i sat test paper 2

8/13/2019 I SAT Test Paper 2

http://slidepdf.com/reader/full/i-sat-test-paper-2 1/16

Duration : 2 Hours Max. Marks : 120

INDIAN INSTITUTE OF SPACE SCIENCE AND TECHNOLOGY

(ISAT-2010)

ENTRANCE TEST FOR ADMISSION TO B.TECH PROGRAMMESACADEMIC YEAR 2010-11

INSTRUCTIONS

1. This question paper is in the form of test booklet with 40 questions.

2. A separate OMR answer sheet is provided.

3. Each equations is provided with a text and figure wherever applicable with multiple answers

(A), (B), (C) and (D). Only one of them is correct.

4. Read the instructions on the OMR sheet carefully. Use HB pencil for marking your answers.

5. All questions carry equal marks of THREE for a correct answer and minus ONE for a wrong answer.

6. Multiple answers for a question will be regarded as a wrong answer.

7. Question book-lets have been marked with A or B or C or D or E on the right hand top corner, which

shall be written on the OMR sheet in the box and bubbled appropriately.

8. Enter your name and bubble the roll number correctly on the OMR sheet.

9. Space available in the booklet could be used for rough work, if required. No separate sheet will be

provided.

10. At the end of the test, the OMR sheet should be returned to the invigilator.

PAPER-II (HINTS & SOLUTIONS)(PHYSICS, CHEMISTRY, MATHEMATICS)

CONSTANTS AND UNITS FOR PHYSICS & CHEMISTRY :

Charge of an electron (e) = 1.6 × 10 –19 C Mass of an electron (me) = 9.11 × 10 –31 kg ;

Mass of a nucleon (mnucleon

) = 1.67 × 10 –27 kg Plank's constant (h) = 6.63 × 10 –34 J.s

Stefan-Boltzmann constant (!) = 5.67 × 10 –8 W/m2 . K4 Velocity of light (C) = 3 × 108 m/s

Gravitational constant (G) = 6.67 × 10 –11 m3 /s2 . kg Acceleration due to gravity (g) = 10 m/s2

Boltzmann constant (k) = 1.38 × 10 –23 J/K 1 fermi (fm) = 1 × 10 –15 m

R = 8.314 Jmol –1 K –1 Atomic mass of Fe = 56 g mol –1

1bar = 105 Pa Faraday constant = 96500 Jmol –1



Page # II

PAPER II

PHYSICS

1. An equilateral prism ABC is made of a material of refractive index 2 . A narrow parallel beam of monochro-

matic light falls on the face AB. The beam coming out from the face AC has minimum deviation from the

original path. A part of the beam also comes out of the face BC after undergoing one reflection at AC. The

angle between the beams coming out from these two surface is

(A) 150º (B) 90º (C) 135º (D) 120º

Ans. (D)

Sol.

As the ray suffers minimum deviation the refracted ray through the prism is parallel to base

" r = r# =2

º60 = 30º

The incident angle i is

rsin

isin = % & sin i = %2

× sin 30º & i = 45º

" The emergent angle e = 45º

The reflected ray is shown the required angle

' = 180 – 45º – 15º = 120º

2. An ideal gas undergoes two successive processes A and B. In the process A, the values of the increase (U

in internal energy and the work W done by the gas are (U = 75 J and W = – 72 J, respectively. For the

process B, (U = 0.

(A) Process A is adiabatic, process B is isocaloric

(B) Process A is adiabatic, process B is isothermal

(C) Process A is isothermal, process B is adiabatic

(D) Process A is isobaric, process B is adiabatic

Ans. (B)



Page # III

Sol. First process

(U + W = 0 & (Q = 0 & adiabatic process

Second process

(U = 0 & (T = 0 & isothermal process

3. A thermally conducting piston can move freely in a thermally insulated cylindrical vessel, separating two

compartments. Compartment I contains 14 mg of N2 gas and compartment 2 contains 20 mg of He gas.

When the piston attains its equilibrium position, the length of compartment 1 becomes L1 and that of com-

partment 2 becomes L2 (see figure). The molecular weight of nitrogen is 28 and that of helium is 4. Then the

ratio2

1

L

L is.

