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XII Part Test-7 Page # 1 Q.1 A ray of light is reflected by two mirrors placed normal to each other. The incident ray makes an angle of 22° with one of the mirrors. At what angle does the ray emerge? (A) 22° (B) 68° (C) 44° (D) None [Geometrical Optics / PM / Moderate ] Q.2 The light source S is over the center of a circular opaque plate of radius 1m at a distance a = 1 m from it. The distance from the plate to the screen is b = 0.8 m. Find the diameter of the shadow of the plate on the screen. S 2r d a b (A) 1.8 m (B) 2.9 m (C) 3.6 m (D) 5.4 m [Geometrical Optics / PM / Easy] Q.3 A concave mirror of radius of curvature 40 cm forms an image of an object placed on the principal axis at a distance 45 cm in front of it. Now if the system is completely immersed in water ( = 1.33) then (A) the image will shift towards the mirror (B) the magnification will reduce (C) the image will shift away from the mirror and magnification will increase. (D) the position of the image and magnification will not change. [Geometrical Optics / SM /Easy] Q.4 The co-ordinates of the image of point object P formed by a concave mirror of radius of curvature 20cm (consider paraxial rays only) as shown in the figure is O y P x (40 cm, 3 cm) (A) (13.33 cm, –1 cm) (B) (13.33 cm, +1 cm) (C) (–13.33 cm, +1 cm) (D) (–13.33 cm, –1 cm) [Geometrical Optics / SM / Tough]
Transcript
  • XII Part Test-7 Page # 1

    Q.1 Aray of light is reflected by two mirrors placed normal to each other.The incident raymakes an angle of 22 with one of the mirrors.At whatangle does the ray emerge?(A) 22 (B) 68(C) 44 (D) None

    [Geometrical Optics / PM / Moderate ]

    Q.2 The light source S is over the center of a circular opaque plate of radius 1m at a distance a = 1 m fromit. The distance from the plate to the screen is b = 0.8 m. Find the diameter of the shadow of the plate onthe screen.

    S

    2r

    d

    a

    b

    (A) 1.8 m (B) 2.9 m (C) 3.6 m (D) 5.4 m[Geometrical Optics / PM / Easy]

    Q.3 A concave mirror of radius of curvature 40 cm forms an image of an object placed on the principal axisat a distance 45 cm in front of it. Now if the system is completely immersed in water (= 1.33) then(A) the image will shift towards the mirror(B) themagnification will reduce(C) the image will shift awayfrom the mirror and magnification will increase.(D) the position of the image and magnification will not change.

    [Geometrical Optics / SM /Easy]

    Q.4 The co-ordinates of the image of point object P formed bya concave mirror of radius of curvature 20cm(consider paraxial rays only) as shown in the figure is

    O

    y P

    x(40 cm, 3 cm)

    (A) (13.33 cm, 1 cm) (B) (13.33 cm, +1 cm)(C) (13.33 cm, +1 cm) (D) (13.33 cm, 1 cm)

    [Geometrical Optics / SM / Tough]

  • XII Part Test-7 Page # 2

    Q.5 The observer 'O' sees the distanceAB as infinitely large. If refractive index of liquid is1 and that of glassis 2, then1/2 is:

    (A) 2 (B) 1/2 (C) 4 (D) None of these[Geometrical Optics / RCS / Tough]

    Q.6 Which of these actions will move the real image point P' farther from the boundary?

    RC PP

    S S

    nAir

    (1) Decrease the index of refraction n.(2) Increase the distance S.(3) Decrease the radius of curvature R.(A) 1, 2, 3 (B) 1 only (C) 2 & 3 only (D) 2 only

    [Geometrical Optics / RCS / Tough]

    Q.7 A rectangular blockof glass is placedon a printed page lyingon a horizontal surface.The minimum valueof the refractive index of glass for which the letters on the page are not visible from any of the verticalfaces of block is(A) >3 (B) >2 (C) > 1.55 (D) > 1.38

    [Geometrical Optics / RPS / Moderate]

    Q.8 A ray incident at a point B at an angle of incidence enters into aglass sphere and is reflected and refracted at the farther surface ofthe sphere, as shown. The angle between the reflected and refractedrays at this surface is 90. If refractive index of material of sphere is

    3 , the value of is

    (A) /3 (B) /4(C) /6 (D) /12

    [Geometrical Optics / RPS / Tough]

  • XII Part Test-7 Page # 3

    Q.9 A light ray hits the pole of a thin biconvex lens as shown in figure. The anglemade by the emergent raywith the optic axis will be(A) 0 (B) (1/3) (C) (2/3) (D) 2

    [Geometrical Optics / Lens / Moderate]

    Q.10 Choose the correct ray diagram of an equi convex lens which is cut as shown

    (A) (B)

    (C) (D)

    [Geometrical Optics / Lens / Moderate]

    Q.11 Consider the four different cases of dispersion of light ray which has all the wave lengths from1 to2(1 > 2). The dotted represents the light ray of wave length avg. Which ray diagram is showingmaximum dispersive power?

    (A) (B) (C) (D)

    [Geometrical Optics / Prism / Easy]

    Q.12 A rayof light strikes a plane mirror at an angle of incidence 45 as shown in thefigure.After reflection, the ray passes through a prism of refractive index 1.5,whose apex angle is 4. The angle through which the mirror should be rotated ifthe total deviation of the ray is to be 90 is(A) 1 clockwise (B) 1 anticlockwise(C) 2 clockwise (D) 2 anticlockwise

    [Geometrical Optics / Prism / Moderate]

  • XII Part Test-7 Page # 4

    Q.13 Anastronomical telescopehas anangular magnificationofmagnitude 5fordistantobjects.Theseparationbetween the objective and the eyepiece is 36 cm. The final image is formed at infinity.The focal lengthsfo of the objective and fe of the eyepiece are(A) 45 cm and 9 cm respectively(B) 50 cm and 10 cm respectively(C) 7.2 cm and 5 cm respectively(D) 30 cm and 6 cm respectively

