IIT-09-STS7-Paper1 Solns.pdf Jsessionid=DNIPNGLEGLCG (2).PDF Jsessionid=DNIPNGLEGLCG

8/13/2019 IIT-09-STS7-Paper1 Solns.pdf Jsessionid=DNIPNGLEGLCG (2).PDF Jsessionid=DNIPNGLEGLCG

1/18

1

BRILLIA N TS

HO M E BA SED FULL-SYLLA BUS SIM ULATO R TEST SERIES

FOR OUR STUDENTS

TOWARDS

IIT-JO INT EN TRA N C E EXA M INA TIO N , 2009

PART A : MATHEMATICS

SECTION I

1. (C) Origin, z, z, z + zar e the vertices of a rh ombus OACB , with an gleAOB = 120

Area of ACB = a rea o f O AB

=1

2OA OB sin 120

= r2

2

3

2

r2

3

4= 8 3

r 2= 32

r = 4 2

2. (B) For logx + 2

(6 + x x2) to exist x + 2 > 0, x + 2 1

a nd 6 + x x 2> 0 x > 2, x 1 an d

(3 x) (x + 2) > 0 x > 2, x 1 and 2 < x < 3

x ( 2, 1) ( 1, 3) ... (1)

Brilliant Tutorials Pvt. Ltd. IIT/STS VII/MPC/P(I)/Solns-1

PAPER I SOLUTIONS

MATHEMATICS PHYSICS CHEMISTRY

IIT-JEE 2009

STS V II/ M PC / P(I) / SO LN S


2/18

2

The given equa tion becomes (x + 2)log x + 2(6 + x x2

) = (x 3) | x |

6 + x x2= (x 3) | x |

6 + x x 2= (x 3)x if x > 0

2x24x 6 = 0

x 22x 3 = 0

(x 3) (x + 1) = 0

x = 3 or x = 1

Both values of x do not satisfy (1)

If x < 0, 6 + x x 2= (x 3) ( x)

2x = 6 x = 3 wh ich is impossible.

th ere is no solution.

3. (A) n = 0

5050

Cn

x 250 n

3n

= x 2 350

= x 150

coefficient of x27= 50C27

=50

C23

.

4. (C) E qu at ion of a n y lin e t hr ou gh A(2, 1) w h ich is a t t he g rea t est d ist a nce fr omB(1, 3) is the line perpendicular to AB passing through A.

its equat ion is y 1 =1

2x 2

i.e. , x = 2y.

5. (C) f (0) exists f(x) is cont inuous a t x = 0

and f (x) is continu ous at x = 0

a 02+ b 0 + s in (0 + ) = b 02+ a

a = s in

and 2a 0 + b + cos = 2b 0

b = cos

a 2+ b2= 1.



3/18

3

6. (D) I = 10

20

cos x

1 x10

dx


4/18

4

9. (A), (B), (D)

Let r = x i y j z k ; r 3 i j k = j k

i j k

x y z

3 1 1

= j k

y z i 3z x j x 3y k = j k

y + z = 0, 3z x = 1 , x + 3y = 1

y = z, x = 3z 1

Thes e condit i ons a re s a t is fi ed b y the v ectors g iv en in the op t ions g iv en in (A),

(B), (D).

10. (B), (C)

I f l, m, n ar e the d irection cosines of th e required line,

l m n = 0

2 l 3m n = 0

l

2=

m

1=

n

1... (1)

B y put t ing z = 0, in x + y + z 2 = 0 a nd 2x + 3y + z 5 = 0, t he poin t (1, 1, 0)lies on t he line. Simila rly ( 1, 2, 1) lies on t he lin e. ( 1, 2, 1) does not lieon the line.

(B) an d (C) ar e th e corr ect options beca use of (1).

SECTION II I

11. (D) t23t 4 = 0

(t 4) (t + 1) = 0

2x= 4 ( Q2x= 1 which is impossible)

stat ement 1 is false and sta tement 2 is true.

