part test ii (main) test - 2 main...ts-part test – 2, jee(main) champ square, sunrise forum, 2nd...

CHAMP SQUARE, Sunrise Forum, 2nd Floor (Debuka Nursing Home Lane), Circular Road, Ranchi – 834 001, Ph.: 7360016500

PART TEST –II (MAIN)

PPAARRTT –– AA ((PPHHYYSSIICCSS))

1. The adjacent diagram shows a charge +Q held on an

insulating support S and enclosed by a hollow spherical

conductor. The point O represents the centre of the

spherical conductor and P is a point such that OP = x and

Sp = r. The electric field at point P will be

(A) 2

0

Q

4 x

(B) 2

0

Q

4 r

(C) 0

(D) none of the above

1.Sol. A

Using Gauss’s law, we have

in0

1E dA q

2

0

20

1E 4 x Q Q Q

1 QE

4 x

2. If a point charge is placed at vertex of cube then, flux linked to surface

shaded in figure.

(A) 0

q

8 (B)

0

q

3

(C) 0

q

12 (D) zero

2.Sol. C

Flux through face 1 is zero and through each of the two faces 2 and 3 is

0

q

24 . So, total flux through shaded surface,

0 0

q q0 2

24 12

3. A neutral conducting spherical shell is kept near a charge q as shown.

The potential at point P due to the induced charges is

(point C represents the centre of the shell)

(A) kq

r (B)

kq

r

(C) kq kq

r r

(D)

kq

CP

q

1

2

3

Q x

r

SP = r

OP = x

- Q

+ Q

E

Charge +Q on

insulating support

O x

r

P

SP = r

OP = x

S

q

C

P r’

r q

TS-PART TEST – 2, JEE(Main)

CHAMP SQUARE, Sunrise Forum, 2nd Floor (Debuka Nursing Home Lane), Circular Road, Ranchi – 834 001, Ph.: 0651-6999023

2

3.Sol. C

Inside conducting shell, C PV V

or

0 0

1 q q q 1 qV

4 r R R 4 r

0

1 q qV

4 r r

4. Consider the shown uniform solid insulating sphere with a short and

light electric dipole of dipole moment ˆp j (embedded at its centre)

placed at rest on a horizontal surface. An electric field ˆE i is suddenly

switched on in the region such that the sphere instantly begins rolling

without slipping. Speed of the sphere when the dipole moment

becomes horizontal for the first time is (m = Mass of the sphere)

(A) 5pE

m (B)

10pE

7m

(C) 5pE

2m (D) zero

4.Sol. B

W K

2 21 1or PE cos 90 cos0 m I

2 2

22 2

2

1 1 2or PE m mR

2 2 5 R

10 PE

7m

5. In the figure below, what is the potential difference between

the point A and B between B and C respectively in steady

state.

(A) AB BCV V 100 V

(B) AB BCV 75 V, V 25 V

(C) AB BCV 25V, V 75 V

(D) AB BCV V 50 V

5.Sol. C

The equivalent circuit is shown in figure

1 2V V 100 and 1 22V 6 V

On solving above equations, we get V1 = 75 V, V2 = 25 V

100 V 20 Ω

1 µF

1 µF 3 µF

3 µF

B

10 Ω

C A

V2 V1

r

r’

q

P

C

R +

+

+

+

− − − −

1 µF

100 V 20 Ω 10 Ω

1 µF

1 µF 3 µF

3 µF

B

p

x

y



3

6. Two capacitors C1 and C2 = 2 C1 are connected in a circuit with a switch

between them as shown in the figure. Initially the switch is open and C1

holds charge Q. The switch is closed. At steady state, the charge on

each capacitor will be

(A) Q, 2Q

(B) Q / 3, 2Q / 3

(C) 3Q / 2, 3Q

(D) 2Q / 3, 4Q / 3

6.Sol. B

Common potential,

1 1 1

Q QV

C 2C 3C

Now charge, 1 1 2 2

Q 2QQ C V and Q C V

3 3

7. In following circuit, key is closed at time t = 0, then what will be

current through battery and charge on capacitor of 2F at that time

(A) 3 A, 12C

(B) 1. 5 A, 6C

(C) 2 A, 3C

(D) 6 A, 6C

7.Sol. A

At t = 0, capacitors offer zero resistance for dc and so effective resistance of the circuit is R 2

