FST Paper 1 Code 0 Revision Class 14-04-2013

7/28/2019 FST Paper 1 Code 0 Revision Class 14-04-2013

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PAPER - 1INSTRUCTIONSA. General :

1. This Question Paper contains 72 questions.

2. The question paper CODE is printed on the right hand top corner on this sheet of this booklet.

3. No additional sheets will be provided for rough work.

4. Blank paper, clipboard, log tables, slide rules, calculators, cellular phones, pagers and electronic gadgets

in any form are not allowed.

5. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is provided separately.

6. Do not Tamper / mutilate the ORS or this booklet.

7. Do not break the seals of the question-paper booklet before instructed to do so by the invigilators.

8. W rite your Name, Roll No. and Sign in the space provide on the back page of this booklet.

B. Filling the Top-half of the ORS :

Use only Black ball point pen only for filling the ORS. Do not use Gel / Ink / Felt pen as it might smudge the ORS.

9. Write your Roll no. in the boxes given at the top left corner of your ORS with black ball point pen. Also, darken

the corresponding bubbles with Black ball point pen only. Also fill your roll no on the back side of your ORS

in the space provided (if the ORS is both side printed).

10. Fill your Paper Code as mentioned on the Test Paper and darken the corresponding bubble with Black ballpoint pen.

11. If student does not fill his/her roll no. and paper code correctly and properly, then his/her marks will not be

displayed and 5 marks will be deducted (paper wise) from the total.

12. Since it is not possible to erase and correct pen filled bubble, you are advised to be extremely careful while

darken the bubble corresponding to your answer.

13. Neither try to erase / rub / scratch the option nor make the Cross (X) mark on the option once filled. Do not

scribble, smudge, cut, tear, or wrinkle the ORS. Do not put any stray marks or whitener anywhere on the

ORS.

14. If there is any discrepancy between the written data and the bubbled data in your ORS, the bubbled data will

be taken as final.

C. Question paper format and Marking scheme :

15. The question paper consists of 3 parts (Physics, Chemistry & Mathematics). Each part consists of Three

Sections.

16. For each question in SectionI, you will be awarded 3 marks if you darken the bubble(s) corresponding to

the correct choice for the answer and zero mark if no bubbled is darkened. In case of bubbling of incorrect

answer, minus one (1) markwill be awarded.

17. For each question in SectionII, you will be awarded 4 marks if you darken the bubble corresponding to the

correct answer and zero marks if no bubble is darkened. In case of bubbling of incorrect answer, minus

one (1) markwill be awarded.

18. For each question in SectionIII, you will be awarded 4 marks if you darken the bubble corresponding to

the correct choice for the answer and zero mark if no bubbled is darkened. There is no negative marking

for incorrect answer(s) in this section.

DONOTBREAKTHE

SEALSWITHOUTBEING

INSTRUCTEDTODO

SOBYTHEINVIGILATOR

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Date : 14-04-2013 Duration : 3 Hours Max. Marks : 249

CODE0

FULL SYLLABUS TEST (FST)-XI

TARGET : JEE (ADVANCED)-2013 COURSE : REVISION CLASSES


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Space for Rough Work


PART- I - PHYSICS

SECTION - IStraight Objective Type

This section contains 13 multiple choice questions. Each question has choices (A), (B), (C), and (D)out of which ONLY ONE is correct.

1. When a gas is compressed adiabatically by moving a piston down, the temperature of the gas increases. Atthe microscopic level, the average kinetic energy of the molecule increase. This increase in kinetic energyoccurs due to :(A) Collision of gas molecules with entire walls(B) Collision of gas molecules with moving piston(C) The decrease in contact area(D) Because of elastic nature of collisions

2. A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased byT, theweight becomes 150 gm. T should be :

(A) 27 C (B)4

27C (C) 300C (D) 327C

3. A wire AB of length 1m, has a sliding support at one end A and fixedsupport at end B. This wire is vibrated by a tuning fork of frequency 100 Hzdue to which the wire AB vibrates in its fundamental tone. The particle atpoint A is vibrating with an amplitude of 10 mm and at t = 0, the photo graphof the wire is as shown in the figure below. (all the particles are at mean position at t = 0)The transverse displacement of the particles as a function of x and t will be :

(A) y (x, t) = (10 mm) sin

x

2cos (200 t) (B) y (x, t) = (10 mm) sin

x

2sin (200 t)

(C) y (x, t) = (10 mm) cos

x

2sin (200 t) (D) y (x, t) = (10 mm) cos

x

2cos (200 t)


