isat-2011 test paper

8/3/2019 ISAT-2011 Test Paper

1/22

ISAT- 1

ISAT-2011 TEST PAPER

1. A projectile is fired at an angle 60 with some velocity u. If the angle is changed infinitesimally, let thecorresponding fractional changes in the range and the time of flight be x and y, respectively, Then y is

(A)3

2x (B)

3

2 x (C) 2x (D) 2x

Sol. R =g

2sinu2

dR = d2cosg

u2

2

R

dR=

d

g

2sinu

2cos

g

u22

2

R

dR= 2cot2d = x

T =g

sinu2

dT = dcos

g

u2

dcotT

dT= y

120cot2

60cot

2cot2

cot

x

y

=2

1

3

12

3

1

y =2

x

Correct Answer is not in the options.


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ISAT- 2

2. A ball is dropped down vertically from a tall building. After falling a height h it bounces elastically from a tableinclined at an angle and hits a wall at a distance d from the point of earlier impact horizontally, then

(A) = (1/2) sin1 (2d/h) (B) = (1/2) tan1 (d/h)(C*) = (1/4) sin1 (d/h) (D) = (1/2) tan1 (2d/h)

Sol. u = gh2

d =g2

)290(2sinu2

d =

g2

4singh2

=

h

dsin

4

1 1

3. A photon with an initial frequency 1011 Hz scatters of an electron at rest. Its final frequency is 0.91011 Hz.The speed of the scattered electron is close to (h = 6.63 1034 Js, m

e= 9.11031 kg)

(A*) 4 103 ms-1 (B) 3 102 ms-1 (C) 2 106 ms-1 (D) 30 ms-1

Sol. h0

= h + 2evm2

1

h(1011 0.9 1011) =2

evm2

1

vm

10h2

e

10

v = 3.8 103 4 103 m/s

Comprehensive (4 - 5)An -particle experiences the following force due to a nucleus (k > 0)

rr

kF

2

if r > R

= rR

k3

if r < R

4. The correct potential energy diagram for the above force is

(A) (B)

(C) (D*)


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ISAT- 3

Sol. rr

kF

2

if r > R

= rR

k3

if r < R

Uf U

i=

R

rd.F

=

r

2dr

r

k=

r

k

assuming U

= 0

for r > R, U =r

k

r < R

Ur U

R= rd.F

= )Rr(R2

kdr

R

krdr.F 22

3

r

R

3

Ur = UR + )Rr(R2k 223

= ]rR[R2

k 223

R2

kU 0r

U =r

kr > R

U = 3R2

k

(R2

+ r2

) r < R

5. Suppose the particle starts from r = with a kinetic energy just enough to reach r = R. Its kinetic energy atr = R/2 will be(A) k/R (B) (5/8) (k/R) (C*) (3/8) (k/R) (D) 0

Sol. The particle is projected from infinity such that is just reaches r = R i.e. its speed at r = R is zero. Nowapplying energy conservation from r = R to r = R/2.

UR

= UR/2

+ K.E.

R

k=

4

RR

R2

k 223 + KE

R8

k5

R

k = K.E.

.E.KR8

k3

6. Let a particle have an instantaneous position )t(r

, velocity )t(v

and acceleration )t(a

. Necessary con-

ditions for it to be considered as an instantaneous circular motion about the origin are

(A) 0r.a;0v.a;0v.r

(B) 0r.a;0v.a;0v.r

(C) 0r.a;0v.a;0v.r

(D*) 0r.a;0v.a;0v.r


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ISAT- 4

Sol. 0r.v

, 0v.a

, 0r.a

7. A large parallel plate capacitor is made of two metal plates of size 2m 1m. It has a dielectric slab made oftwo dielectrics of permeability K

1and K

2as shown in the figure, the distance between them being 0.1m . It is

charged by a battery of 1000V after which the battery is disconnected. Now the dielectric slab is pulled outby 10cm. The work done in doing so is (ignore the gap between the plates and the dielectric slab)

(A) 2.5 105 0

J (B) -2.5 1050

J(C) 5 105

0J (D*) -5 105

0J

Sol.

Ceq = C1 + C2 = 1.0

2

1.0

3 00

Ceq

= 50 0

Q = 500 1000

energy stored =2

1C

eqV2 =

2

150

0(1000)2

= 250 106J

Finally Ceq

= C1 + C

2

=1.0

)9.0(2

1.0

1.13 00

= 51 0

Energy final =0

62

0

2

51

1050

2

1

Energy final =6

0

2

1051

50

2

1

E = 602

1051

50

2

1 25

0 106J

= 5 105 0

J.

