1. Four point masses, each of mass m, are fixed at the corners of a square of side I. The square is rotating with angular frequency ωω , about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is:

[image]

Sol. (3)

[image]

L=I0L=I0? I=m(a2)2×2+m(2a)2I=m(2?a?)2×2+m(2?a)2 =ma2+2ma2=ma2+2ma2 ∴L=I0=3mI02∴L=I0?=3mI02?

1. A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5mm0.5mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively :

(1) Positive, 0.1mm0.1mm (2) Positive, 0.1μm0.1μm (3) Positive, 10μm10μm (4) Negative, 2μm2μm

1. (3)

=L.C=0.550mm=1×10−5m=10μm=L.C=500.5?mm=1×10−5m=10μm

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1. An electron, a doubly ionized helium ion (He++) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths λeλe? , λHe+λHe+? and λpλp? is :

(1)λe>λp>λHe+(2)λe>λHe+>λp(1)λe?>λp?>λHe+?(2)λe?>λHe+?>λp? (3)λe<λp<λHe+(4)λe<λHe+=λp(3)λe?<λp?<λHe+?(4)λe?<λHe+?=λp?

Sol. (1)

λ=hp=h2mK⋅ECp=1221275Eλ=ph?=2mK⋅E?h?Cp?=21?12?527?E mHe>mp>memHe?>mp?>me? λHe<λp<λeλHe?<λp?<λe?

1. For the given input voltage waveform Vin(t)Vin?(t) , the output voltage waveform Vo(t)Vo?(t) , across the capacitor is correctly depicted by :

[image]

Sol. (2)

Answer is (2) because capacitor is charging then discharging then again charging. But during discharging not possible to discharge 100%100% .

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1. Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is :

[image]

(1) MR222MR2?

(2) MR233MR2?

(3) M(R2+H2)44M(R2+H2)?

Sol. (1)

[image]

1. A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is:

(1) 1:2
(2) 1:3
(3) 1:6
(4) 3:4

Sol. (3)

[image]

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1. You are given that Mass of 3Li=7.0160u3Li=7.0160u Mass of 4He=4.0026u4He=4.0026u and Mass of 1H=1.0079u1H=1.0079u When 20 g of 3Li3Li is converted into 4He4He by proton capture, the energy liberated, (in kWh), is: [Mass of nucleon =1GeV/c2=1GeV/c2 ] (1) 6.82×1056.82×105 (2) 4.5×1054.5×105 (3) 8×1068×106 (4) 1.33×1061.33×106 Sol. (4) 3Li+1e+→24He3Li+1e+→24He Δm⇒[mLi+mH]−2[MHe]Δm⇒[mLi?+mH?]−2[MHe?] →Δm=(7.0160+1.0079)−2×4.0003→Δm=(7.0160+1.0079)−2×4.0003 =0.0187=0.0187 Energy released in 1 reaction ⇒Δmc2⇒Δmc2 In use of 7.016 u Li energy is Δmc2Δmc2 In use of 1gm Li energy is Δmc2mLimLi?Δmc2? In use of 20gm energy is ⇒Δmc2mLi×20gm⇒mLi?Δmc2?×20gm 0.0187×931.5×106×1.6×10−19×207×6.023×102336×105=1.33×10636×1050.0187×931.5×106×1.6×10−19×720?×6.023×1023?=1.33×106

2. If the potential energy between two molecules is given by U=−Ar6+Br12U=−r6A?+r12B? , then at equilibrium, separation between molecules, and the potential energy are :

(1)(2BA)1/6,−A24B(2)(2BA)1/6,−A22B(3)(BA)1/6,0(4)(B2A)1/6,−A22B(1)(A2B?)1/6,−4BA2?(2)(A2B?)1/6,−2BA2?(3)(AB?)1/6,0(4)(2AB?)1/6,−2BA2?

Sol. (1)

F=−dUdr=−ddr(−Ar−6+Br−12)F=dr−dU?=dr−d?(−Ar−6+Br−12) for equation F=0for equation F=0 for equation F=0for equation F=0 for equation =A(−6)r7+B⋅12r13=0for equation =r7A(−6)?+r13B⋅12?=0 for equation =12Br13−6Ar7for equation =r1312B?−r76A? for equation =(2BA)1/6for equation =(A2B?)1/6

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1. A clock has a continuously moving second's hand of 0.1m0.1m length. The average acceleration of the tip of the hand (in units of ms−2ms−2 ) is of the order of: (1) 10−310−3 (2) 10−110−1 (3) 10−210−2 (4) 10−410−4

Answer (1)

1. A clock has a continuously moving second's hand of 0.1m0.1m length. The average acceleration of the tip of the hand (in units of ms−2ms−2 ) is of the order of: (1) 10−310−3 (2) 10−110−1 (3) 10−210−2 (4) 10−410−4

(1) 10−310−3 (2) 10−110−1 (3) 10−210−2 (4) 10−410−4

Sol. (1)

a=ν2Rν=2πR60a=Rν2?ν=602πR? =4π2⋅R2(60)2R=4π2R(60)2=4(60)2×10×0.1≈10−3=(60)2R4π2⋅R2?=(60)24π2R?=(60)24?×10×0.1≈10−3

1. Identify the correct output signal Y in the given combination of gates (as shown) for the given inputs A and B.

[image]

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1. An electron is moving along +x+x direction with a velocity of 6×106ms−16×106ms−1 . It enters a region of uniform electric field of 300 V/cm300 V/cm pointing along +y+y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the xx direction will be: (1) 3×10−4T3×10−4T , along −z−z direction (2) 5×10−3T5×10−3T , along −z−z direction (3) 5×10−3T5×10−3T , along +z+z direction (4) 3×10−4T3×10−4T , along +z+z direction

2. An electron is moving along +x+x direction with a velocity of 6×106ms−16×106ms−1 . It enters a region of uniform electric field of 300 V/cm300 V/cm pointing along +y+y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the xx direction will be:

(1) 3×10−4T3×10−4T , along −z−z direction
(2) 5×10−3T5×10−3T , along −z−z direction
(3) 5×10−3T5×10−3T , along +z+z direction
(4) 3×10−4T3×10−4T , along +z+z direction

Sol. (3)

B must be in +z+z axis.