(A) 10/1 (B) 1/10 (C) 49/10 (D) 10/49

Ans. (B)

Sol. PV1 = n1 RT

PV2 = n2RT

10

1

420

28

14

Mm

M

m

L

L

n

n

V

V

2

2

1

1

2

1

2

1

2

1 )))))

4. A solid rectangular parallelepiped has sides of lengths x, y and z, respectively. The solid is pulled along the

z - direction which produces an extension (z in this direction. The relative lateral contractions in the x and y

directions are given by z

zv –

y

y

x

x ()

()

(, where v is a constant. The relative charge in the volume of the

solid given by.

(A)z

z)v2 –1( ((B)

z

z)v1( (*(C)

z

z)v21( (*(D)

z

z3(

Ans. (A)

Sol. V = x × y × z

V

dv =

x

dx + y

dy +

z

dz

V

V( = – v

z

z( –

z

zv( +

z

z(

= )v2 –1(z

z(



Page # IV

Comprehensive : Question no. 5-7

Newton’s law, f = ma, for a particle of mass m is applicable only in inertial frames. If a frame moves linearly

with an acceleration a0 with respect to the earth (assumed to be an inertial frame), the acceleration of the

particle is given by ma = F!

– 0am –F

!!

. The term )am( 0

!

+ is known as the pseudoforce acting on the particle

in this frame. Now consider the situation shown in the adjacent figure. A rectangular box ABCD (as shown

below in the picture) of mass M is made to fall vertically with an acceleration of 2g, as seen from the ground.

The sides AB and CD remain horizontal. The sides AD and BC are of length H. A robot firmly attached to the

box holds an object of mass m at the center of the box.

5. The pseudoforce on the object as seen from the box frame is

(A) 2mg upwards (B) 2mg downwards (C) 2Mg downwards (D) 2Mg upwards

Ans. (A)

Sol. When the rectangular box falls with acceleration 2g, the pseudo force acting on the mass m attached to the

box is 2 mg upwards.

6. The net force (pseudoforce + all real forces) on the object as seen from the box frame is

(A) mg downwards (B) 2mg downwards (C) 0 (D) mg upwards

Ans. (C)

Sol. The mass is at rest with respect to the box. Hence the net force is zero.

7. Now the robot releases the object. It will hit

(A) CD in time square root of 2H/g (B) AB in time square root of H/g

(C) AB in time square root of 2H/g (D) CD in time square root of H/g

Ans. (D)

Sol. When released, the relative acceleration of the mass is (2 g – g) = g upwards.

Assuming the mass is the centre (2

H from CD), time required to hit CD is given by

2

Hgt

2

1 2 ) & g

Ht )



Page # V

8. A square loop and an electric dipole P!

are fixed on a light plastic plate. The loop carries a current i as shown

in the figure. A dipole is at the centre of the loop in the direction shown. Electric and magnetic fields exist in

the region such that there is no net torque on the plate due to them. If the magnetic field is in the plane of the

plate as shown in the figure the electric field must be

(A) Along negative y direction (B) Along positive y direction

(C) Along negative z direction (D) Along positive z direction

Ans. (C)

Sol. , -EpBm .+).

E) j(p) j(B)k(m .+).+

)k(EE +)

9. Positive electric charge is distributed uniformly on the surface of a thin spherical shell made of a non con-

ducting material. The sphere is cut into two hemispheres and the top hemisphere is removed. P is a point

that lies in the plane of the rim of the bottom hemisphere, while O is the centre of the rim. If E!

is the electric

field at P, then

(A) E!

is normal to the plane of the rim, pointing downwards

(B) E!

is directed along OP!

(C) 0E )!

(D) E!

is normal to the plane of the rim, pointing upwards

Ans. (D)



Page # VI

Sol. Consider a similar charge hemisphere placed over the net field due to two hemisphere is zero. It is possible

only when the field due to one hemisphere is perpendicular to the plane of rim. Since the charge is positive,

the field will be pointing upward.

