merination notes

Page # 1

BITSAT PAPER

09.09.2012 (PT-02)

PHYSICAL CHEMISTRY

ORGANIC CHEMISTRY TEST SYLLABUSReaction Mechanism : SN1 of R�X, R�O�R & R�OH, E1 in Alcohols, S

N2 of R�X, R�O�R & R�OH, Sni, Intramo-

lecular SN2, Sayt Zeff product, Hoffmann product, E1 CB, Comparison of E2/E1CB Physical Properties

of Alkyl halide, Alcohol & Ether.

Physical SCQ (32)1. A solution is a mixture of 0.06 M KCl and 0.06 M KI. AgNO

3 solution is being added drop by drop till AgCl

starts precipitating (Ksp

AgCl = 1 x 1010 and Ksp

AgI = 4 x 1016). The concentration of iodide ion at this instantwill be nearly equal to :(A) 4.0 x 105M (B) 4 x 108M (C) 2.4 x 108M (D*) 2.4 x 107M

Sol. When AgCl Starts precipitating [Ag+] = 06.0

10 10

at that time conc of [S�] = 10

16

10

104

= 2.4 × 10�7

2. Heat of neutralization of NH4OH and HCl is

(A) 13.7 kcal/mole (B*) < 13.7 kcal/mole (C) > 13.7 kcal/mole (D) ZeroSol. NH

4OH is a weak base. Heat of netralisation < 13.7 kcal.

3. The solubility of AgCl will be minimum in

(A) 0.001 M AgNO3

(B) Pure watera (C*) 0.01 M CaCl2

(D) 0.01 M NaCl

Sol. 0.01 M CaCl2 gives maximum Cl� ions. To keep K

sp of AgCl constant, decrease in [Ag+] will be maximum

4. A weak base BOH (0.1 mole) is titrated with strong acid HCl (0.08 mole) than the number of H+ ion is (Kb for

BOH = 10�4)(A*) 24.08 20�3 (B) 4 10�10 (C) 6.02 1013 (D) None

Sol. BOH + HCl BCl + H2O

t = 0 0.1 mole 0.08 molet= eq. 0.02 mole � 0.08 mole as solution is bufferso

[OH�] = 10�4 08.0

02.0

[OH�] = 10�4 41

mole

[H+] = 4

14

10

410

[H+] = 4 10�10

No. of H+ ion = 4 10�10 6.02 1023

= 24.08 10+13

Page # 2

5. A certain buffer solution contains X� and HX with their concentrations related as ]X[

]HX[ = 0.2

If the value of Kb at 25°C for X� is 10�9, pH of the buffer at 25°C is : (log 2 = 0.3)

(A*) 5.7 (B) 8.3 (C) 9.7 (D) 4.3Sol. K

b for X� = 10�9 K

a for HX = 10�5 pK

a = 5

pH = pKa + log

10

]HX[]X[

= 5 + 0.7 = 5.7

6. The pH of a 20M

solution of a weak base, if its Kb value at 25°C is 2.5 10�3, will be : [Given log 11.18 = 1.05]

(A) 11.95 (B*) 12 (C) 12.05 (D) 1.95

Sol. Kb =

1C 2

= 0.2

[OH�] = C= 0.05 0.2 = 0.01 pOH = 2 pH = 14 � 2 = 12

7. If equal volume of following solutions are mixed, precipitation of Hg2I

2 (K

sp = 2.5 10�26) will occur only with :

(A) 10�4 M 22Hg + 10�111 M I� (B) 10�5 M 2

2Hg + 10�10 M I�

(C) 10�11 M 22Hg + 10�7 M I� (D*) 10�6 M 2

2Hg + 10�9 M I�Sol. 10�6M Hg

22+ + 10�9M I�

IP = 2

10 6

29

210

= 8

10 24

= 1.25 10�25 > Ksp

precipitation of Hg2I2 will occur.

8. In which of the following solutions, the degree of dissociation of H2O is less than 1.8 10�7 % at 25° C :

(A) 10�6 M HCl (B) 10�7 M NaOH (C) 10�8 M HCl (D*) All of theseSol. The degree of dissociation of pure water at 25°C = 1.8 10�7%

any H+ or OH� ions from an external source will suppress the dissociation of H2O.

9. Three sparingly soluble salts M2X, MX and MX

3 have their solubility product in the ratio of 4: 1 : 27. Their

solubilities will be in the order : (A) MX

3 > MX > M

2 X (B*) MX

3 > M

2X > MX (C) MX > MX

3 > M

2X (D) MX > M

2X > MX

3

Sol. For M2X , 4S

13 = 4x ; S

1 = x1/3

For MX , 4S2

2 = x ; S2 = x1/2

For MX3 , 27S

34 = 27x ; S

3 = x1/4

S3 > S

1 > S

2

10. Calculate the pH of a 0.1 M K3PO

4 solution. The third dissociation constant of phosphoric acid is 10�12.

Given (0.41)1/2 = 0.64 ; log 3 = 0.48 (A) 12.5 (B*) 12.44 (C) 12.25 (D) 12

Page # 3

Sol. Kh =

3a

w

KK

= 12

14

10

10

= 10�2 Kh =

)h1(Ch2

as 1 � h = 1, h = C

Kh = 1.0

10 2

= 0.316

as h > 0.1 1 � h 1

10�2 = )h1(h10 21

or 0.1 (1 � h) = h2

or, 0.1 � 0.1 h = h2

or, h + 0.1 h � 0.1 = 0

or, h = 2

1.04)1.0(1.0 2

= 0.27

as, PO4

3� + H2O HPO

42� + OH�

c(1 � h) ch ch [OH�] = ch

= 0.1 0.27 = 27 10�3

pOH = 3 � log 27 = 3 log33 = 3 � 3 log 3

= 3 � 3 0.48 = 1.56pH = 14 � 1.56 = 12.44

11. The pKa of HCN is 9.3. The pH of a solution prepared by mixture 2.5 mole of KCN and 2.5 mole of HCN in

water and making up the total volume to 500 ml is(A*) 9.3 (B) 7.3 (C) 10.3 (D) 8.3

Sol. pH = pKa + log ]Acid[

]Salt[ = 9.3

12. Calculate the molar solubility of AgCl in 2.5 M NH3 solution. [Given : KspAgCl = 10�10 , K

f[Ag(NH

3)

2]+ = 106]

(A*) 0.025 mol/L (B) 0.2 L mol�1 (C) 0.4 L mol�1 (D) None of these

Sol. AgCl(s) Ag+(aq) + Cl� (aq)

Ag+ + 2NH3 [Ag(NH

3)

2]+

[Ag+] = s

Ksp.... (1)

Kf = 2)s25.2](Ag[

s

or, Kf = 2

sp

2

)s25.2(K

s

or, K

f K

sp = 2

2

)s25.2(

s

or, 10�2 =

s25.2s

or, s25.2

s

= 10�12 or, s = 0.025 � 0.02 s or, 1.02 s = 0.025 or, s = 02.1025.0

0.025 mol/L

13. What is the concentration of acetic acid which can be added to 0.5 M formic acid so that the % dissociationof neither acid is changed by the addition. K

a for acetic acid is 1.85 × 10-5, K

a for formic acid = 2.4 × 10-4.

