1. Which of the following is the energy of a possible excited state of hydrogen?
(1) -6.8 eV
(2) -3.4 eV
(3) +6.8 eV
(4) +13.6 eV
Answer (2)
Sol. Energy of excited state is negative and correspond to n>1n>1 .
2. In the following sequence of reactions :
Toluene KMnO4→A→SOCl2B→H2/PdC,KMnO4?→ASOCl2??BH2?/Pd?C, the product C is (1) C6H5CH3C6?H5?CH3? (2) C6H5CH2OHC6?H5?CH2?OH (3) C6H5CHOC6?H5?CHO (4) C6H5COOHC6?H5?COOH
Answer (3)
Sol.
3. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?
(1)
(2)
(3)
(4)
Answer (1)
Sol. 5- keto- 2- methylhexanal is
4. The ionic radii (in A?A? ) of N3−N3− O2−O2− and F−F− are respectively
(1) 1.36, 1.71 and 1.40
(2) 1.71, 1.40 and 1.36
(3) 1.71, 1.36 and 1.40
(4) 1.36, 1.40 and 1.71
Answer (2)
Sol. Radius of N3−N3− O2−O2− and F−F− follow order N3−>O2−>F−N3−>O2−>F− As per inequality only option (2) is correct that is 1.71 A?A? , 1.40 A?A? and 1.36 A?A?
5. The color of KMnO4KMnO4? is due to
(1) d-d transition
(2) L→ML→M charge transfer transition
(3) σ−σ∗σ−σ∗ transition
(4) M→LM→L charge transfer transition
Answer (2)
Sol. Charge transfer spectra from ligand (L) to metal (M) is responsible for color of KMnO4KMnO4?
6. 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) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion
(2) The assertion is incorrect, but the reason is correct
(3) Both the assertion and reason are incorrect
(4) Both assertion and reason are correct, and the reason is the correct explanation for the assertion
Answer (4)
Sol. N2+O2→2NON2?+O2?→2NO
Required temperature for above reaction is around 3000?C3000?C which is a quite high temperature. This reaction is observed during thunderstorm.
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7. Which of the following compounds is not an antacid?
(1) Cimetidine
(2) Phenelzine
(3) Ranitidine
(4) Aluminium Hydroxide
Answer (2)
Sol. Phenelzine is not antacid, it is anti- depressant.
8. In the context of the Hall-Heroult process for the extraction of Al, which of the following statement is false?
(1) Al2O3Al2?O3? is mixed with CaF2CaF2? which lowers the melting point of the mixture and brings conductivity
(2) Al3+Al3+ is reduced at the cathode to form Al
(3) Na3AlF6Na3?AlF6? serves as the electrolyte
(4) CO and CO2CO2? are produced in this process
Answer (3)
Sol. In Hall- Heroult process Al2O3Al2?O3? (molten) is electrolyte.
9. Match the catalysts to the correct processes :
<table>CatalystProcessa. TiCl3(i) Wacker processb. PdCl2(ii) Ziegler-Natta polymerizationc. CuCl2(iii) Contact processd. V2O5(iv) Deacon's process(1) a(ii), b(i), c(iv), d(iii)(2) a(ii), b(iii), c(iv), d(ii)(3) a(iii), b(i), c(ii), d(iv)(4) a(iii), b(ii), c(iv), d(ii)</table>
Answer (1)
Sol. TiCl3TiCl3? - Ziegler Natta polymerisation V2O5V2?O5? - Contact process PdCl2PdCl2? - Wacker process CuCl2CuCl2? - Deacon's process
10. In the reaction
the product E is
(1)
(2)
(3)
(4)
Answer (2)
Sol.
11. Which polymer is used in the manufacture of paints and lacquers?
(1) Glyptal
(2) Polypropene
(3) Poly vinyl chloride
(4) Bakelite
Answer (1)
Sol. Glyptal is used in manufacture of paints and lacquers.
12. The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3)(NH2OH)]+(NH3?)(NH2?OH)]+ is (py = pyridine)
(1) 3
(2) 4
(3) 6
(4) 2
Answer (1)
Sol.
as per question a=Cla=Cl b=pyb=py c=NH3c=NH3? and d=NH2OHd=NH2?OH are assumed.
13. Higher order (>3)(>3) reactions are rare due to
(1) Increase in entropy and activation energy as more molecules are involved
(2) Shifting of equilibrium towards reactants due to elastic collisions
(3) Loss of active species on collision
(4) Low probability of simultaneous collision of all the reacting species
Answer (4)
Sol. Higher order greater than 3 for reaction is rare because there is low probability of simultaneous collision of all the reacting species.
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14. Which among the following is the most reactive?
(1) Br2Br2? (2) I2I2? (3) IClICl (4) Cl2Cl2?
Answer (3)
Sol. Because of polarity and weak bond interhalogen compounds are more reactive.
15. Two faraday of electricity is passed through a solution of CuSO4CuSO4? . The mass of copper deposited at the cathode is (at. mass of Cu=63.5Cu=63.5 amu)
(1) 63.5g63.5g (2) 2g2g (3) 127g127g (4) 0g0g
Answer (1)
Sol. Cu+2+2e→CuCu+2+2e→Cu
So, 2 F charge deposite 1 mol of Cu. Mass deposited =63.5g=63.5g
16. 3 g of activated charcoal was added to 50 mL50 mL of acetic acid solution (0.06N)(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) 36mg36mg (2) 42mg42mg (3) 54mg54mg (4) 18mg18mg
Answer (4)
Sol. Number of moles of acetic acid adsorbed
=(0.06×501000−0.042×501000)=(0.06×100050?−0.042×100050?)
Weight of acetic acid adsorbed =0.9×60mg=0.9×60mg
=54mg=54mg
Hence, the amount of acetic acid adsorbed per g of
charcoal=543mgcharcoal=354?mg =18mg=18mg
Hence, option (4) is correct.
17. The synthesis of alkyl fluorides is best accomplished by
(1) Sandmeyer's reaction
(2) Finkelstein reaction
(3) Swarts reaction
(4) Free radical fluorination
Answer (3)
Sol. Swart's reaction
CH3−Cl+AgF→ΔCH3F+AgClCH3?−Cl+AgFΔ?CH3?F+AgCl
18. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3NaC8?H7?SO3?Na mol. wt. 206). What would be the maximum uptake of Ca2+Ca2+ ions by the resin when expressed in mole per gram resin?
Answer (3)
Sol. Ca2++2C8H7SO3Na+→Ca(C8H7SO3)2+2Na+Ca2++2C8?H7?SO3?Na+→Ca(C8?H7?SO3?)2?+2Na+ 1 mol 2 mol
The maximum uptake=1206×2=1412mol/gThe maximum uptake=206×21?=4121?mol/g
19. Which of the vitamins given below is water soluble?
(1) Vitamin D
(2) Vitamin E
(3) Vitamin K
(4) Vitamin C
Answer (4)
Sol. Vitamin C is water soluble vitamin.
20. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is
(1) Ion-dipole interaction
(2) London force
(3) Hydrogen bond
(4) Ion-ion interaction
Answer (3)
Sol. H-bond is one of the dipole-dipole interaction and dependent on inverse cube of distance between the molecules.