[image]

qE=qVBqE=qVBE=300v10−2mE=30010−2mv?=30000v/m=30000v/mB=EV=3×1046×106=5×10−3TB=VE?=6×1063×104?=5×10−3T

[image]

1. In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20m20m . The separation between P and Q is 5m5m and the phase of P is ahead of that of Q by 90?90? . A, B and C are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio:

[image]

(1) 4 : 1 : 0
(2) 2 : 1 : 0
(3) 0 : 1 : 2
(4) 0 : 1 : 4

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2

[image]

\left\{ \begin{array}{ll}\Delta x = \lambda /2 Δ?=πΔ?=π I_{\mathrm{c}} = 0 \end{array} \right.\quad \therefore \mathrm{at~A~}\Delta x_{\mathrm{effective}} = 0\mathrm{~or~phase~difference~} = 0  {Same logic as A point but opposits} {Same logic as A point but opposits} \therefore \mathrm{~Answer~is~2.} \end{array} \right.

1. A point like object is placed at a distance of 1 m in front of a convex lens of focal length 0.5 m0.5 m . A plane mirror is placed at a distance of 2 m behind the lens. The position and nature of the final image formed by the system is:

(1) 1 m from the mirror, virtual
(2) 2.6 m from the mirror, virtual
(3) 1 m from the mirror, real
(4) 2.6 m from the mirror, real

Sol. (1, 2 Both are correct)

[image]

for IIIrdIIIrd Refraction, u=−3u=−3

1V+13=21V1?+31?=12?

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1. An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is: (g=10ms−2)(g=10ms−2) (1) 0.45 m (2) 0.60 m (3) 0.20 m (4) 0.80 m

Sol. (3)

[image]

f=mgsin?θf=mgsinθ f=μmgcos?θf=μmgcosθ μmgcos?θ=mgsin?θμmgcosθ=mgsinθ tan?θ=μtanθ=μ

tan?θ=34tanθ=43?cos?θ=416+9=45cosθ=16+9?4?=54?h=1(1−cos?θ)=1−45=15h=1(1−cosθ)=1−54?=51?h=15=0.2mh=51?=0.2m

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1. Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ(CpCv)γ(Cv?Cp??) are given, respectively by:

U=52RTandγ=75(2)U=5RTandγ=65U=25?RTandγ=57?(2)U=5RTandγ=56? U=5RTandγ=75(4)U=52RTandγ=65U=5RTandγ=57?(4)U=25?RTandγ=56?

Sol. (1)

U=f2nRT=52nRT(Cp−Cv=RCv=f2R),γ=CpCd⇒1+2f=1+25=75U=2f?nRT=25?nRT(Cv?=2f?RCp?−Cv?=R?),γ=Cd?Cp??⇒1+f2?=1+52?=57?

1. An object of mass m is suspended at the end of a massless wire of length L and area of cross- section A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:

f=12πYAmL(2)f=12πmLYA(3)f=12πYLmA(4)f=12πmAYLf=2π1?mLYA??(2)f=2π1?YAmL??(3)f=2π1?mAYL??(4)f=2π1?YLmA??

Sol. (1)

Y=F/AΔL/LY=ΔL/LF/A? Y=FLAΔLY=AΔLFL? F=YAΔLLF=LYAΔL? f=12πYALmf=2π1?LmYA?? ? T=2πmk? T=2πkm??

1. An AC circuit has R=100ΩR=100Ω , C=2μFC=2μF and L=80mHL=80mH , connected in series. The quality factor of the circuit is:

(1) 20
(2) 2
(3) 0.5
(4) 400

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1. A sound source S is moving along a straight track with speed v, and is emitting, sound of frequency ν0ν0? (see figure). An observer is standing at a finite distance, at the point O, from the track. The time variation of frequency heard by the observer is best represented by: ( t0t0? represents the instant when the distance between the source and observer is minimum)

[image]

Sol. (4)

[image]

fobserved⇒(VsoundVsound−Vcos?θ)f0fobserved?⇒(Vsound?−VcosθVsound??)f0?

initially θθ will be less ⇒⇒ cosθ more ∴fobserved∴fobserved? more, then it will decrease. Ans.4

1. A particle of charge q and mass m is moving with a velocity −vi?(v≠0)−vi?(v?=0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field B?=B0k^B=B0?k^ , the minimum value of v for which the particle will not hit the screen is:

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1. qdB0mmqdB0?? (2) qdB03m3mqdB0?? (3) 2qdB0mm2qdB0?? (4) qdB02m2mqdB0??

Sol. (1)

[image]

1. Two bodies of the same mass are moving with the same speed, but in different directions in a plane. They have a completely inelastic collision and move together thereafter with a final speed which is half of their initial speed. The angle between the initial velocities of the two bodies (in degree) is ______.

