FST Solutions Code 0 Revision Class 14-04-2013

8/22/2019 FST Solutions Code 0 Revision Class 14-04-2013

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SOLSCT140413 - 1

PAPER-1PART-I (Physics)

1. When a gas is ........................................Sol. (B)

When the particle collides with moving piston, they strike withless speed and bounce with greater speed.

2. A rigid container ........................................Sol. (C)

Initially mass of air is 200 100 = 100 gm, finally mass of air is150 100 = 50 gm. As there is a hole in the wall, pressureinside the container will remain constant = P

0

PV = nRT T n

1

as number of moles of gas is halved, the temperature should bedoubled (in K)

Ti= 300 K So T

f= 600 K T = 300 K = 300 C

3. A wire AB of ........................................Sol. (C)

at x = 0, A = maximum, so A(x) = A cos (kx)at t = 0, all the particles are passing through mean position so y(x, t) = (A coskx) sin (t)

4. Two particle A ............................... .........Sol. (B)

vrel

2 = urel

2 + 2 arel

Srel

0 = (15)2 + 2(3) Srel

Srel

=2

75m.

5. Four Students perform ........................................Sol. (C)

1+

0f4

V

2+

0f4

V3

3+ =

0f4

V5

2

= 31

+ 2 3

= 51

+ 4

so second resonance length is more than 31

and third

resonance length is more than 51.

6. In a cylindrical ........................................Sol. (D)

dp = gdh

PP

PP

10h

0h

2

0

gdh)h6100(dp

Pf P

0= 10h 0h3h20h1000

P

f= 105 + 0.3 105 = 1.3 105 P

a

7. For a cyclic process ........................................Sol. (A)

from B C T = constantfrom C A, P = constant

HINTS & SOLUTIONS

FULL SYLLABUS TEST (FST)-XITARGET : JEE (ADVANCED)-2013

V T T1 2

rmsC

1

8. A canon of mass ........................................Sol. (B)

Initially n

=m

k=

1

100= 10 sec1

and v = 22n xA = 10

22 )2/3(1

= 5 m/sec.Velocity of the cannon after firing the bullet can be foundusing momentum conservation

(m) (5) =

2m

(20) +

2m

V

v = 10m/sec.

v = 221n xA 10

=

2

21

2

3A

)2/1(

100

A

1=

2

5m.

9. A uniform solid ........................................

Sol. (C)

Whether the cylinder rolls up or rolls down, the static friction

will always act up the incline and its value will be fr=

2mR1

sinmg

which is less than mg sinSo about point C which is above cm. torque can be balanced.

So about point C, its angular momentum will be conserved.

We can also find the distance of point C from cm. suppose it is

x. Applying torque balance about point C.

(mg sin) (x) =

2

11

sinmg

(R + x)

Solving we get x = 2R

10. On a horizontal ........................................Sol. (A)

R = 22 Nfr R = 222 mgrm = 109 N

Date : 14-04-2013 COURSE : REVISION CLASSES


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SOLSCT140413 - 2

11. Two identical ........................................

Sol. (B)

For equilibrium of upper shpere 2N cos 30 = mg

N =3

mg

For the equilibrium of lower sphere

Breaking force welding strength (limiting friction)N sin 30 (

s) (mg + N cos 30)

Solving we get s

s

33

1

12. A man squatting ........................................

Sol. (D)

Initially he is sitting so Vi= 0 and finally be becomes stationery

so Vf= 0. So graph of his velocity v/s time graph will be roughly

as shown below

As slope of Vt graph will be initially positive and then negative,

so acceleration of the man will also be initially positive and then

negative. Initially when acceleration is +ve, N = m (g + a) so

reading will be greater than mg, then acceleration will become-ve, N = m (g a) so reading of weighing machine will be less

than mg. And finally man becomes stationery so reading of

weighing machine will be equal to mg.

13. In a container ........................................Sol. (B)

To convert 0 ice into 0 water heat required = 1 80 = 80 calTo convert 0 water into 100 water heat required= 1 1 100 = 100 calTo convert 100 water into 100 steam heat required= 1 540 = 540 calTo convert 100 steam into 190 steam heat required

= ncpT =

18

1(4R) 90 = 40 cal

Total heat given = 80 + 100 + 540 + 40 = 760 cal.

14. Velocity of particle ........................................Sol. (AD)

For t (2, 2.5) speed so motion will be acceleratingfor t (2.5, 3) speed so motion will be retarding

15. On a triangular ........................................

Sol. (ABD)

(A) If ma cos = mg sin a = g tan then no friction will act onthe block.

(B) If acceleration towards x direction is more than g cot,then the block will loose the contact with the wedge. So again

no friction will act on the block.

(D) If downward acceleration is greater than g then the block

will again loose the contact with the wedge. So again fr= 0

16. A wedge of mass ........................................

Sol. (BD)

When the particle reaches diametrically opposite point, sup-

pose displacement of the wedge is x till that time.

(Sx)cm

=21

2x21x1

mm

)S(m)S(m

0 =2/mm

)R2x()2/m()x()m(

x =3

2R which will be maximum displacement of the wedge

because at that instant both the particles and the wedge will

come at rest for a moment. In this way the wedge will perform

oscillatory motion.