(A) Any concentration (B) There can not be any concentration(C*) 6.66 M (D) 3.33 M

Sol. C1

1 = C

2

2

1a CK1 = 2a CK

2

1.8 × 10�5 × C1 = 2.4 × 10�4 × 0.5

C1 = 6.66 M

Page # 4

14. CH3NH2 (Kb = 5 × 10�4) 0.1 mole of CH3NH2 (Kb = 5 × 10�4) is mixed with 0.08 mole of HCl and diluted to onelitre. What will be the H+ concentration in the solution? is mixed with 0.08 mole of HCl and diluted to one litre.What will be the H+ concentration in the solution?

(A) 8 × 10�2 M (B*) 8 × 10�11 M (C) 1.6 × 10�11 M (D) 8 × 10�5 M

Sol. CH3NH2 + HCl CH3 3HN

lC

.1 .08 .08

.02 0 .08

For buffer sol. |OH�| = Kb x ]Salt[]Base[

= 5 x 10�4 x 08.02.

|OH�| = 45

x 10�4

so |H+| = 4

14

10x5410

= 54

x 10�10 = 8 x 10�111 M Ans.

15. 10�2 mole of NaOH was added to 10 litre of water. The pH will change by

(A*) 4 (B) 3 (C) 11 (D) 7Sol. Initially pH = 7

finally [NaOH] = 10�3 so pOH = 3pH = 11

So, (pH) = 4

16. The sum of negative logarithm of hydrogen ion and hydroxide ion concentration at 37ºC : [Kw = 2.5 × 10�4]

(A) 14 (B*) Less than 14 (C) greater than 14 (D) Data insufficient.

Sol. pH + pOH = pKW

= 15 � log (5)2

= 15 � 2 × .699

= 13.6

17. In the reaction : [Ag(CN)2]� + Zn the complex formed will be :

(A*) Tetrahedral (B) square planar (C) octahedral (D) triangal bipyramidal

Sol. 2 [Ag(CN)2]� + Zn [Zn(CN)

4]2� + 2 Ag

Tetrahedral

18. All the following complexes show a decreases in their weights when placed in a magnetic balance. Thenwhich of the these has square planar geometry :(A) Ni(CO)

4(B*) K[AgF

4] (C) Na

2[Zn(CN)

4] (D) None of these

Sol. K [AgF4] is square planar because Ag() is 4d8 and complex is diamagnetic.

19. It is an experiment fact that :

DMG + Ni()salt + NH4OH Red ppt.

Which of the following is wrong about this red ppt :(A) It is a non�ionic complex (B) It involves intra molecular H�bonding

(C*) Ni() is sp3 hybridised (D) It is a diamagnetic complex

Sol. The complex is

Page # 5

20. Sodium nitroprusside is a diamagnetic substance and a important laboratory reagent for the testing ofsulphide ions. The metal involved in the complexation in this is present in which of the following hybridisationstate :(A) sp3 (B) dsp2 (C*) d2sp3 (D) sp3d2

Sol. Sodium nitroprusside is Na2 [Fe(CN)

5 )ON(

] ; a diamagnetic complex.

21. All the following complex ions are found to be paramagnetic :P : [FeF

6]3� ; Q : [CoF

6]3�

R : [V(H2O)

6]3+ ; S : [Ti(H

2O)

6]3+

The correct order of their paramagnetic moment (spin only) is :(A*) P > Q > R > S (B) P < Q < R < S (C) P = Q = R = S (D) P > R > Q > S

Sol. On the basis of number of electrons the correct order is P > Q > R > S.22. When the complex K

6 [(CN)

5 Co�O�O�Co(CN)

5] is oxidised by bromine into

K5[(CN)

5 Co�O�O�Co(CN)

5]. Then which of the following statements will be true about this change:

(A) Co() is oxidised in Co() (B) The O�O bond length will increase

(C*) The O�O bond length will decrease (D) �A� & �B� both are correct

Sol. In the first complex ligand is O2

2� which is oxidised into O2

1� .henceO � O bond length decreases.

23. The octahedral complex [Rh(NO2) (SCN) (en)

2]+ can exist in a total number of isomeric forms including

stereoisomers :(A) 2 (B) 4 (C) 8 (D*) 12

Sol.

(1) NO2 / SCN (5) NO

2 / SCN (9)

(2) ONO / SCN (6) ONO / SCN (10)(3) NO

2 / NCS (7) NO

2 / NCS (11)

(4) ONO NCS (8) ONO / NCS (12)

24. For the reaction Ni2+ + 4NH3 [Ni(NH

3)

4]2+

at equilibrium, if the solution contains 1.6 × 10�4% of nickel in the free state, And the concentration of NH3 at

equilibrium is 0.5 M. Then the instability constant of the complex will be approximately equal to :(A) 1.0 × 10�5 (B) 1.5 × 10�16 (C*) 1.0 × 10�7 (D) 1.5 × 10�17

Sol. Ni2+ + 4 NH3 [Ni(NH

3)

4]2+

k = 43

2

243

]NH[]Ni[

])NH(Ni[

But

243

2

2

])NH(Ni[]Ni[

]Ni[ = 1.6 × 10�6

or

243

2

])NH(Ni[

Ni 1.6 × 10�6

k = 4

6

)5.0(6.1

10

= 107

Hence instability constant = 10�7

Page # 6

25. In which of the following complex ion, the metal ion will have 0g

6g2 e,t configuration according to CFT::

(A) [FeF6]3� (B) [Fe(CN)

6]3� (C*) [Fe(CN)

6]4� (D) None of these

Sol. In [Fe(CN)6]4� ; Fe() is t

2g6 , eg0 due to strong ligands.

26. Spin only magnetic moment of a complex having CFSE = � 0.6 0 and surrounded by weak field ligands can

be(A) 1.73 BM (B) 4.9 BM (C*) both (A) & (B) (D) None of these

Sol. The options can give CFSE = � 0.6 0 with weak field ligands d4 and d9.

27. Which of the following statements is not correct?(a) [Ni(H

2O)

6]2+ and [Ni(NH

3)

6]2+ have same value of CFSE

(b) [Ni(H2O)

6]2+ and [Ni(NH

3)

6]2+ have same value of magnetic moment

(A*) Only a (B) Only b (C) Both a and b (D) None of theseSol. Ammonia is a stronger field ligand than water.

28. The correct IUPAC name of the complex:

C = N

OH

OHH3C

C = NH3C

��

��CoCl2

is :

(A*) Dichlorodimethylglyoximatecobalt (II) (B) Bis(dimethyglyoxime)dichlorocobalt (II)(C) Dimethylglyoximecobalt(II) chloride (D) Dichlorodimethylglyoxime-N, N-cobalt (II)

29. Which of the following pair of complexes have the same EAN of the central metal atoms/ions?(A) [Cu(NH3)4]SO4 and K3[Fe(CN)6] (B) K4[Fe(CN)6] and [Co(NH3)6]Cl3(C) K3[Cr(C2O4)3] and [Ni(CO)4] (D*) all of the above

Sol. (A) [Cu(NH3)

4]2+ = 29 � 2 + 8 = 35

[Fe(CN)6]3� = 26 � 3 + 12 = 35

(B) [Fe(CN)6]4� = 26 � 2 + 12 = 36

[Co(NH3)6]3+ = 27 � 3 + 12 = 36

(C) [Cr(C2O

4)

3]3� = 27 � 3 + 12 = 36

[Ni(CO)4] = 28 + 8 = 36

30. In the reaction [CoCl2(NH

3)

4]+ + Cl� [CoCl

3(NH

3)

3] + NH

3 only one isomer of product is obtained .