21. The following reaction is performed at 298K298K
2NO(g)+O2(g)?2NO2(g)2NO(g)+O2?(g)?2NO2?(g)
The standard free energy of formation of NO(g)NO(g) is 86.6kJ/mol86.6kJ/mol at 298K298K . What is the standard free energy of formation of NO2(g)NO2?(g) at 298K298K ?
(Kp=1.6×1012)(Kp?=1.6×1012)86600+R(298)ln?(1.6×1012)86600+R(298)ln(1.6×1012)86600−ln?(1.6×1012)R(298)86600−R(298)ln(1.6×1012)?0.5[2×86,600−R(298)ln?1.6×1012]0.5[2×86,600−R(298)ln1.6×1012]R(298)ln?(1.6×1012)−86600R(298)ln(1.6×1012)−86600
Answer (3)
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24. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29A?4.29A? . The radius of sodium atom is approximately
(ΔG?)reaction=[(ΔG?)formation]product(ΔG?)reaction?=[(ΔG?)formation?]product? [−(ΔG?)formation]reactant[−(ΔG?)formation?]reactant? ⇒−RTln?KP=2×(ΔG?)NO2−2(ΔG?)NO⇒−RTlnKP?=2×(ΔG?)NO2??−2(ΔG?)NO? ⇒(ΔG?)NO2=2(ΔG?)NO−RTln?KP⇒(ΔG?)NO2??=2(ΔG?)NO?−RTlnKP? ⇒(ΔG?)NO2=2×86600−R(298)ln?KP2⇒(ΔG?)NO2??=22×86600−R(298)lnKP?? ⇒2×86600−R(298)ln?1.6×10122⇒22×86600−R(298)ln1.6×1012? ⇒0.5[2×86,600−R(298)ln?1.6×1012]⇒0.5[2×86,600−R(298)ln1.6×1012]
22. Which of the following compounds is not colored yellow?
(1) K3[Co(NO2)6]K3?[Co(NO2?)6?] (2) (NH4)3[As(Mo3O10)4](NH4?)3?[As(Mo3?O10?)4?] (3) BaCrO4BaCrO4? (4) Zn2[Fe(CN)6]Zn2?[Fe(CN)6?]
Answer (4)
Sol. (NH4)3[As(Mo3O10)4](NH4?)3?[As(Mo3?O10?)4?] BaCrO4BaCrO4? and K3[Co(NO2)6]K3?[Co(NO2?)6?] are yellow colored compounds but Zn2[Fe(CN)6]Zn2?[Fe(CN)6?] is not yellow colored compound.
23. In Carius method of estimation of halogens, 250mg250mg of an organic compound gave 141mg141mg of AgBrAgBr . The percentage of bromine in the compound is (at mass Ag=108Ag=108 ; Br=80Br=80 )
(1) 36
(2) 48
(3) 60
(4) 24
Answer (4)
Sol. Percentage of Br
Ω=Weight of AgBrMoL mass of AgBr×MoL mass of BrWeight of O.C.×100Ω=MoL mass of AgBrWeight of AgBr?×Weight of O.C.MoL mass of Br?×100 Ω=141188×80250×100Ω=188141?×25080?×100 Ω=24%Ω=24%
(1) 3.22 A
(2) 5.72 A
(3) 0.93 A
(4) 1.86 A
Answer (4)
Sol. Edge length of BCC is 4.29A?4.29A? .
In BCC,
edge length=43redge length=3?4?r4.29=43r4.29=3?4?rr=4.2933≈1.86A?r=3?4.29?3?≈1.86A?
25. Which of the following compounds will exhibit geometrical isomerism?
(1) 3 - Phenyl - 1 - butene
(2) 2 - Phenyl - 1 - butene
(3) 1, 1 - Diphenyl - 1 propane
(4) 1 - Phenyl - 2 - butene
Answer (4)
Sol. For geometrical isomerism doubly bonded carbon must be bonded to two different groups which is only satisfied by 1 - Phenyl - 2 - butene.
26. The vapour pressure of acetone at 20?C20?C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100g100g of acetone at 20?C20?C , its vapour pressure was 183 torr. The molar mass (gmol−1)(gmol−1) of the substance is
(1) 64
(2) 128
(3) 488
(4) 32
Answer (1)
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Sol. Vapour pressure of pure acetone PAo=185PAo?=185 torr
Vapour pressure of solution, PS=183PS?=183 torr
Molar mass of solvent, MA=58MA?=58 g/mole
as we know PAo−PSPS=nBnAPS?PAo?−PS??=nA?nB??
⇒185−183183=WBMB×MAWA⇒183185−183?=MB?WB??×WA?MA??⇒2183=1.2MB×58100⇒1832?=MB?1.2?×10058?⇒MB=1.22×58100×183⇒MB?=21.2?×10058?×183=63.68g/mole=63.68g/mole
27. From the following statement regarding H2O2H2?O2? choose the incorrect statement
(1) It decomposes on exposure to light
(2) It has to be stored in plastic or wax lined glass bottles in dark.
(3) It has to be kept away from dust
(4) It can act only as an oxidizing agent
Answer (4)
Sol. H2O2H2?O2? can be reduced or oxidised. Hence, it can act as reducing as well as oxidising agent.
28. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?
(1) BeSO4BeSO4? (2) BaSO4BaSO4?
(3) SrSO4SrSO4? (4) CaSO4CaSO4?
Answer (1)
Sol. BeSO4BeSO4? has hydration energy greater than its lattice energy.
29. The standard Gibbs energy change at 300 K for the reaction 2A?B+C2A?B+C is 2494.2 J. At a given time, the composition of the reaction mixture is [A]=12,[B]=2[A]=21?,[B]=2 and [C]=12[C]=21? . The reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]
(1) Reverse direction because Q>KcQ>Kc?
(2) Forward direction because Q<KcQ<Kc?
(3) Reverse direction because Q<KcQ<Kc?
(4) Forward direction because Q>KcQ>Kc?
Answer (1)
Sol. 2A?B+C,ΔGo=2494.2J2A?B+C,ΔGo=2494.2J
As we know ΔGo=−2.303RTlog?KcΔGo=−2.303RTlogKc?
⇒2494.2=−2.303×8.314×300log?Kc⇒2494.2=−2.303×8.314×300logKc?⇒−0.434=log?Kc⇒−0.434=logKc?⇒Kc=anti log(−0.434)⇒Kc?=anti log(−0.434)⇒Kc=0.367⇒Kc?=0.367
Now [A]=12,[B]=2[A]=21?,[B]=2 and [C]=12[C]=21?
Now Qc=[C][B][A]2=(12)(2)(12)2=4Qc?=[A]2[C][B]?=(21?)2(21?)(2)?=4
as Qc>KcQc?>Kc? hence reaction will shift in backward direction.
30. Which one has the highest boiling point?
(1) Ne
(2) Kr
(3) Xe
(4) He
Answer (3)
Sol. Down the group strength of van der Waal's force of attraction increases hence Xe have highest boiling point.