As extreme to extreme displacement of wedge is3

R2so its

amplitude is3

R

17. A horizontal cylinder ........................................Sol. (AD)

N =R

mu2+ mg (3 cos 2), at = 120,

N = 0 N =R

mu2+ mg(3cos120 2) = 0

u= gR5.3

If the block is moving clockwise, then to cross the point P, theblock has to cross the highest point, so to cross the highest

point u = gR5

18. A Uniform verticle ........................................

Sol. (AD)

The ball will strike the rod with a velocity of

V = )cos1(gR2 V = g .

Applying angular momentum conservation about the hinge point

gm + 0 = 0 +

3

)2(m 2() get =

4

3

g


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SOLSCT140413 - 3

also e = collisionbeforeapproachofVelocity

collisiontheaftersaperationofVelocity

e =

g

0 where =

4

3

g

so e =4

3

19. An elastic rod ........................................Sol. ( A)

If we increase the temperature by T,Compressive force generated is F = YA TFor buckling F = YA T critical load of buckling

(Y) (r2) T 2

43

L

Yr T 2

22

L

r

.

20. An isolated part ........................................Sol. (B)

N1+ N

2= mg ..................(i)

(N1)

2

= (N2) 4

..................(ii)

Solving we get,

N1=

3

mg, N

2=

3

mg2

To avoid buckling,

N2= 2

43Yr

3

mg2

L/2 L/4

cm

mg

N2

N1

21. Velocity of water ........................................Sol. (B)

difference in water level

2

1=

2

=

0f2

V=

1602

320

= 1 m

Pressure difference (P1 P

2) =

wgy

= (103) (10) (1) = 104 Paapplying Bernoulli equation between section (1) and section(2)

P1+

2

1V2 = P

2+

2

1(2V)2

P1 P

2=

2

3V2 104 =

2

3(103)V2 V =

3

20

22. The difference ........................................Sol. ( A)

Balancing pressure at level (1) and (2)

P1

+ wg (h

0+ h) = P

2+

wgh

0+

gh

(P1 P

2) = (

+

w) gh

104 = (11 103 103) (10) h h = 10 cm

23. The horizontal ........................................Ans. (B)

24. The value of ........................................Ans. (A)

Sol.(23 to 24)

Horizontal range =g

h2u

R =g

)y1(2gy2

R = )y1(y2 R will be maximum when R2 will be maximum and R2 will be

maximum when 0dy

)R(d 2

0dy

)R(d 2 = 4(1 2y) = 0 y =

2

1

PART-II (Chemistry)

25. Consider the balanced...............

Sol. (D)

It follows directly form definition of stoichiometry.

26. Given : LiCl . 3NH3(s) LiCl . NH

3(s) + 2NH

3(g) ;

KP

= 9 atm2 ..............

Sol. (D)

Moles of NH3

in vessel =RT

PV=

3000821.0

1.82atm3

= 10 mol

These moles must be present in the vessel before the equilibrium

begins to move backwards and conversion of LiCl . NH3(s) to

LiCl . 3NH3

(s) even begins.

Moles of NH3required for conversion of LiCl . NH

3(s) to LiCl

. 3NH3(s) is 12.

27. The work done in adiabatic compression................

Sol. (B)

q = 0, U = WnC

V(T

2T

1) = P

ext(V

2V

1)

n2

3R (T

2 300) = 2

1

300nR

2

nRT2

2

3(T

2300) = (600T

2) T2 = 420 K

W =nCV(T

2T

1) = 2

2

3 2 (420 300) = 720 cal.

28. Which of the following is the best ...............

Sol. (D)


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SOLSCT140413 - 4

29. Which of the following molecule ...............

Sol. (A)

30. A new flurocarbon of molar mass 102 g mol1 .....................Sol. (A)

Molar mass = 102 gram/mole

P = 650 torr ; T = 77 + 273 = 350 K

Q = i t = 0.25 600 = 150 C

E = Q V = 150 12 = 1800 J

This heat is supplied to the system at constant pressure thats

why this is used for change in enthalpy For vaporisation of 1.8 gram, amount of heat required q =

1800 J

For vaporisation of 102 gram, amount of heat required q

=8.1

1800 102 J

= 102 103 J = 102 KJ/mole

H = U + PVH = U + n

gRT

For determination of U per mol (ng

= 1)

102 (KJ/mol) = U + (1 8.3 350) 103 U = 102 2.9 = 99.1 KJ/mole

31. The correct order of increasing ...............

Sol. (C)

32. Two mole of an ideal gas is expanded .............

Sol. (B)

Ssystem

= nR In1

2

V

V= 2 R In 2 = 11.52 J/K

Ssurrounding

= 310

100041.3 = 11 J/K

Stotal

= + 11.52 11 = + 0.52 J/K

33. ................

Sol. (B)

(A) 2NHC

||O

, when attached to a ring, is a named as

carboxamide.