8. A current I is flowing in a long straight wire along the z-axis. A particle with mass m and charge q has an initial

position x0 i and velocity v0 k The z-component of its velocity after a very short time interval t is

(A*) z(t) =

0[1

2

12

o

0x2

I

2

2

m

q(t)2] (B)

z(t) =

0[1 +

2

12

o

0x2

I

2

2

m

q(t)2]

(C) z(t) =

0[1

2

32

o

0x2

I

2

2

m

q(Dt)2] (D)

z(t) =

0

Sol.


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ISAT- 5

r = )x2(q

mv

qB

mv0

0

00

=r

tv0

)cos(vv 0z

= v0

2

)(1

2

( for small angle cos = 1 2

2

)

=

2

220

0r2

)t(v1v

=

20

2

22

0

022

00

vm

q

x22

)t(v1v

2

2

22

0

00z t

m

q

x22

11vv

9. A non-conducting sphere of radius R has a charge Q distributed uniformly over its volume. The sphere issurrounded by a thin metal shell of radius b (b > R) with a charge Q. The space between the shell and thesphere is filled with air. Which of the following graphs correctly represents the corresponding electric field.

(A*) (B)

(C) (D)

Sol.

3R

kQrE

r R

2r

kQE

R < r < b

E

= 0 r > b


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ISAT- 6

10. The magnetic field at the center of a loop carrying a current I in the circuit shown is given by

(A*) k8

7

R3oI

(B) k8

5

R3oI

(C) k8

7

R3

oI (D) k8

5

R3oI

Sol. k3

5

)R2(4k

3R4B 00

= k

2

51

R2

0

= kR24

7 0

11. A current I is flowing in a wire of length l. The total momentum carried by the charge carrier of mass m andcharge q is

(A*)q

mIl (B)

q

m2Il (C)

m

qIl (D)

m

q2Il

Sol. = nqAvd

P = nAmvd

P =m

q

12. In an oil drop experiment, charged oil drops of mass m and charge q are released at a height h, one at a time,

at intervals t > g/h2 . The drops are collected in a large metal sphere of radius R with a small opening at

the top. The total number of drops that are able to enter the sphere will be.

(A) 20

q

)Rh(4mg (B*) 2

20

q

)Rh(4mg

(C)q

)Rh(4mg 20 (D)

q)Rh(

4mg 0

Sol. Let n number of drops have fallen into the sphere and the charge of the sphere has become Q = nq, andthe next drops just reaches the top of the sphere, applying energy conservation

R

kqQ)R2(mgmgh

)Rh(

kQq

mg(h 2R) =Rh

kQq

R

kqQ

mg(h 2R) =)Rh(R

)R2h(kQq

nq

)Rh(mgR4

2

0

None of the answers is matching. Dimensionally B is correct.


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ISAT- 7

13. Two lenses, one biconvex of focal length f1and another biconcave of focal length f

2are placed along the same

axis. They are separated by a certain distance such that a parallel beam of light incident on the convex lensalso emerges parallel from the concave lens subsequently. The magnification of the combination is given by

(A) M = 2221 f/f (B*) M = f2/ f1 (C) M = f1/ f2 (D) M = (f1 f2)(

22

21 ff )

Sol.

m =1

2

f

f

14. The central fringe in a Youngs double slit experiment with the He-Ne laser ( = 632.8 nm) has intensity I0. If

one of the slits is covered by a 5m thick film of plastic (refractive index = 1.4), the intensity becomes I1. The

ratio I1/I

0is close to

(A) 3/16 (B) 1/4 (C) 1/2 (D*) 3/4Sol. Path difference = ( 1) t

= (1.4 1) 5 106

= 2 106 m

Phase difference =2

x

= 19.86 rad

I1

= I0

cos2

2

4

3

I

I

0

1

15. A polarizer is introduced in the path of a beam of unpolarized light incident on a block of transparent material

(refractive index = 3 ). The polarizer can be placed such that its axis is parallel (P) or normal (N) to the plane

of incidence. The incident beam makes an angle with the surface of the block. The light will be completelytransmitted if(A*) = 30 and the polarizer is placed in P (B) = 30 and the polarizer is placed in N(C) = 60 and the polarizer is placed in P (D) = 60 and the polarizer is placed in N

Sol. = tan1 () = 60

= 30 = 30 & polarizer is placed in P.