PART-B : MATHEMATICS
31. The sum of coefficients of integral powers of x in the binomial expansion of (1−2x)50(1−2x?)50 is
(1) 12(350)21?(350) (2) 12(350−1)21?(350−1)
(3) 12(250+1)21?(250+1) (4) 12(350+1)21?(350+1)
Answer (4)
Sol. (1−2x)50=50C0−50C1(2x)1+50C2(2x)2+?+50C50(−2x)50(1−2x?)50=50C0?−50C1?(2x?)1+50C2?(2x?)2+?+50C50?(−2x?)50
Sum of coefficient of integral power of x
=50C020+50C2⋅22+50C4⋅24+?+50C50⋅250=50C0?20+50C2?⋅22+50C4?⋅24+?+50C50?⋅250
We know that
(1+2)50=50C0+50C1⋅2+?+50C50⋅250(1+2)50=50C0?+50C1?⋅2+?+50C50?⋅250
Then,
50C0+50C2⋅22+?+50C50⋅250=350+1250C0?+50C2?⋅22+?+50C50?⋅250=2350+1?
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32. Let f(x)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]=3limx→0?[1+x2f(x)?]=3 , then f(2)f(2) is equal to
(1)- 4
(2)0
(3)4
(4)- 8
Answer (2)
Sol. Let f(x)=a0+a1x+a2x2+a3x3+a4x4f(x)=a0?+a1?x+a2?x2+a3?x3+a4?x4
Usinglim?x→0[1+f(x)x2]=3⇒lim?x→0f(x)x2=2⇒lim?x→0a0+a1x+a2x2+a3x3+a4x4x2=2So,a0=0,a1=0,a2=2i.e.,f(x)=2x2+a3x3+a4x4Now,f′(x)=4x+3a3x2+4a4x3=x[4+3a3x+4a4x2]Given,f′(1)=0andf′(2)=0⇒3a3+4a4+4=0and6a3+16a4+4=0Solving,a4=12,a3=−2i.e.,f(x)=2x2−2x3+12x4i.e.,f(2)=0(1)?Usinglimx→0?[1+x2f(x)?]=3⇒limx→0?x2f(x)?=2⇒limx→0?x2a0?+a1?x+a2?x2+a3?x3+a4?x4?=2So,a0?=0,a1?=0,a2?=2i.e.,f(x)=2x2+a3?x3+a4?x4Now,f′(x)=4x+3a3?x2+4a4?x3=x[4+3a3?x+4a4?x2]Given,f′(1)=0andf′(2)=0⇒3a3?+4a4?+4=0and6a3?+16a4?+4=0Solving,a4?=21?,a3?=−2i.e.,f(x)=2x2−2x3+21?x4i.e.,f(2)=0?(1)
33. 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) 16.0
(2) 15.8
(3) 14.0
(4) 16.8
Answer (3)
Sol. Mean =16=16
Sum=16×16=256Sum=16×16=256 Newsum=256−16+3+4+5=252Newsum=256−16+3+4+5=252 Mean=25218=14Mean=18252?=14
34. The sum of first 9 terms of the series
131+13+231+3+13+23+331+3+5+……………………………113?+1+313+23?+1+3+513+23+33?+……………………………
(1) 96
(2) 142
(3) 192
(4) 71
Answer (1)
Sol.tn=⌈n(n+1)⌉2n2Sol.tn?=n2⌈n(n+1)⌉2?=(n+1)24=14[n2+2n+1]=14[n(n+1)(2n+1)6+2(n)(n+1)2+12]=14[9×10×196+9×10+9]=96(1)=4(n+1)2?=41?[n2+2n+1]=41?[6n(n+1)(2n+1)?+22(n)(n+1)?+21?]=41?[69×10×19?+9×10+9]=96?(1)
35. Let O be the vertex and QQ be any point on the parabola, x2=8yx2=8y . If the point PP divides the line segment OQOQ internally in the ratio 1:31:3 , then the locus of PP is
(1) y2=x(3) x2=2y(4)(1) y2=x(3) x2=2y?(4)
Answer (3)
Sol. x2=8yx2=8y
Let QQ be (4t, 2t)
∴P=(t,t22)Let P be (h,k)∴h=t,k=t22∴2k=h2 Locus of (h,k) is x2=2y.(1)∴P=(t,2t2?)Let P be (h,k)∴h=t,k=2t2?∴2k=h2 Locus of (h,k) is x2=2y.?(1)
36. Let αα and ββ be the roots of equation x2−6x−2=0x2−6x−2=0
If an=αn−βnan?=αn−βn , for n≥1n≥1 , then the value of a10−2a82a92a9?a10?−2a8?? is equal to
is equal to
(1) -6
(2) 3
(3) -3
(4) 6
Answer (2)
Sol. From equation,
α+β=6α+β=6 αβ=−2αβ=−2
The value of a10−2a82a9=α10+β10+αβ(α8+β8)2(α9+β9)=α9(α+β)+β9(α+β)2(α9+β9)=α+β2=62=3(1)The value of 2a9?a10?−2a8??=2(α9+β9)α10+β10+αβ(α8+β8)?=2(α9+β9)α9(α+β)+β9(α+β)?=2α+β?=26?=3?(1)
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37. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one the boxes contains exactly 3 balls is
55(23)10(2)220(13)1255(32?)10(2)220(31?)12 22(13)11(4)553(23)1122(31?)11(4)355?(32?)11
Answer (4)\*
Sol. Question is wrong but the best suitable option is (4).
Required probability =12C329312=553(23)11=12C3?31229?=355?(32?)11
38. A complex number zz is said to be unimodular if ?z?=1?z?=1 . Suppose z1z1? and z2z2? are complex numbers such that z1−2z22−z1z22−z1?z2?z1?−2z2?? is unimodular and z2z2? is not unimodular. Then the point z1z1? lies on a (1) Straight line parallel to yy -axis (2) Circle of radius 2 (3) Circle of radius 22? (4) Straight line parallel to xx -axis
Answer (2)
Sol.(z1−2z22−z1z2)=1Sol.(2−z1?z2?z1?−2z2??)=1 (z1−2z22−z1z2)(z1−2z22−z1z2)=1(2−z1?z2z1?−2z2??)(2−z1?z2?z1?−2z2??)=1 ⋅z1z1−2z1z2−2z2z1+4z2z2⋅z1?z1?−2z1?z2?−2z2?z1?+4z2?z2? ⋅=4−2z1z2−2z1z2+z1z1z2z2⋅=4−2z1?z2?−2z1?z2?+z1?z1?z2?z2? ⋅z1z1+4z2z2=4+z1z1z2z2⋅z1?z1?+4z2?z2?=4+z1?z1?z2?z2? ⋅zz1(1−z2z2)−4(1−z2z2)=0⋅zz1?(1−z2?z2?)−4(1−z2?z2?)=0 ⋅(z1z1−4)(1−z2z2)=0⋅(z1?z1?−4)(1−z2?z2?)=0 ⇒z1z1=4⇒z1?z1?=4 ?z?=2?z?=2 i.e.zliesoncircleofradius2.i.e.zliesoncircleofradius2.
39. The integral ∫dxx2(x4+1)3/4∫x2(x4+1)3/4dx? equals
(1)(x4+1)14+c(2)−(x4+1)14+c(1)(x4+1)41?+c(2)−(x4+1)41?+c (3)−(x4+1x4)14+c(4)(x4+1x4)14+c(3)−(x4x4+1?)41?+c(4)(x4x4+1?)41?+c
Answer (3)
Sol.I=∫dxx2(x4+1)3/4=∫dxx5(1+1x4)3/4Sol.I=∫x2(x4+1)3/4dx?=∫x5(1+x41?)3/4dx?
