jee main test 9
Short Description
paper...
Description
Q.1
A ray ray of ligh lightt is refl reflec ecte tedd bytw by twoo mirr mirror orss plac placed ed norm normal al to each each othe otherr. The The inci incide dent nt ray ray make makess anan an angl glee of22 of 22°° with with one one of the the mirr mirror ors.At s.At what what angle does does the rayem ray emerg erge? e? (A) 22° (B) 68° (C) 44° (D) None [Geometrical Optics / PM / Moderate ]
Q.2 Q.2
The lig light sourc ourcee S is ove over the the cent center er of a circ circul ulaar opaq opaque ue pla plate of radi radius us 1m at a dis distanc tancee a = 1 m from from it.Thedistancefromtheplatetothescreenisb=0.8m.Findthediameteroftheshadowoftheplateon the screen. screen. S a 2r b d
(A) 1.8 m
(B) 2.9 m
(C) 3.6 m
(D) 5.4 m [Geo [G eome metr tric ical al Opt Optic icss / PM / Easy Easy]]
Q.3
A conc concav avee mirr mirror or of radi radius us of curv curvat atur uree 40 cmfo cm form rmss an imag imagee of an obje object ct plac placed ed on the the prin princi cipa pall axis axis at a dist distan ance ce 45 cm in fron frontt of it. it. Now Now if the the syst system em is comp comple lete telyimm lyimmer erse sedd in wate waterr ( = 1.33 1.33)) then then (A) (A) the the image imagewi will ll shif shiftt towa toward rdss the the mirr mirror or (B)the (B) themag magnifi nificat cation ionwil willl reduce reduce (C)the (C) theima image gewil willl shift shift away awayfro from m themir the mirror roran andd magnif magnifica icatio tionn will will increa increase se.. (D)the (D) thepos positi ition on of theima the image geand andmag magnif nifica icatio tionn will will notcha not chang nge. e. [Geo [G eome metr tric ical al Opt Optic icss / SM /Eas /Easy] y]
Q.4
The The co-o co-ord rdin inat ates esof ofth thee imag imagee ofpo of poin intt obje object ctP P form formed edbya bya conc concaave mirr mirror or ofra of radi dius usof of curv curvat atur uree 20cm 20cm (cons (conside iderr parax paraxia iall rays rayson only ly)) as show shownn in thefig the figur uree is y
O
(A) (1 (13.33 cm cm, –1 –1 cm cm) (C) (– ( –13.33 cm cm, +1 +1 cm cm)
P (40 cm, 3 cm) x
(B) (1 (13.33 cm cm, +1 +1 cm cm) (D) (– (–13.33 cm cm, ––11 cm cm) [Geo [G eome metr tric ical al Opt Optic icss / SM / Tough] ough]
XII Part Test-7
Page # 1
The The obse observ rver er'O 'O'' sees seesth thee dist distan anceAB ceABas asin infi fini nite tely lyla larrge. ge. If refr refrac acti tive vein inde dexx ofli of liqu quid idis is1 andthatofglass is 2, then then 1/2 is:
Q.5
(A) 2
(B) 1/2
(C) 4
(D) None of these [Geom [Ge ometr etrica icall Opt Optic icss / RCS RCS / Tough] ough]
Q.6
Whic Whichh of thes thesee acti action onss will will move moveth thee real real imag imagee poin pointt P'fa P' fart rthe herr from from the the boun bounda dary ry?? n
Air
P •
C R
S
(1) (1) Decr Decrea ease se the the inde indexx of refr refrac acti tion on n. (2) Incr Increa ease se the distan distance ce S. (3) (3) Decr Decrea ease se thera the radi dius us of curv curvat atur uree R. (A) 1, 2, 3 (B) 1 only
•
P
S
(C) 2 & 3 only
(D) 2 only
[Geom [Ge ometr etrica icall Opt Optic icss / RCS RCS / Tough] ough]
Q.7
A rect rectan angu gula larr bloc blockk ofgl of glas asss ispl is plac acedon edon a prin printe tedd page pagely lyin ingg ona on a hori horizo zont ntal alsu surf rfac ace.The e.Themi mini nimu mum m valu valuee of the the refr refrac activ tivee index index of glas glasss for for whic whichh thele the lett tter erss on the the page page are are not not visib visible le from from any any of the the vert vertic ical al face facess of bloc blockk is (A) > 3 (B) > 2 (C) > 1.55 (D) > 1.38 [Geometric [Geometrical al Optics / RPS / Moderate] Moderate]
Q.8 Q.8
A ray ray inci inciddent at a poin oint B at an ang angle of inc incide idence nce ente enters rs into into a glas glasss sphe sphere re and and is refle reflecte ctedd and and refra refract cted ed at the fart farther her surfa surface ce of the the sphe sphere re,, as show shown. n. The The angl anglee betw betwee eenn there the refle flect cted ed and and refr refrac acte tedd rays raysat at this this surf surfac acee is 90°. 90°. If refr refrac acti tive vein inde dexx of mate materi rial al of sphe sphere re is 3 ,thevalueof is (A) /3 (C) /6
(B) /4 (D) /12 [Geom [Ge ometr etrica icall Opt Optic icss / RPS RPS / Tough] ough]
XII Part Test-7
Page # 2
Q.9 Q.9
A lig light ray ray hits hits the the pole pole of a thin thin bico biconv nveex lens lens as show shownn in figu figure re.. The The ang angle made made byth by thee emer emerge gent nt ray ray with with the the opti opticc axis axis will will be (A) 0° (B) (1/3)° (C) (2/3)° (D) 2° [Geometric [Geometrical al Optics / Lens / Moderate]
Q.10 Q.10 Choo Choose se the the corr correc ectt ray ray dia diagra gram of aneq an equi ui conve onvexx lens lens whic whichh is cut as show shownn
(A)
(B)
(C)
(D)
[Geometric [Geometrical al Optics / Lens / Moderate]
Q.11 Q.11
Cons Consid ider er the the four four diff differ eren entt case casess of disp disper ersi sion on of ligh lightt ray ray whic whichh has has all all the the wave wavele leng ngth thss from from1 to 2 (1 > 2). The dotted represents the light ray of wave length avg. Which ray diagram is showing maximumdispe maximum dispersive rsivepower power??