16. A submarine traveling at 10 ms1 is chasing another one in front of it. It locates its position and speed bysending Sonar (ultrasonic sound) towards it and recording the time of its travel and return frequency. Thefrequency of the Sonar is 25000 Hz and the frequency of the reflected signal is 24900 Hz. If the speed ofsound in water is 1500 ms1 , the speed of the submarine being chased is(A) 16 ms1 (B) 14 ms1 (C*) 13 ms1 (D) 11 ms1

Sol. f = f

10v

vv

s

s

vv

10v

s

sv

s velocity of sound

v = 13 m/s


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ISAT- 8

17. When light of intensity Ireflects from a surface separating two media with refractive index 1and

2(

2>

1),

the intensity of the reflected light is (2

1)2 /(

2+

1)2 . To make reflection zero a thin layer of a material of

refractive index of thickness t is inserted between the two media. The value of and t such that wavelengthof light is not reflected at all is(A) = (

2+

1)/ 2; 2t = (2n+1)/2 (B) = (

1+

2)/ 2; 2t = 2n

(C*) = 21 ; 2t = (2n+1)/2 (D) = 21 ; 2t = 2n

Sol.

For destructive interference of ray 1 and 2

2t = (2n + 1)2

for zero intensity of reflected light

1

= 2

2

1

1

2

2

2

1

1

2

2 = 21

18. A point object is placed below a wide glass plate of refractive index n. As an observer moves from left to rightabove the glass plate, the angle subtended by the apparent object is

(A) 2 tan11n

n2

(B) tan11n

n

2

(C*) 2 tan11n

n2

(D) tan11n

n22

Sol. (C)

19. A light sensor is fixed at one corner of the bottom of a rectangular tank of depth 10 m full of a liquid of

refractive index 3/2 . The illuminated area through which the light can exit at the top of the tank is

(A) a quarter of a circle of radius > 10 m (B) a quarter of a circle of radius < 10 m

(C*) a quarter of a circle of radius = 310 m (D) a quarter of a circle of radius > 310 m

Sol.

sinc

=2

3

60tan10

r

r = 310

Area =4

r2


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ISAT- 9

20. The average pressure on a sphere submerged in water is the pressure at the depth of its center. A sphere ofradius 10 cm made of steel is held in water as shown in the figure. The force that water applies on the surfaceof the shaded hemisphere is (g = 10ms2 ;

water= 103 kgm3)

(A) 0 N (B*) 63 N (C) 126 N (D) 252 NSol. F

Hz= P R2

= g (0.2) (0.1)2 = 63 NtF

vertical= Vg

= 103 1010

1

3

23

=3

20= 21Nt

Fres

= 1021

None of the options is matching so the most appropriate answer is corresponding to the Horizontal thrustforce i.e. 63 Newton.

21. Laplace correction to the speed of sound is made only for gases and not for solids and liquids. This isbecause, in comparison to gases, liquids and solids have(A) larger thermal conductivity(B) much smaller compressibility(C) much smaller coefficient of thermal expansion

(D*) much smaller relative pressure change when the wave is passing through them.Sol. (D)

22. Three rods of equal lengths and cross sectional areas are joined as shown in the figure, with respectivethermal conductivities K, 2Kand K. The left end is at a temperature T

Aand the right end at a temperature T

B.

In steady state, the temperatures T1and T

2at the junctions are given by.

(A*) T1=

5

3T

A+

5

2T

B; TT

2=

5

2T

A+

5

3T

B(B) T

1=

5

4T

A+

5

1T

B; TT

2=

5

1T

A+

5

4T

B

(C) T1=

5

3T

A+

5

2T

B; TT

2=

5

1T

A+

5

4T

B(D) T

1=

5

4T

A+

5

1T

B; TT

2=

5

2T

A+

5

3T

B

Sol.

KA

TT A1

+

KA2

3

TT B1

= 0

T1 = BA T5

2

T5

3


10/22

ISAT- 10

0

KA

TT

KA2

3

TT B2A2

T2

= BA T5

3T

5

2

23. The diameter of a metal wire is measured using a screw gauge, whose circular scale has 50 divisions. Twofull rotations of the circular scale move two main scale divisions of 0.5 mm each. When it is used to measurethe diameter of a wire of length 3.14 m, and resistance 10 , its reading is as shown in the figure. Theresistivity of the wire is

(A) 4.84 105m (B) 2.42 105m (C*) 1.21 105m (D) 2.42 106 mSol. Pitch = 0.5 mm

L.C. =50

5.0= 0.01 mm

P.S.R. = 4 0.5 = 2mmC.S.R. = 20 L.C. = 0.2 mdiameter = 2.2 mm

=14.3

)101.1(10RA 23

= 1.21 105

24. Which of the following quantities has the least number of significant digits?(A) 0.80760 (B*) 0.08765 (C) 5.7423 102 (D) 80.760