Let1+1x4=t⇒−4x3dx=dtLet1+x41?=t⇒x3−4?dx=dt So,I=−14∫dtt3/4=−14∫t−3/4dtSo,I=4−1?∫t3/4dt?=4−1?∫t−3/4dt =−14(t1/41/4)+c=4−1?(1/4t1/4?)+c =(−1+1x4)1/4+c=(−1+x41?)1/4+c
So, option (3).
40. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0), is
(1) 861
(2) 820
(3) 780
(4) 901
Answer (3)
Sol.
Total number of integral coordinates as required
=39+38+37+?+1=39+38+37+?+1 ⋅=39×402=780⋅=239×40?=780
41. The distance of the point (1, 0, 2) from the point of intersection of the line x−23=y+14=z−2123x−2?=4y+1?=12z−2? and the plane x−y+z=16x−y+z=16 , is
Answer (3)
Sol.x−23=y+14=z−212=λSol.3x−2?=4y+1?=12z−2?=λ P(3λ+2,4λ−1,12λ+2)P(3λ+2,4λ−1,12λ+2) Lies on plane x−y+z=16Lies on plane x−y+z=16 Then,Then,
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3+2- 4+1+12+2=16 11+5=16 λ=1λ=1 P(5,3,14) Distance =16+9+144=169=13=16+9+144?=169?=13
42. The equation of the plane containing the line 2x−2x− 5y+z=3;x+y+4z=5,5y+z=3;x+y+4z=5, and parallel to the plane, x+3y+6z=1,x+3y+6z=1, is
∴x+3y+6z=−7(2)x+3y+6z=7∴x+3y+6z=−7(2)x+3y+6z=7(3)2x+6y+12z=−13(4)2x+6y+12z=13(3)2x+6y+12z=−13(4)2x+6y+12z=13
Answer (2)
Sol. Required plane is
(2x−5y+z−3)+λ(x+y+4z−5)=0(2x−5y+z−3)+λ(x+y+4z−5)=0
It is parallel to x+3y+6z=1x+3y+6z=1
∴2+λ1=−5+λ3=1+4λ6∴12+λ?=3−5+λ?=61+4λ?
Solving λ=−112λ=2−11?
Required plane is
(2x−5y+z−3)−112(x+y+4z−5)=0(2x−5y+z−3)−211?(x+y+4z−5)=0∴x+3y+6z−7=0∴x+3y+6z−7=0
43. The area (in sq. units) of the region described by {(x,y):y2≤2x{(x,y):y2≤2x and y≥4x−1}y≥4x−1} is
(1)564(3)932(2)1564(4)732(1)645?(3)329??(2)6415?(4)327?
Answer (3)
Sol.
After solving y=4x−1y=4x−1 and y2=2xy2=2x
y=4⋅y22−1y=4⋅2y2?−12y2−y−1=02y2−y−1=0y=1±1+84=1±34y=1,−12y=41±1+8??=41±3?y=1,−21?A=∫−1/21(y+14)dy−∫−1/21y22dyA=∫−1/21?(4y+1?)dy−∫−1/21?2y2?dy=14[y22+y]−1/2−1[y33]−1/21=41?[2y2?+y]−1/2−1?[3y3?]−1/21?=14[4+8−1+48]−12[8+124]=41?[84+8−1+4?]−21?[248+1?]=14[158]−948=41?[815?]−489?=1532−632=932=3215?−326?=329?
44. If mm is the A.M. of two distinct real numbers ll and n(l,n>1)n(l,n>1) and G1,G2G1?,G2? and G3G3? are three geometric means between ll and nn , then G14+2G24+G34G14?+2G24?+G34? equals.
(1)4lm2n(2)4lm2(3)4l2m2n2(4)4l2mn(1)4lm2n(2)4lm2(3)4l2m2n2(4)4l2mn?
Answer (1)
Sol. l+n2=mSol. 2l+n?=m l+n=2ml+n=2m ⋅ ⋅ G1=l(nl)14G1?=l(ln?)41? ⋅ ⋅ G2=l(nl)34G2?=l(ln?)43? ⋅ ⋅ G3=l(nl)34G3?=l(ln?)43?
NowG14+2G24+G33NowG14?+2G24?+G33?l4⋅nl+2⋅(l2)(nl)2+l4(nl)3l4⋅ln?+2⋅(l2)(ln?)2+l4(ln?)3=nl3+2nl2+nl3=nl3+2nl2+nl3=2nl2l2+nl(n2+l2)=2nl2l2+nl(n2+l2)=2nl2l2+nl((n+l)2−2nl)=2nl2l2+nl((n+l)2−2nl)=nl(n+l)2=nl(n+l)2=nl⋅(2m)2=nl⋅(2m)2=4nlm2=4nlm2
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45. Locus of the image of the point (2, 3) in the line (2x−3y+4)+k(x−2y+3)=0,k∈R(2x−3y+4)+k(x−2y+3)=0,k∈R , is a (1) Straight line parallel to yy -axis (2) Circle of radius 22? (3) Circle of radius 33? (4) Straight line parallel to xx -axis
Answer (2)
Sol. After solving equation (i) & (ii)
2x−3y+4=0…(i)2x−3y+4=0…(i)x=1andy=2x=1andy=2
Slope of AB×AB× Slope of MN=−1MN=−1
b−3b+3a−22a+2=−1b−3a−2b+3?a+22?=−1(y−3)(y−1)=−(x−2)x(y−3)(y−1)=−(x−2)xy2−4y+3=−x2+2xy2−4y+3=−x2+2xx2+y2−2x−4y+3=0x2+y2−2x−4y+3=0
Circle of radius =2=2?
46. 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=19x2?+5y2?=1 , is
(1)18(2)272(3)27(4)(1)18(2)227?(3)27?(4)
Answer (3)
Sol. Ellipse is x29+y25=19x2?+5y2?=1
i.e.,a2=9,b2=5i.e.,a2=9,b2=5So,e=23So,e=32?
As, required area =2a2e=2×9(2/3)=27=e2a2?=(2/3)2×9?=27
47. 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) 192
(2) 120
(3) 72
(4) 216
Answer (1)
Sol. 4 digit numbers
5 digit numbers
Total number of integers =72+120=192=72+120=192
48. Let AA and BB be two sets containing four and two elements respectively. Then the number of subsets of the set A×BA×B , each having at least three elements is
(1) 256
(2) 275
(3) 510
(4) 219
Answer (4)
Sol. n(A)=4n(A)=4 n(B)=2n(B)=2
n(A×B)=8n(A×B)=8
Required numbers=8C3+8C4+…+8C8Required numbers=8C3?+8C4?+…+8C8? =28−(8C0+8C1+8C2)=28−(8C0?+8C1?+8C2?) =256−37=256−37 =219=219
49. Let tan?−1y=tan?−1x+tan?−1(2x1−x2)tan−1y=tan−1x+tan−1(1−x22x?)
where ?x?<13?x?<3?1? . Then a value of yy is
(1) 3x+x31−3x2(3) 3x+x31+3x2(4) 3x−x31−3x2(4)(1) 1−3x23x+x3?(3) 1+3x23x+x3?(4) 1−3x23x−x3??(4)
Answer (4)
Sol. tan?−1y=tan?−1x+tan?−1(2x1−x2)tan−1y=tan−1x+tan−1(1−x22x?)
3tan?−1x=tan?−1(3x−x31−3x2)3tan−1x=tan−1(1−3x23x−x3?)y=3x−x31−3x2y=1−3x23x−x3?