(A)
(B)
(C)
(D)
[Geom [G eomet etric rical al Opt Optics ics / Prism Prism / Easy] Easy]
Q.12 A ray ray ofli of ligh ghtt stri strike kess a plan planee mirr mirror orat atan anan angl glee of inci incide denc ncee 45º 45º as show shownn inth in thee figure.After figure.After reflection reflection,, the ray ray passes passes through through a prism prism of refracti refractive ve index index 1.5, whos whosee apex apex angl anglee is 4º.Th 4º. Thee angl anglee thro throug ughh whic whichh the the mirro mirrorr shou should ld be rota rotate tedd if thetotaldeviationoftherayistobe90ºis (A) 1° clockwise (B) 1°an °anticlockwise (C) 2° clockwise (D)2°a )2°annticlockwise [Geom [G eomet etric rical al Opt Optics ics / Prism Prism / Moder Moderate ate]]
XII Part Test-7
Page # 3
Q.13
Anast An astron ronomi omical caltel teles escop copee has has anang an angula ularr magnif magnifica icatio tionn ofmag of magnit nitude ude55 fordis for distan tantt object objects. s.The Thesep separa aratio tionn betwe between en the object objective ivean andd the eyepi eyepiece eceis is 36 cm.The cm. Thefin final al image image is formed formed at infinit infinityy. Thefoc The focal al length lengthss f o of the the obje object ctiv ivee and and f e of the the eyep eyepie iece ce are are (A) (A) 45 cm and and –9 cm resp respec ecti tive vely ly (B) (B) 50 cm and and 10 cm resp respec ecti tive vely ly (C) (C) 7.2 7.2 cm and and 5 cm resp respec ecti tive vely ly (D) (D) 30 cm and and 6 cm resp respec ecti tive vely ly [Geom [Ge omet etric rical al Opt Optics ics / Opt Optics ics Instru Instrume ments nts / Easy] Easy]
Q.14 Ina In a compo compound undmic micro rosco scope pe,, maxim maximum um magnif magnifica icatio tionn is obtai obtained nedwh when en thefin the final al image image (A)isformedatinfinity (B) (B) is forme formedd at the the leas leastt dista distance nceof of disti distinc nctt visio visionn (C) coinc coincide idess with with theobj the objec ectt (D) (D) coinc coincide idess with with the the obje object ctiv ivee lens lens [Geom [Ge omet etric rical al Opt Optics ics / Opt Optics ics Instru Instrume ments nts / Easy] Easy]
Q.15 Yel Yello low w ligh lightt is used used in a sing single le slit slit diff diffra ract ctio ionn expe experi rime ment nt with with a slit slit of widt widthh 0.6 0.6 mm mm.. If yello ellow w ligh lightt is replac replaced ed byX-r by X-ray ays, s, then then the observ observed ed patter patternn will revea reveall (A) (A) that that thecentra thecentrall maxi maximu mum m is narro narrowe wer r (B) (B) more morenu numb mber er of fring fringes es (C) (C) less less numb number er of fring fringes es (D)no (D) no diffra diffracti ction on patter patternn [Wav [W avee Op Opti tics cs / Diffr Diffrac acti tion on / Easy Easy]]
Q.16 In a YDSE YDSEex expe peri rime ment nt =540nm,D=1m,d=1mm.Athinfilmispastedonupperslitandthecentral maxi maxima ma shif shifts ts to the the poin pointt just just in fron frontt of the the uppe upperr slit slit.. What What is the the path path diff differ eren encce at the the cent centre re of the the scre screen en ? (A) 540 nm (B) 270 nm (C) 500 nm (D) 810 nm [Wave [W ave Optics / Interference Interference / Moderate]
Q.17 White light light is incident incident normally normally on a film which which has has µ = 1.5 and thickness thickness of 5000 Å. The wavewaveleng length thss in visi visibl blee spec spectr trum um (400 (40000 Å-70 Å-7000 00 Å) for for whic whichh inte intens nsit ityy of refl reflec ecte tedd ligh lightt be maxim maximum um is ( A) 5 0 00 Å
( B) 3 0 00 Å
(C)
10000 Å 3
( D) 6 00 0 Å
[Wave Optics / Interference / Moderate]
XII Part Test-7
Page # 4
Q.18 Cross section ofa solar cell with reflective SiO coating (thickness,t) is shown. What isthe conditionfor least reflection fora givenwavelength of light in air.Assumethat light is incident normally ?