    [Geometrical Optics / Optics Instruments / Easy]

    Q.14 In a compound microscope, maximum magnification is obtained when the final image(A) is formed at infinity(B) is formed at the least distance of distinct vision(C) coincides with the object(D) coincides with the objective lens

    [Geometrical Optics / Optics Instruments / Easy]

    Q.15 Yellow light is used in a single slit diffraction experiment with a slit of width 0.6 mm. If yellow light isreplaced byX-rays, then the observed pattern will reveal(A) that the central maximum is narrower(B) more number of fringes(C) less number of fringes(D) no diffraction pattern

    [Wave Optics / Diffraction / Easy]

    Q.16 In aYDSE experiment = 540 nm, D = 1m, d = 1mm.Athin film is pasted on upper slit and the centralmaxima shifts to the point just in front of the upper slit. What is the path difference at the centre of thescreen ?(A) 540 nm (B) 270 nm (C) 500 nm (D) 810 nm

    [Wave Optics / Interference / Moderate]

    Q.17 White light is incident normally on a film which has = 1.5 and thickness of 5000 . The wave-lengths in visible spectrum (4000 -7000 ) for which intensity of reflected light be maximum is

    (A) 5000 (B) 3000 (C) 310000

    (D) 6000

    [Wave Optics / Interference / Moderate]

  • XII Part Test-7 Page # 5

    Q.18 Cross section of a solar cell with reflective SiO coating (thickness, t) is shown. What is the condition forleast reflection for a given wavelength of lightin air.Assume that light is incident normally ?

    tSiO

    Air

    Si

    n = 1.45

    n = 1.00

    n = 3.50

    1 2

    (A) t =)n2( SiO

    (B) t =

    )n4( SiO

    (C) t = 2

    (D) t = 4

    [Wave Optics / Interference / Moderate]

    Q.19 Unpolarized light travels through 2 linear polarizers.What (minimum) angle should the second polarizerbe relative to the first polarizer (the difference between1 and2) so that the final intensityof light is 3/8

    th

    of theoriginal value?

    (A) 3

    (B) 6

    (C) 4

    (D) 2

    [Wave Optics / Polarization / Easy]

    Q.20 A young's doubleslit experiment is conducted in water (1) as shown in thefigure, and a glass plate of thickness t and refractive index 2 is placed inthe path of S2. Wavelength of light in water is. Find the magnitude of thephase difference between waves coming from S1 and S2 at 'O'.

    (A)

    2t11

    2 (B)

    2t12

    1 (C)

    2t)( 12 (D)

    2t)1( 2

    [Wave Optics / Tough]

    Q.21 A sphere of brass released in a long liquid column attains a terminal speed v0. If the terminal speedattained by the sphere of marble of the same radius and released in the same liquid is nv0, then the valueof n will be.Given: The specific gravities of brass, marble and the liquid are 8.5, 2.5 and 0.8 respectively.

    (A) 175

    (B) 7717

    (C) 3111

    (D) 517

    [Fluid Mechanics / Viscosity / Moderate]

  • XII Part Test-7 Page # 6

    Q.22 Asoap bubble 50 mm in diameter contains air at a pressure (in excess of atmospheric) of 2 bar. Find thesurface tension in the soap film.(A) 1.25 102 N/m (B) 12.5 102 N/m(C) 125 102 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 followingcurves represents power required to increase surface energy of 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,Acontaining water. Some air is pumpedout through the top of the tube C. The liquids in the left hand limbAand the right hand limb B rise toheights 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 # 7

    Q.26 An iceberg is floating in ocean. What fraction of its volume is above the water ? (Given : densityof ice =900 kg/m3 and density of ocean water = 1030 kg/m3)

    (A) 10390

    (B) 10313

    (C) 10310

    (D) 1031

    [Fluid Mechanics / Statics / Moderate]

    Q.27 The figure shows a conical container of half-apex angle 37ofilled with certain quantities of kerosene andwater. The force exerted by the water on the kerosene is approximately.(Take atmospheric pressure = 105 Pa)

    8 m

    10 mKerosenesp.gr.= 0.8

    Watersp.gr.= 1.0

    P = 10 Pa05

    (A) 3 107 N (B) 4 107 N (C) 2 107 N (D) 5 107 N[Fluid Mechanics / Statics / Tough]

    Q.28 A solid sphere of mass M and radius R is kept on a rough surface. The velocities of air (density) around

    the sphere are as shown in figure.Assuming R to be small and M = kggR4 2

    , what is the minimum

    value of coefficient of friction so that the sphere starts pure rolling? (Assume force due to pressuredifference is acting on centre of mass of the sphere)

    Horizontal

    RM

    m/sm/s

    (A) 0.25 (B) 0.50 (C) 0.75 (D) 1.0[Fluid Mechanics / Dynamics / Tough]

  • XII Part Test-7 Page # 8

    Q.29 Rank in order, from highest to lowest, the liquid heights ha to hc. The air flow is from left to right. Theliquid columns are not drawn to scale but pipe diameter is drawn to scale.

    Suctionpump

    Direction of airflow

    ha hb hc hd

    (A) hb = hd > ha > hb (B) hb = hd > hc > ha (C) hd > hc > hb > ha (D) hb > hd > hc > ha[Fluid Mechanics / Dynamics / Moderate]

    Q.30 Water is pumped through the hose shown below, from a lower level to an upper level. Compared to thewater 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 # 9

    Q.31 Acurrent of 9.65Ais placed for 3 hr between nickel cathode and Pt anode in 0.5 Lof a 2 M solution ofNi(NO3)2. The molarity of Ni

    2+ after electrolysis would be(A) 0.46M (B) 0.92 M (C) 1.08 M (D) none

    [Electrochemistry / Easy]

    Q.32 The resistance of 0.5 M solution of an electrolyte in a cell was found to be 50. If the electrodes in thecell are 2.2 cm apart and have an area of 4.4 cm2 then the molar conductivity(in S m2 mol1) of the solution is(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 various electrolytes (HCl, KCl and CH3COOH)and root of their concentrations in mole per litre.