50. The integral ∫24log?x2+log?(36−12x+x2)dx2∫4?logx2+log(36−12x+x2)dx is equal to
(1) 4
(2) 1
(3) 6
(4) 2
Answer (2)
I=∫24log?x2dx2log?x2+log?(36−12x+x2)I=2∫4?2logx2+log(36−12x+x2)logx2dx?
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I=∫24log?(6−x)2dx2log?x2+log?(6−x)2I=2∫4?2logx2+log(6−x)2log(6−x)2dx?2I=∫241dx2I=2∫4?1dx2I=22I=2I=1I=1
51. The negation of ~ s ∨ ( ~ r ∧ s) is equivalent to
(1) s ∧ (r ∧ ~ s) (2) s ∨ (r ∨ ~ s)
(3) s ∧ r (4) s ∧ ~ r
Answer (3)
Sol. ~(~s∨(~r∧s)) = s∧(r∨~s) = (s∧r)∨(s∧~s) = s∧r
52. 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°, 45° and 60° respectively, then the ratio, AB : BC, is
(1) 3:23?:2? (2) 1:31:3?
(3) 2:32:3 (4) 3:13?:1
Answer (4)
Sol. AO=hcot?30?=h3AO=hcot30?=h3?
BO=hBO=h
CO=h3CO=3?h?
∴ABBC=AO−BOBO−CO=h3−hh−h3=3∴BCAB?=BO−COAO−BO?=h−3?h?h3?−h?=3?
53. lim?x→0(1−cos?2x)(3+cos?x)xtan?4xlimx→0?xtan4x(1−cos2x)(3+cosx)? is equal to
(1) 3 (2) 2
(3) 1221? (4) 4
Answer (2)
Sol. lim?x→02sin?2x⋅(3+cos?x)x2tan?4x4x×4x×x2x=2limx→0?x24xtan4x?×4x2sin2x⋅(3+cosx)?×xx2?=2
54. Let a?,b?a?,b? and c?c? be three non-zero vectors such that no two of them are collinear and (a?×b?)×c?=13?b???c??a?(a?×b?)×c?=31??b???c??a? . If θ is the angle between vectors b?b? and c?c? , then a value of sin θ is
(1) −233−2?? (2) 2332?
(3) −2333−23?? (4) 223322??
Answer (4)
Sol. (a?⋅c?)b?−(b?⋅c?)a?=13?b???c??a?(a?⋅c?)b?−(b?⋅c?)a?=31??b???c??a?
∴−(b?⋅c?)=13?b???c??∴−(b?⋅c?)=31??b???c??∴cos?θ=−13∴cosθ=−31?∴sin?θ=223∴sinθ=322??
55. If A=[12221−2a2b]A=?12a?212?2−2b?? is a matrix satisfying the equation AAT=9IAAT=9I, where II is 3×33×3 identity matrix, then the ordered pair (a, b) is equal to
(1) (-2, 1) (2) (2, 1)
(3) (-2, -1) (4) (2, -1)
Answer (3)
Sol.
[12221−2a2b][12a2122−2b]=[900090009]?12a?212?2−2b???122?21−2?a2b??=?900?090?009??
a+4+2b=0a+4+2b=0
2a+2−2b=02a+2−2b=0
a+1−b=0a+1−b=0
2a−2b=−22a−2b=−2
a+2b=−4a+2b=−4
3a=−63a=−6
a=−2a=−2
−2+1−b=0−2+1−b=0
b=−1b=−1
a=−2a=−2
(−2,−1)(−2,−1)
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56. If the function.
is differentiable, then the value of k+mk+m is
165(2)103(3)4(4)2516?(2)310?(3)4(4)2
Answer (4)
Sol.g(x)={kx+1,0≤x≤3mx+2,3<x≤5Sol.g(x)={kx+1?,mx+2,?0≤x≤33<x≤5?
R.H.D.
=lim?h→0(3m−2k)+mh+2h=m=limh→0?h(3m−2k)+mh+2?=m
L.H.D.
lim?h→0k(3−h)+1−2k−hh→0lim?−hk(3−h)+1?−2k?lim?h→0−k[4−h−2]hh→0lim?h−k[4−h?−2]?lim?h→0−k×4−h−4h(4−h+2)=k4h→0lim?−k×h(4−h?+2)4−h−4?=4k?
From above,
k4=mand3m−2k+2=04k?=mand3m−2k+2=0m=25andk=85m=52?andk=58?k+m=85+25=105=2k+m=58?+52?=510?=2
Alternative Answer
g(x) = \left\{ \begin{array}{ll}k\sqrt{x + 1}, & 0\leq x\leq 3 mx + 2, & 3< x\leq 5 \end{array} \right.
gg is constant at x=3x=3
k4=3m+2k4?=3m+22k=3m+2…(i)2k=3m+2…(i)Also(k2x+1)x=3=mAlso(2x+1?k?)x=3?=mk4=m4k?=mk=4m…(ii)k=4m…(ii)8m=3m+28m=3m+2m=25,k=85m=52?,k=58?m+k=25+85=2m+k=52?+58?=2
57. The set of all values of λλ for which the system of linear equations
2x1−2x2+x3=λx12x1?−2x2?+x3?=λx1?2x1−3x2+2x3=λx22x1?−3x2?+2x3?=λx2?−x1+2x2=λx3−x1?+2x2?=λx3?
has a non- trivial solution
(1) Is a singleton
(2) Contains two elements
(3) Contains more than two elements
(4) Is an empty set
Answer (2)
Sol.x1(2−λ)−2x2+x3=0Sol.x1?(2−λ)−2x2?+x3?=02x1+x2(−λ−3)+2x3=02x1?+x2?(−λ−3)+2x3?=0−x1+2x2−λx3=0−x1?+2x2?−λx3?=0(2−λ)(λ2+3λ−4)+2(−2λ+2)+(4−λ−3)=0(2−λ)(λ2+3λ−4)+2(−2λ+2)+(4−λ−3)=02λ2+6λ−8−λ3−3λ2+4λ−4λ+4−λ+1=02λ2+6λ−8−λ3−3λ2+4λ−4λ+4−λ+1=0⇒−λ3−λ2+5λ−3=0⇒−λ3−λ2+5λ−3=0⇒λ3+λ2−5λ+3=0⇒λ3+λ2−5λ+3=0λ3−λ2+2λ2−2λ−3λ+3=0λ3−λ2+2λ2−2λ−3λ+3=0λ2(λ−1)+2λ(λ−1)−3(λ−1)=0λ2(λ−1)+2λ(λ−1)−3(λ−1)=0(λ−1)(λ2+2λ−3)=0(λ−1)(λ2+2λ−3)=0(λ−1)(λ+3)(λ−1)=0(λ−1)(λ+3)(λ−1)=0⇒λ=1,1,−3⇒λ=1,1,−3
Two elements.