6

i 3A2

8. A mass spectrometer is a device which select particle of equal

mass. A particle with electric charge q > 0 starts at rest from a

source S and is accelerated through potential difference V. It

passes through a hole into a region of constant magnetic field

B perpendicular to the plane of the paper as shown in the

figure. The particle is deflected by the magnetic field and

emerges through the bottom hole at a distance d from the top

hole. The mass of the particle is

(A) q Bd

V (B)

2 2q B d

4V (C)

2 2q B d

8 V (D)

q Bd

2V

8.Sol. C

Radius d m

r2 qB

Also 21 2qV

qV m or2 m

2 2

2qVm

md q B dor m

2 qB 8V

Q C1

R

C2 = 2 C1

S

2 Ω

2 Ω

2 Ω K 6 V

2 µF

1 µF

B

V

S



4

9. A circular current loop of radius a is placed in a radial field B

as shown. Then net force acting on the loop is

(A) zero

(B) 2Balcos

(C) 2alBsin

(D) none

9.Sol. C

Loop experiences the magnetic force due to radial component of

magnetic field. Thus,

F B sin I perimeter Bsin I 2 a

10. A conducting wire bent in the form of a parabola y2 = x

carrying a current i = 1 A as shown in the figure. This wire is

placed in a magnetic field ˆB 2k tesla. The unit vector in

the direction of force is

(A) ˆ ˆ3i 4 j

5

(B)

ˆ î j

2

(C) ˆ î 2 j

5

(D) none of these

10.Sol. B

ˆ ˆ ˆ ˆ ˆ âb Ob Oa 4 i 2 j i j 3 i 3 j

ˆ ˆ ˆ ˆ ˆ ˆF i i B 1 3i 3 j 2k 6 j 6 i

ˆ ˆ ˆ ˆ6 j 6 i i jF̂

6 2 2

11. Consider three quantities 0 0x E/B, y 1/ and zCR

. Here l is the length of a wire, C is

a capacitance and R is a resistance. All other symbols have standard meanings. Then incorrect

option is

(A) x, y have the same dimensions

(B) y, z have the same dimensions

(C) z, x have the same dimensions

(D) none of the three pairs have the same dimensions

11.Sol. D

E / B has dimensions of velocity. CR has dimensions of time and so CR

has dimensions of velocity . So, x,

y and z all the three have dimensions of velocity.

x =1

x =4 x

b

a (1,1)

O

y

(4,-2)

I a

B sin θ

x =1

x =4 x

b

a

O

y

a

I B

θ



5

12. A flexible wire bent in the form of a circle is placed in a uniform

magnetic field perpendicular to the plane of the coil. The radius of the

coil changes as shown in figure. The graph of induced emf in the coil

is represented by

(A)

(B)

(C)

(D)

12.Sol. B

The flux through the coil is given by 2BAcos0 B r

Now d dr

e 2 Brdt dt

Initially dr

0dt

and so e = 0. After that dr

dt

is of constant value, also r increases, so e is also

increases.

Finally dr

dt

becomes zero and so e becomes zero.

13. A rectangular loop of wire shown below is coplanar with a long wire

carrying current I. The loop is pulled to the right as indicated. What

are the directions of induced current in the loop and the magnetic

forces on the left and right sides of the loop?

Induced Current Force on left side Force on right side

(A) Counter clockwise to the left to the right

(B) Clockwise to the left to the right

(C) Counter clockwise to the right to the left

(D) Clockwise to the right to the left

13.Sol. B

As loop move s away, the flux in the loop decreases and so current will induce clockwise to compensate this

flux.

I

t O

e

1 2 t

O

e

1 2

t O

e

1 2 t

O

e

1 2

t (s) O

r



6

14. A cylindrical region of radius 1m has instantaneous homogenous

magnetic field of 5T and it is increasing at a rate of 2T/s. A regular

hexagonal loop ABCDEFA of side 1 m is being drawn in to the

region with a constant speed of 1 m/s as shown in the figure. What

is the magnitude of emf developed in the loop just after the shown

instant when the corner A of the hexagon is coinciding with the

centre of the circle?