    Which of the following is 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 at anode during electrolysis ?(A) Dilute H2SO4 solution with Cu electrodes.(B) Dilute H2SO4 solution with inert electrodes.(C) Fused NaOH with inert electrodes.(D) Dilute NaCl solution with inert electrodes.

    [Electrochemistry / Easy]

  • XII Part Test-7 Page # 10

    Q.35 During discharging of 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 anodeover water at 27C at a total pressure of 580 torr. Determine the current that was used.(Vapour pressure of H2O at 27C = 10 torr) (R = 0.082 atm lit./ mol / K)(A) 0.1Amp (B) 1.25Amp (C) 2.23Amp (D) 1.61Amp

    [Electrochemistry / Medium]

    Q.37 If II2E = + 0.50 V and 32 IIE = + 0.20 V, then II3E is

    (A) 0.30 V (B) 0.65 V (C) 1.30 V (D) 0.067 V[Electrochemistry / Medium]

    Q.38 Choose the incorrect statement(A) Cell constant values of conductivitycells are independent of the solution filled into the cell.(B) Kohlrausch law is valid for strong electrolyte but not for weak electrolyte(C) In general conductivity decreases on dilution whereas equivalent and molar conductivityincrease ondilution.(D)Salt bridgeis employed tomaintain theelectrical neutralityand to minimize the liquid - liquid junctionpotential.

    [Electrochemistry / Easy]

    Q.39 The standard reduction potentials of half cell OCl/Cl, OH and Cl2/Cl are 0.94 volt and +1.36 volt

    respectively. What is the reduction potential of half cell whose cell reaction is represented as2OCl + 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) | PtThe cellpotential :(A) becomes twice on increasing concentration of Cl by10 times(B) becomes half on increasing concentration of Cl by10 times(C) remains unaffected on increasing concentration of Cl by10 times(D) gets affected by change in amount ofAgCl(s)

    [Electrochemistry / Difficult]

  • XII Part Test-7 Page # 11

    Q.41 EMF of a cell is given by E = )T1005.1( 24 V, where T is temperature in Kelvin. Which of the

    following options are correct w.r.t. the galvanic cell.

    (A) RxnH of the cell reaction will be temperature independent.

    (B) The cell reaction involves increase in randomness.(C)At all the temperatures, the cell 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 : Zn/Zn2E = 0.76 Volt,

    )g(Cl/Cl 2E = 1.36 Volt

    From these data one can deduce that :(A) Zn + Cl2 Zn

    2+ + 2Cl is a non-spontaneous reaction at standard conditions.(B) Zn2+ + 2Cl Cl2 + Zn is a spontaneous reaction at standard conditions with

    cellE = 2.12 volt .

    (C) Zn + Cl2 Zn2+ + 2Cl is a spontaneous reaction at standard conditions with

    cellE = 2.12 volt

    (D) Zn + Cl2 Zn2+ + 2Cl is a spontaneous reaction at standard conditions with

    cellE = 0.60 volt

    [Electrochemistry / Medium]

    Q.43 Calculate solubilityofAgBr in 0.1M KBr solution from the followingcell potential data:Pt | H2(g) | H

    +(103M) || KBr(102M) | AgBr | Ag Ecell = 0.26 V

    Given : V8.0EoAg/Ag

    , F298R303.2

    = 0.06

    (A) 1014 M (B) 1015 M (C) 107 M (D) 1013 M[Electrochemistry / Difficult]

    Q.44 Statement-1: Standard reduction potential of hydrogen electrode is independent of temperature.Statement-2: Standard reduction potential of all other electrodes except hydrogen electrode depend

    upon temperature(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is trueandstatement-2 isNOTthecorrectexplanation for statement-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 # 12

    Q.45 Statement-1: The equilibrium constant of a reaction occuring in concentration cell is alwaysequal to unity.

    Statement-2: The equilibrium constant of a reaction occuring in anygalvanic cell depends on the finalactive masses of components present at equilibrium.

    (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is trueandstatement-2 isNOTthecorrectexplanation for statement-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 following compound has aromatic character in it's enol form.

    (A)

    O

    O

    (B)O

    (C)

    O

    O

    (D)O

    NH

    [Isomerism / Easy]

    Q.47 Which of followingcompound doesnot show geometrical isomerism.

    (A) (B)

    (C) H2C = CHCH=CHCH=CH2 (D)

    [Isomerism / Easy]

    Q.48 Write correct order of reactivityof following halogen derivatives.

    (I) (II) CH2=CHCl (III) Me3CCl (IV) PhCH2Cl (V) Ph3CCl

    (A) I > V > IV > III > II (B) V > IV > I > III > II(C) V > I > IV > III> II (D) I > V > III > IV > II

    [GOC / Easy]

  • XII Part Test-7 Page # 13

    Q.49 The correct order of basicities of the following compounds is

    (I) CH3 (II) CH3CH2NH2

    (III) (CH3)2NH (IV) 23 NHCCH||

    O

    (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 followinggroups,OCOCH3 OMe OSO2Me OSO2CF3

    I II III IVthe order of leaving group ability is (A) I > II > III > IV (B) IV > III > I > II (C) III > II > I > IV (D) II > III > IV > I

    [GOC / Easy]

    Q.51 Which of following resonating structure is most stable?

    (A)

    OH

    (B)

    OH

    (C)

    OH

    (D)

    OH

    [GOC / Easy]

    Q.52 Give the correct order of acidic strength of mentioned groups.

    OH

    OHNH2

    C CH(a)

    (b)

    (c)(d)

    (A) a > b > c > d (B) b > c > a > d (C) c > b > a > d (D) b > a > c > d[GOC / Medium]

  • XII Part Test-7 Page # 14

    Q.53 Correct order of HOC of the given compounds are(I) CH3 CH2 CH = CH CH2 CH2 CH3

    (II)CH3

    CH3C= CHCH2CH2 CH3

    (III)

    3

    33

    3

    CH|

    CHCCHCHCH|

    CH

    (IV)CH3

    CH3C

    3

    3

    CH|

    CHCHCH

    (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 of Cl is most difficult?