2. 120

[image]

In Horizontal Direction By Momentum conservation.

mv+mvcos?θ=2mv2cos?αmv+mvcosθ=2m2v?cosα1+cos?θ=cos?α1+cosθ=cosα

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1. Suppose that intensity of a laser is (315π)W/m2(π315?)W/m2 . The rms electric field, in units of V/m associated with this source is close to the nearest integer is (?0=8.86×10−12C2Nm−2;c=3×108ms−1)(?0?=8.86×10−12C2Nm−2;c=3×108ms−1) Sol. 275 I=12?0CErms2I=21??0?CErms2? 3.15π=12×8.86×10−12×3×108×Erms2π3.15?=21?×8.86×10−12×3×108×Erms2? Erms=275Erms?=275 23. The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is (x100)%(100x?)% . If the relative errors in measuring the mass and the diameter are 6.0%6.0% and 1.5%1.5% respectively, the value of xx is Sol. 1050 ρ=m43π(d2)3ρ=34?π(2d?)3m? ρ=k⋅md3ρ=k⋅d3m? log?ρ=log?k0+log?m−3log?dlogρ=logk0+logm−3logd diff. dρρ=dmm−3⋅dddρdρ?=mdm?−3⋅ddd? =6.0+3×1.5=10.5%=6.0+3×1.5=10.5% =x=1050=x=1050

maximum error in the density of the sphere is (x100)%(100x?)% . If the relative errors in measuring the mass and the diameter are 6.0%6.0% and 1.5%1.5% respectively, the value of xx is

Sol. 1050

ρ=m43π(d2)3ρ=34?π(2d?)3m?ρ=k⋅md3ρ=k⋅d3m?log?ρ=log?k0+log?m−3log?dlogρ=logk0+logm−3logddρρ=dmm−3⋅dddρdρ?=mdm?−3⋅ddd?

=6.0+3×1.5=10.5%=6.0+3×1.5=10.5% =x=1050=x=1050

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1. Initially a gas of diatomic molecules is contained in a cylinder of volume V1V1? at a pressure P1P1? and temperature 250 K250 K . Assuming that 25%25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K2000 K , when contained in a volume 2V12V1? is given by P2P2? . The ratio P2/P1P2?/P1? is ______.

Sol. 5

[image]

1. A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1A and it is decreasing at a rate of 102102 A s−1s−1 . The value of the potential difference Vp−VQ′Vp?−VQ′? (in volts) at that instant, is ______.

[image]

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QUESTION PAPER WITH SOLUTION

CHEMISTRY _ 6 Sep. _ SHIFT - 1

1. The INCORRECT statement is :
   (1) Cast iron is used to manufacture wrought iron.
   (2) Brass is an alloy of copper and nickel.
   (3) German silver is an alloy of zinc, copper and nickel.
   (4) Bronze is an alloy of copper and tin

Sol. 2
Brass - (copper Zinc)
Bronze - (copper tin)

1. The species that has a spin-only magnetic moment of 5.9 BM, is : (Td = tetrahedral)
   (1) [Ni(CN)4]2- (square planar) (2) Ni(CO)4(Td)
   (3) [MnBr4]2- (Td) (4) [NiCl4]2- (Td)

Sol. 3
[MnBr4]2-
Mn+2 = 3d 4s 4p
sp3(tetrahedral)
μ = √5(5+2) = 5.9 BM

1. For the reaction
   Fe2N(s) + 3/2 H2(g) ? 2Fe(s) + NH3(g)
   (1) Kc = Kp(RT)1/2 (2) Kc = Kp(RT)-1/2
   (3) Kc = Kp(RT)3/2 (4) Kc = Kp(RT)

Sol. 1
Fe2N(s) + 3/2 H2(g) ? 2Fe(s) + NH3(g)
Δng = 1 - 3/2 = -1/2
Kp/Kc = (RT)Δng = (RT)-1/2
Kc = Kp/(RT)-1/2 = Kp.(RT)1/2

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1. Consider the following reactions :

[image]

'A' is :

[image]

Sol. 1

[image]

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1. Arrange the following solutions in the decreasing order of pOH : (A) 0.01 M HCl (B) 0.01 M NaOH (C) 0.01 M CHCOONa (D) 0.01 M NaCl (1) (A) > (C) > (D) > (B) (2) (B) > (D) > (C) > (A) (3) (B) > (C) > (D) > (A) (4) (A) > (D) > (C) > (B) Sol. 4 (i) 10−2MHCl⇒[H+]=10−2M→pH=210−2MHCl⇒[H+]=10−2M→pH=2 (ii) 10−2MNaOH⇒[OH−]=10−2M→pOH=210−2MNaOH⇒[OH−]=10−2M→pOH=2 (iii) 10−2MCH3COO−Na+⇒[OH+]>10−7⇒pOH<710−2MCH3?COO−Na+⇒[OH+]>10−7⇒pOH<7 (iv) 10−2MNaCl⇒NeutralpOH=710−2MNaCl⇒NeutralpOH=7 (i) > (iv) > (iii) > (ii) 6. The variation of equilibrium constant with temperature is given below : Temperature Equilibrium Constant T1=25?CT1?=25?C K1=10K1?=10 T2=100?CT2?=100?C K2=100K2?=100 The value of ΔH?ΔH? ΔG?ΔG? at T1T1? and ΔG?ΔG? at T2T2? (in Kj mol- 1) respectively, are close to [use R=8.314JK−1mol−1R=8.314JK−1mol−1 ] (1) 28.4, - 7.14 and - 5.71 (2) 0.64, - 7.14 and - 5.71 (3) 28.4, - 5.71 and - 14.29 (4) 0.64, - 5.71 and - 14.29 Sol. 3 ln?[k2k1]=ΔH?R[1T1−1T2]ln[k1?k2??]=RΔH??[T1?1?−T2?1?] ln?(10)=ΔH?R[1298−1373]ln(10)=RΔH??[2981?−3731?] 373×298×8.314×2.30375=ΔH?=28.37kJmol−175373×298×8.314×2.303?=ΔH?=28.37kJmol−1 ΔGT1?=−RT1ln?(K1)=−298Rln?(10)=−5.71kJmol−1ΔGT1???=−RT1?ln(K1?)=−298Rln(10)=−5.71kJmol−1 ΔGT2?=−RT2ln?(K2)=−373Rln?(100)ΔGT2???=−RT2?ln(K2?)=−373Rln(100) =−14.283kJ/mol=−14.283kJ/mol 7. Consider the following reactions A→P1;B→P2;C→P3;D→P4,A→P1;B→P2;C→P3;D→P4, The order of the above reactions are a,b,c and d, respectively. The following graph is obtained when log[rate] vs. log[conc.] are plotted :