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4. Two particle A and B are initially at a distance x. Initial velocity of particles A and B are 10 m/s and 25 m/srespectively in the direction shown in figure and their constant accelerations are 1 m/s2 and 2 m/s2 respectivelyin the direction shown in figure. What should be the minimum value of x, so that these particle can just avoidcollision:

(A)2

125m (B)

2

75m (C)

4

75m (D)

4

125m

5. Four Students perform resonance tube experiment. The first, second and the third resonant length observedare respectively

1,

2and

3. Whose readings are most appropriate.

(A) 1

= 20 cm, 2

= 58 cm, 3

= 98 cm (B) 1

= 20 cm, 2

= 58 cm, 3

= 96 cm(C)

1= 20 cm,

2= 62 cm,

3= 104 cm (D)

1= 20 cm,

2= 60 cm,

3= 100 cm

6. In a cylindrical container, open to the atmosphere from the top, a liquid is filled upto 10 m depth. Density ofthe liquid varies with depth from the surface as (h) = 100 + 6h2 where h is in meter and is in kg/m3.Thepressure at the bottom of the container will be : (atmospheric pressure = 105 Pa, g = 10 m/sec2)(A) 1.7 105 Pa (B) 1.4 105 Pa (C) 1.6 105 Pa (D) 1.3 105 Pa


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7. For a cyclic process, pressure (P) V/s volume (V) graph is as shown in the figure. Here for process B to C,

P V

1. The correct graph of root mean square speed (C

rms) V/s density () is :

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

8. A canon of mass 1 kg (including the bullet) is connected

with a spring of a spring constant k = 100 N/m, and isperforming SHM with an amplitude of 1 m. When the

cannon reaches x =2

3m from the equilibrium, moving

along +xdirection, a bullet of2

1kg is suddenly fired from

it with a velocity of 20 m/s relative to the ground. What will be the new amplitude of the cannon ?

(A)2

3(B)

2

5(C)

2

7(D)

2

11


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9. A uniform solid cylinder is given a linear velocity and angular velocity, so

that it rolls without sliding up the incline. Out of the four points A,B,C and

centre of mass, about which point its angular momentum will be conserved?

(A and C are taken at appropriate distance)

(A) Point A (B) Point B

(C) Point C (D) Centre of mass

10. On a horizontal disc, a block of mass 1kg is placed at a distance 3cm from the centre of the disc. Thecontact surface between the disc and the block is rough having

s= 1/2.

.Now the disc is rotated with a

constant angular velocity of 10 rad/sec about its own axis The contact force applied by the disc on the block

is :

(A) 109 N (B) 125 N (C) 3 N (D) 10 N

11. Two identical spheres, touching each other, are placed on a rough horizontal surface of static frictionco-efficient

sas shown in figure . An another identical sphere of same mass and same radius is also placed

on the spheres symmetrically. The contact surface between sphere C, with other spheres is smooth. Whatshould be the minimum value of

s ,so that the system can stay in equilibrium.

(A)32

1(B)

33

1(C)

3

1(D)

22

1


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12. A man squatting (sitting) on a weighing machine, gradually stands up and then becomes stationary. thereading of weighing machine during this process will be :(A) constant and equal to mg in magnitude(B) constant and greater than mg in magnitude(C) variable but always greater than mg

(D) Initially greater than mg, then less than mg, and then becomes constant equal to mg.

13. In a container, 1 gm 0C ice is covered by a very light piston which is in the contact with atmosphere fromabove. Now it is slowly heated. How much heat is required to convert it into 190C superheated steam ?Assume the steam to be an ideal gas. L

ffor ice = 80 cal/gm, L

vfor water = 540 cal/gm, S

water= 1 cal/gmC

and for the steam = 4/3, R = 2 cal/mol.K(A) 755 cal (B) 760 cal (C) 765 cal (D) 770 cal

SECTION - IIMultiple Correct Answers Type

This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A),(B), (C) and (D), out of which ONE OR MORE is/are correct.

14. Velocity of particle varies with time as v = t2 5t + 6. The motion of the particle will be :(A) Accelerating for t (2, 2.5)(B) Retarding for t (2, 2.5)

(C) Accelerating for t (2.5, 3)(D) Retarding for t (2.5, 3)

15. On a triangular wedge a block is placed. The contact surface between them is rough having friction

co-efficients=

3

1. With how much acceleration should we accelerate the wedge, so that neither static nor

kinetic friction acts on the block.