    (A) (B) (C) (D)

    [Halogen Derivatives / Easy]

    Q.55 Calculate the % of R-configuration in product, if 96% Racemisation takes place for given opticallypurecompound. Cl

    HOH ?

    (A) 48% (B) 50% (C) 52% (D) 96%[Halogen Derivatives / Easy]

  • XII Part Test-7 Page # 15

    Q.56 H C3 ONa

    Et

    D

    ICH3 P,,

    P is

    (A) H C3

    OCH3

    Et

    D (B) H C3 OCH3

    Et

    D

    (C) H CO3 CH3

    Et

    D

    (D)

    CH3

    OCH3

    Et

    D

    [Halogen Derivatives / Easy]

    Q.57

    F

    O N2 NO2

    Cl HO A,A, A is

    (A)

    OH

    O N2 NO2

    Cl

    (B)

    OH

    H N2 NO2

    F

    (C)

    OH

    O N2 NO2

    F

    (D)OH

    H N2 NH2

    F

    [Halogen Derivatives / Medium]

  • XII Part Test-7 Page # 16

    Q.58 Which of the following can be used as laboratory test 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) It gives silver mirror withTollen's reagent

    [Halogen Derivatives / Easy]

    Q.59

    Br|

    CHCHCHCH 323 KOH.alc X (major)

    4

    2

    CClBr Products.

    Last products of the reaction.(A) Racemic mixture (B) Meso compound(C) Diastereomers (D) Opticallyactive

    [Halogen Derivatives / Medium]

    Q.60 Which reaction is incorrect. (Only organic products are given)(A) H3COCH2CH3 + PCl5 H3C Cl + Et Cl

    (B)

    O||

    OHCCH3 + PCl5

    O||

    ClCCH3

    (C)

    O||

    CHCCH 33 + PCl5

    Cl|

    CHCHCH 33

    (D)

    OH

    + PCl5

    Cl

    [Halogen Derivatives / Easy]

  • XII Part Test-7 Page # 17

    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 4th term from the end in the expansion of (2x -1/x2) 10 is(A) 960 x11 (B) 960 x12 (C) 960 x12 (D) 960 x11

    [Binomial Theorem / Moderate]

    Q.63 The greatest term in the expansion of (2x + 7)10, when x = 3 is-(A) T5 (B) T6 (C) T7 (D) None of these

    [Binomial Theorem / Moderate]

    Q.64 If in the expansion of (1+ y)n, the coefficient of 5th, 6th and 7th terms are in A.P., then n isequal 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 requals-(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) a2 = 2b (D) 2a = b2

    [Parabola/ Easy]

    Q.67 The sum of the rational terms in the expansion of 2 31 510

    /e j is equal to(A) 40 (B) 41 (C) 42 (D) 0

    [Binomial Theorem / Easy]

    Q.68 If (1+ x)n = C0+ C1x + C2x2 +....+ Cnx

    n, thenn21

    n1n2110C...CC

    )CC)...(CC)(CC( equals-

    (A) )!1n(nn

    (B) ( )

    !n

    n

    n1 (C) nn

    n

    !(D) None of these

    [Binomial Theorem / Moderate]

  • XII Part Test-7 Page # 18

    Q.69 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-

    (A) 3, 1 (B) , 0 (C) 1, 3 (D) None of these[Complex Number / Moderate]

    Q.71 If z1, z2, z3 are 3 distinct complex numbers such that32 zz

    3

    =13 zz

    4

    =21 zz

    5

    ,

    then the value of211332 zz

    25zz

    16zz

    9

    equals

    (A) 0 (B) 3 (C) 4 (D) 5[Complex Number / Moderate]

    Q.72 The value of m for which y = mx + 6 is a tangent to the hyperbola49y

    100x 22

    = 1, is :

    (A)2017

    (B)1720

    (C)203

    (D)320

    [ Conic Section/ Easy]

    Q.73 If z = 4

    (1 + i)4

    ii

    ii

    11

    then

    zamp|z|

    equals

    (A) 1 (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 # 19

    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]

    Q.76 The middle term of the expansion8

    x2x

    is-

    (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

    (A)

    2571,

    50119

    (B)

    2571,

    50119

    (C)

    50119,

    2571

    (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) 90 (B) 30 (C) tan1

    21

    (D) 45

    [Parabola / Moderate]

    Q.79 The co-efficient of x39 in the expansion of (x4 1/x3) 15 is(A) 455 (B) 455 (C) 105 (D) None of these

    [Binomial Theorem / Moderate]

    Q.80 If tangent drawn from a point 32,1 to the ellipse 222

    by

    9x

    = 1 are at right angles, then

    value of b is(A) 1 (B) 4 (C) 2 (D) None of these

    [Ellipse / Easy]

  • XII Part Test-7 Page # 20

    Q.81 Locus of all such points so that sum of its distances from (2, 3) and (2, 5) is always 10, is

    (A)9

    )1y(25

    )2x( 22

    = 1 (B)

    16)1y(

    25)2x( 22

    = 1

    (C)25

    )1y(16

    )2x( 22

    = 1 (D)

    25)1y(

    9)2x( 22

    = 1

    [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 parabolay2 = 4x,.then the point of intersection of these tangents is

    (A)

    25,

    57

    (B)

    27,

    25

    (C)

    27,

    25

    (D)

    25,

    27

    [Conic section / Easy]

    Q.84 The equation 16x2 = 3y2 32x + 12y 44= 0 represents a hyperbola

    (A) the length of whose transverse axis is 34

    (B) the length of whose conjugate axis is 4(C) whose centre is ( 1,2)