Hence the initial complex must be

(A) cis isomer (B*) trans isomer (C) both (D) mixture of bothSol. Moderate

Cl

symmetrical only single product

Page # 7

Cl two isomers product

replacable positions

31.3AgNO

.Aq

)Major('X' product :

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

Sol. Aqueous AgNO3 catalyse SN1 reaction.

32. Consider the following reaction.

ether

SOCl2

In the above reaction which phenomenon will take place :

(A) Inversion (B*) Retention (C) Racemisation (D) IsomerisationSol. It is SNi reaction so retention takes place

33. Which one of the following has maximum nucleophilicity ?

(A*) SCH3 (B) (C) Et3N (D) Sol. Nucleophilicity size (in a group).

34. NaCN

In the given reaction rate is fastest, when (X) is :

(A) �OH (B)�NH2

(C)

O||

CHOS||O

3 (D*)

O||

CHSO||O

3

Sol. Leaving group ability Stability of anion.

35. In the following reaction the most probable product will be :

Page # 8

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

Sol.

36. When the concentration of alkyl halide is tripled and the concentration of ion is reduced to half, the rateof SN2 reaction increases by:(A) 3 times (B) 2 times (C*) 1.5 times (D) 6 times

Sol. Rate of SN2 [R � X ] [Nu�]

1

2

rr

=

]OH[]RX[

OH]RX3[�

�

21

r2

= 1.5 r1

37. In which of the following reaction the product obtained is t-butyl methyl ether ?

(A) CH3OH + HO �CH2�CH3 42SOH.conc (B)

(C*) (D)

Sol. t-butyl methyl ehter is a mixed ether and for the preparation of mixed ethers in high yield the essentialcondition is the use of primary alkyl halide.Thus,

NaBr

This reaction is williamson's synthesis.

38. CH3CH2CH2OH 5PCl A

KOHalc BB is identified as :(A) propanal (B) propane (C) propyne (D*) propene

Sol. ROH

HCl/ZnCl or I) Br, ,ClX(PX

or SOCl or PCl

23

25

RCl

CH3CH2CH2OH HCl

PCl5

2

Alkene3 CHCHCH

B is an alkene (propene)

Page # 9

39. The only alcohol that cannot be prepared by the indirect hydration of alkene is :(A) ethyl alcohol (B) propyl alcohol (C) isobutyl alcohol (D*) methyl alcohol

Sol. Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbons so alcohol ofat least two carbon atoms can be formed.

H2C = CH2 )SOH(

H

42

CH

| CH

2

3

4HSO

HOSOCH | CH

32

3

HOH Ethanol

242 OHCHCHSOH

H2C = CH � CH3

rule) s'ffMarkowniko(

SOH

HOH

42

product Major

33

OH |

CHCHCH

product Minor322 CHCHCHHO

40. Lucas reagent reacts fastest with :(A) butanol�1 (B) butanol�2 (C*) 2�methyl�propanol�2 (D) 2�methyl�propanol�1

Sol. The order of reactivity with alcohols with lucas reagent is -3º > 2º > 1º

Lucas reagent reacts fastest with 3º alcohol.

(a)

(b)

(c)

(d)

choice (C) is the answer as it is 3º alcohol and rate of reaction is fastest for 3º alcohol.

Page # 1

BITSAT- PT - 2 - XII - (09-09-12)

Syllabus : Sequence&Series, P&C, Binomial Theorem, Mathematical Induction, Determinant, Straight lines

1. For every natural number n, n(n + 3) is always(A*) even (B) odd (C) multiple of 4 (D) multiple of 5

Sol. Let P(n) = n(n + 3), thenP(1) = 1(4) = 4 which is even and multiple of 4.P(2) = 2(5) = 10 which is even and multiple of 5.P(3) = 3(6) = 18 which is even.Hence it is clear that P(n) is even n N

2. The greatest positive integer which divides 32n � 2n � 1 n N is(A) 1 (B*) 2 (C) 4 (D) 8

Sol. Let P(n) = 32n � 2n � 1, then

P(1) = 32 � 2 � 1 = 6

P(2) = 34 � 4 � 1 = 76

P(3) = 36 � 6 � 1 = 722

Obviously 2 is the greatest positive integer which divides P(n) n N.

3. If the 9th terms of an A.P. be zero, then the ratio of its 29th and 19th term is(A) 1 : 2 (B*) 2 : 1 (C) 1 : 3 (D) 3 : 1

Sol. Given that 9th term = a + (9 � 1) d = 0 a + 8d = 0Now ratio of 29th and 19th terms

= d18ad28a

= d10)d8a(

d20)d8a(

=

d10d20

= 12

4. The solution of the equation (x + 1) + (x + 4) + (x + 7) + ....+ (x + 28) = 155 is(A*) 1 (B) 2 (C) 3 (D) 4

Sol. We have (x + 1) + (x + 4) + (x + 7) + ....+ (x + 28) = 155Let n be the number of terms in the A.P. on L.H.S.Then x + 28 = (x + 1) + (n � 1)3 n = 10 (x + 1) + (x + 4) +.....+ (x + 28) = 155

2

10[(x + 1) + (x + 28)] = 155 x = 1

5. If the arithmetic and geometric means of a and b be A and G respectively, then the value of A � G will be

(A) a

b�a(B)

2ba

(C*)

2

2

b�a

(D)

baab2

Sol. Arithmetic mean of a and b = A = 2

ba

and geometric mean G = ab

Then A � G = 2

ba � ab =

2ab2�ba

= 2

)b)(a(2�)b()a( 22

=

2

2

b�a

6. In series 1,2,2,2,2,3,3,3,3,3,3,3,3,3, 4,...........the 400th term is(A) 9 (B) 10 (C*) 11 (D) 12

Sol. Number 11 starts at th222 1)10.....21( position

i.e. 386th position.

7. The sum of the series .........331

3

221

2

111

1424242

to n terms is

(A) 1nn

)1n(n2

2

(B*)

)1nn(2

)1n(n2

(C)

)1nn(2

)1n(n2

2

(D)

2nn2

Page # 2

Sol. Let Tn be the nth term of the series

.........331

3

221

2

111

1424242

Then Tn = 42 nn1

n

= 222 n�)n1(

n

=

)1n�n)(1nn(

n22

=

1nn

1�

1n�n

121

22

=

)1n(n11

�n)1�n(1

121

Now 21

Tn

1rr

2.111

�11

+21

3.211

�2.11

1+

21

4.311

�3.21

1+...+

)1n(n11

�n)1�n(1

121

=

)1n(n11

�121

= )1nn(2

)1n(n2

8. The number of common terms to the sequence 17, 21, 25, .......417 and 16, 21, 26,.....466 is(A) 21 (B) 19 (C*) 20 (D) 22

Sol. Common terms are 21,41,61,......(d = LCM of 4,5 = 20)tn 417

21 + (n � 1)20 417 n 20.8 max value of n = 20

9. If a, b, c are in A.P. as well as in G.P., then(A) a = b c (B) a b = c (C) a b c (D*) a = b = c

Sol. As given b = 2

ca ......(i)

and b2 = ac (a + c)2 = 4ac (a � c)2 = 0 a = cputting a = c in (i) we get b = c a = b = c