(B) has higher priority than C N

(C) C N has prefix cyano

34. Which of the following pairs ..............................

Sol. (B)

35. Silane, SiH4, cannot be ..................

Sol. (C)

36. Which of the following pairs consists ................

Sol. (D)

37. The enthalpy of neutralisation of .................

Sol. (C)

Hionization = Hneutralization of WA + SB Hneutralizationof SA + SB

= 56.1 (57.3) = 1.2 KJ eq1

Enthalpy of ionization for making 100% ionization when there is

no ionization at all = 1.5 KJ eq1

% of ionization =5.1

2.15.1 100 = 20%

38. Point out the correct statement(s) ..................

Sol. (BC)

39. Among B2H

6, CH

4, C

2H

6and SiH

4, ...............

Sol. (AD)

40.

Consider an isothermal previously .........................

Sol. (BD)

41. 1 mole each of H2(g) and I2(g) are ...................

Sol. (BCD)

42. Which of the following statement (s) is/are true ..................

Sol. (A,B,C,D)

(A) Due to the formation of metal ion clusters

(B) M + (x + y) NH3 M+(NH3)X + e(NH3)y

(C) due to the formation of metal clusters.

(D) M (NH3)6 true statement

43. 266HB has ................

Sol. (A)

[6 3 + 6 + 2 = 26 e or 13 e pairs (total)] [6 2 = 12 e = 6e

pairs] = 7 e pairs

n B atoms have (n +1) e pairs for cluster bonding.closo structure (wades rules)

44. B4H

10has ............

Sol. (C)

[4 3 + 10 = 22 electron 11 electron pairs] [4] = 7 electronpairs for cluster bonding.

n = 4, (n + 3) electrons pairs are available for cluster bonding,hence arachno structure.

closo structure (wades rules)

45. Lycopene is expected to absorb ..................

Sol. (D)

46. Look at the structure of -carotene ................

Sol. (A)

47. When the bird is vertical, the ..............

Sol. (A)


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SOLSCT140413 - 5

Due to cooling, low pressure created in bird. So due to

capaliary action liquid flows upward.

48. A person purchases the toy and keeps ............

Sol. (A)

Again alcohol is more volatile and high vapour density. so this

process again repeat.

PART-III (Mathematics)

49. If equation sin4x + asin2x + 1 = 0....................Sol. (A) Let t = sin2x t [0, 1]

f(0).f(1) 0 2 + a 0 a (, 2]

50. Given Sn

=

n

0rr2

1, ....................

Sol. (C) Sn

=

2

11

2

11

1n= 2 n2

1, S =

1

11

2

= 2

S Sn

=n2

1 1000

now, 512 < 1000 < 1024 29 < 1000 < 210

if 2n > 1000, then least value of n is 10.

51. The remainder obtained ........................ .Sol. (A) 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

and n!/15 for n 5 required remainder is same as the remainder obtained bydividing 33 with 15, i.e. equal to 3.

52. Total number of polynomials ........... ..............Sol. (A) We must have

i3 + ai2 + bi + c = 0

and (

i)

3

+ a(

i)

2

+ b(

i) + c = 0 c a + (b 1) i = 0 b = 1, a = cThus total number of such polynomials = 10C

1= 10.

53. If

113

010

121

z

y

x

=

1

2

0

.........................

Sol. (B)

zyx3

y

zy2x

=

1

2

0

y = 2, x + 2y + z = 0 x z = 4, 3x y + z = 1

3x + z = 3 x =4

7, z =

4

9

54.0x

lim [sinx], where [.] .........................

Sol. (C) LHL at x = 0 is 1RHL at x = 0 is 0

so limit doesn't exist

55. A variable line L is drawn.........................Sol. (D) x = rcos, y = rsin

putting in L1

we getL1

rsin= rcos+ 10

OA

1=

10

cossin

Similarly putting the general point of drawn line in the equation of

L2, we get

OB1 =

20cos

sin

rcos= h, rsin= k

we haver

2=

10

cossin +

20

cossin

40 = 3rsin 3rcos 3y 3x = 40

56. In triangle ABC equation ...................... ...Sol. (B) Reflection of P(5, 8) in BC will lie on circumcircle

Clearly P1 (8, 5)

Thus equation of circumcircle is (x 2)2 + (y 3)2

= (8 2)2 + (5 3)2

57. If two different tangents of ..................... ....

Sol. (B) Tangent to y2 = 4x in terms of m is y = mx +m

1

normals to x2 = 4by in terms of m is y = mx + 2b + 2m

b

If these are same lines, thenm

1= 2b + 2m

b

2bm2 m + b = 0For two different tangents, we must have

1 8b2 > 0 |b| a

Similarly3

2m

a+

3

2m

b> c,

3

2m

a+

3

2m

c> b

3

4(ma + mb + mc) > a + b + c ma + mb + mc > 43

(a + b + c)

62. In an arithmetic progression.........................

Sol. (BC) Let the successive terms of the A.P. be a1,a2 ...., a9 a10.