    (D) whose eccentricity is319

    [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)| < /4(B) z : |z - 1| > 2, |arg(z 1)| < /4(C) z : |z + 1| < 2, |arg(z + 1)| < /2(D) z : |z - 1| < 2, |arg(z - 1)| < /2

    [Complex Number / Easy]

  • XII Part Test-7 Page # 21

    Q.86 If n625 = I + f ; n , I N and 0 f < 1, then I equals

    (A) ff1 (B) ff1

    1

    (C) ff1

    1

    (D) ff1

    1

    [Binomial Theorem / Moderate]

    Q.87 2121 zzzz is possible if :

    (A) 12 zz (B) 12 z

    1z

    (C) arg(z1) = arg (z2) (D) 21 zz

    [Complex Number / Easy]

    Q.88 If a > 0, and the equation a2zaz 2 = 3 represents an ellipse, then a lies in

    (A) (1, 3) (B) )3,2( (C) (0, 3) (D) )3,1(

    [Complex Number/ Moderate]

    Q.89 If

    2sini21

    2cos

    2sinitan

    is purely imaginary then general value of is -

    (A) n + 4

    (B) 2n + 4

    (C) n + 2

    (D) 2n + 2

    [Complex Number / Moderate]

    Q.90 The locus of the middle point of the intercept of the tangents drawn from an external point tothe ellipse x2 + 2y2 = 2 between the co-ordinates axes, is

    (A) 1y21

    x1

    22 (B) 1y21

    x41

    22 (C) 1y41

    x21

    22 (D) 1y1

    x21

    22

    [Ellipse/ Easy]

  • XII Part Test-7 Page # 22

    ANSWER KEY

    PHYSICSQ.1 BQ.2 CQ.3 DQ.4 AQ.5 AQ.6 BQ.7 BQ.8 AQ.9 CQ.10 BQ.11 BQ.12 BQ.13 DQ.14 BQ.15 AQ.16 CQ.17 DQ.18 BQ.19 BQ.20 AQ.21 BQ.22 DQ.23 AQ.24 BQ.25 BQ.26 BQ.27 CQ.28 AQ.29 DQ.30 B

    CHEMISTRYQ.31 BQ.32 CQ.33 AQ.34 AQ.35 CQ.36 DQ.37 BQ.38 BQ.39 BQ.40 CQ.41 BQ.42 CQ.43 DQ.44 BQ.45 CQ.46 CQ.47 DQ.48 AQ.49 BQ.50 BQ.51 DQ.52 CQ.53 CQ.54 DQ.55 CQ.56 BQ.57 CQ.58 BQ.59 BQ.60 C

    MATHEMATICSQ.61 BQ.62 DQ.63 BQ.64 BQ.65 BQ.66 DQ.67 BQ.68 BQ.69 DQ.70 AQ.71 AQ.72 AQ.73 DQ.74 BQ.75 BQ.76 CQ.77 AQ.78 AQ.79 BQ.80 CQ.81 DQ.82 DQ.83 BQ.84 DQ.85 AQ.86 DQ.87 CQ.88 CQ.89 AQ.90 C

  • XII Part Test-7 Page # 23

    SOLUTION

    Q.1

    From a corner reflector, reflected ray is antiparallel to incident ray.

    Q.2 2/d1

    8.11

    d = 3.6 m

    Q.3 In mirrors focal length is independentof surrounding medium.

    Q.4 101

    401

    v1

    4014

    401

    101

    v1

    v = 340

    u = 40 ; f = 10

    v = fuuf

    = 10401040

    = 3

    40cm

    40340m

    3h I

    = 31

    hI = 1cm

    Q.5 Image of B must be at infinity

    A BLiquid

    12

    2 13

    3 u1 =

    1

    12

    R

    +2

    23

    R

    3 )R2(1

    = )R(

    12

    +

    23

    R21 = R

    12

    1 = 22 211 = 22

    2

    1

    = 2

  • XII Part Test-7 Page # 24

    Q.6 R1n

    S1

    vn

    S1

    R1n

    vn

    v > 0 S1

    R1n

    S > 1n

    R

    v1v

    nS1

    n11

    R1

    nS1

    nR1

    R1

    v1

    Q.7 For no emergence > cosec A/2

    > 2/90sin1

    > 2

    Q.8 + 90 + = 180

    = 90

    1 sin = 3 sin (90 ) = 60

    Q.9 Refraction at first surface 2 = 2 r

    r = 1

    Refraction at second surface 1 = 3 r'

    r' =

    32

    Ans.

    Q.10 Focal length of planoconvex > focal length of equiconvex lens.

    Q.11 = 1RV

    1 = 0 for B, so B is showing maximum dispersive power.

  • XII Part Test-7 Page # 25

    Q.12

    45

    = A(H 1) = 4 21

    = 2

    total deviation = 90 (due to reflection) + 2 (due to prism) = 92but net deviation should be 90 due to reflection = 88 = 2i i = 46i.e. mirror mut lurotated by1 anticlockwise.

    Q.13 5Ffm

    e

    o & fo + fe = 36 = L

    fo = 30 cmfe = 6 cm

    Q.14 Information based.

    Q.15 sin = a

    or

    Q.16 Infront of upper slint

    On screen = x = d

    D2/d

    ( 1)t = 0

    x = d D)2/d(

    ( 1) t = 0

    at centre on the screen

    x = ( 1)t =D2

    d2

    Q.17 2t 2 = n

    2t =

    21n

    1n2t4

    = = 1n230000

  • XII Part Test-7 Page # 26

    Q.18Air

    Phase change by due to reflection

    SiO

    SiOPhase change by due to reflection

    Si

    Ultimatelyno phasechange

    Optical path diff. = 2nSiOt = 2

    Q.19 83

    I0 = I = 2I0 cos2

    Q.20 x = 1(S2O t) + 2t 1S1Oas S2O = S1O x = (2 1) t

    =air

    2

    (2 1)t & water

    air

    = 1

    = t12

    1

    2

    Q.21 v0 = g)8.05.8(r

    g2 2

    nv0 = g)8.05.2(r

    g2 2

    n = 17/77

    Q.22 P P0 = RT4

    P = 2 105 N/m2

    Q.23 Surface energy = (8r2)T

    p = )Tr8(dtd 2 = 8T(2r dt

    dr)

    p r

    Q.24 R = grcosS2

    = gd

    cosS4

    =

    101010075.04

    33

    = 10

    3.0m = 3 cm

  • XII Part Test-7 Page # 27

    Q.25 Pressure at p & that at q are equal as they are both equal to atmospheric pressure.