(A) g tan i (B) (2g cot ) ( i ) (C) 2g cot j (D) (g sec) (j )


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18. A Uniform verticle rod of mass m and length 2 is hinged at its highest point and is initially at rest. A smallpendulum of mass m and string length , is also suspended from the same point. Initially it is at an angle 60with vertical and then it is released. The pendulum strikes the rod, just after collision, the pendulum comes

to rest, the angular speed of the rod becomes and co-efficient of restitution of this collision is e then

(A) =

g

4

3(B) =

g

3

2(C) e =

3

2(D) e =

4

3

SECTION - III

Comprehension TypeThis section contains 3 paragraphs. Based upon each paragraph, 2 multiple choice questions have tobe answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.


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Paragraph for Question Nos. 19 to 20Buckling of a columnSuppose a rod fixed at both the ends, is subjected to a compressive force. When the compressive force is

too much, the rod may fail due to lateral deflection which is called buckling.

According to Eulers formula, if the compressive force F exceed the critical load FCr

= 2

43

L

Yr, the rod starts

buckling. Here Y = youngs modulus, r = cross section radius of the rod and L = length fo the rod. Thus short,fat and stiff column has a very low tendency to buckle. While long and thin column has a very high tendency

to buckle.

19. An elastic rod is just fit between the walls. Now the temperature of the rod is increased by T. What shouldbe the minimum value of T, so that the rod starts buckle ? (linear expansion coefficient of the rod is )

(A) T = 2

22

L

r

(B) T = 2

23

r

L

(C) T =

L

2

(D) T =r

L2


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20. An isolated part of a bridge of mass m is supported by two identical columnseach of length , crosssectional radius r and young's modulus Y. Whatshould be the minimum crosssection radius r, so that the beam bearingmore load, can escape from buckling ?

(A)

4/1

3

2

Y

mg

(B)

4/1

3

2

Y3

mg2

L/2 L/4

cm

(C)

4/1

3

2

Y3

mg

(D)

4/1

3

2

Y2

mg3

Paragraph for Question Nos. 21 to 22

In a ventury meter tube of cross section area A, water (w

= 103

kg/m3) is flowing steadily. In between the tube, a narrow section

is made whose cross section area is2

A. To measure the

pressure difference at both the sections, two provisions are used.1st provision: Two vertical tube are fitted above the sections. Tomeasure the difference in water level, both the air column are

vibrated by a tuning fork of frequency 160 Hz. The air column offirst tube vibrates in fundamental tone and the air column of thesecond tube vibrates in first overtone.IInd provision: Now a U-tube is arranged between the twosections as shown in the figure, in which a heavy liquid (

= 11

103 kg/m3) is used and the difference in its level is h. (Speed ofsound in air = 320 m/sec, neglect end correction)

21. Velocity of water flowing in the ventury tube of cross section A is :

(A)3

10m/sec. (B)

3

20m/sec (C)

3

40m/sec. (D)

3

50m/sec.

22. The difference in the liquid level in the U-tube is :(A) h = 10 cm (B) h = 20 cm (C) h = 30 cm (D) h = 40 cm


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Paragraph for Question Nos. 23 to 24

An elastic ball is released from a height of 1m above the ground as shown in figure. It falls freely for a distanceof y, and then strikes a smooth fixed incline plane of angle 45. Due to an elastic collision, the ball bounceshorizontally and then performs projectile motion, it reaches the ground at a horizontal distance 'R'.

23. The horizontal distance (R) at which the ball strikes the ground is :

(A) )y1(y2 (B) )y1(y2 (C) )y2(y (D) 2

)y2(y

24. The value of y, so that the range R becomes maximum is :

(A) y =2

1(B) y =

3

1(C) y =

4

1(D) y =

4

3


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PART- II - CHEMISTRY

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,

P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,

Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]

SECTION - I

Straight Objective Type

This section contains 13 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

25. Consider the balanced reaction

2Cl2O7 4ClO2 + 3 O2 (Cl = 35.5)

What can be concluded f rom the coefficients of species in this balanced equation?

(A) For this reaction, exactly 2 g of Cl2O7 must be taken to start the reaction

(B) For this reaction, exactly 2 mol of Cl2O7 must be taken to start the reaction

(C) Mole ratio of Cl2O7ClO2 and O2 during a chemical reaction at any instant (excluding any negative

sign) are 2, 4 and 3 respectively

(D) The ratio of change in number of moles of Cl2O7ClO2 and O2 is 2 : 4 : 3 (excluding any negative

sign)

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

3(s) + 2NH

3(g) ; K

P= 9 atm2

1 mole of LiCl . NH3(s) is placed in an 82.1 L vessel that contains Ne at 3 atm pressure. As some NH

3is

added to the system maintained at 300 K :

(A) LiCl . 3NH3(s) begins to form due to backward reaction as per given equation.