(A) 5/ 3 V (B) 2 /3V

(C) 5 3 2 /3 V (D) 5 3 V

14.Sol. C

The induced emf across the ends B and F due to motion of the loop,

1e B BF 5 1 2 sin60 5 3 V

The induced emf across the loop due to change in magnetic field

22

2

1dB R dB 2e A 2 V

dt 3 dt 3 3

So, 1 2

2e e e 5 3 V

3

15. Key is in position 2 for long time. Thereafter, it is in position

1 at t = 0, Resistances of the bulb and inductance of

inductor are marked in the figure choose the incorrect

alternative

(A) Bulb 2 dies as soon as key is switched into position 1

(B) Time in which brightness of bulb 1 becomes half of its

maximum brightness does not depend on t.

(C) If t = ∞, total heat produced in bulb 1 is 2

22

L

2R

(D) Ratio of maximum power consumption of bulbs is 1 : 1

15.Sol. D

The maximum current in the inductor 02

iR

Energy stored in the inductor

2 220 2

2 2

1 1 LU Li L

2 2 R 2R

When key is in position 1, this energy will convert into heat energy through resistor

16. Consider the circuit shown with respective specifications of

elements marked in the figure. Capacitor – 1 is charged

such that charge on it is Q0 and its left plate is positively

charged. While capacitor – 2 is uncharged. The switch is

closed at t = 0

(A) Frequency of oscillation of charge on left plate of

capacitor – 1 is 1

2 2L C

(B) Frequency of oscillation of charge on left plate of

capacitor – 1 is 1 2

LC

1 m/s A

1 m

1 m

B C

D

E F

key ε

R2

Bulb 2

Bulb 1

R1

1 2

L

Capacitor -2 Capacitor -1

C C

L

Inductor Switch



7

(C) Maximum current through the inductor is 0Q

2L C

(D) Maximum current through the inductor is 0Q

LC

16.Sol. C

eq

1 1f

2 LC 2LC

When i is max, di

0dt

Charge on two capacitors is same

2

02

2 0 00 0

Q

Q Q21Li 2 i

2 2C 4C 2LC

17. A metallic wire is folded to form a square loop of side a. It carries a current i and is kept

perpendicular to the region of uniform magnetic field B. If the shape of the loop is changed from

square to an equilateral triangle without changing the length of the wire and current. The amount

of work done in doing so is

(A) 2 4 3

Bia 19

(B)

2 3Bia 1

9

(C)

22Bia

3 (D) Zero

ANS. A 18. A charged particle of mass m & charge q enters a zone of uniform magnetic

field B with a velocity v making an angle θ with the boundary of the zone,

having width d, when the particle penetrates half – way into the zone, the

change in energy of the particle is

(A)

2mvsin

2B (B)

2

Bqdsin

mv

(C) zero (D) mv d

sinqB

18.Sol. C

Work done by magnetic field is zero.

19. A non conducting ring of radius R1 is charged such that the linear charge density is 2

1cos

where θ is the polar angle. If the radius is increased to R2 keeping the charge constant, the linear

charge density is changed to 2

2 cos . The relation connecting 1 2 1 2R , R , and

(A) 1 1 2 2/R /R (B) 1 2 2 1/R /R (C) 1 2 1 2R R (D) 2 21 1 2 2R R

ANS. B

θ

d

d/2



8

20. In the circuit shown the cells are ideal and have equal emf, the

capacitance of the capacitor is C and the resistance of the

resistor is R. The switch X is first connected to Y and then to Z.

After a long time, the total heat produced in the resistor will be

(A) equal to the energy finally stored in the capacitor

(B) half the energy finally stored in the capacitor

(C) twice the energy finally stored in the capacitor

(D) four times the energy finally stored in the capacitor

ANS. D

Numerical Value 1. In the given circuit, the potential difference across the capacitor is

12 V. Each resistance is of 3. The cell is ideal. The emf (in V) of

the cell is

Ans. 15

2. If energy stored in the capacitors C1 and C2 are same, then the

value of 1

2

C

C is given by

25

K . Find K.

Ans. 36

2.Sol.

At steady state the current, i3R

In close loop A H D E F A, 1

1

q0

C or 1 1q 2 C

Now in close loop BGCE FB

2

2

q2R i 0

C

G i 2R B

A ε

ε

i

F

H q1

q2

C1

R

C2

C

D

R 2R

E

E

C1

C2

ε

R R R

R

R C = 3µF

C

A

B R

Z

Y X

− +

+ −



9

Or 2

2

q2R 0.