10. Two equal positive charges A and B are kept fixed at the positions shown in the figure. A negative charge is

released from the origin. When ' is equal to a certain angle '1, the negative charge starts moving along the

X-axis, whereas when ' is equal to another angle '2, it starts moving along the positive Y-axis. The value of '

1

and '2 are,

(A) 1 2

345

6 74

3sin,

2

1 –(B) 1

2

345

6 79

4sin,

2

1 –(C) 1

2

345

6 73

2sin,

4

1 –(D) 1

2

345

6 75

4sin,

2

1 –

Ans. (B)

Sol. For the released charge to move along Y-axis, the forces along X-axis must balance.

" 210

Kq sin ' = 215

Kq

" sin ' = 2

2

15

10 =

9

4

The only choice is (B) which meets condition for motion along x.

11. An electric charge +q is located at each of the points ±a, ±as2, ±as4 ....ad infinitum on the x-axis, and a

charge -q is located at each of the points ±as, ±as3, ±as5 ......ad infinitum on the same axis. Here s is a

constant > 1. The electrostatic potential at the origin due to the array of charges is

(A) 1 2

345

6 *78 1s

s

a2

q

0(B) 1

2

345

6 78 1 –s

s

a2

q

0(C) 1

2

345

6 *78 1s

1

a2

q

0(D) 1

2

345

6 78 1 –s

1

a2

q

0

Ans. (A)



Page # VII

Sol. The electrostatic potential at the origin is

V = 2 × 1 2

345

6 ** ......

as

kq –

as

kq

as

kq –

a

kq32 = 2

a

kq 1

2

345

6 * .............

s

1

s

1 –1

2 =a

kq2

)s / 11(

1

* =)1s(a

kqs2

*

12. An electron (magnitude of charge e. mass m) is moving in a circular orbit in a magnetic field of magnitude B.

If the orbit contains an integer number n of de Broglie wavelength, the energy of the electron is (h = Planck’s

constant)

(A) n2 1 2

345

6 m4

heB

ð(B) n 1

2

345

6 m4

heB

ð(C) 1

2

345

6 m4

heB

n

1

ð(D) 1

2

345

6 m4

heB

n

12

ð

Ans. (B)

Sol. Radius of orbit =qB

mv

de-Broglie wavelength 9 = n

r2ð

= nqB

mv2ð

& mv

h =

nqB

mv2ð

" 2

1 mv2 =

m4

qB

ð

. nh = n 1 2

345

6 m4

heB

ð

13. A source emits sound having a range

of frequencies ,the distribution of intensity

:(v) sketched in the figure.

An observer moves with a speed u towards the source. Which of the following represents the intensity

distribution as determined by the observer?

(A) (B)

(C) (D)

Ans. (B)

Sol. As the listener moves towards the source apparent frequency increases. Intensity remains the same.



Page # VIII

CHEMISTRY

14. Standard molar enthalpies of several substances are summarised pictorially below. The correct representations

is :

(A) (B)

(C) (D)

Ans. (B)

Sol. (Hºƒ – H2(g) = 0, H+

(aq) = 0

Br2(g) = 31 kJ

H2O(g) = –241.8 kJ

D2O(g) = –249.2 kJ

H2O(l) = –285.2 kJ

15. The observed rate of a chemical reaction is substantially lower than the collision frequency. One or more of

the following statements is/are true to account for this fact.

(a) the reactants do not have the required energy.

(b) the partners do not collide in the proper orientation.

(c) collision complex exists for a very short time.

(d) collision frequency overstimates the number of effective collision.

(A) A, B & C (B) A, B & D (C) B, C & D (D) A, C & DAns. (B)



Page # IX

16. The correct statement(s) for alkali halides is/are :

(a) metal excess defects make NaCl yellow.

(b) metal excess defects make LiCl, NaCl and KCl coloured.

(c) metal excess defects make NaCl yellow but has no effect of LiCl.

(d) metal excess defects make both NaCl and KCl coloured.

(A) A & B (B) A & C (C) A, B & D (D) A

Ans. (C)

Sol. Metal excess defect makes NaCl-yellow,k LiCl-red and KCl-violet.