12. (D) Event hough becau se of stat ement 2 which is true x2+ 4x + 2 2x 2+ 2x 1

w here x is an int eger x = 1, 0, 1, 2, 3,

x = 1 makes x2+ 4x + 2 negat ive and x = 0, makes 2x2+ 2x 1 negat ive.

13. (D) U nit digit in 3100

is 1 but t he digit in t he t ent h pla ce is not zer o a nd hen ce

3100

w hen divided by 100 would not give the r emainder 1.

14. (B) Sta tements 1 a nd 2 a re t rue. B ut st a tement 2 is not sufficien t to just ifys ta tement 1.



5/18

5

SECTION IV

15. (C)cos A

a 2

cos B

b

cos C

c=

a

bc

b

a c

b2 c

2 a

2

2abc 2

c2 a

2 b

2

2abc

a2 b

2 c

2

2abc=

a2 b

2

a bc

3a 2+ b2+ c2= 2(a 2+ b2)

a 2+ c2= b2

B =

2

16. (B) Inradius r =Ar ea of AB C

s=

1

2a c

1

2a b c

=a c a c b

a c2 b

2

=a c a c b

2a c

=a c b

2

( Qa2+ c

2= b

2)

17. (B) Now A, C, B ar e in A.P . 2C = A + B = C

C =

3; A =

6, B =

2

Hence sin 2A = sin

3 sin sin

2

3.

=3

2

0 3

2

= 3

18. (C) b2(c + a) a 2(b + c) = c(b2a 2) + ab (b a ) = (b a ) a b.

c2(a + b) b 2(c + a) = a (c2b2) + bc (c b) = (c b ) a b

But b a = c b .

a 2(b + c), b2(c + a ), c2(a + b) ar e in A.P.



6/18

6

19. (C) Dividing bc, ca, a b by a bc,

we get1

a,

1

b,

1

cw hich a r e in H.P .

bc, ca, ab a re also in H .P .

20. (A) Since a, b, c ar e in A.P .,

a = b d , c = b + d

sin (b + c) = a sin (2b + d) = b d ,

s in (c + a) = b sin 2b = b

sin (a + b) = c sin (2b d) = b + d .

sin (2b + d) + sin (2b d) = 2 sin 2b

2 sin 2b cos d = 2 sin 2b cos d = 1.

d = 2n.

t an a = tan (b 2n ) = tan b , tan c = tan (b + 2n) = tan b .

21. (C) f x = 1 e

t a n x

t

1 t2

dt

f (x) = sec2x t a n x

1 t a n2

x

= t a n x

22. (D) Lt

x 0

0

x2

sin t dt

x3

= Lt

x 0

2x sin x

3x2

Lt

x 0

2x sin x

3x2

= 2

3 a nd Lt

x 0

2x sin x

3x2

=2

3

Hence the limit does not exist.

23. (C) f x = x

3

x3 2

e t

dt

f x = 3x2

e x

3 2

3x2

e x

3

= 3x2

e x

3

e2

1 < 0 V x R

Q x2

e x

3

> 0 a nd e2

< 1 .

f(x) decreases in R.



7/18

7

PART B : PHYSICSSECTION I

24. (A) Th e gr avit a t ion al pot en tia l en er gy (G P E ) in t he ca se of a sph er ica l sh ellelement dr, is

d=G

r

4

3 r

2 4 r

2dr

=16

3

2G

2r

4dr

Tota l G.P .E = = d

= 16

3

2G

20

R

r4

dr

=16

15

2G

2R

5

Volume V =4

3R

3

Mass M =4

3R

3

=3

5

G M2

R= 3

5

4

3

13

G M2

V1 3

gravitational pressure Pg=

d

dV

=1

5

4

3

1 3

G M2

V 4 3

= 4

375

13

G M2

V 4 3

PgV

4/3

Aliter solution

G ravita t ional potentia l energy [] = G m

r

1

r



8/18

8

volume V = 43

r3

1

r3

1

V

1

r

3

1

V

1

r V

1 3

[P V] = [En ergy ] Pg= d

dV V

1

3

V

V

1

3 1

V

4

3

25. (C) -r a y s pen et r a t e m et a l (pipe) m u ch bet t er t h a n or -ra y s . E mis sion of ca nbe cut off by th in sheet of paper an d can be cut off by lead.