    10012gppQ

    10010gppP

    2f

    f equating pw10 = p2 12 p2 = 0.83 g/cm3

    Q.26 Let V be the volume of iceberg and let x be the fraction of volume above water. Using law of floatation,weightof floatingbody=weightof liquiddisplacedbypartofthe floatingbodyinsidethe liquid.Therefore,Viceg = (1 x) Vwaterg. Using the value of ice and water, we get x = (13/103).

    Q.27 37tan8r

    = 43

    r = 6mF = (P0 + hg) r

    2

    r

    37= (105 + 10 800 10) 361.8 36 105= 2 107

    Q.28 Force due to pressure difference is F = 21R2 (v2

    2 v12) =

    2R7 2

    Now, F f = Ma

    fR = 52

    MR2 Ra

    M F

    Mgf

    N

    f = 52

    Ma f = 72

    F = R2 Mg = R2

    = 25.041

    MgR2

    Q.29 hb > hd > hc > ha. The liquid level is higher where the pressure is lower. The pressure is lower where theflow speed is higher. The flow speed is highest in the narrowest tube, zero in the open air.

    Q.30 A1V1 = A2V2 V2 > V1

    P2 + 21

    eV22 + gh = P1 + eV1

    2 + 0

    P2 < P1

  • XII Part Test-7 Page # 28

    SOLUTION

    Q.31 EW

    = 965003600365.9

    = 1.08 ; MW

    = 0.54

    [Ni2+] = 5.054.05.02

    = 0.92 M

    Q.32 conductivity, K = 501

    4.42.2

    = 0.011cm1

    molar conductivity, ^m = K C1000

    = 0.01 5.01000

    = 201cm2 mol1 = 0.0021m2 mol1

    Q.33 Molar conductance of HCl will be maximum and CH3COOH will be minimumbecause (i) [H+] haveexceptional ionic mobility

    (ii) CH3COOH is a weak electrolyte

    Q.36 21

    H2O 41

    O2 + H+ + e

    2On = RTPV

    = 33

    105.2300082.0

    1082760

    )10580(

    i = tq

    = tF.ne = 6010

    965004n2O

    = 1.61 AmpAmp

    Q.37 (1) 2I23

    + 3e 3I

    (2) e + 2I23

    3I31

    (3) 2e +

    3I31 3I

    (3) = (1) (2)G3 = G1 G22FE3 = (3FE1) (1 FE2)

    E3 = 2EE3 21 = 2

    2.05.03 = 0.65 V

  • XII Part Test-7 Page # 29

    Q.38 Kohlrausch Law is valid for strong as well as weak electrolyte.

    Q.39 2Cl Cl2 + 2e o1G = 2 F (1.36)

    2OCl + 2H2O + 4e 2Cl + 4OH o2G = 4 F (0.94)

    2OCl + 2H2O + 2e Cl2(g) + 4OH

    o3G = 2 F o3E

    2 F o3E = 2 F ( 1.36) 4F(0.94)

    o3E = 1.36 + 2 0.94 = 0.52 V Ans.

    Q.40 Ag(s) | AgCl(s) | KCl(aq) (0.1M) | Hg2Cl2(s) | Hg(l) | PtCell can be represented as

    Ag |Ag+(C1M) || Hg2+2(C2M) | Hg(l) | Pt(s)

    ]Cl[KspCAg 2Hg ]Cl[

    KspC 22

    Anode : (Ag Ag+(an) + e ) 2Cathode: Hg2+2 (aq) + 2e 2Hg(l)

    Overall Rxn: 2Ag(s) + Hg+22(aq) 2Ag+(aq) + 2Hg(l)

    Q = ]Hg[]Ag[2

    2

    2

    Q =

    2

    2sp

    2

    2sp

    ]Cl[

    k]Cl[

    k

    2Cl2Hg

    AgCl

    =22ClspHg

    2spAgCl

    KK

    = constant

    If concentration of Cl will change the Ecell will remain unaffected.

    Q.41 E = ( 1.05 + 104 T2)V

    (A)PT

    E

    T2100 4

    H = nFE + nFTPT

    E

    H will depends on temperature

    (B) S =PT

    EnF

    > 0 Thus entropy will increase.

    (C) G = nFE as E = 1.05 + 104 T2It can be less then zero at certain temperature so reaction will not be spontanous at all temperature.

  • XII Part Test-7 Page # 30

    (D) At 200 KE = 1.05 + 104 (200K)2

    = 1.05 + 0.02E = 1.03Vthus reaction is non-spontanous.

    Q.42 ECell =

    Zn/Zn2E + )g(Cl/Cl 2E

    = 0.76 + 1.36 = 2.12 V

    Q.43 Anode : 21

    H2 H+ + e

    103 MCathode : e + AgBr(s) Ag + Br (aq)

    102 M

    Ecell =cellE 1

    06.0log (102 103)

    cellE =

    2H/HBr,AgBr,AgEE

    = 0.80 + 0.06 log Ksp 00.26 = 0.80 + 0.06log Ksp 0.06 log 1050.26 0.80 0.3 = 0.06 log Ksp 0.84 = 0.06 log KspKsp = 1014Let in 0.1M KBr solubility ofAgBr be ss(s + 0.1) = 1014 ; s 0.1 = 1014s = 1013M

    Q.46 (A)

    O

    O

    do not tautomerize (B)O

    OH

    Non Aromatic

    (C)

    O

    O

    OH

    OH

    10 e Aromatic

    (D)O

    NH

    4 e Anti-aromatic

    OH

    N

  • XII Part Test-7 Page # 31

    Q.47 (B) allene will not show geometrical isomerism.