(B) 2 mole NH3is required to be added for complete conversion of LiCl . NH

3(s) to LiCl . 3 NH

3(s)

(C) Pressure in the vessel remains constant at 3 atm.

(D) No LiCl . 3NH3(s) is formed till the pressure in vessel increases to 6 atm.

27. The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure

of 2 atm starting from intial pressure of 1 atm and initial temperature of 300 K (R = 2 cal/mol-degree)

(A) 360 cal (B) 720 cal (C) 800 cal (D) 1000 cal



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28. Which of the following is the best reducing agent amongst the following ?

(A) H3BO

3(B) NaOH (C) NaHCO

3(D) NaH

29. Which of the following molecule contains shortest NO bond ?

(A) NOF (B) NO2 (C) NO3

(D) NH2OH

30. A new flurocarbon of molar mass 102 g mol1 was placed in an electrically heated vessel. When the

pressure was 650 torr, the liquid boiled at 770C. After the boiling point had been reached, it was found that

a current of 0.25 A from a 12.0 volt supply passed for 600 sec vaporises 1.8g of the sample. The molar

enthalpy & internal energy of vaporisation of new flourocarbon will be :(A)H = 102 kJ/mol, E = 99.1 kJ/mol (B) H = 95 kJ/mol, E = 100.3 kJ/mol

(C) H = 107 kJ/mol, E = 105.1 kJ/mol (D) H = 92.7 kJ/mol, E = 97.4 kJ/mol

31. The correct order of increasing bond order :

(A) CO < CO2< CO

32 (CO bond) (B) CN < NCN2 < (CN bond)

(C) ClO < ClO2< ClO

3 < ClO

4 (Cl-O bond) (D) SO

2< SO

42< SO

32 (SO bond)

32. Two mole of an ideal gas is expanded irreversibly and isothermally at 37C until its volume is doubled and

3.41 kJ heat is absorbed from surrounding. Stotal

(system + surrounding) is :

(A) 0.52 J/K (B) 0.52 J/K (C) 22.52 J/K (D) 0


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33.

has correct IUPAC name as :

(A) 3-Carbamoylbenzene-1-carbonitrile (B) 3-Cyanobenzene-1-carboxamide

(C) 3-Cyanobenzamide (D) 3-Aminocarbonylcyanobenzene

34. Which of the following pairs contain functional isomers?

(A) , (B) ,

(C) , (D) ,

35. Silane, SiH4

, cannot be obtained by treating :

(A) Mg2Si with H2O (B) SiCl4 with NaH

(C) SiCl4 with PH3 (D) all the above options would produce SiH4

36. Which of the following pairs consists of only +M groups?

(A) (B) (C) Cl, NO2

(D)


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37. The enthalpy of neutralisation of a weak acid in 1 M solution with a strong base is 56.1 kJ mol1. If the

enthalpy of ionization of the acid is 1.5 kJ mol1 and enthalpy of neutralization of the strong acid with a strong

base is 57.3 kJ equiv1 , what is the % ionization of the weak acid in molar solution (assume the acid to be

monobasic)?

(A) 10 (B) 15 (C) 20 (D) 25

SECTION - II

Multiple Correct Answer Type

This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of

which ONE OR MORE is/are correct.

38. Point out the correct statement(s) :

(A) Condensation occurs at minima of Z vs P graph when Z < 1.

(B) Each of these lines represents isotherms

(C) T1critical temperature

(D) T4critical temperature


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39. Among B2H

6, CH

4, C

2H

6and SiH

4, which gases burn to produce a glassy solid?

(A) B2H

6(B) CH

4(C) C

2H

6(D) SiH

4

40.

Consider an isothermal previously evacuated vessel containing two separate vessels with H2O() and Na

2SO

4(s)

as shown in the figure. Given that

H2O () H

2O (g) ; K

p1= P

1atm

Na2SO

4. 10H

2O(s) Na

2SO

4(s) + 10H

2O(g) ; K

P2= (P

2)10 atm10

Then,

(A) P1

must be equal to P2

(B) If P1< P

2no Na

2SO

4would be converted to Na

2SO

4.10 H

2O at all

(C) If P1

> P2, all the H

2O () must dry up eventually.