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3

Among the following the correct sequence for the order of the reactions is :

(1) c>a>b>dc>a>b>d (2) d>a>b>cd>a>b>c

(3) d>b>a>cd>b>a>c (4) a>b>c>da>b>c>d

Sol. 3

A→P1B→P2C→P3D→P4A→P1B→P2C→P3D→P4

Rate == K (conc.)order

log(rate) == log(K) ++ order log (case)

[image]

Slope == order According graph d>b>a>cd>b>a>c order of slope

1. The major product obtained from the following reactions is :

[image]

[image]

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1. Which of the following compounds shows geometrical isomerism ?
   (1) 2-methylpent-1-ene (2) 4-methylpent-2-ene
   (3) 2-methylpent-2-ene (4) 4-methylpent-1-ene

Sol. 2
4-Methylpent-2-ene
Can show G.I.

1. The lanthanoid that does NOT shows +4 oxidation state is :
   (1) Dy (2) Ce
   (3) Tb (4) Eu

Sol. 4
Fact

1. The major products of the following reactions are :

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Sol. 1

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1. The major product of the following reaction is :

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Sol. 2

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1. The increasing order of pKbpKb? values of the following compounds is :

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(1) I<II<III<IVI<II<III<IV (3) I<II<IV<IIII<II<IV<III

(2) II<IV<III<III<IV<III<I (4) II<I<III<IVII<I<III<IV

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3 Order of pKb

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1. kraft temperature is the temperature :

(1) Above which the aqueous solution of detergents starts boiling
(2) Below which the formation of micelles takes place.
(3) Above which the formation of micelles takes place.
(4) Below which the aqueous solution of detergents starts freezing.

Sol. 3

Tk+Tk?+ temp. above which formation of micelles takes place.

1. The set that contains atomic numbers of only transition elements, is ?

(1) 9, 17, 34, 38
(2) 21, 25, 42, 72
(3) 37, 42, 50, 64
(4) 21, 32, 53, 64

Sol. 2

Transition elements =21=21 to 30 37 to 48 57 & 72 to 80

Ans.21,25,42&72

1. Consider the Assertion and Reason given below.

Assertion (A) : Ethene polymerized in the presence of Ziegler Natta Catalyst at high temperature and pressure is used to make buckets and dustbins.

Reason (R) : High density polymers are closely packed and are chemically inert.

Choose the correct answer from the following :

(1) (A) and (R) both are wrong.
(2) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(3) (A) is correct but (R) is wrong
(4) Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Sol. 2

From ziegler - Natta catalyst HDPE is produced, HDPE is closely packed and are chemically inert, so used to make bucket and dustbin.

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1. A solution of two components containing n1n1? moles of the 1st1st component and n2n2? moles of the 2nd2nd component is prepared. M1M1? and M2M2? are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in g mL-1, C2C2? is the molarity and x2x2? is the mole fraction of the 2nd2nd component, then C2C2? can be expressed as :

C2=dx1M2+x2(M2−M1)(2)C2=1000x2M1+x2(M2−M1)(3)C2=dx2M2+x2(M2−M1)(4)C2=1000dx2M1+x2(M2−M1)(5)C2?=M2?+x2?(M2?−M1?)dx1??(2)C2?=M1?+x2?(M2?−M1?)1000x2??(3)C2?=M2?+x2?(M2?−M1?)dx2??(4)C2?=M1?+x2?(M2?−M1?)1000dx2??(5)

Sol. 4

C2=x2[x2M1+(1−x2)M2]/d×1000C2?=[x2?M1?+(1−x2?)M2?]/dx2??×1000 C2=1000dx2M1+(M2−M1)x2C2?=M1?+(M2?−M1?)x2?1000dx2??

1. The correct statement with respect to dinitrogen is ?

(1) Liquid dinitrogen is not used in cryosurgery.
(2) N2N2? is paramagnetic in nature
(3) It can combine with dioxygen at 25?C25?C (4) It can be used as an inert diluent for reactive chemicals.

Sol. 4

(1) Liquid nitrogen is used as a refrigerant to preserve biological material food items and in cryosurgery.
(2) N2N2? is diamagnetic, with no unpaired elctrons.
(3) N2N2? does not combine with oxygen, hydrogen or most other elements. Nitrogen will combine with oxygen, however ; in the presence of lightning or a spark.
(4) In iron and chemical Industry inert diluent for reactive chemicals.

1. Among the sulphates of alkaline earth metals, the solubilities of BeSO4BeSO4? and MgSO4MgSO4? in water, respectively, are :

(1) Poor and high
(2) High and high
(3) Poor and poor
(4) High and poor

Sol. 2

Order of solubility of sulphate of Alkaline earth metals BeSO4>MgSO4>CaSO4>SrSO4>BaSO4BeSO4?>MgSO4?>CaSO4?>SrSO4?>BaSO4?