1
Air n
= 1.00
n
= 1.45
n
= 3.50
2
SiO
Si
(A) t =
(2n SiO )
(B) t =
(4n SiO )
t
(C) t =
(D) t =
2
4
[Wave Optics / Interference / Moderate]
Q.19 Unpolarized light travelsthrough 2 linear polarizers.What (minimum)angle should thesecond polarizer berelativeto thefirst polarizer (the differencebetween1 and 2)sothatthefinalintensityoflightis3/8th oftheoriginalvalue? (A)
3
(B)
6
(C)
(D)
4
2
[Wave Optics / Polarization / Easy]
Q.20 A young'sdoubleslit experiment isconducted in water (1)asshowninthe figure, and a glass plate of thickness t and refractive index 2 is placed in thepathofS2. Wavelength of light in water is . Find the magnitude of the phase difference between waves coming from S1 andS2 at 'O'.
2 2 (A) 1 t 1
1 2 (B) 1t 2
(C) ( 2 1 ) t
2
(D) ( 2 1) t
2
[Wave Optics / Tough]
Q.21 A sphere of brass released in a long liquid column attains a terminal speed v0. If the terminal speed attained bythe sphere of marble ofthe same radiusand releasedin the sameliquid is nv0,thenthevalue ofnwillbe. Given:Thespecific gravities of brass, marble and the liquidare8.5, 2.5 and0.8 respectively. (A)
5 17
(B)
17 77
(C)
11 31
(D)
17 5
[Fluid Mechanics / Viscosity / Moderate]
XII Part Test-7
Page # 5
Q.22 Asoapbubble 50 mm indiameter contains air ata pressure (in excessof atmospheric) of 2 bar. Find the surface tensionin the soapfilm. (A) 1.25 × 10 –2 N/m (B) 12.5 × 10 –2 N/m (C) 125 × 10 –2 N/m (D) 1250 N/m [Fluid Mechanics / Surface Tension / Easy]
Q.23 A spherical soap bubble is blown such that its radius increases at a constant rate. Which of following curves represents power required to increase surface energyof the bubble versus radius of drop.
(A)
(B)
(C)
(D)
[Fluid Mechanics / Surface Tension / Easy]
Q.24 The height risen by a liquid in a capillary tube having diameter of its bore as 1 mm is given by. (Take S = 0.075 N/m, density = 1 gm/cc, contact angle = 0°) (A) 6 cm (B) 3 cm (C) 2 cm (D) 4 cm [Fluid Mechanics / Surface Tension / Easy]
Q.25 The limbs of a glass U-tube are lowered into vesselsAand B,A containing water. Some air is pumped out through the top of the tube C.The liquids in the left hand limbA and the right hand limb B rise to heights of 10 cm and 12 cm respectively. The density of liquid B is :
(A) 0.75 gm/cm3
(B) 0.83 gm/cm3
(C) 1.2 gm/cm3
(D) 0.25 gm/cm3 [Fluid Mechanics / Statics / Easy]
XII Part Test-7
Page # 6
Q.26 Aniceberg isfloatinginocean. Whatfractionof its volumeisabove the water? (Given: density ofice = 900 kg/m3 and density of ocean water = 1030 kg/m3) (A)
90 103
(B)
13 103
(C)
10 103
(D)
1 103
[Fluid Mechanics / Statics / Moderate]
Q.27 The figureshows a conical container of half-apex angle 37o filledwith certain quantities of keroseneand water. The force exerted bythewater on the kerosene is approximately. (Take atmospheric pressure = 105 Pa) 5
P0 = 10 Pa
10 m
Kerosene sp.gr.= 0.8 Water sp.gr.= 1.0
8m
(A) 3 × 107 N
(B) 4 × 107 N
(C) 2 × 107 N
(D) 5 × 107 N [Fluid Mechanics / Statics / Tough]
Q.28 A solid sphereofmassM and radiusR iskeptona rough surface. The velocitiesofair (density)around 4R 2 kg , what isthe minimum the sphere are as shown in figure.Assuming R to be small and M = g value of coefficient of friction so that the sphere starts pure rolling? (Assume force due to pressure differenceis actingon centre of mass of the sphere)
m/s
m/s M
(A) 0.25
(B) 0.50
R
Horizontal (C) 0.75
(D) 1.0 [Fluid Mechanics / Dynamics / Tough]
XII Part Test-7
Page # 7
Q.29 Rank in order, from highest to lowest, the liquid heights ha to hc.Theairflowisfromlefttoright.The liquid columnsare not drawn to scale but pipe diameter is drawn to scale.
Direction of airflow
Suction pump
ha
(A)h b = hd > ha > h b
h
hc
(B) h b = hd > hc > ha
hd
(C) hd > hc > h b > ha (D) h b > hd > hc > ha [Fluid Mechanics / Dynamics / Moderate]
Q.30 Wateris pumpedthrough the hose shown below, from a lower level to an upper level. Compared to the water at point 1, the water at point 2: (A) has greater speed and greater pressure (B) has greater speed and less pressure (C) has less speed and less pressure (D) has less speed and greater pressure
[Fluid Mechanics / Dynamics / Easy]
XII Part Test-7
Page # 8
Q.31 Acurrent of 9.65A is placed for 3 hrbetween nickel cathode and Pt anode in 0.5 Lof a 2 M solution of Ni(NO3)2. The molarityof Ni2+ after electrolysis would be (A) 0.46M (B) 0.92 M (C) 1.08 M (D) none [Electrochemistry / Easy]
Q.32 The resistanceof 0.5 M solution of an electrolyte in a cell was found tobe 50. If the electrodesin the cell are 2.2 cm apart and have an area of 4.4 cm 2 then the molar conductivity (inSm2 mol –1)ofthesolutionis (A) 0.2 (B) 0.02 (C) 0.002 (D) None of these [Electrochemistry / Easy]
Q.33 A graph was plotted between the molar conductance of variouselectrolytes (HCl,KCl and CH3COOH) and root of their concentrations in mole per litre.