Sol. 0.08765

25. In an experiment designed to determine the universal gravitational constant, G, the percentage errors inmeasuring the appropriate mass, length and time variables are given by a, b, c respectively. The total error indetermining G is then(A*) (a + 3b + 2c) (B) ( a + 3b 2c) (C) (2a + 3b + 2c) (D) (a + 9b + 4c)

Sol. [G] = M1L3T2

100T

T2

L

L3

M

M100

G

G

100G

G

= a + 3b + 2c

26. The relative stability of the octahedral complexes of Fe(III) over Fe(II) with the bidentate ligands,(i) HO-CH

2-CH

2-OH, (ii) HO-CH

2-CH

2-NH

2, (iii) H

2N-CH

2-CH

2-NH

2, (iv) H

2N-CH

2-CH

2-SH, follows the order

(A) (i) > (ii) > (iii) > (iv) (B) (ii) > (i) > (iv) > (iii)(C*) (iii) > (ii) > (iv) > (i) (D) (iv) > (i ) > (iii ) > (ii)

Ans. It is fact (A/c to spectrochemical series of ligand strength).

27. Number of isomers that [Pt(Cl)(Br)(NO)(NH3)] exhibit,

(A*) 3 (B) 6 (C) 4 (D) 8Ans. In Mabcd compounds total geometical isomer are three.


11/22

ISAT- 11

28. When a metal is in its low oxidation state, the metal-carbon bond in M-CO is stronger than the metal-chloridebond in M-Cl, because,(A) chloride is a donor and carbon monoxide is a acceptor(B) chloride is a acceptor and carbon monoxide is a donor(C*) chloride is a donor and the carbon monoxide is both a donor as well as a acceptor(D) chloride is both the donor as well as acceptor and the carbon monoxide is both and accepter

Ans. Due to synergic effect in metal carbonyl complexes.

29. Freshly prepared, bright blue coloured, dilute solution of sodium in liquid ammonia can be used to reduce theorganic functional moieties. In this, the actual reducing species is, (e is an electron)(A) [Na(NH

3)n]+ (B) [H

2(NH

3)n]

(C) [NaNH2(NH

3)n] (D*) [e(NH

3)

n]

Sol. Dilute solution of Na & NH3forms ammonated e which act as reducing agent.

Na + (x + y) NH3 [Na(NH

3)

x]+ + [e(NH

3)

y]or [e(NH

3)

n]

ammonated e

30. The statement that is NOT correct in case of silicates is,(A) cement is a silicate(B) the Si-O bond is 50% covalent and 50% ionic(C) silicate structures could have holes to fit cations in tetrahedral and octahedral geometries(D*) silicates are mainly built through SiO

2 units

Sol. Silicate cantions SiO4

4 units in their lattice.

31. The (SiO32)

nare

(A) pyrosilicates (B) orthosilicates (C) cyclic silicates (D) sheet silicatesSol. The normal formula of cyclic silicate anion is (Si

nO

3n2n)

n. So, the ans is cyclic silicate.

32. The oxo-acid of sulphur that does NOT contain S-S bond is,(A) pyrosulphurous acid (H

2S

2O

5) (B) dithionus acid (H

2S

2O

4)

(C) dithionic acid (H2S

2O

6) (D) pyrosulphuric acid (H

2S

2O

7)

Sol. Structure of oxyacid are follows

33. The reason for the formation of H+ when B(OH)3is dissolved in water is,

(A) acidic nature of B(OH)3

(B) high polarizing power of B3+

(C) hydrogen bonding between B(OH)3and water

(D) high electronegativity of oxygenSol. Boric acid is behave as levies acid with H

2O

34. An optically active alcohol (X) on catalytic hydrogenation gives an optically inactive alcohol. The alcohol (X)is(A) 3-ethyl-3-buten-2-ol (B) 3-methyl-3-penten-2-ol(C*) 2-ethyl-3-buten-1-ol (D) 4-methyl-4-penten-2-ol

Sol.


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ISAT- 12

35. The major product formed in the following reaction is

NaOH,aq)ii

Zn/3O)i

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

Sol. OH

Zn/O

2

3

36. The following transformation is effected by

(A) alkaline KMnO4

(B) NaOH/CHCl3

(C*) NaOH/2

(D) peracetic acidSol. I

2in alkaline medium doesnot add to C = C but removes C

1by iodoform reaction.