17. For the cell reactions, Mg(s) + 2Ag+ (aq) —; Mg2+ (aq) + 2Ag(s), Eº(cell)

is 3.17 V at 298 K. The values of

E(cell), (Gº and Q at Ag+ and Mg2+ concentrations of 0.001 and 0.02 M, respectively are :

(A) 3.04 V, –605.8 kJ mol –1, 20000 (B) 3.04 V, 611.8 kJ mol –1, 20000

(C) 3.13 V, –604.0 kJ mol –1, 20 (D) 3.04 V, –611.8 kJ mol –1, 20000

Ans. (D)

Sol. Ecell = Eº(cell) + 206.0 log

)Mg(

)Ag(

2

2

*

*

Ecell = 3.17 + 0.03 log02.0

)001.0( 2

Ecell

= 3.04 V

(Gº = –nFEº = – 2 × 96500 × 3.17 J mol –1

(Gº = –611.81 kJ mol –1

18. Amongst the following, the most thermally stable polymer is

(A) Polyethylene (B) Polypropylene

(C) Polystyrene (D) Poly a-methylstyrene

Ans. (A)

Sol. Linear chain and hence effective packing.

19. Pick the group which does not contain a neutral oxide :

(A) NO2, P4O10, Al2O3, NO (B) MgO, N2O5, SO3, N2O

(C) CO2, SO

3, CaO, XeO

3(D) CO, SiO

2, SnO

2, Na

2O

3

Ans. (C)

Sol. CO2, SO3, XeO3 ..... acidic (non metallic oxides). CaO - basic (metallic oxide)

20. The three p-orbitals of a p-block element E, combine to form p3 hybride orbitals in compound EX3. The

X-E-X bond angle in EX3 is :

(A) 109º (B) 120º (C) 134º (D) 90º

Ans. (D)

Sol. 3p orbitals are mutually perpendicular to each other.



Page # X

21. The species with metal ion having d5 configuration is :

(A) K4[Fe(CN)]

6(B) [Co(NH

3)4]PO

4(C) K

4[Mn(CN)

6] (D) [Co(NH

3)5(SO

4)]NO

3

Ans. (C)

Sol. Mn is in +2 oxidation state and has d5 configuration.

22. The monobasic acid among the following is :

(A) H2PO3 (B) H2S2O7 (C) H3PO2 (D) H4P2O7

Ans. (C)

Sol. H3PO2 is a monobasic acid as there is only one –OH group in it.

23. The best explosive among the following is :

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

Ans. (D)

Sol. The most unstable structure.

24. An organic compound on treatment with chromic acid / H2SO4 gave a clear arange solution which turned

greenish and opaque immediately. The compound is :

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

Ans. (D)

Sol. Secondary alcohols are oxidized to ketones by chromic acid.

25. Among the following, the homo polymer is :

(A) (B)

(C) (D)

Ans. (A)

Sol. Structure (B), (C) and (D) are copolymers.



Page # XI

26. The correct IUPAC nomenclature of the given compound is :

(A) Ethyl-3-aminomethyl-5-cyano-2-hydroxypentanoate

(B) 4-Aminomethyl-5-ethoxycarbonyl-5-hydroxypentanenitrile

(C) 2-Aminomethyl-4-cyano-1-ethoxycarbonylbutanol

(D) Ethyl 4-amino-3-cyanoethyl-2-hydroxybutanoate

Ans. (D)

Sol.

Ethyl-3-aminomethyl-5-cyano-2-hydroxypentanoate

MATHEMATICS

27. Let z1, z

2, z

3 be complex numbers of modulus 1 satisfying |z

1 – z

2|2 + |z

1 – z

3|2 = 4. Then

(A) z1 + z2 = 0 (B) z2 + z3 = 0 (C) z1 + 2z3 = 0 (D) z1 + z2 + z3 = 0

Ans. (B)

Sol. Put z2 = – z3

|z1 + z

3|2 + |z

1 – z3

|2 = 4

indeed 2 × (|z1|2 + |z3|

2) = 4

" z2 + z

3 = 0

28. The number of ways in which 7 balls can be put in 7 bags such that atmost 5 bags are empty is

(A) 77 (B) 77 – 7 (C) 77 – 5 (D) 77 – 2

Ans. (B)

Sol. 7 balls in 7 bags; Atmost 5 bags empty.