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1. The presence of soluble fluoride ion upto 1ppm concentration in drinking water, is : (1) Harmful to skin (2) Harmful to bones (3) Safe for teeth (4) Harmful for teeth Sol. 3 Environmental chemistry - safe for teeth 21. A spherical balloon of radius 3cm containing helium gas has a pressure of 48×10−348×10−3 bar. At the same temperature, the pressure, of a spherical balloon of radius 12cm containing the same amount of gas will be. ×10−6×10−6 bar. Sol. 750 moles=48×10−3×43π(3cm)3R×Tmoles=R×T48×10−3×3π4?(3cm)3? moles=P×43π(12cm)3RTmoles=RTP×3π4?(12cm)3? P×144×12=48×9×3×10−3P×144×12=48×9×3×10−3 P=2736×10−3P=3627?×10−3 P=2700036×10−6P=3627000?×10−6 P=30004×10−6P=43000?×10−6 P=750×10−6P=750×10−6 bar 22. The elevation of boiling point of 0.10m0.10m aqueous CrCl3⋅xNH3CrCl3?⋅xNH3? solution is two times that of 0.05m0.05m aqueous CaCl2CaCl2? solution. The value of xx is. [Assume 100%100% ionisation of the complex and CaCl2CaCl2? , coordination number of CrCr as 6, and that all NH3NH3? molecules are present inside the coordination sphere] Sol. 5 ΔTb=i×Kb×mΔTb?=i×Kb?×m i×0.1×Kb=3×0.05×Kb×2i×0.1×Kb?=3×0.05×Kb?×2 i=3i=3 [Cr(NH3)5.Cl]Cl2→[Cr(NH3)5Cl]+2+2Cl−[Cr(NH3?)5?.Cl]Cl2?→[Cr(NH3?)5?Cl]+2+2Cl− x=5x=5 23. Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6OH−+Cl−?ClO3−+3H2O+6e−6OH−+Cl−?ClO3−?+3H2?O+6e− If only 60%60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10g10g of KClO3KClO3? using a current of 2A is. (Given : F=96,500CF=96,500C mol- 1; molar mass of KClO3=122gKClO3?=122g mol- 1)

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11

10122×6=2×t(hr)×3600×60%9650012210?×6=965002×t(hr)×3600×60%? t(hr)=96500122×72=10.98hrt(hr)=122×7296500?=10.98hr =11hours=11hours

1. In an estimation of bromine by Carius method, 1.6g1.6g of an organic compound gave 1.88g1.88g of AgBr. The mass percentage of bromine in the compound is ...... (Atomic mass, Ag=108Ag=108 , Br=80g mol−1Br=80g mol−1 )

Sol. 50%

Carius method

%ofBr=wt of AgBrwt of organic compound×100×molar mass of BrAgBr%ofBr=wt of organic compoundwt of AgBr?×100×AgBrmolar mass of Br? =1.881.6×80188×100=15040300.8=50%=1.61.88?×18880?×100=300.815040?=50%

1. The number of Cl=OCl=O bonds in perchloric acid is, "......"Sol. 3

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QUESTION PAPER WITH SOLUTION

MATHEMATICS _ 6 Sep. _ SHIFT - 1

Q.1 The region represented by {z = x + iy ∈ C : |z| − Re(z) ≤ 1} is also given by the inequality:
{z = x + iy ∈ C : |z| − Re(z) ≤ 1}
(1) y² ≤ 2(x + 1/2) (2) y² ≤ x + 1/2 (3) y² ≥ 2(x + 1) (4) y² ≥ x + 1

Sol. 1
{z = x + iy ∈ c : |z| − Re(z) ≤ 1}
|z| = √(x² + y²)
Re(z) = x
|z| − Re(z) ≤ 1
⇒ √(x² + y²) − x ≤ 1
⇒ √(x² + y²) ≤ 1 + x
⇒ x² + y² ≤ 1 + x² + 2x
⇒ y² ≤ 2(x + 1/2)

Q.2 The negation of the Boolean expression p ∨ (∼p ∧ q) is equivalent to:
(1) p ∧ ∼q (2) ∼p ∨ ∼q (3) ∼p ∨ q (4) ∼p ∧ ∼q

Sol. 4
p ∨ (∼p ∧ q)
(p ∧ ∼p) ∨ (p ∧ q)
t ∧ (p ∨ q)
p ∨ q
∼ (p ∨ (∼p ∧ q)) = ∼ (P ∨ q)
= (∼ P) ∧ (∼ q)

Q.3 The general solution of the differential equation √(1 + x² + y² + x²y²) + xy dy/dx = 0 is:
(where C is a constant of integration)
(1) √(1 + y²) + √(1 + x²) = 1/2 log?( (√(1 + x²) − 1)/(√(1 + x²) + 1) ) + C
(2) √(1 + y²) − √(1 + x²) = 1/2 log?( (√(1 + x²) − 1)/(√(1 + x²) + 1) ) + C
(3) √(1 + y²) + √(1 + x²) = 1/2 log?( (√(1 + x²) + 1)/(√(1 + x²) − 1) ) + C
(4) √(1 + y²) − √(1 + x²) = 1/2 log?( (√(1 + x²) + 1)/(√(1 + x²) − 1) ) + C

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3

1+x2+y2+x2y2+xydydx=01+x2+y2+x2y2?+xydxdy?=0 (1+x2)(1+y2)+xydydx=0(1+x2)(1+y2)?+xydxdy?=0 (x+x2)dx=−y1+y2dy(x+x2)?dx=−1+y2?y?dy

Integrate the equation

∫1+x2xdx=−∫y1+y2dy∫x1+x2??dx=−∫1+y2?y?dy 1+x2=t21+x2=t2 2xdx=2tdt2xdx=2tdt dx=txdtdx=xt?dt ∫tdtt2−1=−∫zdxz∫t2−1tdt?=−∫zzdx? ∫t2−1+1t2−1dt=−z+c∫t2−1t2−1+1?dt=−z+c ∫dt+∫1t2−1dt=−z+c∫dt+∫t2−11?dt=−z+c t+12ln?(t−1t+1)=−z+ct+21?ln(t+1t−1?)=−z+c 1+x2+12ln?(1+x2−11+x2+1)=−1+y2+c1+x2?+21?ln(1+x2?+11+x2?−1?)=−1+y2?+c 1+y2+1+x2=12ln?(x2+1+1x2+1−1)+c1+y2?+1+x2?=21?ln(x2+1?−1x2+1?+1?)+c