Which of thefollowingis correct match? (A) I (CH3COOH) ; II (KCl) ; III (HCl) (B) I (HCl) ; II (KCl) ; III (CH3COOH) (C) I (CH3COOH) ; II (HCl) ; III (KCl) (D) I (KCl) ; II (CH3COOH) ; III (HCl) [Electrochemistry / Medium]
Q.34 Which of the following arrangement will not produce oxygen atanode during electrolysis ? (A)DiluteH2SO4 solution withCu electrodes. (B)DiluteH2SO4 solution with inert electrodes. (C) Fused NaOH with inert electrodes. (D)Dilute NaCl solution with inert electrodes. [Electrochemistry / Easy]
XII Part Test-7
Page # 9
Q.35 During dischargingof lead storage battery, which of the following is/are true ? (A)H2SO4 is produced (B)H2O is consumed (C) PbSO4 is formed at both electrodes (D) Densityof electrolytic solution increases [Electrochemistry / Easy]
Q.36 An aqueous solution of Na2SO4 was electrolysed for 10 min. 82 ml of a gas was produced at anode over water at 27°C at a total pressure of 580 torr. Determine the current that was used. (Vapour pressure of H2Oat27°C=10torr)(R=0.082atmlit./mol/K) (A) 0.1 Amp (B) 1.25 Amp (C) 2.23 Amp (D) 1.61 Amp [Electrochemistry / Medium]
Q.37 If E I
2
I
= + 0.50 V and E I
(A) 0.30 V
2
I 3
= + 0.20 V, then E I
(B) 0.65 V
3
I
is
(C) 1.30 V
(D) – 0.067 V [Electrochemistry / Medium]
Q.38 Choosetheincorrect statement (A) Cell constant values of conductivity cells are independent ofthesolution filled into the cell. (B) Kohlrausch lawis valid for strong electrolyte but notfor weak electrolyte (C)In general conductivity decreases on dilutionwhereas equivalent andmolar conductivityincrease on dilution. (D) Salt bridgeis employedtomaintaintheelectrical neutralityandtominimize the liquid- liquid junction potential. [Electrochemistry / Easy]
Q.39 The standard reduction potential’s of half cell OCl – /Cl – , OH – and Cl2/Cl – are 0.94 volt and +1.36 volt respectively. What is the reduction potentialof half cell whosecell reaction is represented as 2OCl – + 2H2O + 2e Cl2(g) + 4OH – (A) 0.21 V (B) 0.52 V (C) 1.04 V (D) 2.1 V [Electrochemistry / Medium]
Q.40 Consider the cell : Ag(s) |AgCl(s) |KCl(aq) (0.1M) | Hg2Cl2(s) | Hg(l ) | Pt The cellpotential : (A) becomes twice on increasing concentration of Cl – by10 times (B) becomes half on increasing concentration of Cl – by 10 times (C) remainsunaffected on increasing concentration of Cl – by10 times (D) gets affected bychange in amount ofAgCl(s) [Electrochemistry / Difficult] XII Part Test-7
Page # 10
Q.41 EMF of a cell is given by E = (1.05 10 4 T 2 ) V, where T is temperature in Kelvin. Which of the following optionsare correct w.r.t. thegalvanic cell. (A) H Rxn of the cell reaction will be temperatureindependent. (B)Thecell reaction involves increase in randomness. (C)At all thetemperatures, thecell reaction will be spontaneous. (D) The cell reaction will be non-spontaneous at T = 200K. [Electrochemistry / Medium]
Q.42 For the electrochemical cell : Zn(s) | Zn2+(aq)||Cl – (aq)|Cl2(g)|Pt(s)
Given : E Zn 2 / Zn = – 0.76 Volt, E Cl / Cl ( g ) = – 1.36 Volt 2
From these data one can deduce that : (A)Zn+Cl2 Zn2+ +2Cl – is a non-spontaneous reaction at standard conditions. (B) Zn2+ +2Cl – Cl2 + Zn is a spontaneous reaction at standard conditions with
E cell = 2.12 volt . (C)Zn+Cl2 Zn2+ +2Cl – is a spontaneous reaction at standard conditions with
E cell =2.12volt (D)Zn+Cl2 Zn2+ +2Cl – is a spontaneous reaction at standard conditions with
E cell =0.60volt [Electrochemistry / Medium]
Q.43 Calculate solubilityofAgBr in 0.1MKBr solutionfrom the followingcell potential data: Pt | H2(g) | H+(10 –3M) || KBr(10 –2M) | AgBr | Ag Ecell = 0.26 V 2.303 R 298 Given : E o , = 0.06 0 . 8 V Ag / Ag F (A) 10 –14 M
(B) 10 –15 M
(C) 10 –7 M
(D) 10 –13 M [Electrochemistry / Difficult]
Q.44 Statement-1: Standard reduction potential of hydrogen electrode is independent of temperature. Statement-2: Standard reduction potentialof all other electrodes excepthydrogen electrode depend upon temperature (A) Statement-1 is true, statement-2 is true andstatement-2 is correct explanation forstatement-1. (B) Statement-1istrue,statement-2is true andstatement-2 isNOTthecorrectexplanation forstatement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. [Electrochemistry / Easy] XII Part Test-7
Page # 11
Q.45 Statement-1: The equilibrium constant of a reaction occuring in concentration cell is always equalto unity. Statement-2: The equilibrium constantof a reactionoccuringin any galvaniccell depends onthe final activemassesof components present at equilibrium. (A) Statement-1 is true, statement-2 is true andstatement-2 is correct explanation forstatement-1. (B) Statement-1 istrue,statement-2istrue andstatement-2 isNOTthecorrectexplanation forstatement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. [Electrochemistry / Easy]
Q.46 Which of the followingcompound has aromatic character in it'senol form. O (A)
O
O
(B)
O
(C)
O
(D)
NH
O [Isomerism / Easy]
Q.47 Whichof following compounddoesnot show geometrical isomerism. (A)
(B)
(C) H2C = CH–CH=CH–CH=CH2
(D) [Isomerism / Easy]
Q.48 Write correct order of reactivityof followinghalogen derivatives.
