b.mat full syllabus test vii main solns

8/12/2019 B.mat Full Syllabus Test VII Main Solns

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1

Brilliant Tutorials Pvt. Ltd. IIT/BMAT7/PMC(MAIN)/Solns - 1

PART A : PHYSICS

1. (3) Half the sum of all charges given to

the plates appears on each of the

two outer most surfaces of the

system of plates. Total charge given

to the system is 10 C 4 C = 6 C

and half of 6 C is 3 C. Hence on

the outer faces of A and B 3 C

each appears. By law of

conservation of charge, the charge

on the inner face of A is 10 C

3 C = 7 C. By virtue of the fact

that the pair of apposite faces

carries equal and apposite charges,

the charge on the inner face of B is

7 C. When the plate B is earthed

3 C of charge flows from the

earth to the plate B, because 3 C

of positive charge resides on B. Thecharges on the inner faces being

bound they would remain intact.

Therefore choice (3) is correct.

2. (2) The given circuit can be redrawn as

follows:

A

B

The circuit can be further reduced

to

A B

C

C2C 2C

Hence equivalent capacitance is

3C

2

BRILLIANTS

FULL SYLLABUS TEST 7

FOR OUR STUDENTS

TOWARDS

JOINT ENTRANCE EXAMINATION, 2013

JEE 2013

B.MAT 7 (MAIN) SOLNS

PHYSICS

MATHEMATICS

CHEMISTRY

SOLUTIONS


2/17

2


3. (1) The charge configuration can be

viewed as the sum of an isolatedcharge and a dipole.

Setting V = 0 at infinity, we have

potential dV due to the dipole at p

is given by

= =

d 2 20

2qd k2qdV (r)

4 r rwhen r >> d

The potential at p due to the point

charge

= = +p 0

q q

V (r) k4 (r d) r when r >> d

The resultant potential at p is

= +

q 2dV(r) k 1

r r

4. (4) The distance moved by A in time t

is given by

= + 2A1

S 3 t 1 t2

(1)

The distance moved by B in time t

is given by

= + +2B1

S 1 t 2 t 102

(2)

The distance between B and A is

= = + +2

BA B At

S S S 2t 102

For BAS to be a minimum

Now ( )= +BAd

S 2 tdt

(3)

( )=2

BA2d S 1dt (4)

From (4), it is obvious that BAS is

minimum.

From (3), it is obvious that at t = 2,

the distance BAS is minimum.

From (1) = =AS (t 2) 8 m

From (2) = =BS (t 2) 16 m

=B AS S 8 m

5. (1) Comparing the given equation

namely y = bx 2cx of the trajec-

tory with the standard equation

namely

=

2

22gxy x tan sec2u

, we get

b = tan and

( )= = + 2 22 2g g

c sec 1 tan2u 2u

( ) = + =22a

c 1 b [ a g]2u

( ) = +2 2a

u 1 b2c

i.e., ( ) ( )+

= + =

2 10 1 25a

u 1 b2c 2 13

= 110 ms

6. (2) The acceleration of the system is

= =+ +

F Fa

2m m 2m 5 m

Hence the force acting on c of

mass 2 m is 2 ma = 2m =F 2

F5 m 5

,

consequently the reaction due to

mass C on B is2

F5 .

The F,B,D, of B is as follows


3/17

3


B

mg (weight)

2F (normal

5reaction)

rf (friction force)

Hence = r2

f F5

For the body not to slip in the down-

ward direction the condition is

=2

F mg5

=

5F mg

2

7. (3) In accordance with the principle of

superposition, the wave function

describing the resultant wave pulse is

obtained by adding the two wave

pulses. At t = 0, the distance

between the wave pulses is 30 cm.

After two seconds the wave pulse

ABCD would cover a distance of

20 cm towards right and the wave

pulse PQRS, would cover a distance

of 20 cm towards left. Superposition

takes place between the two pulses

and overlapping occurs as indicated

below.

A

P Q

S R

CB

D

pulse 1

pulse 2

Hence the resultant pulse will be as

follows.