Total number of ways = 77

Let 6 bags be empty

& 7C1 = 7

" Atmost 5 bags empty is possible in (77 – 7) ways



Page # XII

29. tan –1 11

2+ 2 tan –1

7

1is

(A) tan –1 77

36(B) tan –1

2

1(C) tan –1

7

1(D)

73

36

Ans. (B)

Sol. tan –1

1111

2

3

4444

5

6

1111

2

3

4444

5

6

+*1

2

345

6 +

49

11

7

1.2

tan11

2 1

= tan –1 1 2

345

6 11

2+ tan –1

149

29.7

1.2

+

1 2

345

6

= tan –1 1 2 34

5 6 11

2+ tan –1 1

2 34

5 6

48

14

= tan –1 11

2 + tan –1 1

2

345

6 24

7 = tan –1

1111

2

3

4444

5

6

.+

*

24

7

11

21

24

7

11

2

= tan –1 1 2

345

6 +.

*142411

7748 = tan –1 1

2

345

6 250

125 = tan –1 1

2

345

6 2

1

30. A traffic police reports that 20 percent of the vehicles passing through a check point are from outside the

state. Then the probability that more than 8 of the next 10 vehicles are from the state is

(A)

9

5

4

5

14 –1 1

2

345

6 (B)

9

5

1

5

411 2

345

6 (C)

9

5

4

5

141 2

345

6 (D) 1 –

9

5

1

5

411 2

345

6

Ans. (C)

Sol. P(outside the state) =5

1

P(inside the state) = 1 – 5

1 =

5

4

" there can be 9 inside state or 10 inside state vehicles.

Required probability = 10C9 .

9

5

41 2

345

6 .

5

1 + 10C10

10

5

41 2

345

6 = 10

9

5

4 (10 + 4) = 10

9

5

414 .



Page # XIII

31. Let c,b,a!

!

!

be three non-zero vectors such that c is not orthogonal to b. If r!

= x i + y j + zk ( i , j , k are unit

vectors ) satisfies br!

!

. = ba!

!

. and c.r!!

= 0, then ar!!

. is

(A) , -bac.b

c.a !!

!!

!!

.11 2

3445

6 (B) , -cb

c.b

c.a !

!

!

!

!!

.11 2

3445

6 (C) , -ac

c.b

c.a !!

!

!

!!

.11 2

3445

6 (D) , -ab

c.b

c.a !

!

!

!

!!

.11 2

3445

6

Ans. (A)

Sol. )ba(c)br(c!

!!!

!!

..)..

b)ac(a)b.c(b)r.c(r)b.c(!

!!!!

!!

!!!!

!

++)+

" b)a.c(a)b.c(r)b.c(!

!!!!

!!!

!

+)

Since 0r.c )!!

" )ab)(a.c(aa)b.c()ar)(b.c(!

!!!!!

!!!!

!!

.+.).

)c.b(

)a.c(ar

!!

!!

!!

). = )ba(!

!

.

)ba(c.b

c.a !!

!!

!!

.11 2

3445

6

32. Let an object be placed at some height h cm and let P and Q be two points of observation which are at adistance 10 cm apart on a line inclined at an angle 15º to the horizontal. If the angles of elevation of the object

from P and Q are 30º and 60º, respectively, then h is

(A) 25 (B) 2 / 5 (C) 65 (D) 10 2

Ans. (D)

Sol.

In ( PQO

º45sin

h=

º30sin

10

h = 10 2



Page # XIV

33. An unbiased die is rolled until two consecutive trials result in even numbered faces. The probability that

exactly six trials are required to get two consecutive even numbered faces is

(A)

6

2

15 1

2

345

6 (B)

6

2

16 1

2

345

6 (C)

6

2

14 1

2

345

6 (D)

6

6

16 1

2

345

6

Ans. (A)

Sol. 5th and 6th trials will have even numbered faces ------ E E

The remaining 4 trials can be filled only as

follows :

2 E 2O ; 1 way

3 E 1O ; 3 way

4O ; 3 way

5 ways to fill and P(E)/ P(O) =2

1" 5

6

2

11 2

345

6

34. A student is allowed to select at most n books from a collection of 2n + 1 books. If the total number of ways

in which a student can select atleast one book is 63, then n is

(A) 9 (B) 3 (C) 8 (D) 4

Ans. (B)