37. Among the following halides, the one that is least reactive towards solvolysis

(I) (II) (III) (IV)

(A) I (B) II (C*) III (D) IV

Sol. All solvolysis reaction are SN1 and passes, through carbocation formation. The carbocation formed in III is

antiaromatic hence it is least reactive.

38. Isopropanol can be converted to methylacetate using(A*) pyridinium chlorochromate followed by peracetic acid(B) peracetic acid followed by pyridinium chlorochromate(C) pyridinium chlorochromate followed by NaOH/

2

(D) NaOH/2followed by pyridinium chlorochromate

Sol.

39. Among the isomeric butylbenzenes, the one that is NOT oxidized by alkaline KMnO4

to benzoic acid is

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

Sol. Only alkyl benzene with at least one benzylic H is oxidised.


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ISAT- 13

40. The following reaction is effected by

(A*) i. (CH3)

2CHCOCl/AlCl

3; ii. Br

2/FeBr

3; iii. NH

2.NH

2/KOH

(B) i. (CH3)2CH-CH

2Cl/AlCl

3; ii. Br

2/FeBr

3

(C) i. (CH3)2CHCOCl/AlCl3; ii. NH2.NH2/KOH; iii. Br2/FeBr3(D) i. Br

2/FeBr

3; ii. (CH

3)

2CH-CH

2Cl/AlCl

3

Sol.

41. The major product of the following reaction is

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

Sol.

42. Conversion of benzene into 1,3-dibromobenzene is accomplished through(A) i. Br

2/FeBr

3; ii. HNO

3/conc. H

2SO

4; iii. Sn/HCl; iv. NaNO

2/HCl, 0-5C ; v. CuBr

(B) i. Br2/FeBr

3; ii. HNO

3/conc. H

2SO



2

(C*) i. HNO3/conc. H

2SO

4; ii. Br

2/FeBr



(D) i. HNO3/conc. H

2SO

4; ii. Br

2/FeBr



2

Sol.

43. Liquid oxygen and liquid nitrogen are allowed to flow between the poles of an electromagnet. Choose thecorrect observation(A) Both will be attracted but to opposing pole pieces(B) Both will be attracted to the same pole(C) Liquid oxygen will be attracted and liquid nitrogen will be repelled to the same degree(D*) Liquid oxygen will be attracted but liquid nitrogen unaffected

Sol. On the basic of MOT, O2

is paramagnetic while N2

is diamagnetic. So, liquid oxygen will be attracted butliquid nitrogen unattracted.

44. The highest transition energy in the Balmer series in the emission spectra of hydrogenis (R

H=109737 cm1)

(A) 4389.48 cm

1 (B) 2194.74 cm

1 (C) 5486.85 cm

1 (D*) 27434.25 cm

1

Sol. For balmer series of hydrogen n1

= 2, z = 1and for highest energy, n

2=


14/22

ISAT- 14

1

= RH

Z2

22

21 n

1

n

1

1

= 109737 (1)222

1= 27434.25 cm1

45. A one litre glass bulb is evacuated and weighed. The weight is 500 g. It is then filled with an ideal gas at 1 atm

pressure at 312.5 K. The weight of the filled bulb is 501.2 g. The molar weight of the gas is(R = 8 102L.atm.K1.mol1)(A) 28 (B) 34 (C*) 30 (D) 24

Sol. V = 1, wgas

= 501.2 500 = 1.2 gm

using Pv=nRT 1 1 =M

2.1 8 102 312.5

M = 30

46. The van der Waals coeffcient of the inert gases He, Ar and Xe are given below

16.5137.4Xe

20.3337.1Ar38.234.0He

moldm10(b)mol.dm.atm(agasInert 13226

Choose the appropriate pair to complete the following statement.

The increase in the value of a signifies the increasing importance of- interaction while increase

in value in b is due to -(A) Ion-ion; increased atomic volume(B) Induced dipole-induced dipole; increased atomic volume(C) Induced dipole-dipole; dipole-dipole interaction(D) Dipole-dipole; decreasing ionization energies

Sol. On going down the group, size increases due to which atomic volume increases and with increases of sizevanderwall force increases which is induced dipole - induced dipole.

47. Assuming H0 and S0 do not change with temperature, calculate the boiling point of liquid A using thethermodynamic data given below

Thermodynamic data A(liq) A(gas)H

0(kJ/mol) 130 100

S0

(J/K/mol) 100 200

(A*) 300 K (B) 130 K (C) 150 K (D) 50 KSol. G = H TS,

Aliq

AAgas

.