10. If G be the geometric mean of x and y, then 2222 y�G

1

x�G

1 =

(A) G2 (B*) 2G

1(C) 2G

2(D) 3G2

Sol. As given G = xy

2222 y�G

1

x�G

1 = 2x�xy

1 + 2y�xy

1

= y�x1

y1

x1

� = xy1

= 2G

1

11. If a1, a

2, a

3,........ are in A.P. such that a

1 + a

5 + a

10 + a

15 + a

20 + a

24 = 225, then

a1 + a

2 + a

3 + ..... + a

23 + a

24 is equal to

(A) 909 (B) 75 (C) 750 (D*) 900Sol. a

1 + a

5 + a

10 + a

15 + a

20 + a

24 = 225

3 (a1 + a

24) = 225 (sum of terms equidistant

from beginning and end are equal) a1 + a

24 = 75

Now a1 + a

2 + ........ + a

23 + a

24

= 2

24 [a

1 + a

24] = 12 × 75 = 900

Page # 3

12. There are n distinct points on the circumference of a circle. The number of pentagons that can be formedwith these points as vertices is equal to the number of possible triangles. Then the value of n is-(A) 7 (B*) 8 (C) 15 (D) 30

Sol. 5nC = 3

nC

5nC = 3�n

nC

5 = n � 3

n = 8

13. A question paper consists of two parts A and B. Part A has 4 questions in which each question has analternative and part B has 3 questions without any alternative. The number of ways to attempt paper whenat least one question must be attempted for each part is (are)(A) 561 (B*) 560 (C) 648 (D) 127

Sol. Required ways = (34 �1)(23 � 1) = 560

14. Number of ways such that 6 boys and 3 girls can be seated such that there is exactly one boy in betweenany two girls(A) 50400 (B*) 21600 (C) 10800 (D) 36000

Sol. Number of ways = 6! × 5 × 3!

15. All letters of the word 'RACHIT' are permuted in all possible ways and the words so formed (with or withoutmeaning) are written as in dictionary, then the 484th word is-(A) RACHIT (B*) RACITH (C) RACTHI (D) RACIHT

Sol. ACHIRT 5!CAHIRT 5!HACIRT 5!IACHRT 5!RACHIT is 481th word RACHITRACHTI is 482th word

RACIHI is 483th wordRACITH is 484th word

16. The number of ways in which 6 different red roses and 3 different white roses can form a garland so that allthe white roses come together is(A*) 2160 (B) 2165 (C) 2155 (D) 4320

Sol. ways = 2

!3!)1�7( = 2160

17. There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The numberof different circles that can be drawn through at least 3 of these points is(A) 116 (B*) 117 (C) 120 (D) 115

Sol. Total number of solutions = 10C3 � 4C

3 + 1 = 117.

18. How many different arrangements can be made out of the letters in the expansion A2B3C4, when writtenin full length ?

(A*) !4!3!2!9

(B) 4.3.2

!9(C) 2 ! 3 ! � 4 (D) !4!3!2

!9

Sol. Here A, B, C are repeated twice, thrice and four times respectively

No. of arrangements = !4!3!2)!432(

= !4!3!2!9

19. Number of positive integral solutions of x1. x

2 x

3 = 210 is-

(A) 25 (B) 26 (C) 27 (D*) 81Sol. We have

x1 x

2 x

3 = 210 = 2.3.5.7

Total no. of solutions of the equation x1x

2x

3 = 210 is 3 × 3 × 3 × 3 = 81

Page # 4

20. Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each ofthem gets at least two blankets, equal to(A*) 10C

3(B) 9C

3(C) 11C

3(D) none of these

Sol. Let 4 persons recieve B1, B

2 , B

3, B

4 number of blankets

B1 + B

2 +

B

3 +

B

4 = 15 {B

1,B

2 , B

3 ,B

4 2}

Thus number of ways = 10C3

21. Let S (k) = 1 + 3 + 5 +.......+ (2k � 1) = 3 + k2. Then which of the following is true ?(A) S(1) is correct(B*) S(k) S (k + 1)(C) S(k) S(k + 1)(D) Principle of mathematical induction can be used to prove the formula

Sol. S(k) = 1 + 3 + 5 + ..... + (2k � 1) = 3 + k2

put k = 1 in both sides, we get LHS = 1 and RHS = 3 + 1 = 4 LHS RHSPut (k + 1) in both sides in the place of kLHS = 1 + 3 + 5 + .... + (2k � 1) + (2k + 1)

RHS = 3 + (k + 1)2 = 3 + k2 + 2k + 1Let LHS = RHS1 + 3 + 5 + .......... + (2k � 1) + (2k + 1)

= 3 + k2 + 2k + 1 1 + 3 + 5 + ...... + (2k � 1) = 3 + k2

If S(k) is true, then S(k + 1) is also true.Hence, S(k) S(k + 1).

22. If the coefficients of second, third and fourth terms in the expansion of (1 + x)2n are in A.P., then which of thefollowing is TRUE.(A) n2 � 9n + 7 = 0 (B) 3n2 � 9n + 7 = 0 (C) 3n2 + 9n + 7 = 0 (D*) 2n2 � 9n + 7 = 0

Sol. 2nC1 , 2nC

2 , 2nC

3 are in A.P.

2n2 � 9n + 7 = 0

23. The term that is independent of x in the expansion of

92

x31

x23

is

(A)

45

69

31

23

C

(B*)

3

39

61

C

(C)

54

49

31

23

C

(D)

66

69

313

23

C

Sol.9

2

x31

x23

=

9

0r

rr92

r9

x31

�x23

C

For the term that is independent of x18 � 2r � r = 0 r = 6

Required term = 69 C

3

23

6

31

= 6

9 C

3

61

24. In the expansion of (21/5 + 3 )20, the sum of all rational terms is equal to

(A) 21 (B) 84 (C) 97 (D*) none of these

Sol. Tr+1

= r20 C (21/5)20�r ( 3 )r

= r20 C . 5

r4

2 . 2

r

3

As 2 and 3 are relatively primes. Tr+1

is rational, if 5r

and 2r

are integer�s�s

r is multiple of 10 0 r 20r = 0 ,10, 20Thus sum of rational terms = T

1 + T

11 + T

21 = 20C

0 24 + 20C

10 22. 35 + 20C

20 . 310

This is more than 21, 84, 97

Page # 5

25. The value of

2832003

where {.} denote the fractional part, is equal to

(A) 2815

(B) 285

(C*) 2819

(D) 289

Sol. 32003 = 32001.32

= 9(27)667

= 9(28 � 1)667

= 9[ 0667C 28667 + 1

667C (28)666 + ...... + 667667C (�1)667]

that means if we divide 32003 by 28, remainder is 19

Thus

2832003

= 2819

26. The middle term in the expansion of 10

x1

x

is

(A) 10C1

x1

(B*) 10C5

(C) 10C6

(D) 10C7x .

Sol. Middle term = 2

210T = T5 + 1

= 10C5x10 � 5

5

x1

= 10C

5.