Q.4 Let L1L1? be a tangent to the parabola y2=4(x+1)y2=4(x+1) and L2L2? be a tangent to the parabola y2=8(x+2)y2=8(x+2) such that L1L1? and L2L2? intersect at right angles. Then L1L1? and L2L2? meet on the straight line:

(1)x+2y=0(2)x+2=0(3)2x+1=0(4)x+3=0(1)x+2y=0(2)x+2=0(3)2x+1=0(4)x+3=0

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4 Let tangent of y2=4(x+1)y2=4(x+1) L1:t1y=(x+1)+t12…………………………L1?:t1?y=(x+1)+t12?…………………………? (i) and tangent of y2=8(x+2)y2=8(x+2) L2:t2y=(x+2)+2t22L2?:t2?y=(x+2)+2t22?? L1⊥L2L1?⊥L2? 1t1⋅1t2=−1t1?1?⋅t2?1?=−1 t1t2=−1t2(i)−t1(ii)t1?t2?=−1t2?(i)−t1?(ii)? t1t2y=t2(x+1)+t2.t12t1t2y=t1(x+2)+2t22.t1t1?t2?y=t2?(x+1)+t2?.t12?t1?t2?y=t1?(x+2)+2t22?.t1?? (t2−t1)x+(t2−2t1)+t2t1(t1−2t2)(t2−t1)x+(t2−2t1)−(t1−2t2)=0(t2−t1)x+3t2−3t1=0(t2?−t1?)x+(t2?−2t1?)−(t1?−2t2?)(t2?−t1?)x+(t2?−2t1?)+t2?t1?(t1?−2t2?)?=0(t2?−t1?)x+3t2?−3t1?=0? ⇒x+3=0⇒x+3=0

Q.5 The area (in sq. units) of the region A={(x,y):?x?+?y?≤1,2y2≥?x?}A={(x,y):?x?+?y?≤1,2y2≥?x?}

(1) 1661? (2) 5665? (3) 1331? (4) 7667?

Sol. 2

[image]

Total area=4∫01/2[(1−x)−(x2)]dxTotal area=4∫01/2?[(1−x)−(2x??)]dx=4[x−x22−12x3/23/2]?1/2?=4[x−2x2?−2?1?3/2?x3/2?]?1/2?=4[12−18−23(12)3/2]=4[21?−81?−32??(21?)3/2]=4×524=56=4×245?=65?

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6 The shortest distance between the lines x−10=y+1−1=z10x−1?=−1y+1?=1z? and x+y+z+1=0x+y+z+1=0 , 2x−y+z+3=02x−y+z+3=0 is:

(1)1(2)12(3)13(4)12(1)1(2)2?1?(3)3?1?(4)21?

Sol. 3

Plane through line of intersection is x+y+z+1+λ(2x−y+z+3)=0x+y+z+1+λ(2x−y+z+3)=0 It should be parallel to given line 0(1+2λ)−1(1−λ)+1(1+λ)=0⇒λ=00(1+2λ)−1(1−λ)+1(1+λ)=0⇒λ=0 Plane: x+y+z+1=0x+y+z+1=0 Shortest distance of (1,−1,0)(1,−1,0) from this plane

=?1−1+0+1?12+12+12=13=12+12+12??1−1+0+1??=3?1?

Q.7 Let a, b, c, d and p be any non zero distinct real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0.(a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0. Then:

(1) a, c, p are in G.P.
(2) a, b, c, d are in G.P.
(3) a, b, c, d are in A.P.
(4) a, c, p are in A.P.

Sol. 2

(a2+b2+c2)p2−2(ab+bc+cd)p+b2+c2+d2)=0(a2+b2+c2)p2−2(ab+bc+cd)p+b2+c2+d2)=0 (a2p2−2abp+b2]+[b2p2−2bcp+c2]+[c2p2−2cdp+d2](a2p2−2abp+b2]+[b2p2−2bcp+c2]+[c2p2−2cdp+d2] (ap−b)2+(bp−c)2+(cp−d)2=0(ap−b)2+(bp−c)2+(cp−d)2=0

ap=bba=cb=dc=pap=bab?=bc?=cd?=p

bp=c

cp=d a,b,c,d are in G.P.

Q.8 Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

(1) 2! 3! 4!
(2) (3!)^3.(4!)
(3) 3! (4!)^3
(4) (3!)^2.(4!)

Sol. 2

F1→3F1?→3 members

F2→3F2?→3 members

F3→4F3?→4 members

No. of ways can they be seated so that the same family members are not separated

=3!×3!×3!×4!=(3!)3.4!=3!×3!×3!×4!=(3!)3.4!

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9 The values of λλ and μμ for which the system of linear equations x+y+z=2x+y+z=2 x+2y+3z=5x+2y+3z=5 x+3y+λz=μx+3y+λz=μ has infinitely many solutions are, respectively: (1) 6 and 8 (2) 5 and 8 (3) 5 and 7 (4) 4 and 9

Sol. 2

x+y+z=2x+y+z=2 x+2y+3z=5x+2y+3z=5 x+3y+λz=μx+3y+λz=μ

has infinitely many solutions

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10 Let m and M be respectively the minimum and maximum values of

Then the ordered pair (m, M) is equal to:

(1) (−3,−1)(−3,−1) (2) (−4,−1)(−4,−1) (3) (1, 3) (4) (−3,3)(−3,3)

Sol. 1

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Equation of P′B→y−23=tan?120?(x−0)P′B→y−23?=tan120?(x−0) 3×+y=233?×+y=23? (3,−3)(3,−3?) satisfy the line

Q.12 Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:

(1)1099(2)533(3)15101(4)5101(1)9910?(2)335?(3)10115?(4)1015?