3R C

22

5 Cq

3

Given,

2 21 2

1 2

q q

2C 2C

221 1 1 1

22 2 22

C q 2C C 25or

C 5 /3C C 36q

3. Consider a wire of non uniform cross section. If the area of

cross–section at point A is double of the area of cross – section

at point B. What is ratio of heat energy dissipated in a unit

volume at points A and B?

Ans. 0.25

3.Sol. 21 1

22 2

H i R R 1

H 4R 4i R

4. An electron and a positron are projected in a transverse

magnetic field (of strength B) with speed v each. If width of

magnetic field region is kr (where k is a constant, r is radius

of revolution of each of the particle). Time spent by each of

the particles in magnetic field is one–fourth of the time

period of revolution of particles for k equal to (Assume

elastic collision, if particles collide).

Ans. 1.414

4.Sol. krx 2sin or sinr 4 r

k 2

kr / 2 kr / 2

π / 4

4R R

i

B A

Positron

1.5 r

kr

Electron



10

5. In figure, there is a four way key at the middle. If key is

thrown from situation BD to AC, then how much charge

(in C) will flow through point O?

Ans. 72

5.Sol. When key is on the position BD, the situation is shown in the figure.

When key is on the position AD, the situation is shown in the figure

The charge flows to each capacitor is 36 C and so total charge flows through point O is 72 C

O

12 C

- 24 C

3 V

6 V

- 24 C

24 C

- 12 C

O

6 V

- 24 C

24 C

- 12 C

12 C

3 V

3 V

4 µF 4 µF

6 V

A B

O D

K

C



11

PPAARRTT –– BB ((CCHHEEMMIISSTTRRYY))

1. When an aromatic compound (A) was treated with Br2 / aq. KOH gives compound (B). Compound

(B) upon treatment with alcoholic potash and another compound (Y) gives a foul smelling gas with

formula C6H5NC was formed. Y was formed by reacting a compound (Z) with Cl2 in the presence

of slaked lime. The compound (Z) is

(A) C6H5NH2 (B) C2H5OH (C) CH3OCH3 (D) CHCl3

1.Sol. B

C

O

NH2Br2 / KOH

NH2

CHCl3 / alc. KOHNC

(A)

H3 CH2 OHCCl2 / Ca (OH)2

CHCl3 + HCOO

(Y)(Z)

2. Acid – catalysed hydration of an unknown compound X (C6H12) yielded Y (C6H14O) as the major

product a racemic mixture. Which (if any) of the following is (are) to be X ?

CH3CH3

CH3

(1)

CH3CH3

CH3(2)

CH2CH3

CH3

(3)

CH2CH3

CH3

CH3

(4)

(A) (1) and (3) (B) only (2) (C) only (4) (D) none of above

2. Sol. D

After hydration non of the compound (1) to (4) form an optically active compound.

3. Bakelite is obtained from phenol and ……………

(A) acetaldehyde (B) acetal (C) formaldehyde (D) chlorobenzene

Ans. C 4. Examine the following statements pertaining to a SN

2 reaction.

(a) The rate of reaction is independent of the concentration of the nucleophile

(b) The nucleophile attacks the carbon atom on the side of the molecule opposite to the group

being displaced

(c) The reaction proceeds with simultaneous bond formation and bond rupture amongst the

following which of the above were true?

(A) a, b (B) a, c (C) a, b, c (D) b, c

4.Sol. D

For SN2 reaction, Rate k R x Nu

Me

Et D

Cl

Nu

SN2

Me

Et D

ClNu- Cl

Me

Et Nu

D



12

5. Choose the compound which can react with [Ag(NH3)2]+ and on treatment with alk. KMnO4 gives

(CH3)3C – COOH

(A) CH3CH2CH2 – C C – CH3 (B) (CH3)2 CHCH2 – C CH

(C) (CH3)3 C – C CH (D) (CH3)3 C – C C – CH3

Ans. C 6. Methyl alcohol can be distinguished from ethyl alcohol using

(A) Fehling solution (B) Schiff’s reagent

(C) Sodium reagent (D) Sodium hydroxide and iodine

Ans. D 7. CH3

NH2

(CH3CO)2O

What is X?