(D) Partial pressure of water vapour in the vessel finally (P) must lie in the range P2 P P

1if P

1> P

2.


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41. 1mole each of H2(g) and I2(g) are introduced in a 1L evacuated vessel at 523K and equilibrium

H2(g) + I2(g) 2HI (g) is established. The concentration of HI(g) at equilibrium

(A) Changes on changing pressure(B) Changes on changing temperature(C) Is same even if only 2 mol of HI (g) were introduced in the vessel in the begining.(D) Is same even when a platinum gauze is introduced to catalyse the reaction.

42. Which of the following statement (s) is/are true for the solutions of alkali metals and alkaline earth metals in

ammonia () ?

(A) Concentrated solutions of alkali metals in ammonia are copper - bronzed coloured and have a metallic

lusture.(B) Dilute solutions of alkaline earth metals are bright blue/deep bule black in colour due to the spectrum

from the solvated electron.(C) Concentrated solutions of the alkaline earth metals in ammonia are bronze coloured.(D) Evaporation of the ammonia from solutions of alkali metals yields the metal, but with alkaline earth

metals evaporation of ammonia gives hexammoniates of the metals.

SECTION - III

Comprehension TypeThis section contains 3 paragraphs. Based upon each paragraph, there are 2 questions. Each question has4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.


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Paragraph for Question Nos. 43 to 44The structure of the boron atom clusters in boron hydrides and related species are usefully summarised by

Wade's rules, which have a foundation in molecular orbital theory but are presented here as an empirical

basis for remembering and predicting structures in the following steps:

(A) Any borane or borane anion can be represented by the formula p

mnnHB ; e.g. for B5H111, n = 5, m = 6,

p = 0 for 21410HB , n = 10, m = 4, p = 2.

(B) Count the number of valence electrons in the species [3n+(n + m) + p]

(C) Assign 2n to n ordinary two-electron BH bonds, the remaining 2R = 2n + m + p electrons (or half that

number of electron pairs, R) are used for bonding the cluster.

(D) if R= n + 1, this is called a closo(closed) cluster, a complete polyhedron.(E) if R = n+2, this is called a nido(nest) cluster, a polyhedron with one vertex missing.

(6) if R = n+3, this is called an arachno (web) cluster, a polyhedron with two vertices missing.

For example, consider B12

H12

2, B5H

9and B

5H

11. An electron count shows that there are n + 1, n +

2 and n + 3 electron pairs available for boron cluster bondings in B12

H12

2, B5H

9and B

5H

11respectively; hence

the description of their boron clusters is given as a dodecahedron (a closostructure), an octahedron with one

position vacant, a square pyramid (a nidostructure), and a pentagonal bipyramid with two positions vacant

(an arachnostructure), respectively.



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It should be noted that in cases where there is choice we have not predicited which positions are

vacant, nor what the polyhedron will be (though symmetric structures would be expected to be favoured). In

practice the polyhedron is found to be a trigonal bipyramid, an octahedron, a pentagonal bipyramid, a

dodecahedron, an octadecahedron and an icosahedron for 5, 6, 7, 8, 11 and 12 vertices respectively.

43. 266HB has -

(A) closo-structure (B) nido-structure (C) arachno-structure (D) unpredictable structure

44. B4H

10has -

(A) closo-structure (B) nido-structure(C) arachno-structure (D) unpredictable structure

Paragraph for Question Nos. 45 to 46If given the right amount of energy, electrons can be promoted from a low-energy atomic orbital to a higher-

energy one. This gives rise to an atomic absorption spectrum. Exactly the same process can occur with

molecular orbitals.

HOMO-LUMO gap

Electrons can be promoted from any filled orbital to any empty orbital. The smallest energy difference

between a full and empty molecular orbital is between the HOMO and the LUMO. The smaller this difference,

the less energy will be needed to promote an electron from the HOMO to the LUMO : the smaller the amount

of energy needed, the longer the wavelength of light needed since E = h. Therefore, an important

measurement is the wavelength at which a compound shows maximum absorbance, max.The energy difference between the HOMO and LUMO for conjugated butadiene is less than that for

ethene. Therefore we would expect butadiene to absorb light of longer wavelength than ethene (the longer the

wavelength the lower the energy, E = hc/). This is found to be the case : butadiene absorbs at 215 nm

compared to 185 nm for ethene. The conjugation in butadiene means it absorbs light of a longer wavelength

than ethene. In fact, this is true generally.