    (D)

    it cannot show geometrical isomerism

    Q.49 I conjugate acid has two equivalent RSII 2 amineIII 1 amineIV amide less basic

    Q.50 Strong base are poor leaving groups

    Q.51

    OH

    octet complete structure

    Q.52 c > b > a > dc (1 - alcohol)b (2 - alcohol)a (Triplybounded carbon is sp-hybridised)b (Nitrogen atom is sp3 hy)

    Q.53 For Hydrocarbons containing same C

    HOC stability1

    Q.55

    Cl

    HO

    +

    HO

    Racemic mixture 96 %

    R + S48% 48%

    Remaining 4 % will be R onlyTotal R = 48 % + 4% = 52 % Ans.

  • XII Part Test-7 Page # 32

    Q.56 O

    Et

    D

    CH3CH I3S 2N

    H C3 OCH3

    Et

    D

    Q.57

    F

    O N2 NO2

    Cl

    Cl

    HO

    OH

    O N2 NO2

    F

    Q.58 Phenol gives a characteristic purple colour with neutral FeCl3 Since it is an enol.

    Q.59CH CHCHCH3 3

    Br

    HOHCH3CH=CHCH3 (Cis/ trans)

    Trans is major

    Br /CCl2 4

    Br BrBrBr

    HHHHH3C

    CH3 CH3

    CH3Antiaddition at transobtained erythreo product(P.O.S., Meso compound)

    Q.60

    O||

    CHCCH 33 + PCl5

    Cl|

    CHCCH|Cl

    33

  • XII Part Test-7 Page # 33

    SOLUTIONQ.61 For rectangular hyperbola

    0, (k 2) k 0 k 2, k 0h2 > ab k(k 2) < 0 0 < k < 2

    Also coff.of x2 + coff. of y2 = 0 k 2 + k = 0 k = 1

    Q.62 Required term = T10-4+2 = T8= 10C7(2x)

    3(-1/x2)7

    = 960 x11 Ans.

    Q.63 Here axa)1n(

    = 76

    7).110(

    = 13

    77= 5 13

    12

    Greatest term = T5+1 = T6 Ans.

    Q.64 As given nC4,nC5,

    nC6 are in AP.

    nC4 +nC6 = 2.

    nC5

    !4)!4n(

    !n

    +!6)!6n(

    !n

    = 2 !5)!5n(!n

    30 + (n 5) ( n 4) = 2.6 (n 4) n2 21n + 98 = 0 (n 7) (n 14) = 0 n = 7, 14 Ans.

    Q.65 We haveTr+1 =

    29 Cr329r (7x) r = (29Cr.3

    29r.7r) xr

    ar = coefficient of (r +1)th term =29 Cr. 3

    29r . 7 r

    Now, ar = ar1 29Cr . 3

    29r.7r = 29 Cr-1 . 330r . 7 r1

    29

    291

    37

    CC

    r

    r

    30 rr

    = 37

    r = 21. Ans.

  • XII Part Test-7 Page # 34

    Q.66 T = S1, then yb 2 (x + a) = b2 4a

    Passing through (0, 0), thenb2 = 2a Ans.

    Q.67 Here Tr+1 =10 Cr ( 2 )

    10r (31/5 )r,where r = 0, 1, 2, ....,10.

    We observe that in general term Tr+1 powers of 2 and 3 are12

    (10r) and15 r respectively

    and 0 r 10. So both these powers will be integers together only when r = 0 or 10 sum of required terms

    = T1 + T11= 10 C0( 2 )

    10 + 10 C10 (31/5) 10

    = 32 + 9 = 41 Ans.

    Q.68 The given expression

    =n21

    1n210C....CCC....CCC

    . 1 10

    FHG

    IKJ

    CC

    1 21

    FHG

    IKJ

    CC

    1 32

    FHG

    IKJ

    CC ... 1 1

    FHG

    IKJ

    CC

    n

    n

    = 1 1FHG

    IKJ

    n 1 12

    F

    HGIKJ

    n 1 23

    F

    HGIKJ

    n... 1

    1

    FHG

    IKJn C Cn0 b g

    =( )

    !n

    n

    n1Ans.

    Q.69 The required complex number is point of contact P as shown in the figure. C(0, 25) isthe centre of the circle and radius is 15.

    Now |z| = OP = OC PC2 2

    = 625 225 = 20

    amp (z) = = XOP = OCP

    (0, 25)

    15 P

    cos PCOC

    = 1525

    = 35

    and sin = OPOC

    = 2025

    = 45

    z = 2035

    45

    FHG

    IKJi = 12 + 16i. Ans.

  • XII Part Test-7 Page # 35

    Q.70 |z 3 + i| = |(z + 2i) ( 3 + i)| |(z + 2i) |+| ( 3 + i)| 1 + 2 = 3

    The greatest value of |z 3 + i | is 3.

    Again |z 3 + i| = | (z + 2i) ( 3 + i)|

    | 3 + i | | z + 2i| 2 1 = 1

    Thus least value of |z 3 + i | is 1. Ans.

    Q.7132 zz

    3

    =13 zz

    4

    =21 zz

    5

    232 |zz|

    9

    = 213 |zz|

    16

    = 221 |zz|

    25

    )zz)(zz(9

    3232 =

    )zz)(zz(16

    1313 = )zz)(zz(

    252121

    =

    )zz()zz(

    932

    32

    , )zz(

    16

    13 = )zz( 13 z, )zz()zz(

    2521

    21

    Hence211332 zz

    25zz

    16zz

    9

    = 0 Ans.