(A)

CH3

Br

NH2

(B)

(C) (D)

Br2/CH3COOH H+/H2O

X

CH3

NH2

Br

CH3

NH2

COCH3

CH3

COCH3

NH2

7.Sol. B

CH3

NH2

(CH3CO)2O

CH3

NHCOCH3

Br2 / CH3COOH

NHCOCH3

Br

CH3

H3O+

NH2

Br

CH3

8. Maltose is made of

(A) two molecules of glucose (B) two molecules of fructose

(C) glucose and fructose molecules (D) two molecules of sucrose

Ans. A



13

9. An ester (A) with molecular formula C9H10O2 was treated with excess of CH3MgBr and the

compound so formed was treated with conc. H2SO4 to form olefin (B). Ozonolysis of B gave

ketone with formula C8H8O which shows positive iodoform test. The ester (A) is

(A) C2H5COOC6H5 (B) C7H7COOCH3

(C) ethyl benzoate (D) both (A) and (C)

Ans. C 10. C N

OCH3

+ CH3MgBr Q PH3O+

The product 'P' in the above reaction is

(A) (B)

COCH3

OH

(C) (D)

COCH3

OCH3

CH3 CH3

OH

OCH3

CH3 CH3

OH

OH

10.Sol. B

CN

OCH3

CH3MgBr

C NMgBr

CH3

OCH3

H3O+

OH

C CH3O

11. Which of the following pairs are correctly matched for the number of mono-chlorinated products

(ignoring stereoisomers) obtained from the listed isomeric hexanes?

Isomer Number of monochlorinated products

I : 2, 3 - dimethyl butane Two

II : 2, 2 - dimethyl butane Three

III : 2 – methyl pentane Five

Select the correct answer using the code given below:

(A) I, II and III (B) I and II only (C) II and III only (D) I and III only

Ans. A



14

12. Consider oxidation of the following alcohols

H2 CHCH2OHC PH2 CHCHOCI :

II : (CH3)3COH H3 C

CH3

CH2C

III : CH3CH2OHR

CH3COOH IV : CH3CH CH CH3

SCH3COOH

The oxidising agents A, B, C and D respectively are

P Q R S

(A) MnO2 Cu/ 4KMnO /OH,

H2CrO4

(B) Cu/ MnO2 H2CrO4 4KMnO /OH,

(C) MnO2 Cu/ H2CrO4

4KMnO /OH,

(D) MnO2 H2CrO4 Cu/

4KMnO /OH,

Ans. C 13. The IUPAC name of following compound is

C CH

CH3

CH

CH3

CH

CHO

CH2 CH3H3

(A) 2 – ethyl – 3, 4 – dimethylpentanal (B) 2, 3 dimethyl – 4 – aldohexane

(C) 2, 3, - dimethyl – 4 – ethyl pentanal (D) 1 – ethyl – 2, 3 - dimethyl pentanal

Ans. A

14. (2R, 3R) – 2,3 – pentandiol is

(A)

C

HHO

OHH

CH3

H2 CH3

(B)

C

OHH

HHO

C

H3

H2 CH3

(C)

C

HHO

OHH

C

H3

H2 CH3

(D) Both (A) & (C)

14.Sol. D

C

HHO

OHH

C

H3

H2 CH3

1

2

3

4 5

(2R, 3R)



15

15. Which comparison is correct as indicated?

(A) OH > H3 OHC (acidic nature)

(B)

NNH2

< ( kb value)

(C) COOH < HCOOH (pka

value)

(D) All

Ans. A

16. CH3

Cl

H2

H

CH3C

aq. KOH

SN2

product ( 2 - butanol )

d - (dextrorotatory)

(A) product is (laevorotatory) (B) product is racemic mixture

(C) product is d (D) none of the above

16.Sol. D

In SN2 reaction, inversion in configuration occurs but optical rotation may or may not be opposite

17. Most stable free radical in the following is

(A)

CH3

(B)

(C)

(D)

CH2

17.Sol. C

Higher the conjugation more will be stability.