Sol. 2

Case- 1E, O, E, O, E, O, E, O, E, O, E 2b=a+c→2b=a+c→ Even ⇒⇒ Both a and c should be either even or odd.

P=6C2+5C211C3=533P=11C3?6C2?+5C2??=335?

Case- 2

O,E,O,E,O,E,O,E,0,E,OO,E,O,E,O,E,O,E,0,E,OP=5C2+6C211C3=533P=11C3?5C2?+6C2??=335?Total probability=12×533+12×533=533Total probability=21?×335?+21?×335?=335?

Q.13 If f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) and ∑x=1∞f(x)=2∑x=1∞?f(x)=2 , x,y∈Nx,y∈N , where N is the set of all natural number, then the value of f(4)f(2)f(2)f(4)? is :

(1)23(2)19(3)13(4)49(1)32?(2)91?(3)31?(4)94?

Sol. 4

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[image]

∑x=1∞f(x)=f(1)+f(2)+f(3)+.......f(∞)=2x=1∑∞?f(x)=f(1)+f(2)+f(3)+.......f(∞)=2⇒f(1)+f((1))2+f((1))3.......=2⇒f(1)+f((1))2+f((1))3.......=2f(1)1−f(1)=21−f(1)f(1)?=2f(1)=2/3f(1)=2/3f(2)=(23)2,f(4)=(23)4f(2)=(32?)2,f(4)=(32?)4f(4)f(2)=(2/3)4(2/3)2=49f(2)f(4)?=(2/3)2(2/3)4?=94?

Q.14 If {p} denotes the fractional part of the number p, then {32008}{83200?}, is equal to :

(1) 5885? (2) 1881? (3) 7887? (4) 3883?

Sol. 2

{32008}={91008}={(8+1)1008}{83200?}={89100?}={8(8+1)100?}{100C01100+100C1(8)199+100C2(82)198+....+100C10081008}{8100C0?1100+100C1?(8)199+100C2?(82)198+....+100C100?8100?}={100C01100+8k8}={8100C0?1100+8k?}={1+8k8}={18+k}K∈I={81+8k?}={81?+k}K∈I=18=81?

Q.15 Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, x24+y22=14x2?+2y2?=1 from any of its foci ?

(1) (−1,3)(−1,3?) (2) (−2,3)(−2,3?) (3) (−1,2)(−1,2?) (4) (1,2)(1,2)

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4 Let foot of perpendicular is (h,k)(h,k)

[image]

x24+y22=1(GivenΦ)Φ4x2?+2y2?=1(GivenΦ)Φ a=2,b=2,e=1−24=12a=2,b=2?,e=1−42??=2?1?

Focus (ae,0)=(2,0)(ae,0)=(2?,0) Equation of tangent

y=mx+a2m2+b2y=mx+a2m2+b2?y=mx+4m2+2y=mx+4m2+2?

Passes through (h,k)(h,k) (k−mh)2=4m2+2(k−mh)2=4m2+2

line perpendicular to tangent will have slope

−1m−m1?y−0=−1m(x−2)y−0=−m1?(x−2?)my=−x+2my=−x+2?(h+mk)2=2(h+mk)2=2Add equation (1) and (2) k2(1+m2)+h2(1+m2)=4(1+m2)Add equation (1) and (2) k2(1+m2)+h2(1+m2)=4(1+m2)h2+k2=4h2+k2=4x2+y2=4(Auxiliary circle)x2+y2=4(Auxiliary circle)

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1. ∴(−1,3)∴(−1,3?) lies on the locus.

Q.16 lim?x→1[∫0(x−1)2tan?(t2)dt(x−1)sin?(x−1)]limx→1?[(x−1)sin(x−1)∫0(x−1)2?tan(t2)dt?]

(1) is equal to 1
(2) is equal to 1221? (3) does not exist
(4) is equal to −12−21?

Sol Bouns

lim?x→1[∫0(x−1)2tan?(t2)dt(x−1)sin?(x−1)]x→1lim??(x−1)sin(x−1)∫0(x−1)2?tan(t2)dt??

Using L- Hopital rule

=lim?x→12(x−1)⋅(x−1)2cos?(x−1)4−0(x−1)⋅cos?(x−1)+sin?(x−1)(00)=x→1lim?(x−1)⋅cos(x−1)+sin(x−1)2(x−1)⋅(x−1)2cos(x−1)4−0?(00?)=lim?x→12(x−1)3⋅cos?(x−1)4(x−1)[cos?(x−1)+sin?(x−1)(x−1)]=x→1lim?(x−1)[cos(x−1)+(x−1)sin(x−1)?]2(x−1)3⋅cos(x−1)4?=lim?x→12(x−1)2cos?(x−1)4(x−1)[cos?(x−1)+sin?(x−1)(x−1)]=x→1lim?(x−1)[cos(x−1)+(x−1)sin(x−1)?]2(x−1)2cos(x−1)4?=lim?x→12(x−1)2cos?(x−1)4cos?(x−1)+sin?(x−1)(x−1)=x→1lim?cos(x−1)+(x−1)sin(x−1)?2(x−1)2cos(x−1)4?

on taking limit

=01+1=0=1+10?=0

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17 If ∑i=1n(xi−a)=n∑i=1n?(xi?−a)=n and ∑i=1n(xi−a)2=na∑i=1n?(xi?−a)2=na , (n,a>1)(n,a>1) then the standard deviation of n observations x1,x2,…,xnx1?,x2?,…,xn? is :

(1) a−1a−1? (2) na−1na−1? (3) a−1a−1 (4) a−1a−1?

