8 Ω1.6\times10^{-8}\,\Omegam

Correct answer: C

Explanation: ρ=Enevd=508×1028⋅1.6×10−19⋅2.5×10−4≈1.6×10−5 Ω\rho=\frac{E}{nev_d}=\frac{50}{8\times10^{28}\cdot1.6\times10^{-19}\cdot2.5\times10^{-4}}\approx1.6\times10^{-5}\,\Omegam.

Question 84

Physics · 4 marks · −1

For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? (Graphs are schematic and not drawn to scale)

Figure for question 84
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: A

Explanation: KE is an inverted parabola (maximum at d = 0), PE an upright parabola (zero at d = 0), both non-negative.

Question 85

Physics · 4 marks · −1

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g=10g = 10 m/s2^2) (The figures are schematic and not drawn to scale)

Figure for question 85
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: B

Explanation: Until the first stone lands (t=8t=8 s) y2−y1=30ty_2-y_1=30t (linear, reaching 240 m); afterwards y2−y1=y2y_2-y_1=y_2 follows a downward parabola reaching 0 at t=12t=12 s.

Question 86

Physics · 4 marks · −1

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:

  1. A.30 μ30\,\mum
  2. B.100 μ100\,\mum
  3. C.300 μ300\,\mum
  4. D.1 μ1\,\mum

Correct answer: A

Explanation: θ=1.22λD=1.22⋅5×10−75×10−3\theta=\frac{1.22\lambda}{D}=\frac{1.22\cdot5\times10^{-7}}{5\times10^{-3}}; x=25 cm⋅θ≈30 μx=25\text{ cm}\cdot\theta\approx30\,\mum.

Question 87

Physics · 4 marks · −1

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle θ\theta with the vertical. If wires have mass λ\lambda per unit length then the value of I is: (g = gravitational acceleration)

Figure for question 87
  1. A.2sin⁡θπλgLμ0cos⁡θ2\sin\theta\sqrt{\frac{\pi\lambda gL}{\mu_0\cos\theta}}
  2. B.2πgLμ0tan⁡θ2\sqrt{\frac{\pi gL}{\mu_0}\tan\theta}
  3. C.πλgLμ0tan⁡θ\sqrt{\frac{\pi\lambda gL}{\mu_0}\tan\theta}
  4. D.sin⁡θπλgLμ0cos⁡θ\sin\theta\sqrt{\frac{\pi\lambda gL}{\mu_0\cos\theta}}

Correct answer: A

Explanation: Separation d=2Lsin⁡θd=2L\sin\theta; μ0I22πd=λgtan⁡θ⇒I=2sin⁡θπλgLμ0cos⁡θ\frac{\mu_0I^2}{2\pi d}=\lambda g\tan\theta\Rightarrow I=2\sin\theta\sqrt{\frac{\pi\lambda gL}{\mu_0\cos\theta}}.

Question 88

Physics · 4 marks · −1

On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam:

  1. A.goes horizontally without any deflection
  2. B.bends downwards
  3. C.bends upwards
  4. D.becomes narrower

Correct answer: C

Explanation: The lower part of the wavefront travels faster (smaller n), so the wavefront tilts and the beam bends upward, toward higher refractive index.

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