H = 100 (130) = 30 kJ/moleS = 200 100 = 100 J/K /mole 0 = 30,000 T 100 T = 300 K

48. A solution of CaCl2was prepared by dissolving 0.0112 g in 1 kg of distilled water (molar mass of Ca = 41 g mol1

and Cl = 35.5 g mol1). The freezing point constant of water is 2 K.kg mol1. The depression in the freezingpoint of the solution is(A) 0.0002 (B) 0.002 (C) 0.003 (D*) 0.0006

Sol. Tf

= i Kf m For CaCl

2, i = 3

= 3 2x1

111/0112.

= .0006


15/22

ISAT- 15

49. Of the four values of pH given below which is the closet to the pH of 0.004 M carbonic acid(K

a1= 4 107 ; K

a2= 2 1012)

(A*) 4.4 (B) 5.0 (C) 5.4 (D) 4.0Sol. Due to large difference in Ka

1and Ka

2values pH is only due to Ka

1

=3

71

104

104

C

Ka

= .01 (1%)

[H+] = C = 4 103 102 = 4 105

pH = 5 log4= 4.4

50. The Habers process for the production of ammonia involves the equilibrium

N2(g)

+ 3H2(g)

2NH3(g)

Assuming that H and S for the reaction does not change with temperature, which of the statementsis true (H = 95 kJ and S = 190 J/K)(A) Ammonia dissociates spontaneously below 500 K(B*) Ammonia dissociates spontaneously above 500 K(C) Ammonia dissociates at all temperatures(D) Ammonia does not dissociates at any temperature

Sol. G = H TS

Spontaneous, G < 0G = 95 103 T (190)

= 95,000 190 T < 0 95000 < 190 T 190 T > 95000

T > 500 K

51. Martin throws two dice simultaneously. If the sum of the outcomes is 12, he offers lunch at a five star hotelwith probability 2/3. If the sum is 7, he offers lunch with probability 1/2. In all the other cases, he offers lunchwith probability 1/3. Given that the lunch was offered, the probability that the sum of the outcomes equals 12is(A) 1/18 (B*) 1/20 (C) 1/24 (D) 1/36

Sol. P(12) =361

P(7) =6

1

P(not 7 not 12) = 1 6

1

36

1=

36

29

P =

3

1.

36

29

2

1.

6

1

3

2.

36

13

2.

36

1

=2992

2

= 201

40

2

52. A species has an initial population 410. At the end of the first day, the population increases by 50% At theend of the second day, it decreases by the same percentage. If the process continues in the samepattern, the number of days for the population to reach 310 is(A) 10 (B*) 20 (C) 50 (D) 100

Sol. end of first day = (1.5) . 410 =2

3. 220 = 3.219

end of second day = 3.219. 2

1= 3.218


16/22

ISAT- 16

3.219 + 3.218 + 9.217 + 9.216 + 27.215 + . . . . . . . .at the end of 20 trial it will become 310

53. If 4 squares are chosen at random on a chessboard (there are 64 square arranged in 8 rows and 8 col-umns in a chessboard), then the probability that all the four squares are in the same main diagonal is

(A)4

64

48

C

C(B*)

464

48

C

C2 (C)

464

48

C

C3 (D)

464

48

C

C4

Sol.4

64

48

C

C.2Two main diagonal have 8 element.

54. A student was calculating the variance of a data that consists of ten observations. By mistake, he used

one of the observations as 1 instead of 10 and found the variance as100

744and mean as

10

46. The actual

variance of the data is

(A*)100

825(B)

100

725(C)

100

625(D)

100

525

Sol.1046x0 (given)

Now, actual mean10

55x

variance =100

744

100

744

)10(

)xx(........)xx()xx( 2102

22

1

so actual variance = 10

)x10(

10

)x1(

100

744 220

= 100

825

10

461

10

5510

10

1

100

74422

55. A fair coin is tossed 6 times. The probability that the head appears in the sixth trial for the third time is

(A)16

5(B*)

32

5(C)

36

5(D)