(I)
(II) CH2=CH–Cl
(A) I > V > IV > III > II (C) V > I > IV > III> II
(III) Me3C–Cl
(IV) PhCH2Cl
(V) Ph3C–Cl
(B) V > IV > I > III > II (D) I > V > III > IV > II [GOC / Easy]
XII Part Test-7
Page # 12
Q.49 The correct order of basicities of the following compounds is (I) CH3 –
(II) CH3CH2 NH2
(III) (CH3)2 NH
O || (IV) CH3 C NH 2
(A) II > I > III > IV
(B) I > III > II > IV
(C) III > I > II > IV
(D) I > II > III > IV [GOC / Easy]
Q.50 In the following groups, –OCOCH 3 –OMe I II the order of leaving group ability is – (A) I > II > III > IV (B) IV > III > I > II
–OSO2Me III
–OSO2CF 3 IV
(C) III > II > I > IV
(D) II > III > IV > I [GOC / Easy]
Q.51 Which offollowing resonatingstructure is most stable? O–H (A)
O–H (B)
O–H
(C)
O–H
(D)
[GOC / Easy]
Q.52 Give the correct order of acidic strength of mentioned groups. (b)
OH
(a)
CCH
(c) (d)
(A) a > b > c > d
NH2 (B) b > c > a > d
OH (C) c > b > a > d
(D) b > a > c > d [GOC / Medium]
XII Part Test-7
Page # 13
Q.53 Correct order of HOC of the given compounds are (I) CH3 – CH2 – CH = CH – CH 2 – CH2 – CH3 (II)
CH3 CH3
(III)
(IV)
C = CH–CH2 –CH2 – CH3
CH 3 | CH 3 CH CH C CH3 | CH 3 CH3 CH3
C CH CH CH 3 | CH 3
(A) IV > I > II > III
(B) I > II > III > IV
(C) III > I > IV > II
(D) I > III > IV > II [GOC / Easy]
Q.54 In which of the following,replacement ofCl – is most difficult? (A)
(B)
(C)
(D) [Halogen Derivatives / Easy]
Q.55 Calculate the % of R-configuration in product, if 96% Racemisationtakes place for given opticallypure compound. Cl
HOH ?
(A) 48%
(B) 50%
(C) 52%
(D) 96% [Halogen Derivatives / Easy]
XII Part Test-7
Page # 14
Et
Q.56
H 3C
H3C I O¯Na P,,
D
P is Et
Et
(A) H3C
(B) H3C
D
OCH3 D
OCH3
Et
Et (C) H3CO
(D) D
CH3
OCH3 CH3
D
[Halogen Derivatives / Easy]
F Cl
OH A,
Q.57 O2N
A is
NO2 OH
F Cl
(A)
OH (B)
O2N
H2N
NO2
F
NO2
F OH
(C)
OH (D)
O2N
NO2
H2N
NH2 [Halogen Derivatives / Medium]
XII Part Test-7
Page # 15
Q.58 Which of the following can be used as laboratorytest for phenol? (A) It gives effervescene of CO2 with NaHCO3 (B) It gives purple colour with neutral FeCl3 (C) It gives yellow ppt. with NaOH I2 (D)Itgives silvermirror with Tollen'sreagent [Halogen Derivatives / Easy]
Q.59
alc. KOH
Br 2 X (major) Products. CH3 CH 2 CH CH3 CCl4 | Br
Last products of the reaction. (A) Racemic mixture (C) Diastereomers
(B) Meso compound (D) Optically active [Halogen Derivatives / Medium]
Q.60 Whichreaction is incorrect. (Only organicproducts are given) (A) H3C–O–CH2 –CH3 + PCl5 H3C –Cl + Et – Cl (B) H 3C C OH + PCl5 H 3C C Cl || || O O (C) H 3C C CH 3 + PCl5 H 3C CH CH 3 || | O Cl OH (D)
Cl + PCl5 [Halogen Derivatives / Easy]
XII Part Test-7
Page # 16
Q.61 If (k – 2)x2 + ky2 = 4 represents a rectangular hyperbola, then k equals (A) 0 (B) 1 (C) 2 (D) 3 [Hyperbola / Easy]
Q.62 The 4 th term from the end in the expansion of (2x -1/x2) 10 is– (A) 960 x –11 (B) 960 x –12 (C) –960 x –12 (D) –960 x –11 [Binomial Theorem / Moderate]
Q.63 The greatest term in the expansion of (2x + 7)10 , when x = 3 is(A) T5 (B) T 6 (C) T 7 (D) None of these [Binomial Theorem / Moderate]
Q.64 If in the expansion of (1+ y)n, the coefficient of 5 th , 6 th and 7 th terms are in A.P., then n is equal to(A) 7, 11 (B) 7, 14 (C) 8, 16 (D) None of these [Binomial Theorem / Easy]
Q.65 If the coefficients of rth and (r +1)th terms in the expansion of (3+7x)29 are equal, then r equals(A) 15 (B) 21 (C) 14 (D) None of these [Binomial Theorem / Easy]
Q.66 If (a, b) is the mid-point of chord passing through the vertex of the parabola y2 = 4x, then (A) a = 2b (B) 2a = b (C) a 2 = 2b (D) 2a = b 2 [Parabola/ Easy]
Q.67 The sum of the rational terms in the expansion of e (A) 40
(B) 41
(C) 42
2
31/ 5 j
10
is equal to (D) 0 [Binomial Theorem / Easy]
Q.68 If (1+ x)n = C 0 + C 1x + C 2 x 2 +....+ C n x n, then nn (A) (n 1)!