18 Select the correct statement (s)

(A) All natural amino acids which are constituents of proteins, are -amino acids

(B) All -amino acids are optically active and have L – configuration

(C) In proteins -amino acids are connected by amide linkage

(D) Both (A) and (C)

Ans. D



16

19. End product of the reaction

Hg (OAc)2

NaBH4

AOH

CH3 CH3

A is

(A) OH

HO

CH3

(B) O

CH3 CH3

(C) OH

HO

CH3

(D) OH

CH3CH3

19.Sol. B

OH

CH2

Hg (OAc)2 OH

CH3CH3

HgOAc

O

CH3

HgOAc

NaBH4 O

CH3 CH3

20. Glycerol can be separated from spent - lye in soap industry by:

(A) Steam distillation (B) fractional distillation

(C) distillation under reduced pressure (D) ordinary distillation

Ans. C Numerical Value 1. Number of chiral center(s) present in hydrolysis product of A (given below) is

O O

Ans. 1

1.Sol.

O O

H3O+

OH OHHO OH- H2O

O HHO OH

2. Me

NMe

EtBu

OH

Major product

n -

The number of possible hyper conjugation form in major product is



17

Ans. 0

2.Sol. There are four β – hydrogens in this quaternary ammonium salt, on heating it will give Hoffman eliminated

product by abstraction of most acidic β – H.

3. The sulphur content of cystine is 32%. If cystine contains two sulphur atoms, then molecular

weight of cystine is

Ans. 200 4. How many of the following is/are correctly matched?

(i) Neoprene CH3 CH2 C

Cl

CH CH2 CH3

n

(ii) Nylon – 66 NHCH3 (CH2)6NH C

O

(CH2)4C

O

CH3

n

(iii) Terylene OCH3 CH2CH2 C

O

C

O

CH3

n

(iv) Teflon CCH3 F2 CF2 CH3

n

Ans. 3

4.Sol.

OC COOCH2 CH2 CH3OCH3

n

Terylene

5. An organic compound ‘A’ having molecular formula C2H3N on reduction gave another compound

‘B’. Upon treatment with nitrous acid, ‘B’ gave ethyl alcohol. On warming with chloroform and

alcoholic KOH ‘B’ forms offensive smelling compound ‘C’. The number of lone pairs of electron in

compound ‘C’ is

Ans. 1

5.Sol.

H3 C NCReduction

CH3CH2NH2

HNO2CH3CH2 OH

(A) (B)

CH3CH2 NH2CHCl3 / KOH

CH3CH2NC

(B)



18

PPAARRTT –– CC ((MMAATTHHEEMMAATTIICCSS))

1. Let z1, z2 be two complex numbers such that z1 + z2 and z1z2 both are real, then

(A) z1 = - z2 (B) 1 2z z (C) 1 2z z (D) z1 = z2

1. Sol. (B)

Let z1 = a + ib, z2 = c + id, then z1 and z2 is real

(a + c) + i(b + d) is real

b + d = 0 d = - b

z1z2 is real

(ac – bd) + i(ad + bc) is real

ad + bc = 0

a(-b) + bc = 0

a = c

z1 = a + ib = c – id = 2z a c and b d

2. The value of

6

k 1

2 k 2 ksin icos

7 7

is

(A) -1 (B) 0 (C) – i (D) i

2. Sol. (D)

Let 2 2

z cos isin7 7

Then by De Moivre’s theorem, we have

k 2 k 2 k

z cos isin7 7

3. The points representing the complex number z for which |z + 5|2 - |z – 5|2 = 10 lie on

(A) a straight line

(B) a circle

(C) a parabola

(D) the bisector of the line joining (5, 0) and (-5, 0)

3. Sol. (A)

z 5 z 5 z 5 z 5 10

or 5 z z 25 5 z z 25 10

12 2x 2 x (a)

2

4. If z 2

2z 3

represents a circle, then its radius is equal to

(A) 1

3 (B)

3

4 (C)

2

3 (D) 1

4. Sol. (C)

2 2

z 22

z 3

20 32x y x 0

3 3



19

Which is a circle with centre 10

,03

and

radius = 100 32 4 2

9 3 9 3

5. How many ways are there to arrange the letters in the word GARDEN with the vowels in

alphabetical order?

(A) 120 (B) 240 (C) 360 (D) 480

5. Sol. (C)

Take A, E as similar letter (eg. BB)

Then six letters can be arranged in 6 !/2 !, i.e., 360 ways.

6. Seven women and seven men are to sit around a circular table such that there is a man on either

side of every women. The number of seating arrangements is

(A) (7 !)2 (B) (6 !)2 (C) 6 ! 7! (D) 7 !