64

3

Sol.5

C2 2

1

.2

15

= 32

5

64

10

56. The sum of the roots of the equation x + 1 2log2(2x+ 3) + 2log

4(10 2x) = 0 is

(A*) log211 (B) log

212 (C) log

213 (D) log

214

Sol. x + 1 log2

(2x + 3)2 + log2

(10 2x) = 0

x + 1 = log2

x

2x

210

)32(

2x.2 = x

2x

210

)32(

20.2x 2 = (2x)2 + 9 + 6.2x , put 2x = yy2 14y + 11 = 0


17/22

ISAT- 17

y1

y2

= 11

21 xx2 = 11 x1 + x2 = log2 111

57. Let z = a

5sini

5cos , a R, |a| < 1. then z2010 + z20111 + . . . equals

(A) a1

z2010

(B) a1

a2010

(C) z1

za2010

(D*) z1

a2010

Sol. z = a ei/5

z2010 + z2011 + . . . . . + =z1

z2010

=z1

e.a 402i2010

=z1

a2010

58. The locus of the point z satisfying arg(z + 1) = and arg(z 1) = , when (0, ) vary subject to the

condition tan1

tan

1= 2, is

(A) two parallel lines (B) a single point(C*) a line parallel to the x-axis (D) two intersecting lines

Sol. arg(z + 1) = arg(z 1) =

tan1 1x

y

= & tan1 1x

y=

tan = 1xy

& tan = 1xy

tan1

tan

1= 2

y

1xy

1x= 2 y = 1

59. For the equation, sin x + cos x =

a

1a

2

1, a > 0,

(A) there is no solution, for any a > 0(B) there is a solution, for infinitely many a > 0(C) there is a solution, for two or more, but finitely many a > 0(D*) there is a solution, for exactly one a > 0

Sol. sin x + cos x = 2a

1a

2

1

solution will only be there if a = 1

60. The number of solution of the equation sin x + cos x = 1 sin 2x in the interval [2, 3] is(A) 2 (B) 4 (C*) 6 (D) 8

Sol. sin x + cos x = 1 sin 2x1 + t = t2 2t + 1t2 3t = 0t = 0, 3sin 2x = 0

x =2

n

x =2

,

2

5,

2

3

, 2, 0, 2


18/22

ISAT- 18

61. Consider the circles C1

: x2 + y2 = 64 and C2

with radius 10. If the center of C2

lies on the line y = x and C2

intersects C1

such that the length of the common chord is 16, then the center of the circle C2

is

(A)

2

3,

2

3(B)

2

4,

2

4(C)

2

6,

2

6(D)

2

8,

2

8

Sol. (x a)2 + (y a)2 = 102

x2 + y2 = 64

equation of common chord is

2ax

2ay + 2a2

100 + 64 = 0It passes through (0, 0)

a2 = 18, a = 23

2

6,

2

6

62. A line segment joining (1, 0, 1) and the origin (0, 0, 0) is revolved about the x-axis to form a right circular cone.If (x, y, z) is any point on the cone, other than the origin, then it satisfies the equation(A) x2 2y2 z2 = 0 (B*) x2 y2 z2 = 0 (C) 2x2 y2 2z2 = 0 (D) x2 2y2 2z2= 0

Sol

Equaiton of plane AB is x = 1

OA = 2

Cone OACB can be find by section of plane x = 1 & sphere x2 + y2 + z2 = 22by homoginizing the equation of sphere by plane equation of required conex2 + y2 + z2 = 2(x)2

x2 y2 z2 = 0

63. Let (x, y, z) be any point on the line passing through (x0, y

0, 0) and parallel to the vector kji

. If

x0

+ y0

= 2, then (x, y, z) lies on the plane normal to the vector

(A*) k2ji

(B) k2ji

(C) kji

(D) kji

Sol. Equation of line1

z

1

yy

1

xx 00

Let point on this line is (r + x0, r + y

0,r)

But x0

+ y0

= 2Point is (r + x

0, r + 2 x

0, r)

by eleminating x0

& r byx = r + x

0, y = r + 2 x

0, z = r

we get x + y 2z = 2

Normal vector to the plane k2ji


19/22

ISAT- 19

64. A tangent to the ellipse16

y

25

x 22 = 1 meets the coordinate axes at A and B. If the points A, B and the origin

are vertices of an isosceles triangle, then the length of AB is

(A) 2/41 (B) 41 (C*) 82 (D) 164

Sol.4

siny

5

cosx

= 1

A

0,

cos

5

B

sin4

,0

sin4

cos

5

tan =5

4

AB = 22 eccos16sec25

=

16

25116

25

16125

25161625 = 82

65. Let an

=n

1[(2n + 1) (2n + 2) ... (2n + n)]1/n, for n N and

nlim a

n= eL, then L is

(A*)

3

2

dxxlog (B)

3

2

dxx2log (C) 3

2

dx)x1log( (D) 3

2

dx)x2log(

Sol. an

=

n/1

n

n2.......