Sol. 4

Q.18 If αα and ββ be two roots of the equation x2−64x+256=0x2−64x+256=0 . Then the value of

(α3β5)1/8+(β3α5)1/8is:(β5α3?)1/8+(α5β3?)1/8is:

Sol. 3

x2−64x+256=0x2−64x+256=0 α+β=64α+β=64 αβ=256αβ=256

(α3β5)1/8+(β3α5)1/8(β5α3?)1/8+(α5β3?)1/8=α+β(αβ)5/8=64(256)5/8=6432=2=(αβ)5/8α+β?=(256)5/864?=3264?=2

Q.19 The position of a moving car at time tt is given by f(t)=at2+bt+c,t>0f(t)=at2+bt+c,t>0 , where a,ba,b and cc are real numbers greater than 1. Then the average speed of the car over the time interval [t1,t2][t1?,t2?] is attained at the point :

(1)(t1+t2)/2(2)2a(t1+t2)+b(3)(t2−t1)/2(4)a(t2−t1)+b(1)(t1?+t2?)/2(2)2a(t1?+t2?)+b(3)(t2?−t1?)/2(4)a(t2?−t1?)+b

Sol. 1

f′(t)=Vav=f(t2)−f(t1)t2−t1f′(t)=Vav?=t2?−t1?f(t2?)−f(t1?)? =a(t22−t12)+b(t2−t1)t2−t1=t2?−t1?a(t22?−t12?)+b(t2?−t1?)? =a(t1+t2)+b=2at+b=a(t1?+t2?)+b=2at+b t=t1+t22t=2t1?+t2??

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2

50505049(2)50505051(3)50515050(4)5049505050495050?(2)50515050?(3)50505051?(4)50505049?

Sol. 2

I1=∫01(1−x50)100dxI1?=∫01?(1−x50)100dx I2=∫01(1−x50)(1−x50)100dxI2?=∫01?(1−x50)(1−x50)100dx =∫01(1−x50)100dx−∫01x50(1−x50)100dx=∫01?(1−x50)100dx−∫01?x50(1−x50)100dx I2=I1−∫01x−x49(1−x50)100dxdxI2?=I1?−∫01?dxx−x49(1−x50)100?dx

By using by parts

I2=I1−[x(−150)(1−x50)101101]01+∫01(−150)(1−x50)101101I2?=I1?−[x(50−1?)101(1−x50)101?]01?+∫01?(50−1?)101(1−x50)101?I2=I1−0+∫01(1−x50)101(−5050)dxI2?=I1?−0+(−5050)∫01?(1−x50)101?dxI2=I1−I25050I2?=I1?−5050I2??50515050I2=I150505051?I2?=I1?I2=50505051I1I2?=50515050?I1?α=50505051α=50515050?

Q.21 If a?a? and b?b? are unit vectors, then the greatest value of 3?a?+b??+?a?−b??3??a?+b??+?a?−b?? is

Sol. 4

3?a?+b??+?a?−b??3??a?+b??+?a?−b?? =3(2+2cos?θ)+2−2cos?θ=3?(2?+2cosθ)+2?−2cosθ =6(1+cos?θ)+2(1−cos?θ)=6?(1?+cosθ)+2?(1?−cosθ)

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2.22 Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If AD=8 m,BC=11 mAD=8 m,BC=11 m and AB=10 mAB=10 m ; then the distance (in meters) of a point M on AB from the point A such that MD2+MC2MD2+MC2 is minimum is ______.

Sol. 5

[image]

(MD)2=x2+82=x2+64(MD)2=x2+82=x2+64 (MC)2=(10−x)2+(11)2=(x−10)2+121(MC)2=(10−x)2+(11)2=(x−10)2+121 f(x)=(MD)2+(MC)2=x2+64+(x−10)2+(2)f(x)=(MD)2+(MC)2=x2+64+(x−10)2+(2) Differentiate f′(x)=0f′(x)=0 2x+2(x−10)=02x+2(x−10)=0 4x=20⇒x=54x=20⇒x=5 f′′(x)=4>0f′′(x)=4>0 at x=5x=5 point of minima

Q.23 Let f : R → R be defined as

The value of λλ for which f′′(0)f′′(0) exists, is ______.

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5

24 The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45?45? . After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30?30? to the horizontal plane, the angle of elevation of the top of the hill becomes 75?75? . Then the height of the hill (in meters) is ______.

Sol. 80

[image]

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1 x=80cos?30?=403x=80cos30?=403? y=80sin?30?=40y=80sin30?=40 In ADC tan 45?=hx+z⇒h=x+z45?=x+zh?⇒h=x+z ⇒h=403+z…(i)⇒h=403?+z…(i) In AEDF tan 75?h−yz75?zh−y? 2+3=h−40z⇒z=h−402+3…(ii)2+3?=zh−40?⇒z=2+3?h−40?…(ii) Put the value of z from (i) h−403=h−402+3h−403?=2+3?h−40? h(1+3)=40(23+3−1)h(1+3?)=40(23?+3−1) h(1+3)=80(1+3)h(1+3?)=80(1+3?) h=80h=80

Q.25 Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m.n is ______.

Sol. 28

A & B are set No. of subset of A=2mA=2m No. of subset of B=2nB=2n 2m=2n+1122m=2n+112 2m−2n=1122m−2n=112 2n(2m−n−1)=1122n(2m−n−1)=112 2n(2m−n−1)=24(23−1)2n(2m−n−1)=24(23−1) n=4n=4 m−n=3m−n=3 m−4=3⇒m=7m−4=3⇒m=7 m.n = 28

 

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