To cut , very t hick lead is necessar y.

Absorption of -ra ys is less compar ed to a nd -rays .

emitt er is to be chosen.

The count ra te of emission of -ra ys is propor t iona l to the ra te ofd isin t eg ra t ion of a t om s a n d t h is in t ur n is pr opor t ion a l t o r a dioa ct iv e d eca yconstant .

1

T12

A h igher count ra te w ill be obt ained if t he -emit ter ha s T1/2

of a few

hours (ra ther t ha n of severa l years).

26. (B) Kinetic energy of neutron =1

2mv

2

1061.6 1019J = 12

1.67 1027v 2

v = 1.384 107m /s

velocity of cent re of ma ss (C.O.M) = Vc

=m

1v

m1 m

2

=1

1 41.384 10

7

= 2.768 106m /s



9/18

9

vn

= velocity of neutr on in th e C.O.M syst em before collision

= v Vc

= 13.84 1062.768 106= 11.072 106m /s

pn= (v

n) (m

n)

pn

= (11.072 106) (1.67 1027) = 18.49 1021N -s

pH e

w ill be equal t o pn

but opposite in sign .

27. (C) D iffer en t ia t in g g iv en expr ession for V w it h r espect t o x a n d eq ua t in g t o z er o(condit ion for V minimu m for eq uilibrium),

we have

dV

dx = a k sin kx = 0

s in kx = 0

kx = m, (m = 0, 1, 2, ....)

x =m

k

For stable equilibrium,d

2V

dx2< 0

a k2cos kx < 0

a k2cos (m) < 0Case (i) a > 0

d2

V

dx2< 0if m is even (i.e., if m is of form m = 2n w here n = 0, 1, 2, ... .)

x =2n

k

Case (ii) a < 0

d2

V

dx

2< 0if m is odd (i.e., if m is of form m = 2n + 1 wh ere n = 0, 1, 2, ... .)

x =2n 1

k

x =2n

kif a > 0

x =2n 1

kif a < 0



10/18

10

28. (B) V = IR 1I

= RV

Sensitivity can be determined by calculating the va lue

V

1

If.s.d

for X =1

40 10 3

= 25 V

1

If.s.d

for Y =1

50 10 3

= 20 V

X is more sensitive tha n Y.

29. (D)

V = kq

1q

2

rw here k =

1

40

= 2 q

2

x

q2

2x

q2

3x . . . . k

w h er e t h e pr e-fa ct or 2 t a kes ca r e of t he fa ct t ha t t he st rin g of ch a rg es

ar e extended along both d irections.

V = 2 kq2

x1

1

2

1

3

1

4 . . . .

= 2kq

2

xl n 2 Q l n 1 y = y

y2

2

y3

3 .. .

V = 2q2

l n 2

40

x

SECTION II

30. (A), (C)

For (A)

As x = 0 a t t = 0 a nd x = R0

a t t = 2 s, t here a re some w hole number of t ur ns

(in clu din g t he possibilit y of t he w hole n um ber bein g zer o) a nd left over

fra ction of tur n.

t h er e is n o d ef in it e in for m a t ion a b ou t per iod (or eq u iv a len t ly t h e n u mberof turns).

cannotbe determined by the data.



11/18

11

For (C)

For this 3-D motion, v =dx

dt

2

dy

dt

2

dz

dt

2

dx

dt= R sin t

dy

dt= R cos t

dz

dt= c

v = c2 R

0

2(Qusing sin

2+ cos2= 1 for a ny a ngle ).