27. If |x| < 1, then the coefficient of xn in the expansion of (1 + x + x2 + x3 .......)2 is(A) n (B) n � 1 (C) n + 2 (D*) n + 1

Sol. (1 + x + x2 + ...)2 =

2

x�11

= (1 � x)�2

Tr + 1

= !r))1�r(�2)......(�3)(�2(�

(�x)r

= !r

)1)(�1r......(3.2)1(� rr

xr

= !r

)1r(r......3.2.1)1(� r2 xr = (r + 1) xr

coefficient of xr = r + 1 coefficient of xn = n + 1

28. If a2m

=

m2

0r rm2 C

1, then

m2

0r rm2 C

requals

(A) (2m � 1) a2m

(B) 2ma2m

(C*) ma2m

(D) ma2m

Sol. E =

m2

0r rm2 C

r =

m2

0r rm2 C

r�m2 2E =

m2

0r rm2 C

r�m2r = 2m

m2

0r rm2 C

1

29. Last three digits of the number N = 7100 � 3100 are(A) 100 (B) 300 (C) 500 (D*) 000

Sol. N = 7100 � 3100

= (10 � 3)100 � 3100 = (100C0. 10100 � ........... � 100C

99. 10.399) + 3100 � 3100

= (1000)N + 3100 � 3100

= (1000) Nlast 3 digits = 000

Page # 6

30. If the lines represented by x2 � 2pxy � y2 = 0 are rotated about origin through an angle , one in clockwisedirection and other in anticlockwise direction, then the equation of the bisectors of the angle between thelines in new position is(A) px2 + 2xy + py2 = 0 (B) px2 � 2xy + py2 = 0 (C) px2 � 2pxy � py2 = 0 (D*) px2 + 2xy � py2 = 0

Sol. Bisectors of x2 � 2pxy � y2 = 0 is 2

y�x 22

= p�xy

px2 + 2xy � py2 = 0Lines in new position will also have same angle bisectors.

31. If 3a + 2b + 6c = 0, then the family of straight lines ax + by + c = 0 passes through a fixed point whosecoordinates are given by

(A*)

31

,21

(B) (2, 3) (C) (3, 2) (D)

21

,31

Sol. ax + by +

6b2�a3�

= 0

6ax + 6by � 3a � 2b = 0

a(6x � 3) + b(6y � 2) = 0

x = 1/2, y = 1/3

32. The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is(A) 3/2 (B*) 3/10 (C) 6 (D) none of these

Sol. Distance = 22 43

215

�9

=

103

33. A ray of light passing through the point A (1, 2) is reflected at a point B on the x-axis and then passesthrough C (5, 3) . Then the equation of AB is-(A*) 5x + 4y = 13 (B) 5x � 4y = � 3 (C) 4x + 5y = 14 (D) 4x � 5y = � 6

Sol.

AB will pass through C.

equation of AB is y + 3 = 4�

5 (x � 5)

4y + 5x = 13

34. If bx + cy = a, where a, b, c are the same sign, be a line such that the area enclosed by the line and the

axes of reference is 81

unit2, then

(A) b, a, c are in G.P. (B) b, 2a, c are in A.P.

(C) b, 2a

, c are in A.P.. (D*) b, �2a, c are in G.P.

Page # 7

]Sol. bx + cy = a

ab

x + ac

y = 1

bax

+

cay

= 1

Area of OAB = 81

(given)

21

. ba

. ca

= ± 81

bca2

= ± 41

4a2 = ± (bc) (2a)2 = ± bc

b, ±2a, c are in G.P.

35. In a ABC if A is (1, 2) and equation of the medians through B and C are x + y = 5 and x = 4 respectivelythen B is(A) (1, 4) (B*) (7, � 2) (C) (4, 1) (D) (�2, 7)

Sol.

point of intersection of x + y = 5 and x = 4G (4, 1)

3

4b1 = 4 b = 7

So B (7, � 2)

36. Let B1 = 3x + 4y � 7 = 0 & B

2 4x � 3y � 14 = 0 are angle bisectors of the angle between the lines L

1 = 0 &

L2 = 0 in which L

1 is passes through the point (1, 2) then

(A*) B1 is acute angle bisector (B) B

2 is acute angle bisector

(C) B1 & B

2 both are right angle bisector (D) Data is insufficient

Sol. Let d1 & d

2 are the distance of point (1, 2) from the bisector B

1 & B

2.

d1 =

5

783 =

54

d2 =

5

1464 =

516

d1 < d

2

B1 is an acute angle bisector

37. If P is a point (x, y) on the line y = � 3x such that P and the point (3, 4) are on the opposite sides of the

line 3x � 4y = 8, then

(A*) y < � 58

(B) y > � 58

(C) y > � 511

(D) y < � 51

Page # 8

Sol. Since 3.3 � 4.4 � 8 = � 15 < 0

3x � 4y � 8 > 0 3

3

y� � 4y � 8 > 0

5y < � 8 y < � 58

38. If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in-(A) H.P. (B) G.P. (C*) A.P. (D) None of these

Sol. Since the given lines are concurrent,

114c

113b

112a

= 0 01b�c

113b

01�b�a

= 0

[Applying R3 R

3 � R

2, R

1 R

1 � R

2]

a � b + c � b = 0or 2b = a + c a, b, c are in A.P.

39. A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is-

(A) 31

(B) 32

(C) 1 (D*) 34

Sol. Equation of the line through the point (2, 2)

and to line (1) is (y � 2) = 31

(x � 2)

3y � 6 = x � 2

x � 3y + 4 = 0

Its y-intercept = 34

. [Putting x = 0]

40. The line segment joining the points (1, 2) and (k, 1) is divided by the line 3x + 4y � 7 = 0 in the ratio

4 : 9, then k is-(A*) � 2 (B) 2 (C) �3 (D) 3

Sol. L : 3x + 4y � 7 = 0

� L (1, 2) : L (k, 1) = 4 : 9

� (3 + 8 � 7) : (3k + 4 � 7) = 4 : 9

� 4 : (3k � 3) = 4 : 9 k = � 2.

41. If px4 + qx3 + rx2 + sx + t = x34x3x

3xx21x

3x1xx3x2

then t is equal to

(A) 33 (B) 20 (C) 15 (D*) 21

Sol. px4 + qx3 + rx2 + sx + t = x34x3�x

3�xx�21x

3x1�xx3x2

Putting x = 0

t =

043�

3�21

31�0

= 21

42. If the system of linear equations x + 2ay + az = 0, x + 3by + bz = 0, and x + 4 cy + cz = 0 has a non-zerosolution, then a, b, c(A) are in AP (B) are in GP (C*) are in HP (D) satisfy a + 2b + 3c = 0

Page # 9

Sol. The system of linear equations has a non-zero solution, then

cc41

bb31

aa21

= 0

Applying R2 R

2 � R

1, R

3 R

3 � R

1

a�ca2�c40

a�ba2�b30

aa21

= 0

(3b � 2a) (c � a) � (4c � 2a) (b � a) = 0

3bc � 3ba � 2ac + 2a2 = 4bc � 2ab � 4ac + 2a2

4ac � 2ac = 4bc � 2ab � 3bc + 3ab

2ac = bc + ab

On dividing by abc, we get

b2

= a1

+ c1

Hence, a, b, c are in HP.