Sol. Atleast one book and atmost

& 2n+1C1 +2n+2C2 + .... + 2n+1Cn = 63

But <*

)

*1n2

0r

r1n2 C = 22n+1 and

2n+1Cr =2n+1C2n+1 –r

" 2[2n+1C1 +....+ 2n+1Cn] 22n+1 – 2

& 2(63) = 22n+1 – 2

& 2n = 6 & n = 3

35. The sum of the series 1 – (3/2) + (5/4) – (7/8) + ......is

(A) 0 (B) 2/9 (C) 2/3 (D) e

Ans. (B)

Sol. S = 1 + 3r + 5r2 + 7r3 + ...........

Sr = r + 3r2 + 7r3 + ...........

_________________

S(1 – r) = 1 + 2r + 2r 2 + 2r3 ....

S(1 – r) = 1 +r1

r2

+



Page # XV

" S 1 2

345

6 *

2

11 = 1 +

32

12

+.

= 1 – 3

2 =

3

1

S ×2

3 =

3

1& S =

9

2

36. A group of 47 students received 27 medals in football, 26 medals in basket ball and 28 medals in cricket, out

of which 8 students got medals in all the three events. Then the number of students who received medals in

exactly two of the events is

(A) 42 (B) 34 (C) 26 (D) 18

Ans. (D)

Sol. n(F) = 27 n(B) = 26

n(C) = 28

n(F = B = C) = 8

n(F > B > C) = n(F) + n(B) + n(C)

– n(A = B) – n(B = C)

– n(F = C) + n(A = B = C)

47 = 27 + 26 + 28 - n(F = B) – n(B = C) – n(E = C) + 8

" n(F = B) + n(B = C) + n(E = C) = 42

" No student received exactly two events = 42 – 3n (A = B = C) = 42 – 24 = 18

37. Let f(x) = 3 ? *

x

0

2 1dt)t(ft, x @ [0, A). Then f(1) is

(A) 1 (B) e (C) e2 (D) e6

Ans. (B)

Sol. f#(x) = 3x2 f(x)

)x(f

)x(f # = 3x2 & log f(x) = x3 + logC

" f(x) =3xCe .....(i)

f(0) = 3 ?0

0

2 )x(f + 1 = 1 & C = 1

" f(x) =3xe & f(1) = e



Page # XVI

38. The general solution of the differential equation ydx – (x2 – 4) dy = 0 satisfies

(A) y4 = C2x

2 –x

*(B) y4 = C

2 –x

2x *(C) y2 = C

2x

2 –x

*(D) y2 = C

2 –x

2x *

Ans. (A)

Sol.

4x

dx2

+

= y

dy& log y =

4

1 log 1

2

34

5

6

*

+

2x

2x

& y4 = C 1 2

345

6 *+

2x

2x

39. If f(x) = [x] denotes the greatest integer function, then , -?5.1

0

22 dx]x[ –]x[ is

(A) 2 –2

3(B) 135/64 (C) 2

2

3* (D) 0

Ans. (A)

Sol. , -? +2 / 3

0

22 dx]x[]x[ = ? ? 11

2

3

44

5

6 +

2 / 3

0

2 / 3

0

22 dx]x[dx]x[ = ? ? ?**1

0

2

1

2 / 3

2

222 dx]x[dx]x[dx]x[

– 44

5

6 *? ?

1

0

2 / 3

1

22 dx]x[dx]x[ = 0 + , -12 + + 2 1 2

345

6 + 2

2

3 – B

C

DEF

G1 2

345

6 +1

2

3 =

2

223 + =

2

3 – 2

40. The value of A;xlim (ex + x)1/x is

(A) 0 (B) 1 (C) e (D) A

Ans. (B)

Sol. A;xlim

xxe

1

x

e

xe

x1

111

2

3

444

5

6

1 2

345

6 * = A;x

lim

xxe

1

x

e

xe

x1

111

2

3

444

5

6

1 2

345

6 * = e0 = 1

i sat test paper 2

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