31. (A), (D)

At a steady sta te, the capacitors will be cha rged to

vC

1

=24

3 9 6 9 = 12 V

a nd vC

2

=24

3 9 6 6 = 8 V

At t > 0,t he circuit w ill be as shown below. Writing KVLs and KCL, we have

24 8 = 3i + 9i124 12 = 3i + 6i

2

i = i1+ i

2

Solving, i1

=36

33A, i

2=

32

33A

Also, we h a ve 8 vs= 6 i

2

8 vs= 6

32

33

v s=24

11V (For t > 0)

Theoretically, it is obvious that as t , vs24 V

32. (B), (C), (D)

Classically: Th e oscilla t in g elect r ic field a s socia t ed w i t h t h e in cid en t lig h tca u ses t h e elect r on s bou n d t o t h e s ur fa ce t o os cilla t e, a b sor bin g en er g y u n t il

they overcome their binding a nd leave th e surface.

Th e ph ot ocu rr en t is pr opor t ion a l t o t h e in t en sit y of in cid en t lig ht , t h etime of exposure and the size of the incident beam.



12/18

12

The photocurr ent is independent of th e frequency of the incident w ave.

Quantum mechanically (a n d a lso exper im en ta lly ): Th e ph ot ocu rr en ta ppea r s a lm ost im media t ely (befor e a n oscilla t in g elect ron sh ou ld h a ve

ov er com e elect r on bin d in g ) a n d t h er e w i ll be a cu t-of f f req uency b elow which

no electr ons a re ejected.

33. (A), (C)

B y da ta T is increased.

U s in g t h e ex pr ess ion f or fu n da m en t a l m od e w h ose fr eq u en cy is g iv en by

f =1

2L

T

, on e ca n sa y t ha t t he w ir e w ill be br ou gh t ba ck t o fun da men ta l

m od e by in cr ea sin g fr eq uen cy or in cr ea sin g len gt h . (H er e is lin ea r ma ss

density of wire).SECTION II I

34. (C) Ferrite cores ha ve only medium h igh flux density (2 kG to 5 kG).

35. (A) (Theory /kn owled ge ba sed).

36. (D) There w ill be tw o peaks in such a gra ph corresponding t o the tw o fragment s.

37. (D) Let, Moment of inertia w ithout t aking ma ss m = I0

Moment of inertia t aking m into considera tion = I

I = I0 m

l

4

l

2

2

wh ere I0= M l

2

12 M

l

2

2

=M l

2

3

I = M l2

3+ m l

4

l

2

2

Using M = 4m a s given

I =91

48

m l2

SECTION IV

38. (C) (Theory /kn owled ge ba sed).

39. (B) Energy is to be conserved

mgh =1

2mv

2w h er e v is velocit y of bob of pen du lu m ju st befor e it h it s t h e

spring.

v = 2gh



13/18

13

As F g iven is conserva tive (being fu nction of x)

we ha ve the potentia l energy as V = F dx

=1

2kx

2

1

4bx

4

B y conservat ion of energy ,

1

2mv

2= mgh =

1

2kx

2

1

4bx

4

x 2 kb

2

= 4mghb

k2

b2

x =4mgh

b

k2

b2

k

b

40. (A) U sing Work-Energy Theorem,

we ha ve, W = EK

But W =

F dx

1

2kx

2

1

4bx

4=

1

2mv

2

x 4+ 2k

bx

2= 2

m

bv

2

x2

k

b

2

= 2m

bv

2

k

b

2

x 2= k

b 2

m

bv

2

k

b

2

x = k

b1

2mv2

b

k2

1

12



14/18

14

41. (B) A gra in (of san d) falling thr ough H w ill acquire a velocity v = 2gHw hen it reaches the bott om and th e whole tr ip ta kes time t

1.