6. Sol. (C)

First we seat seven women, which can be done by (7 – 1) ! = 6 ! ways

In any such arrangement, seven men can be seated in seven places marked in 7P7 = 7 ! ways

Required number of ways = 6 ! 7 !

7. The number of ways in which 10 persons can go in two boats, so that there may be 5 on each

boat, supposing that two particular persons will not go in the same boat is

(A) 105

1C

2 (B) 8

51

C2

(C) 8

42 C (D) 8C4

7. Sol. (C)

First omit two particular persons. Remaining eight persons may be four in each boat. This can be done in 8C4

ways. The two particular persons may be placed in two ways one in each boat

Total number of ways = 2 8C4

8. The number of ways of dividing 52 cards amongst four players so that three players have 17

cards each and the fourth player just one card, is

(A)

3

52! 1

3!17! (B) 52 ! (C)

3

52!

17! (D) none of these

8. Sol. (A)

3

52! 35! 18! 1 ! 52! 1 !

35!17! 18!17! 17!1! 3! 3!17!

W

W

W

W W

W

W

X

X

X

X X

X

X



20

9. There are four balls of different colors and four boxes of colors same as those of the balls. The

number of ways in which the balls, one in each box, could be placed such that a ball does not go

to box of its own colour is

(A) 8 (B) 7 (C) 9 (D) none of these

9. Sol. (C)

1 1 1

4! 12 4 1 92! 3! 4!

10. If the second term in the expansion

n

13

1

xx

x

is 14x5/2 then the value of n n3 2C / C is

(A) 12 (B) 6 (C) 4 (D) 3

10. Sol. (C)

n 1

1n 5 / 2132 1T C x x x 14x

or

n 1 35 / 213 2nx 14x

n

rn

r 1

n3

n2

n 1 3 5

13 2 2

n 14

C n r 1Now, for r 3, n 14

rC

C 14 3 14

3C

11. The number of terms which are free from radical signs in the expansion of 55

1/5 1/10x y is

(A) 6 (B) 7 (C) 8 (D) none of these

11. Sol. (A)

55 r5555 r /10 55 r /5 r /105

r 1 r r

r 0

T C x y x C x y

The terms which are free from radical will correspond to those values of r in which the radical signs

disappear, i.e., r r

and5 10

. Clearly from (i) r should be a multiple of 10 of range 0 to 55.

r = 0, 10, 20, 30, 40, 50

Thus there will be six terms free from radical signs.

12. If the 4th term in the expansion of

n1

axx

is

5

2, then the value of a and n are

(A) 1

,62

(B) 1, 3 (C) 1

,32

(D) cannot be found

12. Sol. (A)

It is given that the fourth term in the expansion

n1 5

of ax isx 2

, therefore



21

3n n 3

3

n n 3 n 63

1 5C (ax)

x 2

5C a x

2

We must have x0 for which n – 6 = 0. Hence,

n = 6 and we get 6 3

35

C a2

3 1 1

a a8 2

13. If (1 + x + 2x2)20 = a0 + a1x + a2x2 + …. + a40x40 then a1 + a3 + a5 + ….. + a39 equals

(A) 219(220 – 21) (B) 230(219 – 19) (C) 219(220 + 21) (D) none of these

13. Sol.: (D)

Putting x = 1 and -1 and subtracting, we get 420 – 220 = 2[a1 + a3 + a5 + …. + a39]

219(220 – 1) = a1 + a3 + … + a37 + a39

14. The sum

m

i 0

10 20 p, where 0 if p q

i m i q

is maximum when m is

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

14. Sol. (C)

m10 20 10 20 10 20 10 20 10 20

i m i 0 m 1 m 1 2 m 2 m 0

i 0

C C C C C C C C ... C C

= Coefficient of xm in the expansion of product (1 + x)10 (x + 1)20

= Coefficient of xm in the expansion of (1 + x)30

= 30Cm

Hence the maximum value 30Cm is 30C15.

15. The largest real value for x such that

4 4 k k

k 0

5 x 8

4 k ! k! 3

is

(A) 2 2 5 (B) 2 2 5 (C) 2 2 5 (D) none of these

15. Sol. (A)

4 4 k k

k 0

4 4 k k

k 0

4 4 4 k kk

k 0

4

5 x

4 k ! k!

5 x 4!

4 k ! k! 4!

C 5 x

4!

5 x

4!