(B)
(n 1)n n!
(C)
(C0 C1 )(C1 C 2 )...(Cn 1 Cn ) equalsC1C 2...Cn nn n!
(D) None of these [Binomial Theorem / Moderate]
XII Part Test-7
Page # 17
Q.6 9 The complex number z having least positive argument which satisfy the condition |z – 25i | 15 is (A) 25i (B) 12 + 25i (C) 16 + 12i (D) 12 + 16i [Complex Number / Moderate]
Q.70 If |z + 2i| 1, then greatest and least value of |z– 3 + i| are(B) , 0
(A) 3, 1
(C) 1, 3
(D) None of these [Complex Number / Moderate]
Q. 71 If z1, z 2 , z 3 are 3 distinct complex numbers such that
then the value of (A) 0
9 z 2 z3
16
z 3 z1
(B) 3
25 z1 z 2
3 z 2 z3
=
4 z3 z1
=
5 z1 z 2
,
equals
(C) 4
(D) 5 [Complex Number / Moderate]
x 2 y2 = 1, is : Q.72 The value of m for which y = mx + 6 is a tangent to the hyperbola 100 49
(A)
17 20
(B)
20 17
(C)
3 20
(D)
20 3
[ Conic Section/ Easy]
Q.73 If z =
4
(1 +
(A) 1
1 i | z | i then equals i 1 i amp z
i) 4
(B)
(C) 3
(D) 4 [Complex Number / Tough]
Q.74 If the line x + y – 1 = 0 touches the parabola y2 = kx, then the value of k is (A) 4 (B) –4 (C) 2 (D) –2 [Parabola / Easy]
XII Part Test-7
Page # 18
Q.75 Coordinates of the focus of the parabola x2 – 4x – 8y – 4 = 0 are (A) (0, 2) (B) (2, 1) (C) (1, 2) (D) (–2, –1) [Parabola / Easy]
8
2 Q.76 The middle term of the expansion x is x (A) 560
(B) –560
(C) 1120
(D) –1120 [Binomial Theorem / Easy]
Q.77 Vertex of the parabola whose directrix is 3x + 4y – 5 = 0 and focus is (4, 5) is
119 , 71 50 25
(A)
119 , 71 50 25
(B)
71 , 119 25 50
(C)
(D) None of these [Parabola / Moderate]
Q.78 The angle between the tangents drawn from the origin to the parabola y2 = 4a(x – a) is (A) 90º
(B) 30º
1 (C) tan –1 2
(D) 45º [Parabola / Moderate]
Q.79 The co-efficient of x39 in the expansion of (x4 –1/x 3) 15 is– (A) 455 (B) – 455 (C) 105
(D) None of these
[Binomial Theorem / Moderate]
x 2 y2 = 1 are at right angles, then Q.80 If tangent drawn from a point 1, 2 3 to the ellipse 9 b2 value of b is (A) 1
(B) 4
(C) 2
(D) None of these [Ellipse / Easy]
XII Part Test-7
Page # 19
Q.81 Locus of all such points so that sum of its distances from (2, –3) and (2, 5) is always 10, is ( x 2)2 ( y 1) 2 (A) =1 25 9
( x 2)2 ( y 1)2 (B) =1 25 16
( x 2)2 ( y 1) 2 (C) =1 16 25
( x 2)2 ( y 1) 2 (D) =1 9 25 [Ellipse / Moderate]
Q.82 If the line 2x – 3y = k touches the parabola y2 = 6x, then the value of k is (A) 27/4 (B) –81/4 (C) – 7 (D) – 27/4 [Conic section / Easy]
Q.83 Tangents are drawn to the points of intersection of the line 7y – 4x = 10 and parabola y2 = 4x,.then the point of intersection of these tangents is 7 5
(A) 5 , 2
5 7
(B) 2 , 2
5
7 5 (D) 2 , 2
(C) 2 , 2 7
[Conic section / Easy]
Q.84 The equation 16x2 = 3y2 – 32x + 12y – 44= 0 represents a hyperbola (A) the length of whose transverse axis is 4 3 (B) the length of whose conjugate axis is 4 (C) whose centre is (– 1,2) (D) whose eccentricity is
19 3 [Conic section / Moderate]
Q.85 The locus of z which lies in shaded region is best represented by (A) z : |z + 1| > 2, |arg(z + 1)| 2, |arg(z – 1)| 0 v
n 1 1 R S > R S n 1
1 v
1 1 1 1 1 1 1 1 v R nR nS R n nS
Q.7
For no emergence > cosec A/2 1 > sin 90 / 2 >
2
+ 90° + = 180°
Q.8
= 90 – 1sin = 3 sin (90 – ) =60° Q.9
Refractionatfirst surface × 2° = 2 × r r = 1° Refraction at second surface × 1° = 3 × r'
2 r' = 3
Ans.