58. The normal to the curve, x2+2xy−3y2=0x2+2xy−3y2=0 at (1,1)
(1) Meets the curve again in the second quadrant
(2) Meets the curve again in the third quadrant:
(3) Meets the curve again in the fourth quadrant
(4) Does not meet the curve again
Answer (3)
Sol. Curve is x2+2xy−3y2=0x2+2xy−3y2=0
Differentiate w.r.t.x,2x+2[dydx+y]−6y⋅dydx=0Differentiate w.r.t.x,2x+2[dxdy?+y]−6y⋅dxdy?=0⇒(dydx)(1,1)=1⇒(dxdy?)(1,1)?=1
So equation of normal at (1, 1) is
y−1=−1(x−1)y−1=−1(x−1)
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59. The number of common tangents to the circles x2+y2−4x−6y−12=0x2+y2−4x−6y−12=0 and x2+y2+6x+18y+26=0x2+y2+6x+18y+26=0 , is (1) 2 (2) 3 (3) 4 (4) 1
(3) 4
(4) 1
Answer (2)
Sol. x2+y2−4x−6y−12=0x2+y2−4x−6y−12=0
C1(center)=(2,3),r=22+32+12=5C1?(center)=(2,3),r=22+32+12?=5x2+y2+6x+18y+26=0x2+y2+6x+18y+26=0C2(center)(−3,−9),r=9+81−26C2?(center)(−3,−9),r=9+81−26?C1C2=13,C1C2=r1+r2C1?C2?=13,C1?C2?=r1?+r2?
Number of common tangent is 3.
60. Let y(x)y(x) be the solution of the differential equation
(xlog?x)dydx+y=2xlog?x,(x≥1).(xlogx)dxdy?+y=2xlogx,(x≥1).
Then y(e)y(e) is equal to
(1) 0
(2) 2
(3) 2e
(4) e
Answer (2)\*
Sol. It is best option. Theoretically question is wrong, because initial condition is not given.
xlog?xdydx+y=2xlog?xIfx=1theny=0xlogxdxdy?+y=2xlogxIfx=1theny=0dydx+yxlog?x=2dxdy?+xlogxy?=2IF,e∫1xlog?xdx=elog?log?x=log?xIF,e∫xlogx1?dx=eloglogx=logxSolution is y⋅log?x=∫2log?xdx+cSolution is y⋅logx=∫2logxdx+cylog?x=2(xlog?x−x)+cylogx=2(xlogx−x)+cx=1,y=0x=1,y=0
Then, c=2,y(e)=2c=2,y(e)=2
PART-C:PHYSICS
61. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion
(1) Kinetic energy, potential energy and total energy decrease
(2) Kinetic energy decreases, potential energy increases but total energy remains same
(3) Kinetic energy and total energy decrease but potential energy increases
(4) Its kinetic energy increases but potential energy and total energy decrease
Answer (4)
Sol. PE=−27.2z2n2eVPE=−27.2n2z2?eV
TE=−13.6z2n2eVTE=−n213.6z2?eVKE=13.6z2n2eVKE=n213.6z2?eVKE=13.6n2eV,Asn decreases,KE↑KE=n213.6?eV,Asn decreases,KE↑PE=−27.2n2eV,asn decreases,PE↓PE=−n227.2?eV,asn decreases,PE↓TE=−13.6n2eV,asn decreases,TE↓TE=−n213.6?eV,asn decreases,TE↓
62. The period of oscillation of a simple pendulum is
T=2πLgT=2πgL?? . Measured value of LL is 20.0cm20.0cm known to 1mm1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s 90 s using a wrist watch of 1 s 1 s resolution. The accuracy in the determination of gg is
(1) 3%3% (2) 1%1%
(3) 5%5% (4) 2%2%
Answer (1)
Sol. g=4π2lT2Sol. g=4π2T2l? ⇒Δgg×100=Δll×100+2ΔTT×100⇒gΔg?×100=lΔl?×100+2TΔT?×100 =Δll×100+2Δtt×100=lΔl?×100+2tΔt?×100 =0.120.0×100+2×190×100=20.00.1?×100+2×901?×100 =100200+20090=12+209≡3%=200100?+90200?=21?+920?≡3%
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63. A long cylindrical shell carries positive surface charge σσ in the upper half and negative surface charge −σ−σ 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)
(1)
(2)
(3)
(4)
Answer (4)
Sol. The field line should resemble that of a dipole.
64. A signal of 5kHz5kHz frequency is amplitude modulated on a carrier wave of frequency 2MHz2MHz . The frequencies of the resultant signal is/are
(1) 2005 kHz and 1995 kHz
(2) 2005 kHz, 2000 kHz and 1995 kHz
(3) 2000 kHz and 1995 kHz
(4) 2 MHz only
Answer (2)
Sol. Frequencies of resultant signal are
(2000+5)kHz,2000kHz,(2000−5)kHz,(2000+5)kHz,2000kHz,(2000−5)kHz,
65. Consider a spherical shell of radius RR at temperature TT . The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u=UV∝T4u=VU?∝T4 and pressure p=13(UV)p=31?(VU?) . If the shell now undergoes an adiabatic expansion the relation between TT and RR is (1) T∝e−3RT∝e−3R (2) T∝1RT∝R1? (3) T∝1R3T∝R31? (4) T∝e−RT∝e−R
Answer (2)
P=13(UV)=13kT4P=31?(VU?)=31?kT4 PV=μRTPV=μRT μRTV=13kT4VμRT?=31?kT4 ⇒V∝T−3⇒V∝T−3
66. An inductor (L=0.03H)(L=0.03H) and a resistor (R=0.15kΩ)(R=0.15kΩ) are connected in series to a battery of 15V15V ,EMF in a circuit shown below. The key K1K1? has been kept closed for a long time. Then at t=0t=0 K1K1? is opened and key K2K2? is closed simultaneously. At t=1mst=1ms the current in the circuit will be (e5≡150)(e5≡150)
(1) 67mA67mA (2) 6.7mA6.7mA (3) 0.67mA0.67mA (4) 100mA100mA
Answer (3)
\displaystyle = \frac{15}{150} e^{-\frac{1\times 10^{-3}}{1 / 5\times 10^{3}}} = 0.67\mathrm{mA
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67. A pendulum made of a uniform wire of cross- sectional area AA has time period TT . When an additional mass MM is added to its bob, the time period changes to TMTM? . If the Young's modulus of the material of the wire is YY then 1YY1? is equal to
g=g= gravitational acceleration)
[(TMT)2−1]MgA[(TTM??)2−1]AMg?[1−(TMT)2]AMg[1−(TTM??)2]MgA?[1−(TTM)2]AMg[1−(TM?T?)2]MgA?[(TMT)2−1]AMg[(TTM??)2−1]MgA?
Answer (4)
Sol.T=2πIgSol.T=2πI+ΔIgSol.Y=FIAΔI⇒ΔI=MgIAYSol.⇒1Y=AMg[(TMT)2−1](3)Sol.T=2πgI??Sol.T=2πgI+ΔI??Sol.Y=AΔIFI?⇒ΔI=AYMgI?Sol.⇒Y1?=MgA?[(TTM??)2−1]?(3)
68. 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 1m1m from the diode is
(1)2.45V/m(2)5.48V/m(1)2.45V/m(2)5.48V/m(3)7.75V/m(4)1.73V/m(3)7.75V/m(4)1.73V/m
Answer (1)
I=P4πr2=Uav×cUav=12?0E02⇒P4πr2=12?0E02×c⇒E0=2P4πr2?0c=2.45V/m(1)I=4πr2P?=Uav?×cUav?=21??0?E02?⇒4πr2P?=21??0?E02?×c⇒E0?=4πr2?0?c2P??=2.45V/m?(1)
69. Two coaxial solenoids of different radii carry current II in the same direction. Let F1‾F1?? be the magnetic force on the inner solenoid due to the outer one and F2‾F2?? be the magnetic force on the outer solenoid due to the inner one. Then
(1) F1‾F1?? is radially inwards and F2‾F2?? is radially outwards
(2) F1‾F1?? is radially inwards and F2‾=0F2??=0 (3) F1‾F1?? is radially outwards and F2‾=0F2??=0 (4) F1‾=F2‾=0F1??=F2??=0
Answer (4)
Sol. Net force on each of them would be zero.