16. A man has 3 pairs of black socks and 2 pairs of brown socks kept together in a box. If he dressed

hurriedly in the dark, the probability that after he has put on a black sock, he will then put on

another black sock is

(A) 1

3 (B)

2

3 (C)

3

5 (D)

2

15



22

16. Sol. (A)

6

11 10

1

C 6 3P B

10 5C

2 1 2

1 2 1 2 1

5P B / B B Black

9

P B B P B P B / B

3 5 1

5 9 3

17. A and B are two events such that P(A) > 0, P(B) 1, then P A /B is equal to

(A) 1 – P(A/B) (B) 1 P A /B (C)

1 P A B

P B

(D)

P A

P B

17. Sol. (C)

P A B P A BP A / B

P B P B

1 P A B

P B

18. A bag contains 5 white and 3 black balls. Four balls are successively drawn out and not replaced.

The probability that they are alternately of different colors is

(A) 1

196 (B)

2

7 (C)

1

7 (D)

13

56

18. Sol. (C)

Required probability = P(WBWB) + P(BWBW)

5 3 4 2 3 5 2 4

8 7 6 5 8 7 6 5

1 1 1

14 14 7

19. A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of

distribution of number of successes is

(A) 8

3 (B)

3

8 (C)

4

5 (D)

5

4

19. Sol. (D)

Here n = 5

p = P (an odd number) = 3 1

6 2

1 1q 1 p 1

2 2

1 1 5Variance npq 5

2 2 4

20. Three of the six vertices of a regular hexagon are chosen at random. The probability that the

triangle with three vertices is equilateral equals

(A) 1/2 (B) 1/5 (C) 1/10 (D) 1/20



23

20. Sol. (C)

Out of 6 vertices 3 can be chosen in 6C3 ways, will be equilateral if it is ACE or BDF (2 ways)

Required probability = 6

3

2 2 1

20 10C

Numerical Value

1. If 1 sin2 cos2

f2cos2

then the value of 0 02f 11 f 34 equals

Ans. 1

1.

2 2 2

1 sin2 cos2f

2cos2

cos sin cos sin

2 cos sin cos sin

2cos 1

2 cos sin 1 tan

0 0

0 0

1 1f 11 f 34

1 tan11 1 tan34

0 0 0

1 1

1 tan11 1 tan 45 11

00

0

0

0

1 1

1 tan111 tan111

1 tan11

1 1 tan11 1

2 21 tan11

2. Value of 1

2r 0

1tan

1 r r

is equal to

k

, then k equals to

Ans. 2

2. Sol.

1 1

2

1 r 1 rtan tan

1 r r 11 r r

= tan-1 (r + 1) – tan-1(r)

n1 1 1 1

r 0

tan (r 1) tan (r) tan (n 1) tan (0)

= tan-1 (n + 1)

A B

C

D E

F



24

n

1 1

2r 0

1tan tan

21 r r

3. A(3, 1), B(6, 5) and C(x, y) are three points such that angle ACB is a right angle and the area of triangle ABC

= 7 sq. unit, then the number of such points C is

ANS. 0

Sol. It is clear that AB = 5, let AC = b, BC = a,

then area of

ab

ABC 72

Thus such a point C is possible if

a2 + b2 = 5, ab = 14

We easily observe that this system does not have real solutions.

4. The area of the triangle formed by joining the origin to the points of intersection of the line 5x 2y 3 5

and circle x2 + y2 = 10 is

ANS. 5

Sol. The equation of lines joining origin and point of intersection of 5x 2y 3 5 and 2 2x y 10,

is

2

2 2 5x 2yx y 10

3 5 or

2 2x y 8 5xy 0

These lines are perpendicular.

Area of triangle 1

base height2

1

radius radius2

1

10 10 52

5. The foci of the ellipse 2 2

2

x y1,

16 b and hyperbola

2 2x y 1

144 81 25 concide. Then the value of b2 is

ANS. 7

Sol. We must have 12

ae a'e' 4e e'5

Where 2 2 281 144b 16 1 e , e' 1

25 25 from last equation

15e'

12 then from, first equation

3e

4

whence from second equation

2 9b 16 1 7

12

C(x, y)

900

A(3, 1) B(6, 5)

part test ii (main) test - 2 main...ts-part test – 2, jee(main) champ square, sunrise forum, 2nd...

Documents