The more conjugated a compound is, the smaller the energy transition between its HOMO and

LUMO and hence the longer the wavelength of light it can absorb. Hence UV-visible spectroscopy

can tell us about the conjugation present in a molecule.


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Both ethene and butadiene absorb in the far-UV region of the electromagnetic spectrum (215 nm is just

creeping in to the UV region) but, if we extend the conjugation further, the gap between HOMO and LUMO will

eventually be sufficiently decreased to allow the compound to absorb visible light and hence be coloured. A

good example is the red pigment in tomatoes. It has eleven conjugated double bond (plus two unconjugated)

and absorbs light.

Approximate wavelengths for different colours

Absorbed frequency, nm Colour absorbed Colour transmitted R(CH=CH)nR, n =

200-400 ultraviolet < 8

400 violet yellow-green 8

425 indigo-blue yellow 9

450 blue orange 10

490 blue-green red 11

510 green purple

530 yellow-green violet

550 yellow indigo-blue

590 orange blue

640 red blue-green

730 purple green

Look at the table above and answer the following questions :


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45. Lycopene is expected to absorb which of the following wavelength to a maximum extent ?

(A) 640 nm (B) 380 nm (C) 420 nm (D) 470 nm

46. Look at the structure of -carotene. Its colour is expected to be

(A) red (B) green (C) orange (D) yellow

Paragraph for Question Nos. 47 to 48 THE DRINKING BIRD

Let us consider the scheme of a Chinese toy

known as the drinking bird (Fig. aand b). Oncetriggered to action, which is done by bringing it toposition b, the bird continues to swing up anddown: in one position it drinks water from theglass . and then returns to the upper position, ina seemingly perpetual motion. A detailed analysisof the action of the toy show, however, that it

corresponds to the laws of thermodynamics.

Construction of Toy: As seen from Fig. the

bird consists oftwoflasks joined by a tube, theseflasks containing a volatile liquid andthermetically sealed. The entire system is capable of rotating about its axis fixed in the stationary support,the position of equilibrium shown on the left of the figure. The head of the bird is covered by a layer of amoistureabsorbing material, cotton.


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CHEMISTRY

P1FST-XI-140413C0-21

Working of Toy: The system is brought into action by moistening the head, say, when the beak is dipped inwater, as in position b. Then, the system acts on its own i.e. water vaporizing from the coating of the headlowers its temperature by a few degrees as compared with the temperature of the body (and of thesurroundings). This lowers the vapour pressure in the head chamber. The pressure difference that arises runsthe liquid in the direction of the head. The centre of gravity is shifted, the head becomes heavier and bendsforward. At the turning point of motion the bird assumes a nearly horizontal position, as shown on the right.The beak dips in water, maintaining the moisture content of the head coating. The end of the connecting tubeemerges from the liquid residue in the left flask. In this position of the system, when the liquid being acted onby the force of gravity flows back into the left flask, theinitial position of equilibrium is restored and the cyclethen repeats itself.The drinking bird acts better at a higher temperature and a lower moisture content. With a relative moisturecontent close to 100 per cent the action discontinues. The working substance must have a high vapourpressure, a higher density and a low heat of vaporization; Freon(CCI

3F) is a suitable substance. The

efficiency of the system can be increased by replacing water in the beaker with a more volatile liquid.

47. When the bird is vertical, the liquid flows up because of :(A) low pressure zone created in birds head due to cooling(B) low pressure zone created in birds head due to vaporisation of liquids within birds head(C) High pressure zone created in birds tail due to latent heat(D) none of these

48. A person purchases the toy and keeps it in a small closed glass cabinet. After a few days, the bird will stopdrinking liquid. Just for fun, the person replaces the water beaker by alcohol. Then, when the beak of bird is

dipped in alcohol and released, the bird :(A) seems to like alcohol, and begins drinking again !(B) shows no change in its behaviour and refuses to take up alcohol (a Gandhian Bird!)(C) does not take up alcohol but rebegins to drink water now (for washing away alcohol : Saintly bird!)(D) shows erratic behaviour and flies away



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P1FST-XI-140413C0-22

MATHEMATICS

PART- III - MATHEMATICS

SECTION - I

Straight Objective Type

This section contains 13 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

49. If equation sin4x + asin2x + 1 = 0 has a solution then a (A) (, 2] (B) [2, 2] (C) (, 2] [2, ) (D) [2, )

50. Given Sn

=

n

0rr2

1, S =

0rr2

1. If S S

n

Documents

FST Paper 1 Code 0 Revision Class 14-04-2013