For 0


15/18

15

Current I = Bl

0

N= 0.25 4

4 10 7

100= 7960 A

Resista nce of coil R =l

A=

N 2r

A2 A

1

=3 10

8100 2 1.5

4 2 2 1 104

= 0.047

V = IR

= (7960) (0.047)

= 375 V

P = VI

= (375) (7960) = 2.99 106

3 103kW

45. (B) Let Rate required = f

density of wa ter = d = 1 kg/litre

specific hea t of wa ter = 4190 (SI u nit s)

T = 40C

clear ly , P = d f C(T)

f =P

d C T

3 103 10

3

1 4190 40

18 lit res/s.

46. (C) Magn etic pressure =1

2

B2

0

=0.25

2

2 4 107

2.5 104N/m2



16/18

16

PART C : CHEMISTRY

SECTION I

47. (B) Volume of an at om4

3 r

3=

4

3 a

2

3

= a

3

6

Number of atoms in unit cell =1

8 8 = 1

Volume of unit cell = a3

P acking fraction = 1 a3

6 a3

= 6

48. (C) SO2(g)

+ NO2(g)

SO3(g)

+ NO(g)

In it ia lly 1 mole 1 mole 1 mole 1 mole

At e qu ilibr iu m (1 x) (1 x) (1 + x) (1 + x)

Kc

=S O

3NO

S O2

NO2

=1 x 1 x

1 x 1 x

16 =1 x

2

1 x2 ;

1 x

1 x= 4

x = 0.6

[NO2] = 1 x = 1 0.6 = 0.4 mol L1

49. (B) F or a n y h yd rog en like species, t he en er gy of t he elect ron in t he n th or bit isgiven by

En

= 21.76 10

19Z

2

n2

For He+

in 3rd orbit Z = 2, n = 3.

E3

= 21.76 10

19 4

9

E = E3E

E = 21.76 1019

4

9= 9.68 10

19J



17/18

17

We know hc

= E

=hc

E=

6.62 10 34

J s 3 108

ms1

9.68 1019

J

= 2.05 107

m

= 2050

50. (D)

51. (B) I n d ich rom at e d ia n ion t he six t er min a l C r O bon ds a re eq uiva len t due t o

resonance.

52. (D)

SECTION II

53. (A), (B)

54. (A), (D)

55. (A), (C), (D)

56. (B), (C), (D)

SECTION II I

57. (A)

58. (D)

59. (D) This is due to the fa ct th at the r eaction is reversibleI

2+ CH

4 CH

3I + H I

60. (D) Cell potential is a n int ensive property.

SECTION IV

61. (B) The in volved rea ction is

H2O

2+ H

++ I

I

2+ 2H

2O

1 mol of H2O

21 mol of I

2

Amount of I2l iberated =

0.508g

2 127= 2 10

3mol



18/18

18

Amount of I2in 5 cm

3

solution is 2 103

m ol

Molar concentra tion of H2O

2=

2 103

mol

5 10 3

L

= 0.4 mol L1

62. (A) The equ a tion is 2 H2O

22H

2O + O

2

2 mol of H2O

21 mol of O

222.4 L of O

2a t STP .

2 molar solut ion of H2O

2ha s a volume strengt h of 22.4

0.4 molar solut ion of H2O

2ha s volume strength of

22.4

2

0.4 = 4.48

63. (C) An a queous solution of H2O

2is a cidic.

H2O

2+ H

2O H

3O

++ H O

2

64. (C) F lu or in e h a s m a xim um elect r on w it h dr a w in g effect , h en ce F C H2C OOH is

the str ongest a cid a nd h ence ionizes readily.

65. (A) sp h ybr id ca r bon is m or e elect ron eg at ive t ha n sp2

h ybr id ca r bon w h ich is

m or e elect r on eg a t iv e t h a n s p3

h ybr id ca rbon . As t he h ybr id st a t e of Cchang es from sp

3to sp, the acid strengt h t ends to increase.

66. (D)

67. (A)

68. (C)69. (D)


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IIT-09-STS7-Paper1 Solns.pdf Jsessionid=DNIPNGLEGLCG (2).PDF Jsessionid=DNIPNGLEGLCG