70. 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 VηVη , where VV is the volume of the gas. The value of qq is
(γ=CpCv)(γ=Cv?Cp??)
Answer (2)
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71. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0Q0? and then connected to the LL and RR as shown below :
If a student plots graphs of the square of maximum charge (QMax2)(QMax2?) on the capacitor with time (t)(t) for two different values L1L1? and L2(L1>L2)L2?(L1?>L2?) of LL then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)
(1)
(2)
(3)
(4)
Answer (4)
Sol. For a damped pendulum, A=A0e−bt/2mA=A0?e−bt/2m
⇒A=A0e−(R2L)t⇒A=A0?e−(2LR?)t
(Since LL plays the same role as mm )
72. From a solid sphere of mass MM and radius RR , a spherical portion of radius R22R? is removed, as shown in the figure. Taking gravitational potential V=0V=0 at r=∞r=∞ , the potential at the centre of the cavity thus formed is
( G=G= gravitational constant)
(1) −GMRR−GM? (2) −2GMRR−2GM? (3) −2GMRR−2GM?
Answer (1)
Sol. V=V1−V2V=V1?−V2?
V1=−GM2R3[3R2−(R2)2]V1?=−2R3GM?[3R2−(2R?)2] ⇒V=−GMR⇒V=R−GM?
73. A train is moving on a straight track with speed 20ms−120ms−1 . It is blowing its whistle at the frequency of 1000Hz1000Hz . The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound =320ms−1=320ms−1 ) close to
(1) 12%12% (2) 18%18% (3) 24%24% (4) 6%6%
Answer (1)
f1=f[vv−vs]=f[320320−20]=f×320300Hzf1?=f[v−vs?v?]=f[320−20320?]=f×300320?Hz f2=f[vv+vs]=f×320340Hzf2?=f[v+vs?v?]=f×340320?Hz ⋅100×(f2f1−1)=(f2−f1f1)×100⋅100×(f1?f2??−1)=(f1?f2?−f1??)×100 ⋅=100[300340−1]=12%⋅=100[340300?−1]=12%
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74. F→AB
Given in the figure are two blocks AA and BB of weight 20 N20 N and 100 N100 N respectively. These are being pressed against a wall by a force FF as shown. If the coefficient of friction between the blocks is 0.1 and between block BB and the wall is 0.15, the frictional force applied by the wall on block BB is
(1) 80 N80 N (2) 120 N120 N (3) 150 N150 N (4) 100 N100 N
Answer (2)
Sol.
Clearly fs=120 Nfs?=120 N (for vertical equilibrium of the system)
75. Distance of the centre of mass of a solid uniform cone from its vertex is z0z0? . If the radius of its base is RR and its height is hh then z0z0? is equal to
(1)3h4(2)5h8(3)3h28R(4)h24R(4)(1)43h?(2)85h?(3)8R3h2?(4)4Rh2??(4)
Answer (1)
Sol. dm=πr2dy?dm=πr2dy?
yCM=∫ydm∫dm=∫πr2dy×ρ×y13πR2hρyCM?=∫dm∫ydm?=31?πR2hρ∫πr2dy×ρ×y? =3h4=43h?
76. A rectangular loop of sides 10 cm10 cm and 5 cm5 cm carrying a current II of 12 A12 A is placed in different orientations as shown in the figures below:
If there is a uniform magnetic field of 0.3 T0.3 T in the positive zz direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?
(1) (a) and (c), respectively
(2) (b) and (d), respectively
(3) (b) and (c), respectively
(4) (a) and (b), respectively
Answer (2)
Sol. Stable equilibrium M‾??B‾M??B
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77. In the circuit shown, the current in the 1Ω1Ω resistor is
(1) 0 A
(2) 0.13 A, from QQ to PP (3) 0.13 A, from PP to QQ (4) 1.3 A, from PP to QQ
Answer (2)
Sol. From KVL,
9=6I1−I2…(1)9=6I1?−I2?…(1) 6=4I2−I1…(2)6=4I2?−I1?…(2)
Solving, I1−I2=−0.13I1?−I2?=−0.13 A
78. A uniformly charged solid sphere of radius RR has potential V0V0? (measured with respect to ∞∞ ) on its surface. For this sphere the equipotential surfaces with potentials 3V02,5V04,3V0423V0??,45V0??,43V0?? and V044V0?? have radius R1,R2,R3R1?,R2?,R3? and R4R4? respectively. Then
(1)R1≠0and(R2−R1)>(R4−R3)(1)R1??=0and(R2?−R1?)>(R4?−R3?) (2)R1=0andR2<(R4−R3)(2)R1?=0andR2?<(R4?−R3?) (3)2R<R4(3)2R<R4? (4)R1=0andR2>(R4−R3)(4)R1?=0andR2?>(R4?−R3?)
Answer (2, 3)
Sol. V0=kQR…(i)Sol. V0?=kRQ?…(i) V1=kQ2R3(3R2−r2)V1?=2R3kQ?(3R2−r2) V=32V0⇒R1=0V=23?V0?⇒R1?=0 5kQ4R=kQ(3R2−r2)2R34R5kQ?=kQ2R3(3R2−r2)? ⇒R2=R2⇒R2?=2?R? ⇒3kQ4R=kQR3⇒4R3kQ?=R3kQ? ⇒R3=4R3⇒R3?=34R? ⇒1kQ4R=kQR4⇒4R1kQ?=R4?kQ? ⇒R4=4R⇒R4>2R⇒R4?=4R⇒R4?>2R
79. In the given circuit, charge Q2Q2? on the 2μF2μF capacitor changes as CC is varied from 1μF1μF to 3μF3μF Q2Q2? as a function of CC is given properly by : (Figures are drawn schematically and are not to scale)
(1)
(2)
(3)
(4)
Answer (1)
Sol.Caq=3C3+C…(i)Sol.Caq?=3+C3C?…(i) Total charges q=(3C3+C)E…(ii)Total charges q=(3+C3C?)E…(ii) Charge upon capacitor 2μF,Charge upon capacitor 2μF,
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80. A particle of mass mm moving in the xx direction with speed 2v2v is hit by another particle of mass 2m2m moving in the yy direction with speed vv . If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
(1) 50%50% (2) 56%56%
(3) 62%62% (4) 44%44%
Answer (2)
Sol. m?2vm?2v
KE loss=12m(2v)2+12(2m)v2KE loss=21?m(2v)2+21?(2m)v2 −12×(3m)(2mv23m)2=53mv2−21?×(3m)(3m2mv2??)2=35?mv2
KE loss=12m(2v)2+12(2m)v2KE loss=21?m(2v)2+21?(2m)v212×(3m)(2mv23m)2=53mv221?×(3m)(3m2mv2??)2=35?mv2
Required %=53mv22mv2+mv2×100=56%%=2mv2+mv235?mv2?×100=56%
81. Monochromatic light is incident on a glass prism of angle AA . If the refractive index of the material of the prism is μμ , a ray, incident at an angle θθ , on the face ABAB would get transmitted through the face ACAC of the prism provided.