By hypothesis a2 + a4 + a6 + a8 + a10 = 15

i.e., (a + d) + (a + 3d) + ...... + (a + 9d) = 15

i.e. , 5a + 25d = 15 .... (i)

and a1 + a3 + a7 + a9 = 2

112

5a + 20d =2

112

From (i) and (ii)., we get 5d =

2

12 or d =

2

1and a =

2

1

Hence the A.P. is2

1, 1, 1

2

1, 2, .....

63. If a, b, c are in H.P., .........................Sol. (ABCD)

(A) ifacb

a

,

bac

b

,

cba

c

are in H.P..

a

acb ,

b

bca ,

c

cba are in A.P..

a

cba

, b

cba

, c

cba

are in A.P..

a

1,

b

1,

c

1are in A.P..

a, b, c are in H.P.

(B)b

2=

ab

1+

cb

1

2(b2 bc ab + ac) = 2b2 b(a + c) b(a + c) = 2ac

b =ca

ac2

a, b, c are in H.P.

(C) a 2

b,

2

b, c

2

bare in G.P..P.

2

bc

2

ba

4

b2

4

b2= ac

2

b(a + c) +

4

b2

ac =2

)ca(b

b =ca

ac2

a,b,c are in H.P.

(D)cb

a

,

ac

b

,

ba

c

are in H.P..

a

cb ,

b

ac ,

c

ba are in A.P..

a

cba ,

b

cba ,

c

cba are in A.P..

a

1

, b

1

, c

1

are in A.P.. a,b,c are in H.P.

64. Total number of ways of.........................Sol. (CD) Required number = number selections of one or more out

of three 25 paise coins and two 50 paise coins= 4 3 1 = 11

65. A(1, 2) and B(7, 10) .........................

Sol. (BC)

P must lie on the perpendicular bisector of AB.Equation of perpendicular of AB is AB

y 6 = 102

)71((x 4) 3x + 4y 36 = 0

also AB = 10 AL = 5

AH = ALcosec60 =3

10

66. If (acosi, asin

i) ; .........................

Sol. (AB) The given points lie on the circle x2 + y2 = a2 for all (i), sincethe points are the vertices of an equilateral triangle, thecircumcentre and the centroid of the triangle are same.

3

1a (sin

1+ sin

2+ sin

3) = 0

cos1+ cos

2+ cos

3= sin

1+ sin

2+ sin

3= 0

67. Latus rectum of .........................Sol. (C) If (x, y) is the centroid of triangle formed by A(0, 0), P(t)

and Q

t

1is

3x = 2a

2

2

t

1t

3y = 4a

t

1t

Eliminating t we get


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SOLSCT140413 - 7

9y2 = 8a(3x 4a)

or y2 =8a

3

4ax

3

\ length of latus rectum is3

a8

68. Vertex of parabola .........................

Sol. (B) Since y2 =3a4 4ax

3

is the equation of parabola P1

hence vertex is

0,

3

a4

69. (C) In the expansion of ...................... ...70. (B) The number of integral .........................Sol. 69 T

r + 1= 6561C

r(71/3)6561 r

(111/9)r

= 6561Cr.72187 r/3

11r/9

Tr + 1

is rational

If9

rand

3

rare integers

r is a multiple of 9.0 r 6561

9

r= 0,1,2,....., 729

Total terms = 730

70. Tr + 1

= 256Cr(31/2)256 r

(51/8)r

= 256Cr.2128 r/2

5r/8

for Tr + 1

is integer, then 128 2

r0 ,

8

r0

r = 0, 8, 16,...., 256

Number of integral terms = 33

71. If1p

1+

2p

1+

3p

1=

2

1.........................

72.

1p

Acos+

2p

Bcos+

3p

Ccos=

Sol.71 (D)1p

1+

2p

1+

3p

1=

2

1

AM G.M.

3

p

1

p

1

p

1

321

3/1

321 p1.

p1.

p1

or

3

6

1

321 ppp

1

p1

p2

p3 216

72. (B)1p

Acos+

2p

Bcos+

3p

Ccos

=

2

acosA +

2

bcosB +

2

ccosC

=

2

)C2sinB2sinA2(sinR

=

R2 a b c. .

2R 2R 2R

=

2R4

R4=

R

1

PAPER-2PART-I (Physics)

1. A Cylindrical rod ......................................Sol. (D)

for equilibrium Buoancy force (B) = mgNow lets displace the rod by an angle

net

= (mg)

4

sin

net

=

4

mg

net

= K K=4

mg

T = mg

4

2

cm

Where cm

+

2

4m

=

0

cm

= 0

16

m 2.

2. In a region, potential ..................................... .Sol (C)

The potential energy graph can be assumed to be equivalent toa mountain. A particle is released from x 10+ with forwardinitial velocity. The particle will move forward and then returnback. And when it will pass through x = 0, its speed will bemaximum. applying energy conservation between x = 10 to x =0 :

Ki+ U

i= K

f+ U

f

21 (2) (2)2 + 10 =

21 (2) v2 + 0

V = s/m14 .

3. A football player ......................................

Sol. (C)

t

VVa if

=t

)jV()jV(

a

= )i(t

V

+ j

t

V


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SOLSCT140413 - 8

As average acceleration is towards north west, the external

force also should be towards north west. This external forcemust be static friction.