Q.10 Focal length ofplanoconvex > focal length of equiconvex lens.
V R Q.11 = 1
– 1 = 0 for B, so B is showing maximum dispersive power.
XII Part Test-7
Page # 24
45°
Q.12
1 = 2° 2 total deviation = 90° (due to reflection) + 2° (due to prism) = 92° but net deviation should be 90° due to reflection = 88° = – 2i i = 46° i.e. mirror mut lurotated by1° anticlockwise.
= A(H – 1) = 4 ×
f o 5 & Q.13 Fe f o = 30 cm f e = 6 cm m
f o + f e = 36 = L
Q.14 Information based. Q.15 sin =
a
or
Q.16 Infrontofupper slint
d / 2 – (µ – 1)t = 0 On screen = x = d D ( d / 2) – (µ – 1) t = 0 D at centre on the screen
x = d
d2 x = (µ – 1)t = 2D
Q.17 2µt – 2 = n
n 1 2µt = 2 4µt 30000 Å = = 2n 1 2n 1
XII Part Test-7
Page # 25
Phase change by due to reflection
Q.18
Air SiO Phase change by due to reflection
Ultimately no phase change
SiO Si
Optical path diff. = 2nSiOt =
Q.19
2
3 I I0 = I = 0 cos2 8 2
Q.20 x = 1(S2O – t) + 2t – 1S1O as S2O = S1O x = (2 – 1) t 2 = (2 – 1)t & air
2 2
= 1 t 1
Q.21 v0 =
air water = 1
2 2 (8.5 0.8)g r g
2 2 (2.5 0.8)g r g n = 17/77 nv0 =
4T R P = 2 × 105 N/m2
Q.22 P – P0 =
Q.23 Surface energy = (8r 2)T d dr (8r 2T ) = 8T(2r ) dt dt p r
p =
2S cos 4S cos 4 0.075 0. 3 Q.24 R = = = 3 = m = 3 cm r g dg 10 103 10 10
XII Part Test-7
Page # 26
Q.25 Pressure at p & that at q are equal as they are both equal to atmospheric pressure. 10 100 equating pw10 = p2 × 12 p2 = 0.83 g/cm3 12 Q p f p 2 g 100 P p f p g
Q.26 LetV be the volume oficeberg and let x bethe fraction of volume above water. Using law of floatation, weight offloatingbody=weight ofliquiddisplaced bypart ofthefloatingbodyinsidetheliquid. Therefore, Viceg = ( 1 – x ) Vwater g. Using the value of ice and water , we get x = (13/103). Q.27
r 3 tan 37 = 8 4 r=6m F = (P0 + hg) r 2 = (105 + 10 × 800 × 10) × × 36 1.8 × 36 × × 105 = 2 × 107
• r 37°
N
1 7R 2 2 2 2 Q.28 Force due to pressure difference is F = R (v2 – v1 ) = 2 2 Now, F – f = Ma
M
2 a fR = MR 2 × 5 R 2 f = Ma 5
=
R 2 Mg
2 f = F = R 2 7
Mg
= R 2
f
F
Mg
1 4
0.25
Q.29 h b > hd > hc > ha. The liquid level is higher where the pressure is lower. The pressure is lower where the flow speed is higher. The flow speed is highest in the narrowest tube, zero in the open air. Q.30 A1V1 = A2V2 V2 > V1 1 eV22 + gh = P1 + eV12 + 0 2 P2 < P1
P2 +
XII Part Test-7
Page # 27
SOLUTION
Q.31
W 9.65 3 3600 W = = 1.08 ; = 0.54 E 96500 M
[Ni2+] =
2 0.5 0.54 = 0.92 M 0.5
Q.32 conductivity, K =
1 2.2 × = 0.01 –1cm –1 50 4.4
molar conductivity,
1000
1000
^m = K × C = 0.01 × 0.5 = 20 –1cm2 mol –1 = 0.002 –1m2 mol –1
Q.33 Molar conductanceof HCl will bemaximum and CH3COOHwillbeminimum because (i) [H+] have exceptional ionic mobility (ii) CH3COOH is a weak electrolyte
Q.36
1 1 O2 + H+ + e¯ H2O 2 4 PV = = RT
n O2
(580 10) 82 10 3 760 2.5 103 0.082 300
q n e .F n O2 4 96500 i= = = = 1.61 Amp 10 60 t t
Q.37 (1) (2)
3 I 2 + 3e – 3I – 2
e – +
1 3 3 I 2 I3 2
—————————— 1 3 – (3) 2e + I 3I – 3 —————————— (3) = (1) – (2) G3 = G1 – G2 –2FE3 = (–3FE1) – (–1 × FE2) E3 =
3E1 E 2 3 0.5 0.2 = = 0.65 V 2 2
XII Part Test-7
Page # 28
Q.38 Kohlrausch Law is valid for strong as well as weak electrolyte.