43. If a, b, c are pth, qth and rth, terms of a G.P., then 1rclog

1qblog

1palog

equals -

(A*) 0 (B) 1 (C) log abc (D) pqr

Sol. If A be the first term and R be the c.r. of G.P., then

a = ARp�1, b=ARq�1, c=ARr�1

log a = logA + (p � 1)log R

= 1rAlog

1qAlog

1pAlog

+ 1rRlog)1r(

1qRlog)1q(

1pRlog)1p(

= 0 + log R 11r1r

11q1q

11p1p

= 0 [by C2 � C1]

44. For positive numbers x, y, z, the numerical value of the determinant

1ylogxlog

zlog1xlog

zlogylog1

zz

yy

xx

is

(A*) 0 (B) 1 (C) 2 (D) None of these

Sol. Value of determinant

1ylogxlog

zlog1xlog

zlogylog1

zz

yy

xx

= zlog1

.ylog

1.

xlog1

zlogylogxlog

zlogylogxlog

zlogylogxlog

= 0

45. The number of values of ' r

' satisfying the equation, 2r

391r3

39 CC = r3

391r

39 CC 2

is

(A) 1 (B*) 2 (C) 3 (D) 4

Page # 10

Sol. r339

1r39

r39

1r339 CCCC 22

r339

1r339 CC

= 22 r39

1r39 CC

r240C = 2r

40 C

r2 = 3r or r = 0, 3

or r2 + 3r = 40 r = 5, �8

BITSAT(XII)_PT-2_Pg.No # 1

BITSAT�XII/XIIIPT�02

1. The average velocity of molecules of a gas of molecular weight M at temperature T is:

(A*) 0 (B) MRT3

(C) 8RT

M(D)

2RT

M

Sol. Average velocity of a molecule at any temperature is zero because of its random motion.

2. The ratio of r.m.s. speed to the r.ms. angular speed of a diatomic gas at certain temperature is:(assume m = mass of one molecule, M = molecular mass, = moment of inertia of the molecules)

(A) 3

2(B)

3

2

I

M (C*)

3

2

I

m(D) 1

Sol. kT23

mV21 2

kT22

21 2

m23V

3. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting allvibrational modes, the total internal energy of the system is:(A) 4 R

T (B) 5 R

T (C) 15 R

T (D*) 11 R

T

Sol. In an ideas gas internal energy = 2f

nRT

U = 25

× RT + 4 × 23

RT = 11 RT.T.

4. Maxwell�s velocity distribution curve is given for the same quantity at two different temperatures. For

the given curves.

(A) T1 > T2 (B*) T1 < T2 (C) T1 T2 (D) T1 = T2Sol. Higher is the temperature greater is the most probable velocity.

5. In a process the density of a gas remains constant. If the temperature is doubled, then the change inthe pressure will be:(A*) 100 % Increase (B) 200 % Increase (C) 50 % Decrease (D) 25 % Decrease

Sol. We have = RTPM

1

1

RTMP

= 2

2

RTMP

1

2

1

1

T2P

TP

P2 = 2P

1

6. 12 gm He and 4 gm H2 is filled in a container of volume 20 litre maintained at temperature

300 K. The pressure of the mixture is nearly :(A) 3 atm (B) 5 atm (C*) 6.25 atm (D) 12.5 atm


Sol. PV = n RT

P = V

nRT =

31020

30031824

412

.

= 6.25 × 105 Pa

7. In an experiment the speeds of any five molecules of an ideal gas are recorded. The experiment isrepeated N times where N is very large. The average of recorded values, is :

(A*)M

RT2(B)

M

RT8

(C)

M

RT3(D)

M

RT

Sol. When speed of 5 molecules which are selected randomly, then the average is most likely to be equal tothe most probable speed.

The average of these values is most likely equal to MRT2

.

8. P-V diagram of a cyclic process A B C A is shown in figure. The temperature of the gas will bemaximum at :

(A) A (B) B(C*) a point between A and B (D) a point between B and C

Sol. Temperature at points A and B are equal. A to B temperature first increases then decrease.

9. On an X temperature scale, water freezes at � 125.0° X and boils at 375.0° X. On a Y temperature

scale, water freezes at � 70.0°Y and boils at � 30.0°Y. The value of temperature on X-scale equal to the

temperature of 50.0°Y on Y-scale is :

&(A) 455.0° X (B) � 125.0° X (C*) 1375.0° X (D) 1500.0° X

Sol.500

)125(X =

40)70(Y

For Y = 50X = 1375.0°X

10. The amount of heat supplied to decrease the volume of an ice water mixture by 1 cm3 without anychange in temperature, is equal to : (ice = 0.9, water = 80 cal/gm)(A) 360 cal (B) 500 cal (C*) 720 cal (D) None

Sol. x gm ice convert into x gm water

9.0x

� x = 1 x = 1.09.0

= 9

Q = 9 × 80 = 720 cal

11. n moles of a gas filled in a container at temperature T is in equilibrium initially. If the gas is compressedslowly and isothermally to half its initial volume, the work done by the atmosphere on the piston is:


(A*) n R T

2(B)

n R T

2(C) n R T n 2

1

2

(D) n R T l n 2

Sol. Work done by atmosphere = Patm V

= Patm 2V

................(i)

As ; Initially gas in container is in thermodynamic equilibrium with its surroundings. Pressure inside cylinder = Patm& PV = nRT

PatmV = nRT or V = atmP

nRT

Putting in (1),

W = 2

nRT

12. In the figure shown the pressure of the gas in state B is:

(A) 2563

P0 (B*) 2573

P0 (C) 2548

P0 (D) none of these

Sol.

AN = 3v0 cos2 37º

PB =

2516

v3vv

P00

0

0

=

2548

1

= P0(73/25) Ans. (B)

13. A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is givento it. Then the work done in expansion is:

(A) Q (B)3

5 Q (C*)

2

5 Q (D)

2

3 Q


.Sol. For process at constant pressure

Q = nCp T = 25

nR T and W= PV = nRT = 52

Q

14. A thermodynamic process of one mole ideal monoatomic gas is shown in f igure. Theefficiency of cyclic process ABCA will be :

(A) 25% (B) 12.5% (C) 50% (D*) 13

100%

Sol. W = 21

P0V0 = 21

RT0 .

Heat absorbed = QAB + QBC = CVT0 + CP2T0 = 2

13 RT0

Efficiency = 00

00

VP2

13

VP21

× 100

000 RT

2

13VP

2

13

=131

× 100 = 7.7 % Ans.

15. 1 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graphshows the volume V versus Q. The gas temperature is nearly equal to : (use R = 8.31 J/K.mole)

(A) 208.4 K (B) 268.2 K (C*) 312.6 K (D) 353.8 KSol. For isothermal process

Q = nRT n 1

2

vv

1800 = 1 × 8.3 T n zget T = 312.6 K

16. Curve in the figure shows an adiabatic compression of an ideal gas from 15 m3 to 12 m3, followed by anisothermal compression to a final volume of 3.0 m3. There are 2.0 moles of the gas. Total heat supplied tothe gas is equal to : (n2 = 0.693)


3 12 15

400

p(Pa)

V(m )3

(A) 4521 J (B) �4521 J (C*) �6653 J (D) �8476 J

Sol. There is no heat transfer in adiabatic compression. In isothermal process

pwafd :)ks"e laihMu esa dksbZ m"ek LFkkukUrj.k ugha gksrk gSA lerkih izfØ;k esa

Q = W = P1V

1 ln

1

2

V

V

= 400 x 12 ln 41

= �6653 J

17. Two bodies A and B have emissivities 0.5 and 0.8 respectively. At some temperatures the two bodieshave maximum spectral emissive powers at wavelength 8000 Å and 4000 Å respectively. The ratio of

their emissive powers at these temperatures is:

(A*) 128

5(B) 10 (C)

165

(D) None of these

Sol. Let the body have temperatures T1 and T2 respectively at wavelength 1 = 8000Å and 2 = 4000Å.