n

32

n

22

n

12

log an

=

n

1rn

r2log

n

1

log an

= 1

0

dx)x2log( Put x + 2 = t

= 3

2

dttlog

66. The value of 4

333

n n3

)n3(...21lim

is

(A) 1/2 (B) 9/4 (C*) 27/4 (D) 81/4

Sol.4.n3

)1n3()n3(lim

4

22

n

=4.3

81=

4

27


20/22

ISAT- 20

67. The value of

2/

0

2/xexcos1

xsin2dx is

(A) e/4 (B) e/2 (C) 2e/2 (D*) 2e/4

Sol. dx)2/xtan()2/x(sece2/

0

22/x

at2

x= t

=

4/

0

2t dt2.ttantsece

= 2/02/x 2/xtan2.e

= 2e/4

68. The differential equation satisfied by the family of curves which cut the curves y = x3, R at right anglesis

(A) xy' 3y = 0 (B*) x + 3yy' = 0 (C) y' 3x2 = 0 (D) 3x2y' + 1 = 0Sol. y = x3

dx

dy= 3x2 = 3x

y3. x2 =

x

y3

x

y3

dy

dx

x dx + 3y dy = 0x2 + 3y2 = C

69. Let f(x) = x (|x |) (2 + cos2x), x R. Then the function f : R R is(A) one-one but NOT onto (B*) onto but NOT one-one(C) both one-one and onto (D) neither one-one nor onto

Sol. f(x) = x(|x |) (2 + cos2x)It is onto Range Rvalue at x = 0, is 0so function is many one.

70. The equation 2x3 3x2 + p = 0 has three distinct real roots for all p belonging to(A*) (0, 1) (B) (2, ) (C) (, 1/2) (D) (, 0) (1, )

Sol. p = 2x3 3x2

dx

dy= 6x2 6x

= 6x(x 1)

1 < p < 0

0 < p < 1


21/22

ISAT- 21

71. For a real number x, let [ x ] denote the greatest integer less than or equal to x. Let f : [1/2, ] R be defined

by f(x) = (x [ x ])[x] cos2

x. Then f is

(A*) Continuous at x = 1 but NOT continuous at x = 2(B) Continuous at x = 2 but NOT continuous at x = 1(C) Continuous at both x = 1 and x = 2(D) Discontinuous at x = 1 and x = 2

Sol. f :

,21 R

f(x) = (x [x])[x] cos2

x

cont. at x = 1but not at x = 2

72. Let f : (0, ) R be defined by f(x) = 2xsin2x cos2x. Then0x

lim

f(x) is

(A) 1 (B*) 2 (C) e (D) 2e

Sol. y = 0xlim 2xsin2x cos 2x

log y =0x

lim

(log2 + sin 2x log x + log cos 2x)

log y = log 2 +x2eccos

xloglim

0x +

0xlim

log cos 2x

= log 2 +x2cotx2eccos2

x

1

lim0x

+ 0

y = 2

73. The distance of the point (1, 2, 3) from the plane )k2ji2(.r

= 5 measured parallel to the line

)kji()j2i3(r

is

(A*) 33 (B) 25 (C) 53 (D) 65

Sol.1

3z

1

2y

1

1x (x, y, z) ( + 1, + 2, + 3)

k2ji2.r = 52( + 1) + ( + 2) + 2( + 3) = 5

5 = 15= 3P (2, 1, 0)A (1, 2, 3)

AP = 999 = 33


22/22

74. If the vector k7j4i3

= 21 vv

, where 1v

is parallel to kji

and 2v

is perpendicular to ki2

, then

21 |v|

+ 22 |v|

is

(A) 68 (B) 70 (C) 88 (D*) 92

Sol. k7j4i3vv 21

1v

= kji

2v = (3 ) i + (4 + )j + (7 ) k

2v

. ki2 = 02(3 ) (7 ) = 06 2 7 + = 0 1 = 0 = 1

kjiv1

3v1

k8j3i4v2

89v2

2

2

2

1 vv

= 92

75. A plane H passes through the intersection of the planes )kji.(r

= 3 and )kji.(r

= 2. If H divides

the line segment joining (3, 0, 2) and (0, 3, 1) in the ratio 2 : 1 internally, then the equation of H is

(A*) )k3ji3.(r

= 1 (B) )k3ji.(r

= 3 (C) )kj3i5.(r

= 1 (D) )k3ji2.(r

= 4

Sol.

P (1, 2, 0)(x + y + z + 3) + (x y + z 2) = 0passes through (1, 2, 0)

6 + (

3) = 0 = 23x y + 3z 1 = 0

isat-2011 test paper

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