4. A uniform rod of ......................................Sol. ( A)

Applying torque balance about the hinge point

(mg)

2

= (T) ()

T =2

mg= 40 N

So wave velocity will be

V = T

=1.0

40 = 20 m/sec

Time taken to reach the topmost point

t =speed

cetandis=

20

1= 5 102 seconds.

5. A river is 240 m ......................................Sol. (C)

Time to cross the river

t =y

y

V

S=

4

240= 60 seconds

Drifting during this timeS

x= V

xt

Sx

= (3 1) (60)S

x= 120 m

Time to walk along the shore

t =1

120= 120 seconds

Total time to go from A to B = 60 + 120 = 180 seconds = 3minutes.

6. 20 gm of water at ......................................

Sol. (C)

Heat available by water till 0C

= msT = 20 1 30 = 600 calHeat required by ice to reach 0C water

= mL + msT = 5 80 + 5 80 + 5 0.5 10 =425 calThus we have 25 gm of water at 0C and extra heat 600 425

= 175 cal

Hence, 175 = 25 1 TT = 7C (temp. of mixture)

7. A cuboidal vessel of ......................................Sol. (B)

dgha2 = ah2

dgh h = 2m.

8. Surface mass ......................................

Sol. (A)dm = 2 rdr () = (Kr2) 2 rdr

= 1

0

2

2

dmr= 1 kg.m2 .

9. A 1m long tube is ....................... ...............

Sol. (B)

Temperature of the gas remains constant so

PiV

i= P

fV

f

(P0) (0.25) = (P

0gh) (1 h)

solving h = 50 cm

10. Three blocks and ......................................Sol. (C)

a =

428

)g4)5.0(g2()g8(

a = 7

g2m/sec

2

to find the tension at the middle point of the rod, draw the FBDof the part :

7g T = (7) (a)

Solving T = 50 N = 5 kg-force

11. A uniform cubical ......................................Ans. (A) q, s ; (B) p, q, s ; (C) s, t ; (D) p, q, r

Sol.


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SOLSCT140413 - 9

If surface is rough, the breaking force in x direction is zero,

while welding strength (limiting friction) will be non-zero. So

sliding cannot occur. But the cube can topple about the corner

point A.

For topping :-

(Fa) > (mg)

2

a F >

2

mg

(A) If f = 3

mg

then the block will neither slide nor topple. But F

force will create an anticlockwise torque which will lift the right

half and press the left half. So at right half pressure is less and

at left half pressure will be more. So net pressing force (Normal

reaction) will be shifted to the left of the middle point.

(B) F =3

mg2: The block cannot slide, but as the force is

greater than2

mg, So it will start topple about the corner point

A.

(C) F =2mg3

, since F > mg, so the block will be lifted up. so it

cannot slide also.

(D) F =4

mg3, and the surface is smooth. As F >

2

mg, so the

block start toppling about the corner point A. To keep the centre

of mass stationary in x direction, the corner point A will have to

slide toward right.

12. In a container, an ........................ ..............Ans. (A) q,s,t ; (B) p,r,t ; (C) p,r ; (D) p,sSol. (A) Initially the gas is compressed slowly so process is

isothermal. Then the gas is expended, so the process will bealmost adiabatic. So, pv graph will be as shown :

So , pf< p

iand T

f< T

iand W

net= negative

(B) Initially the gas is expanded slowly against atmosphericpressure. So the process is isobaric expansion. Then it iscompressed adiabatic. So the pv graph will be as shown :

So, pf> p

i, T

f> T

iand W

net= negative

(C) As the piston moves very slowly its acceleration a 0, Fnet

0 upwards and downwards forces should be almostbalanced. So PA = P

0A + Kx

Pressure of gas will be P = P0

+A

Kx

As the gas is expanded, pressure will increase. As both P andV are increasing. So temperature will also increase. As volumeis increasing, W

gas= positive.

(D) dQ = 2dU

dQ =

RdT

23n2

R3n

dT/dQ

Molar heat capacity C = 3R = constant, So the process must be

a polytropic process. For a polytropic process C = Cv

+x1

R

= 3R x =3

1

C = Cv+

x1

R

= 3R x =

3

1

So, equation of the polytropic process will be

PV1/3 = constant 3/1v

1P So, V P

Now lets convert this equation into vT

3/1vV

nRT

= constant T V2/3

So, V T

and V So, Wgas will be positive.

13. An object is moving ......................................Ans. 9

Sol. V = 2t (12 6) =dt

ds

ss

0sds =

tt

0tt2 (12 t) dt S = 12t2

3

t2 3

Now =t

S

=t

3

t2t12

32

= 12t 3

t2 2

will be maximum whendt

vd = 0

12 3

t4= 0 t = 9 second

14. Helium gas is ......................................Ans. 8Sol. Assume pressure and volume at state B are P

Band V

Brespec-

tively.from A B process is isothermalso (2P

0) (2V

0) = (P

B) (V

B) .....(1)

from B C process is adiabatic

PBVBr

= P0V0r

.....(2)

Solving we get VB

=8

V0.