From Wien�s displacement law

T = constant 1T1 = 2T2 or 8000 × T1 = 4000T2

or2

1

TT

= 21

Emissive power = e AT4

Ratio of emissive powers at these temperature is

= 422

411

Te

Te =

8.05.0

×

4

2

1

=

1285

18. N(< 100) molecules of a gas have velocities 1, 2, 3........ N km/s respectively. Then(A) rms speed and average speed of molecules is same.(B) ratio of rms speed to average speed is (2N + 1)(N + 1)/6N

(C) ratio of rms speed to average speed is (2N + 1)(N + 1)/6

(D*) ratio of rms speed to average speed of molecules is )1N(6

)1N2(2

Sol. Vrms

= N

V..........VV 2N

22

21

= N

N..........21 222

= N6)1N2()1N(N

Vrms

= 6

)1N2()1N(

Vavg

= N

V........VV N21 =

NN........21

= N2)1N(N

= 21N

avg

rms

V

V = )1N(6

)1N2(2


19. A solid spherical black body of radius r and uniform mass distribution is in free space. It emits power�P� and its rate of colling is R then

(A) R P r2 (B*) R P r (C) R P 1/r2 (D) R P r1

Sol. Rate of radiation per unit area is proportional to (T4)

P AT4

P r2.

Also ms dtdT

ATT4 dtdT

= R r1

(because m = (v) r3 and A r2)

20. A black body emits radiation at the rate P when its absolute temperature is T. At this temperature thewavelength at which the radiation has maximum spectral emissive power is

0. If at another temperature

T the power radiated is P and wavelength at maximum spectral emissive power is 20 then

(A*) P T = 32PT (B) P T = 16PT (C) P T = 8PT (D) P T = 4PT

Sol. For a black body, wavelength for maximum intensity :

T1

& P T4

4

1P

P = 16 P. P T = 32PT

21. Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles havingpressure P

0, volume V

0 and temperature T

0 is

(A) 00VP

R(B)

R

VP 00 (C*) 0T1

(D) 0Tn

1

Sol. [Easy]PV = nRTPdV = nRdT

= dTdV

V1

andP

nRdTdV

= T1

For given temperature T0 , =

0T1

22. A solid sphere of iron at 2°C is lying at the bottom of a bucket full of water at 2°C. If the temperature of

the water is increased to 3°C, the buoyant force on the sphere due to water will

(A*) Increase (B) Be unchanged (C) Decrease(D) Increase or decrease depends upon the numerical values of coefficient of expansion of water andiron.

Sol. As the temperature of water is increased from 2°C to 3°C the density of water increases (remember

anamolous behaviour of water), also the volume of sphere increases. Therefore bouyant force on spheredue to water shall increase.

23. The lengths of two metallic rods at temperatures are LA and L

B and their linear coefficient of expansion are

A and

B respectively. If the difference in their lengths is to remain constant at any temperature then

(A) LA/L

B=

A/

B(B*) L

A/L

B=

B/

A(C)

A=

B(D)

A

B=1

Sol. Change in LA = change in L

B

i.e. LA = L

B

AT L

A =

BTL

B

or AL

A =

BL

B .


24. Two identical long, solid cylinders are used to conduct heat from temp T1 to temp T

2. Originally the

cylinder are connected in series and the rate of heat transfer is H. If the cylinders are connected inparallel then the rate of heat transfer would be :(A) H

/4 (B) 2H (C*) 4H (D) 8H

Sol. (B) Initially effective resistance = 2R. In parallel effective resistance = 2R

. It has reduced by a factor of

1/4 so rate of heat transfer would be increased by a factor of 4, keeping other parameters same.

25. A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fasteach day at a temperature of 15°C and 10 seconds slow at a temperature of 30°C. The temperature at

which it is designed to give correct time, is(A) 18°C (B*) 20°C (C) 24°C (D) 25°C

Sol. Fractional loss of time per second = 21

T

Therefore 21

(T0 � 15) × (24 hrs) = 5

and21

(30 � T0) × (24 hrs) = 10

on solving T0 = 20°C

26. A shell, made of material of electrical conductivity

910(-m)�1, has thickness t = 2 mm and radius R = 10 cm.

In an arrangement, its inside surface is kept at a lower potential than its outside surface. The resistanceoffered by the shell is equal to -

+-+-+-

(A) 5 x 10�12 (B) 2.5 x 10�11 (C) 5 x 10�12 (D*) 5 x 10�11

Sol. R =

1 . 2R4

t

Using values R = 5 x 10�11

27. In the given network of four resistances, the equivalent resistance is

(A) 20 (B) 5.4 (C) 12 (D*) 4.5

Hint : Given circuit is equivalent to

solve it can be calculated


28. In the figure shown:

(A) current will flow from A to B(B*) current may flow from A to B(C) current will flow from B to A(D) the direction of current will depend on r.

Sol. current through resistance will be fromA to B if20 � > < 18

and fromB to A if20 � < 2 < 18

29. The terminal voltage across a battery of emf cannot be:

(A) 0 (B) >

(C) < (D*) none of these is correctSol. Termianal potential across battery is :

�ir If battery works as a source + ir If battery works as a local Ideal battery or if i = 0.

30. In the circuit shown the readings of ammeter and voltmeter are 4A and 20V respectively. The meters are non-ideal, then R is

(A) 5 (B) less than 5

(C*) greater than 5 (D) between 4 and 5.

Sol. Effective resistance in the branch of R and voltmeter is ;

Reff

= 4

20 = 5

Also in parallel effective resistance is less than the individual resistance. Value of R must be greater than 5.

31. For an adiabatic process graph between PV & V for a sample of ideal gas will be :

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

Sol. PV Tfor adiabatic process,TV�1 = constant

32. The maximum current in a galvanometer can be 10 mA. It�s resistance is 10. To convert it into an ammeterof 1 Amp. a resistor should be connected in(A) series, 0.1 (B*) parallel, 0.1 (C) series, 100 (D) parallel, 100.

Sol. G = 10 mAG = 10S ( � G) = G G where S is shunt in parallel

S = 3

3

G

G

10101

101010G

= 0.1


33. Battery of internal resistor ' r ' and e.m.f. is connected to a variable external resistance AB. If the slidingcontact is moved from A to B, then terminal potential difference of battery will :

(A) remain constant & is independent of value of external resistance(B*) increase continuously

(C) decrease continuously

(D) first increase and then will decrease.Sol. Terminal potential difference across battery will be

= � ir

If resistance increases then �i� will decrease

So, potential will increase.

34. Two cells of emf 1 and

2 (

2 <

1) are joined as shown in figure :

When a potentiometer is connected between x and y it balances for 300 cm length against 1. On connecting

the same potentiometer between x and z it balances for 100 cm length against 1 and

2. Then the ratio

1

2

is :

(A) 31

(B) 43

(C) 41

(D*) 32

Sol. 1 = 300 ..........(i)

�2 +

1 = 100 ..........(ii)

where, is the potential gradient

1

2

=

32

.