G1o = – 2 × F × (–1.36)
Q.39 2Cl – Cl2 + 2e – 2OCl – + 2H2O + 4e – 2Cl – + 4OH – ————————————————
G o2 = – 4 × F × (0.94)
2OCl – + 2H2O + 2e – Cl2(g) + 4OH –
G 3o = – 2 × F × E3o
– 2 × F × E3o = – 2 × F × (– 1.36) – 4F(0.94) E3o = – 1.36 + 2 × 0.94 = 0.52 V Ans. Q.40 Ag(s) | AgCl(s) | KCl(aq) (0.1M) | Hg2Cl2(s) | Hg(l ) | Pt Cell can be represented as Ag|Ag+(C1M)||Hg2+2(C2M)|Hg(l ) |Pt(s) Ksp Ksp C 2 Hg 2 [Cl ] [Cl ]2 (Ag Ag+(an) + e – ) × 2 Anode : Cathode: Hg2+2 (aq) + 2e – 2Hg(l ) ––––––––––––––––––––––––––––––––– Overall Rxn: 2Ag(s) + Hg+22(aq) 2Ag+(aq) + 2Hg(l ) ––––––––––––––––––––––––––––––––– C Ag
[Ag ]2 Q= [Hg2 2 ] 2 k sp AgCl
2 [ Cl Q= 2 ] k spHg Cl
2 2
2 K spAgCl = K spHg2Cl2
= constant
2
[Cl ]
If concentrationof Cl – will change the Ecell will remain unaffected. Q.41 E = (– 1.05 + 10 –4 T2)V E (A) T 0 104 2T P
E H° = – nFE° + nFT T P H° will depends on temperature (B) (C)
E nF S = T > 0 Thus entropy will increase. P G = – nFE as E = – 1.05 + 10 –4 T2 It can be less then zero at certain temperature so reaction will not be spontanous at all temperature. XII Part Test-7
Page # 29
(D)
At 200 K E = – 1.05 + 10 –4 × (200K)2 = – 1.05 + 0.02 E = – 1.03V thus reaction is non-spontanous.
Q.42 ECell = E Zn 2 / Zn + E Cl / Cl2 ( g )
= 0.76 + 1.36 = 2.12 V 1 H 2 2
Q.43 Anode :
H+
+
10 –3 M AgBr(s)
Cathode :
e –
Ecell = E cell –
0.06 log (10 –2 × 10 –3) 1
+
e –
Ag
Br – (aq) 10 –2 M
+
E cell = E Ag,AgBr ,Br E H / H2
= 0.80 + 0.06 log Ksp – 0 0.26 = 0.80 + 0.06log Ksp – 0.06 log 10 –5 0.26 – 0.80 – 0.3 = 0.06 log Ksp – 0.84 = 0.06 log Ksp Ksp = 10 –14 Let in 0.1M KBr solubilityofAgBr be s s(s + 0.1) = 10 –14 ; s × 0.1 = 10 –14 s = 10 –13M
O
OH
O
Q.46 (A)
do not tautomerize
(B)
Non Aromatic
O
(C)
O
OH (D)
O
OH 10 e Aromatic
XII Part Test-7
OH
O
N
NH
4 e Anti-aromatic
Page # 30
Q.47 (B)
allene willnot show geometricalisomerism.
(D)
it cannotshowgeometricalisomerism Q.49 I conjugate acid has two equivalent R–S II 2° amine III 1° amine IV amide less basic Q.50 Strong base are poor leaving groups
O–H
Q.51
octet complete structure
Q.52 c > b > a > d c (1° - alcohol) b (2° - alcohol) a (Triplybounded carbon is sp-hybridised) b (Nitrogen atom is sp3 hy) Q.53 For Hydrocarbons containing same C 1 HOC stability Cl
HO
Q.55
HO
+
Racemicmixture 96 %
R + S 48% 48% Remaining4%willbeRonly Total R = 48 % + 4% = 52 % Ans.
XII Part Test-7
Page # 31
Et
Et
Q.56
O
CH3
CH3–I S N2
H3C
OCH3 D
D
F
F Cl
OH
OH
Q.57 O2N
Cl
NO2
O2N
NO2
Q.58 Phenol gives a characteristicpurple colour with neutral FeCl3 Sinceitisanenol. OH Q.59
H
CH3 –CH=CH–CH3 (Cis/ trans)
CH3–CH–CH–CH 3 Br
Trans is major Br2/CCl4
H3C
H
Br
Br CH3
H
Q.60
CH 3 H H
Br Br CH 3
Antiaddition at trans obtained erythreo product (P.O.S., Meso compound)
Cl | H 3C C CH 3 + PCl5 CH 3 C CH 3 || | O Cl
XII Part Test-7
Page # 32
SOLUTION
Q.61 For rectangular hyperbola 0, (k – 2) k 0 k 2, k 0 h 2 > ab k(k – 2) < 0 Also coff.of x 2 + coff. of y2 = 0 k–2+k=0 k=1
0
View more...
Comments