15. An astronaut is ......................................Ans. 5

Sol.g2

u2

= 10m

g

u2= 4 sec

Solving , g = 5 m/s2 .

16. Wind entering in .................... ..................

Ans. 2

Sol. Energy entering in the windmill


10/14

SOLSCT140413 - 10

=2

1mv2

Pin

=dt

dE=

2v2

1

dt

dm

Pin

=

2v2

1(AV) =

2

1 A V3

Electrical power output

Pout

=

3AV

2

1

3

1

Pout

=3AV

6

1 =

3)10(102.16

1

Pout

= 2 kW.

17. An aeroplane of ......................................

Ans. 5

Sol.

Applying bernaullis equation for air above and below the wing

P1+

2

1V

12 = P

2+

2

1V

22

(P2 P

1) =

2

1(V

1+ V

2) (V

1 V

2)

Here V1

= V + VV

2= V V

So, (P2 P

1) =

2

1(2V) (2V) = 2V V

For level flight

(P2 P

1) = mg

(2V V)A = mgPutting the values we will get

V = 5 m/sec.

18. A wooden block ......................................

Ans. 2

Sol. Due to the perfectly inelastic collision,

Eloss

= )mm(2mm

21

21 (u1 u2)

2 (1 e2)

Eloss

=)m9m(2

)m9)(m(

(u 0)2 (1 02)

Eloss

=2u

20

m9

Heat produce =

2u20

m9

2

1=

2u40

m9

Due to this heat, the temperature of the bullet will reach its

melting point, and then the bullet will melt, so

Heat produced =2u

40

m9

= mST + mL

f

22. TlI3contains ..........

2u40

m9

= m(2000) (200) + m(5 105)

Solving u = 2 103 m/sec.

19. A small uniform ..................................... .

Ans. 2

Sol. Lets see relative to the truck

For perfect rolling relative to the truck

arel

= R .............(i)ma f

r= (m) (a

rel) .............(ii)

(fr) (R) = )(2

MR2

.............(iii)

Solving we get arel

=3

a2=

3

32

arel = 2 m/sec2

Srel

= urel

+2

1a

relt2

4 = 0 +2

1(2) t2 t = 2 seconds.

20. A particle is moved ......................................Ans. 2Sol. work done in path 2

w2

= )jdyidx(F

= )jdxidx(jxas the line : y = x

= 1

0dxx =

2

1J.

work done in path 1w

1= w

OB+ w

BA

= idxF

+

jdyF

= jdyjxidxjx

= 0 +

1

0y

jdy)j1(= 1 J

2

1

w

w= 2. Ans. 2

PART-II (Chemistry)

21. Which of the following is the hybridisation for the ................

Sol. (A)

P

Cl

||O

Cl Cl

sp3

I

F

||O

F

F

F

sp d3 2

F


11/14

SOLSCT140413 - 11

Sol. (A)

23. Bubbling CO2 through which of the ..................

Sol. (A)

2NaAlO2 + CO2 +3H2O Na2CO3 + 2Al(OH)3

24. Which of the following is a water .................

Sol. (D)

25. Addition of water to which of the ...................

Sol. (D)

26. A certain mass of gas is expanded ..................

Sol. (B)

State 1 State 2

(1 L, 10 atm, 300 K) (4L, 5 atm, 600 K)

Heat given, q = 50 (600 300) = 15000 J

W = Pext

(V2 V

1) = 1 (41) = 3 L atm = 300 J

E = q + W = 15000 300 = 14700 JH = E + P

2V

2 P

1V

1= 14700 J + (20 10) 100 J = 15.7 KJ

27. Which of the compounds does not ..............

Sol. (D)

28.

Which alkyl group is .......................

Sol. (A)

29. Which of the following pairs consists of ................

Sol. (A)

30. The average OH bond energy ......................

Sol. (C)

(A) H2O() H

2O(g) H = 40.6 KJ/mole

(B) 2H(g) H2(g) H = 435.0 KJ/mole

(C) 2O(g) O2(g) H = 489.6 KJ/mole

(D) H = 571.6 KJ/mole

(A) Calculation of Hf

(H2O,)

2H2(g) + O

2(g) 2H

2O() H = 571.6 KJ/mole

Hr= 2H

F(H

2O, ) 2H

F{H

2,(g)} H

F(O

2, g)

Zero Zero571.6 = 2H

F(H

2O, ) so H

F(H

2O, ) = 285.8 KJ/mole

(B) Calculation of HF(H

2O, g)

H2O() H

2O(g) H = 40.6

Hr= H

F(H

2O, g) H (H

2O,)

HF

(H2O, g) = H

F(H

2O, ) + H

r

= 285.8 + 40 = 245.8 KJ/mole

(C) H = 245.8 KJ/mole

Hr = HH

+2

1

OO 2

OH

245.8 = + 435 + 2

1(489.6) 2

OH

2OH

= 435 + 244.8 + 245.8

2OH = 925.6

OH

= 462.8 KJ/mole

31. (A) Na3B

3O

6...................................