35. The equivalent resistance of the circuit across points A and B is equal to :

A B 15 10

15

20

3020

30

10

(A) 22.5 (B) 25 (C*) 37.5 (D) 75 Ans. (C)Sol. Equivalent circuit is


10 15

20 3010

20 30

15A B

= 37.5

36. In the circuit shown in figure find the current in branch AB of the circuit :

20 VB

A

(A*) 5 A (B) 0.5 A

(C) 3

11A (D) None of these

Sol. Here in this circuit its equivalentresistance across battery can be given as

Req = 1140

20 VB

A

1.5A

4A

5.5A

1.5A

Thus current through battery is

I = 11

40

20 = 5.5 A.

Thus current 1.5A (from figure) will be divided in 10 & 5 in inverse ratio thus

[rP] Vivah - 01 - Mujhe Haq Hai.mp3

I5 = 15

105.1 = 1A

Thus current is branch AB isIAB = 1 + 4 = 5A Ans.

37. What should be value of E for which galvanometer shows no deflection :

(A*) 10 V (B) 5 V (C) 15 V (D) 20 V

Sol.20

10E =


10 = E � 20

10E × 5

40 = 4E � E + 10

30 = 3EE = 10 V.

38. In the circuit shown in the figure, the potential difference between B and C is :

(A) 0.1 V (B) 2V (C) 0.5 V (D*) 4.25 VSol. Apply K.V.L.

K.V.L.i.5 + 2 + i.3 + i.6 + i.2 � 4 = 10

16i = 12

i = 43

A

Potential difference between B and C is

2 + 3 × 43

= 4.25 V

39. An ideal gas is taken through cyclic process as shown in the figure. The net work done by the gas is:

(A) zero (B*) PV (C) 2 PV (D) 3 PV

40. Heat energy absorbed by a system in going through a cyclic process is shown in the figure [ V in litresand p in kPa ] is:


(A) 107 J (B) 104 J (C*) 102 J (D) 107 J n2 =

a 2

T2

T =

2/14n

a

11PT-02_XII_BITSAT_PAGE # 1

BITSAT PAPER (PT-02)_DATE : 09-09-12_CLASS-XII

Instructions (Q.1 & 2) Choose the alternative which can replace the italicized word :

1. He is a candid politician(A*) frank (B) faithful (C) soft spoken (D) fearless

2. He was punished for shirking his official work :(A) solving (B*) avoiding (C) delegating (D) postponing

Instructions (Q. 3 & 4) Choose the correct antonym :

3. Quell(A) Anger (B) Query (C) Suppress (D*) Aggravate

4. Soporific(A) Inducing (B) Inciting (C) Consoling (D*) Vigorous

Instructions(Q 5 & 6) Fill in the blanks with the most appropriate word from the given options.

5. They have decided to meet the prime minister in order to have their _______ heard.(A) agony (B) suffering (C) sorrow (D*) woes

6. The pleasures of the world are _________ as they are not permanent.(A) tangible (B) existent (C) corporeal (D*) illusory

Instructions (Q 7 to 9) Pick up the correct synonyms :

7. Facsimile(A) laughter (B) not genuine (C) epithet (D*) exact copy

8. Tenable(A) actual (B) valuable (C*) defensible (D) ever-lasting

9. Stellar(A) glorious (B) stolen (C) outstanding (D*) starry

Instructions(Q 10 to 13) Read the following passage and answer the questions carefully.The world of today has achieved much, but for all its declared love for humanity it has based itself far more

on hatred and violence than on the virtues that make man human. War is the negation of truth and

humanity. Sometimes, war may be unavoidable but its progeny are terrible to contemplate. Not mere

killing, for man must die, but the deliberate and persistent propagation of hatred and falsehood, which

gradually become the normal habits of the people.

It is dangerous and harmful to be guided in our life's course by hatreds and aversions, for they are

wasteful of energy and limit and twist the mind, and prevent if from perceiving the truth.

10. The achievements of the world are not impressive because

(A) there is nothing much to boast of (B) they are mostly in the field of violence

(C*) its love of humanity is a pretence (D) the world hasn't made any achievement

11. War is the negation of truth means

(A) wars do not exist (B) wars are evil

(C) wars kill human beings (D*) wars spread and advertise falsehood

12. The world's declared love of humanity is

(A) false (B) true (C) non-existent (D*) not to be taken seriously

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13. Man should be guided by

(A) scientific discoveries (B) practical wisdom

(C*) generous human feelings (D) materialism

Instruction (Q. 14 & 15) Select the option which has same relation as the given pair of words.

14. �Heart� is related to �Blood� in the same way as �Lung� is related to(A*) Oxygen (B) Chest (C) Purification (D) Air

15. �Face� is related to �Expression� in the same way as �Hand� is related to(A*) Gesture (B) Work (C) Handshake (D) Pointing

Directions (16 to 19): Find the missing numbers/letter/terms :

16. 2.5, 3.5, 15, 72, 352, ?

(A) 1785 (B*) 1885 (C) 1925 (D) 1980

Sol. Even number cube + cube�s digit�s multiplication

17. CK 10 5 JROF 7 3 TXKM ? ? PV(A) 4, 6 (B) 6, 8 (C*) 6, 11 (D) 10, 12

Sol. Sum of alphabets numbers and then add their digits.

18.

10 54 ?

7 45 32

24 144 68

(A) 42 (B) 36 (C) 6 (D*) 4Sol. (D) Half of the difference of top & bottom is the middle number in that column.

19. At a dinner party every two guests used a bowl of rice between them, every three guests used a bowl ofdal between them and every four used a bowl of meat between them. There were altogether 65 dishes.How many guests were present at the party?(A) 60 (B) 65 (C) 90 (D) None of these

Sol. (A) Let the number of guests be x. Then,

number of bowls of rice = 2x

; number of bowls of dal = 3x

; number of bowls of meat = 4x

.

1265x136512

x3x4x665

4x

3x

2x

.6013

1265x

20. If the following scrambled letters are rearranged to form the name of a city, the city so formed is famousfor its :ACGHHIORRTT(A) Locks (B*) Cement Plant (C) Temples (D) Pottery

Sol. (B) The city is CHITTORGARH and it is famous for Cement Plant.

21. DRAMA is coded as 73 and STAGE as 25. How will you code ACTOR ?(A) 56 (B) 50 (C*) 75 (D) 67

Sol. (C) DRAMA = (4 + 18 + 1 + 13 + 1) = 37STAGE = (19 + 20 + 1 + 7 + 5) = 52.ACTOR = (1 + 3 + 20 + 15 + 18) = 57


22. What was the day on 24th October, 1984 ?(A) Monday (B*) Wednesday (C) Thursday (D) Friday

23. Three positions of the same dice are given below. Observe the figures carefully and tell which number willcome in place of ?

5 4 52 1 ?

6 2 4

(i) (ii) (iii)

(A) 1 (B*) 3 (C) 2 (D) 6

24. Ram is to the South-East of Mukesh, Shyam is to the East of Mukesh and North-East of Ram. IfSuresh is to the North of Ram and North-West of Shyam, in which direction of Mukesh is Sureshlocated ?(A) North-West (B) South-West (C*) North-East (D) South-East

Directions : (25) In the following questions, some figures are given in a sequence. Find out the figure from thealternatives, which will come in place of the question mark to continue the sequence.

Problem Figure

25.

(1) (2) (3) (4) (5)

Answer Figure

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

Ans. D

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. A B D D D D D C D C D D C A A

Ques. 16 17 18 19 20 21 22 23 24 25

Ans. B C D A B C B B C D

BITSAT_PT-02 (CLASS-XII)

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