    Q.72 If y = mx + c touches

    1by

    ax

    2

    2

    2

    2

    then c2 = a2m2 b2

    Here, c = 6, a2 = 100, b2 = 49 36 = 100m2 49 100m2 = 85

    m =2017

  • XII Part Test-7 Page # 36

    Q.73 z = 4

    (1 + i)4

    ii

    ii

    11

    z = 4

    (1 + i)4

    i1i1

    i1i

    ii

    ii1

    z = 4

    (1 + i)4

    )1(ii

    )1(ii

    z = 4

    (1 + i)4 (2i) z = 2

    24 i4i4 ee2

    z = 2

    2

    3ie = 2 2ie

    )zarg(|z|

    =2/

    2

    = 4 Ans.

    Q.74 x + y 1 = 0 y = x + 1 (1)

    y2 = kx y2 = 4, 4k

    x (2)

    Since, line (1) is touching the parabola (2),

    c = ma

    1 = )1(4k

    k = 4

    Q.75 x2 4x 8y 4 = 0x2 4x + 4 = 8y + 8 (x 2)2 = 8(y + 1) is of the form X2 = 4aY a = 2 vertex = (2, 1) and focus is (2, 1)

    Q.76 Since (n = 8) is even then there is only one middle term i.e.2

    28T = T5

    T5 =8C4(x)

    4(2/x)4

    = 8C4. (2)4 = 16.8C4

    = 1120 Ans.

  • XII Part Test-7 Page # 37

    Q.77 Axis of parabola is 4x 3y = k(4, 5) lies on it. k = 1 axis is 4x 3y = 1

    If A is the intersection of directrix and axis of parabola then A is

    2517,

    2519

    .

    Since vertex is the mid point of A and the focus (4, 5)

    Vertex is

    2571,

    50119

    Q.78 The combined equation of the tangents drawn from (0, 0) to y2 = 4a(x a) is(y2 4ax + 4a2) (0, 0 + 4a2) = [y 0 2a (x + 0 a)]2

    (y2 4ax + 4a2) (4a2) = 4a2 (x a)2

    y2 4ax + 4a2 = (x a)2

    x2 y2 + 2ax 3a2 = 0Since, coefficient of x2 + coefficient of y2 = 0, therefore, the required angle is 90.

    Q.79 General term Tr+1 =15Cr (x

    4)15rr

    3x1

    =15Cr x607r (1)r

    For coefficient of x39

    60 7r = 39 r = 3

    The required = T4 =15C3 (x

    4)123

    3x1

    coefficient of x39 = 455 Ans.

    Q.80 Since tangents drawn from 32,1 to the ellipse 222

    by

    9x

    = 1 are right angles, therefore,

    32,1 will lie on the director circel of the ellipse, i.e., 32,1 will lie on x2 + y2 = 9 + b2 13 = 9 + b2 b2 = 4 b = 2

  • XII Part Test-7 Page # 38

    Q.81 Since the segment joining (2, 3) and (2, 5) is parallel to y-axis, therefore, ellipse is vertical.

    2be = 8, 2b = 10 b = 5 e = 54

    a2 = b2 (1 e2) = 9 and centre of the ellipse is (2, 1) Equation of required ellipse is

    25)1y(

    9)2x( 22

    = 1

    Q.82 Equation of tangent of parabola y2 = 4ax is ty = x + at2

    Hence t = 23

    and 2k

    =2

    23

    23

    = 4

    27

    Q.83 Here let the point be (x1, y1) chord of contact yy1 = 2 (x + x1)compare with the given line

    10x2 1

    = 7

    y1 =4

    2

    x1 = 5/2, y1 = 7/2

    Point reqd.

    27,

    25

    Q.84 We have, 16(x2 2x) 3 (y2 4y) = 44 16 (x 1)2 3 (y 2)2 = 48

    3

    )1x( 2 16

    )2y( 2 = 1

    This equation represents a hyperbola with eccentricity given

    e =2

    axisTransverseaxisConjugate1

    =

    2

    341

    =

    319

  • XII Part Test-7 Page # 39

    Q.85 1z > 2 and 4)1z(arg

    2

    2

    2( 1, )

    (0, 1)(1, 0)

    4

    2

    Q.86 I + f + f ' = n625 + n625 = 2k (even integer) f + f ' = 1

    Now, (I + f) f ' = n625 n625 = (1)n = 1 (I + f) (1 f) = 1

    or I = f)f1(1

    . Ans.

    Q.87 21 zz = 21 zz

    2

    21 zz =2

    1z +2

    2z + 2 21 zz

    2

    1z +2

    2z + 2Re )zz( 21 =2

    1z +2

    2z + 21 zz2

    2Re )zz( 21 = 2 21 zz

    2 21 zz cos (1 2) = 2 21 zz ( z1 = r1 1ie and z2 = r2 2ie )

    cos (1 2) = 1 or 1 2 = 0 arg (z1) = arg (z2). Ans.

    Q.88 The equation a2zaz 2 = 3 will represent an ellipse if

    3a2a2

    3 < a2 2a < 3 2 < (a 1)2 < 4 (a 1)2 < 4 2 < a 1 < 2 1 < a < 3 a (0, 3) [ a > 0] Ans.

  • XII Part Test-7 Page # 40

    Q.89 Multiply above and below by conjugate of denominator and put real part equal to zero.

    =

    2sini21

    2cos

    2sinitan

    2sini21

    2sini21

    =

    2sin41

    2sintan2

    2cos

    2sini

    2cos

    2sin

    2sin2tan

    2

    tan 2 sin 2

    2cos

    2sin = 0

    cossin

    (1 cos ) sin = 0

    sin

    coscos1

    (1 cos ) = 0

    (1 cos ) (tan 1) = 0cos = 1 = 2n and

    tan = 1 = n +4 Ans.

    Q.90 Let the point of contact be

    R )sin,cos2(

    Equation of tangent AB is

    BQ(h, k)

    R

    O

    Y'

    Y

    XX'

    2y2x 22

    1sinycos2

    x

    A )0,sec2( ; B (0, cosec )Let the middle point Q of AB be (h, k)

    h = 2sec

    , k = 2cosec

    cos =2h

    1, sin =

    k21

    Required locus is 22 y41

    x21

    = 1 Ans.


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