Sol. (A) p, r, t (B) p, s, t (C) p, q, r, s, t (D) p, s

(A) Na3B

3O

6

(B) NaCaAlSi3O

9

(C) Borax [Na2B

4O

7.10H

2O Na

2B

4(OH)

4O

5.8H

2O]

(D) P4O

10

32. (A) Perfectly crystalline solid .................... .....

Sol. (A q) ; (B p, r, s, t) ; (C p, s) ; (D r, t).

(A) IIIrd law states 0TLim

S 0 for perfect solid.

(B) In reversible cyclic process the every net energy change is zero

(i.e. Hsys = 0, Usurr = 0 and net heat exchange = 0) and Ssurr = 0.(C) In irreversible cyclic process Ssys

= 0 but Ssurrounding

> 0 by IInd

law and all state function are zero for it.

(D) In adiabatic process net heat change is zero so Ssurrounding

= 0.

33. Sum of oxidation numbers of all the ..............

Sol. (7)

KO2, K = + 1 On two 'O' atoms, Sum

PbO2, O = 2 2 2 = 4. BaO

2: Ba = + 2

On two 'O' atoms

34. Consider the equilibrium ..................

Sol. (4)

Ni(s) + 4CO(g) Ni(CO)4(g)

P P

For backward reactionQ

p K

p

p4K

P

P


12/14

SOLSCT140413 - 12

3-

3atm125.0

P

1

P3 8 atm3

P 2 atmP

Total= 2 P = 4 atm.

35. ICl3 is an orange colored solid ....................

Sol. (08)

I2Cl6 is a planar molecule.

36. How many of these species are ..................

Sol. (4)

O2, O

2+, O

2, B

2

37. reacts with NH2OH to form ...................

Sol. (4)

38. 1 mole of how many of the following acids ......................

Sol. (4)

(HCl, HNO3, H3PO2, H3BO3) H2SO4, H2SO3, H3PO3, H4P2O5 are

diprotic. HCl, HNO3, H3PO2, H3BO3 are monoprotic.

39. How many of these are hydrolysed .............................

Sol. (6)

SOCl2 + H2O H2SO3 + 2HCl

SO3 + H2O H2SO4

XeO4 + 2H2O H2XeO4

POCl3 + H2O H3PO4 + 3HCl

PCl5 + H2O H3PO4 + 5HCl

SiCl4 + H2O H2SiO3 + 4HCl

40. Intramolecular H-Bonding is observed .................Sol. (2)

,

PART-III (Mathematics)

41. Number of integral .......................................

Sol. (B) 0)5x(.)1x(log

)8x()3x()2e()8x2x(22

2

x2

(x 3) (x 8) 0 and x 5 x [3,8] x2 2x + 8 > 0 xR

ex + 2 > 0 x R(x 5)2 > 0 x R {5}

and log2(x2 + 1) x R

Number of integral values of x is 5 (x = 3,4,6,7,8 x 5).

42. Sum of first 2n terms ...................................Sol. (D) (13+33+53+....+ (2n 1)3 + (1.12 + 2.32 + 3.52 +....+ n(2n 1)2)

=

n

1r

n

1r

22n

1r

3 )1r3()1r2()1r2(r)1r2(

=

n

1r

23 )1r7r16r12( =6

)1n(n (18n2 14n + 5) n

43.

n

0r

rn )rx(sinC is .......................................

Sol. (A) In

=

n

0r

xrir

n eC

= In

= ((1 + eix)n)

= In

=

n2

2

xcos.

2

xsini2

2

xcos2

=

2

nxsin.

2

xcos.2 nn

44. 15 identical balls have to ..............................Sol. (A) Let the balls put in the box are x

1, x

2, x

3, x

4and x

5.

we must havex

1+ x

2+ x

3+ x

4+ x

5= 15, x

i 2

(x1

2) + (x2

2) + (x3

2) + (x4

2) + (x5

2) = 55 + 5 1C5= 9C

5

45. Line ax + by + p = 0 .......................................Sol. (B) Lines xcos + ysin = p, xsin ycos = 0 are mutually

perpendicular. Thus, ax + by + p = 0 will be equally inclined tothese lines and would be the angle bisector of these lines.

Now equations of angle bisectors is

xsin ycos = ( xcos + ysin p) x(cos sin) + y(sin + cos) = por x(sin + cos) y(cos sin) = pcomparing these lines with ax + by + p = 0, we get

a

cos sin =

cossinb

= 1 a2 + b2 = 2

46. If the circles x2 + y2 + 2ax + 2by = 0 ............Sol. (C) Both circles pass through origin. If they touch each other

then tangents drawn to respective circles at origin must beidentical lines.i.e. ax + by = 0 and bx + cy = 0 should represents same line

c

b

b

a b2 = ac

47. Maximum number of.....................................Sol. (D) Normals to y2 = 4ax and x2 = 4by in terms of 'm' are

y = mx 2am am3 and y = mx + 2b + 2

b

mfor a common normal

2b + 2m

b+ 2am + am3 = 0


13/14

SOLSCT140413 - 13

am5 + 2am3 + 2bm2 + b = 0That means there can be atmost 5 common normals

48. If the tangent to the.....................................

Sol. (A)4

y

16

x 22 = 1.The tangent at P() is

4

xcos +

2

ysin = 1.