A D

CB

8. (4) Let f be the frequency of the first

fork. Since two successive forksproduce 3 beats when sounded

together, the common difference is

evidently 3.

Hence the frequency of the 55 forks

forms an arithmetic progression. In

A.P., the sum to n terms is given by

the formula

= + nn

S [2a (n 1)d]2

Here a f, d = 3, and n = 55

13365 =

55

2 [2 f + (55 1) 3]

= + 13365

f 27 355

= 13365

f 27 355

= =243 81 162

Then the frequency of the last fork is

= + lastf f (55 1) 3

= + =162 54 3 324 Hz

9. (4) The resistance of an ideal ammeter iszero.

The resistance of an ideal voltmeter

is infinity. Since current takes the

path of least resistance, no current

flows through the voltmeter and

hence the voltmeter reads 0 V.

10. (3) By law of conservation of momen-

tum, we have

( ) = + 3 38 10 v 2.50 8 10 u (1)

where u is the velocity of bullet plus

block and v initial velocity of thebullet to be determined.

The bullet covers a vertical distance

of 1m.


4/17

4


Using the law of conservation of

energy, we have

u = =21

mu 2gs2

u 2gs=

19.6=

Substituting for u in equation (1) we

get

38 10 v 2.508 19.6 =

3

2.508v 19.6

8 10

=

32.508 4.43 108

=

= 1.39km/ s

11. (2) The circuit pertaining to the given

problem is as follows. Let E be the

equivalent emf of the source

balanced across the potentiometer

wire.

2v R = 790

40 cm

J BA

E2

E1

r1

r2

The current through AB is given by

= =+

2 2i A

790 10 800

Potential difference across AB is=

210V

800

Hence fall of potential per cm.,

lengths of the potentiometer wire is

=2 10 2

V800 100 8000

Hence fall of potential across 40 cm

lengths of the wire is 2

40 V8000

.

This is equal to emf of the source.

= = =2 1

E V 10 mV200 100

But( )

+=

+12 2 1

1 2

E r E rE

r rby theory

+

= 210 1/2 E 3 /2

102

= + 240 10 3E

=2E 10mV

12. (1) Eventhough the loop is closed,

current does not complete the

loop. Hence net force is not zero.

The segment OAB and OCB can be

replaced by OB. Hence the square

loop is equivalent to three wires

each of lengths 2 2 m and

carrying a current of 3.0A. Hence

the magnetic force on the square

loop is given by

( )= F i B Here ( )= i j

( ) ( )= + + + 3.0 2 2 i j 2.0 i j k

= + + + 12 2 i i j j i k j i

+ +

i j j k

= +

12 2 j i

( ) = F 12 2 2 = 24 N


5/17

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13. (2) Since the proton passes without

deviation, we have

Bev = Ee

= =

3

3

E 120 10v

B 50 10

= 62.4 10 m/ s

The number of protons striking the

target is

= =

19 3 1619

0.80 mA 1 110 10 10

2 21.6 10

As soon as the proton strikes the

grounded target, the proton almostloses its velocity. Hence change in

the momentum of the proton per

second is

mv = 27 61.6 10 2.4 10

Total change in momentum of the

beam per second

= 16 27 61

10 1.6 10 2.4 102

= 51.6 1.2 10

=

5

2

m

1.92 10 kg s

But rate of change of momentum =

force

Force exerted on the target= 52 10 N

14. (2) We have( )

= d N1

iR dt

( ) = =1

i dt d N 1R

= 0

1i dt d

R

But i dt , is equal to the area under

the given ( )1 t plot.

Area = 61

40 10 42

= 680 10

And = d ; the total charge in flux

= 680 10R

= 610 80 10

= 30.8 10 wb

15. (1) The induced emf is due to B

changing with respect to space

and time.

Due to B changing in time t, we

have say, 1E .

Due to B changing with space, we

have, say 2E .