θ<sin?−1[μsin?(A−sin?−1(1μ))]θ<sin−1[μsin(A−sin−1(μ1?))]θ>cos?−1[μsin?(A+sin?−1(1μ))]θ>cos−1[μsin(A+sin−1(μ1?))]θ<cos?−1[μsin?(A+sin?−1(1μ))]θ<cos−1[μsin(A+sin−1(μ1?))]θ>sin?−1[μsin?(A−sin?−1(1μ))]θ>sin−1[μsin(A−sin−1(μ1?))]
Answer (4)
Sol.
sin?θ=μsin?r1sinθ=μsinr1?⇒sin?r1=sin?θμ⇒sinr1?=μsinθ?⇒r1=sin?−1(sin?θμ)⇒r1?=sin−1(μsinθ?)r2=A−sin?−1(sin?θμ)r2?=A−sin−1(μsinθ?)⇒r2<sin?−1(1μ)⇒r2?<sin−1(μ1?)A−sin?−1(sin?θμ)<sin?−1(1μ)A−sin−1(μsinθ?)<sin−1(μ1?)⇒A−sin?−1(1μ)<sin?−1(sin?θμ)⇒A−sin−1(μ1?)<sin−1(μsinθ?)⇒sin?(A−sin?−1(1μ))<sin?θμ⇒sin(A−sin−1(μ1?))<μsinθ?⇒μ(sin?(A−sin?−1(1μ)))<sin?θ⇒μ(sin(A−sin−1(μ1?)))<sinθ⇒sin?−1(μsin?(A−sin?−1(1μ)))<θ⇒sin−1(μsin(A−sin−1(μ1?)))<θ
82. From a solid sphere of mass MM and radius RR 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
Answer (2)
Sol. d=2R=a3d=2R=a3?
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⇒a=23R⇒a=3?2?RMM′=43πR3(23R)3=32πM′M?=(3?2?R)334?πR3?=23??π⇒M′=2M3π⇒M′=3?π2M?I=M′a26=2M3π×43R2×16I=6M′a2?=3?π2M?×34?R2×61?I=4MR293πI=93?π4MR2?
83. Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list:
<table><tr><td></td><td>List -I</td><td></td><td>List-II</td></tr><tr><td>(A)</td><td>Franck-Hertz experiment</td><td>(i)</td><td>Particle nature of light</td></tr><tr><td>(B)</td><td>Photo-electric experiment</td><td>(ii)</td><td>Discrete energy levels of atom</td></tr><tr><td>(C)</td><td>Davison-Cermer experiment</td><td>(iii)</td><td>Wave nature of electron</td></tr><tr><td></td><td></td><td>(iv)</td><td>Structure of atom</td></tr></table>
(1) (A) - (ii) (B) - (iv) (C) - (iii)
(2) (A) - (ii) (B) - (i) (C) - (iii)
(3) (A) - (iv) (B) - (iii) (C) - (ii)
(4) (A) - (i) (B) - (iv) (C) - (iii)
Answer (2)
Sol. Franck-Hertz exp.- Discrete energy level.
Photo-electric effect- Particle nature of light
Davison-Germer exp.- Diffraction of electron beam.
84. When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5×10−4ms−12.5×10−4ms−1. If the electron density in the wire is 8×1028m−38×1028m−3, the resistivity of the material is close to
(1) 1.6×10−7Ωm1.6×10−7Ωm
(2) 1.6×10−6Ωm1.6×10−6Ωm
(3) 1.6×10−5Ωm1.6×10−5Ωm
(4) 1.6×10−8Ωm1.6×10−8Ωm
Answer (3)
Sol. V=IR=IρlAV=IR=IρAl?
⇒ρ=VAIl=VAIneAvd=Vl×n×e×vd⇒ρ=IlVA?=IneAvd?VA?=l×n×e×vd?V?⇒ρ=50.1×2.5×10−4×1.6×10−19×8×1028⇒ρ=0.1×2.5×10−4×1.6×10−19×8×10285?=1.6×10−5Ωm=1.6×10−5Ωm
85. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement dd. Which one of the following represents these correctly?
(Graphs are schematic and not drawn to scale)
(1)
(2)
(3)
(4)
Answer (1)
Sol. KE=12mω2(A2−d2)KE=21?mω2(A2−d2)
PE=12mω2d2PE=21?mω2d2
At d=±Ad=±A,
PE = maximum while KE = 0.
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86. Two stones are thrown up simultaneously from the edge of a cliff 240m240m high with initial speed of 10m/s10m/s and 40m/s40m/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=10m/s2g=10m/s2
(The figures are schematic and not drawn to scale)
(1)
(2)
(3)
(4)
Answer (2)
Sol. Till both are in air (From t=0t=0 to t=8t=8 sec)
Δx=x2−x1=30tΔx=x2?−x1?=30t ⇒Δx∝t⇒Δx∝t
When second stone hits ground and first stone is in air ΔxΔx decreases.
87. A solid body of constant heat capacity 1J/?C1J/?C is being heated by keeping it in contact with reservoirs in two ways :
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature 100?C100?C to final temperature 200?C200?C Entropy change of the body in the two cases respectively is
(1)ln?2,ln?2(2)ln?2,2ln?2(1)ln2,ln2(2)ln2,2ln2 (3)ln?2,8ln?2(4)ln?2,4ln?2(3)ln2,8ln2(4)ln2,4ln2
Answer (None)
88. Assuming human pupil to have a radius of 0.25cm0.25cm and a comfortable viewing distance of 25cm25cm , the minimum separation between two objects that human eye can resolve at 500nm500nm wavelength is
Answer (1)
Sol.RP=1.22λ2μsin?θ=1.22×(500×10−9m)2×1×(1100)Sol.RP=2μsinθ1.22λ?=2×1×(1001?)1.22×(500×10−9m)?=3.05×10−5m=3.05×10−5m=30μm=30μm
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89. Two long current carrying thin wires, both with current II , are held by insulating threads of length LL and are in equilibrium as shown in the figure, with threads making an angle θθ with the vertical. If wires have mass λλ per unit length then the value of II is g=g= gravitational acceleration)
2sin?θπλgLμ0cos?θ2sinθμ0?cosθπλg?L?? 2πgLμ0tan?θ2μ0?πgL??tanθ πλgLμ0tan?θμ0?πλg?L??tanθ sin?θπλgLμ0cos?θsinθμ0?cosθπλg?L??
Answer (1)
Sol.
Tcos?θ=λglTcosθ=λg?l Tsin?θ=μ02π⋅I×Il(2Lsin?θ)Tsinθ=2πμ0??⋅(2Lsinθ)I×Il? ⇒I=2sin?θπλgLμ0cos?θ⇒I=2sinθμ0?cosθπλg?L??
90. On a hot summer night, the refractive index of air is smallest near the ground and increases with height form the ground. When a light beam is directed horizontally, the Huygen's principle leads us to conclude that as it travels, the light beam
(1) Goes horizontally without any deflection
(2) Bends downwards
(3) Bends upwards
(4) Becomes narrower
Answer (3)
Sol. Consider a plane wavefront travelling horizontally. As it moves, its different parts move with different speeds. So, its shape will change as shown
⇒⇒ Light bends upward
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