The focus of the parabola x2 = 8(y 6),is (0, 8).

0 +

2

8sin = 1 sin=

4

1.

49. Locus of the point of intersection ..................Sol. (D) Let the normal drawn to the curve at P(t

1) cuts the curve

again at Q(t2), then

If R(h,k) be the point of intersection of tangents drawn to curveat P and Q, then

h =21

21

tt

tct2

, k = 21 ttc2

k

h= t

1t2= 2

1t

1

Also h

31

1 t

1

t = 2ct1.

31t

1

h(t14 1) = 2ct

1

h2(t14 1)2 = 4c2t

12

h2

2

2

2

1h

k

= 4c2

h

k

Thus locus is (x2 y2)2 + 4c2xy = 0

50. Ifa

xcos=

b

xsin,.......................................

Sol. (D)a

xcos =b

xsin tanx =ab

cos2x = 22

22

ba

ba

and sin2x = 22 ba

ab2

so acos2x + bsin2x = 3 22 2a aba b

= a

51. Ans. (A) (q,s), (B) (p,t), (C) (r,s,t), (D) (q)(A) If , are the .................................

Sol. (A) sinx = 2

1 x =

6

7,

6

11

and cosx = 2

3 x =

6

5,

6

7

so + = 3 and =

(B ) cotx = 3 x = 6

5,

6

11

and cosecx = 2 x =6

7,

6

11

so + = 2 and = (C) sinx =

21 x =

67 ,

611

and tanx =1

3 x =

6

,

6

7

so + = 3 and + = 2and = (D) = 0 =

52. Ans. (A) (r,t), (B) (p), (C) (q,s), (D) (q)(A ) The solution of ..............................

Sol. (A) |x + y| = 10 ........(1)

and log10

y log10

|x| = log100

4

y = 2|x| ........(2)

so from (1) and (2) we have x =3

10and y =

3

20

or x = 10 and y = 20

(B) Let log2x = A and log

2y = B

4A2 + 1 = 2B and 2A B

A =2

1

also

2

2

14

+ 1 = 2B B = 1

log2x =

2

1and log

2y = 1

x = 2 and y = 2

(C) log4x = log

2y

xlog 22 = log2y

x = y2 and x2 5y2 + 4 = 0 x2 5x + 4 = 0 x = 1, x = 4 y = 1, y = 2 Let y > 0 Solution set are {1, 1}, {4, 2}

(D) (x 1)2 + (y 1)2 = 0x = 1, y = 1

53. Ifxlog3log 44 3x2 = 27, .... ...................... ...

Sol. (4)xlog3log 44 3x2 = 27 log

4x = 2

x = 4

54. If 3 be the remainder, ................... ...........Sol. (5) 22005 = 2.22004 = 2.(24)501 = 2(17 1)501

= 2.{501C0. 17501501C

1. 17500 +..........}

= 34( 1) + 32= 34( 1) + 17 + 15

3 = 15 = 5

55. Among the 8! permutations .................... ......Sol. (7) Let the arrangement be x

1x

2x

3x

4x

5x

6x

7x

8.

Clearly 5 should occupy the position x4

or x5.

Thus, required number = 2(7!) i.e. = 7

56. If Sn

= 31

1+ 33 21

21

..............................

Sol. (1) Tr= 333 r...21

r.....321

= 2

r

1

1r

1

n

1r)r(T = 2

1n

1

1 = Sn

nlim [S

n] = 1


14/14

SOLSCT140413 14

57. Consider the point ................... ...........Sol. (2) We have |PA PB| AB Thus for |PA PB| to be maximum,

point A,B and P must be collinear. Equation of AB is x + 2y = 2.

Solving it with give line we get P 24 17

,5 5

so a + 2b =24 34

5

= 2

58. Locus of the mid point .................... ..........Sol. (2) Clearly the points A,B and P are concyclic, we have

PD2 = DA2 = BD2 = OA2 OD2

Let D (x, y) x2 + (y b)2 = a2 (x2 + y2) 2x2 + 2y22by + b2 a2 = 0 + + = 2

59. Consider an ellipse with .................... ..........Sol. (2) Equation of ellipse can to take as

2 2x y1

25 16 with e =

5

3

Thus one focus is (3, 0)Let the drawn circle touches the ellipse at point ' P'. Focal distance

of P = a ex1

= 5 5.5

3cos.

= 5 3cosIts minimum value is 2. Thus radius of largest circle is 2 units.

60. Ellipse 1by

ax

2

2

2

2 and ............................

Sol. (2) We have aee

= Aeh

and b = B

ee2 a2 = A2.e

h2

ee2 .

)e1(

b2e

2

= eh2 .

2

2h

B

(e 1)

2e

1

11

e

=

2he

11

1 2

h2e e

1

e

1 = 2

FST Solutions Code 0 Revision Class 14-04-2013

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Transcript of FST Solutions Code 0 Revision Class 14-04-2013