Then = 1dB

E Adt

and

= 2dB

E Adt

= dB dx

A .dx dt

( ) ( ) = 2 31E 0.12 10

= 51.44 10 V

( ) ( ) ( ) = 2 1 22E 0.12 10 0.8 10

= 51.15 10 V

Since the effects are additive

= +1 2E E E

( ) = + 51.44 1.15 10

= 52.59 10 V

52.6 10 V


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16. (2) We have ( ) = t /i 1 e

= 0

q idt

i.e., ( )

=

t /

0

q i 1 e dt

= +

t /e

0i t

=i

qe

(1)

Hence

= = =

34L 1 10

10 sR 10

and = =V V

iR 10

From (1), we have,

=

46 V 103.68 10

2.718 10

= 6

4

3.68 27.18V 10

10

= =

6

4

100 101V

10

17. (3) The resistance R of the lamp is

= = =

V 100R 10

10

In order to make the lamp run at itsnormal condition, an inductancemust be connected in series withthe resistance.

For this = = =

V 200Z 20

10

Power supplied by the battery withinductance is in series is

= = = L

R 10P VI cos VI 200 10

Z 20

= 1000W

When inductance is replaced byadditional resistance the value of

additional resistance to be

connected in series with 10 resistance is given by

= add totalR R R

But = = total200

R 2010

= = addR 20 10 10

Power delivered by the battery is

= =V cos0 200 10 2000 W

Hence the ratio required is

=1000 W

2000W =

1

2

18. (4) As per Einsteins photoelectricequation, we have

= + 201

h h mv2

=21

mv eVs2

= +0h h eVs whose sV is the

stopping potential

= 0s

hV h

e e

From the graphh

e

= slope = 14

1.656

4 10

= 14 s0.414 10 V

= 15 s4.14 10 V

Hence in units of =15 sh

10 V , 4.14e

19. (1) Since the drop carries two units ofelectronic charge the velocity andforce diagrams will be as follows.

+

vx

kvy

qE

mg

kvx

vyv


7/17

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As the drop is falling with a constant

velocity, the net force on the dropmust be zero. Hence the electricforce and the component ofviscous force must balance along xdirection and the weight of thedrop and the component of viscousforce must balance along y-direction.Therefore

=xkv qE

=ykv mg

=y

x

v mg

v qE

From the velocity diagram

= y

x

vcot

v

Hence =mg

cotqE

or

=

1 mgtanqE

=

151

19

2

1.6 10 10tan

10003.2 10

2 10

( )= 1tan 1

= 45

20. (2) Let 0N be the initial number of

atoms of the radioactive source.

Then = t0N N e where N is the

number of atoms that remain intactat time t (in years)

we know = = n2 n2

T 4

( )T 4 years=

Number of radioactive atoms

needed for the treatment is

( )=

=

t 0

dN 10N

dt 4 365 24 60

=

010 NN4 365 24 60

... (1)

Two years later the activity

remained is

=

= =

0

t 2

dNN N 2

dt

If t minutes is needed for the

same treatment, the

t 2

dN tN

dt 4 365 24 60=

=

(2)

comparing (1) and (2) we get

=

20 010 N t N e

4 365 24 60 4 365 24 60

= =

n22 2t 10e 10 e

But

=

n2 n2e

2 n10 10e

n = 2

21. (3)

The diameter of the image of

aperture on the screen when the

screen is at a distance of 35 cm from

the lens is given by

20d 2 cm15

=

40 8cm 2.67 cm

15 3= = =


8/17

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22. (4) The mass rate of flow of water is

given by

P t = change in PE, where P is

power.

mgh =

P dm

gh dt=

6dm 2000 10

dt 10 160

=

6 65 10 1.25 10 kg4

= =

In units of =1dm

kgs , 1.25dt

23. (1) From the given data, the diameter

of the 4 balls together is given by

( )7.0 2.0 = 5.0 cm, Since the radius

of the third ball that is removed is

( )0.83 0.3 the diameter of the

same is ( )1.7 0.3 . Therefore the

sum of diameter of the remaining 3

balls is

( )5.0 1.7 3.3 cm =

Hence the radius of each ball is

3.30.3 cm

6

( )0.55 0.3 cm=

( )0.6 0.3 cm=

24. (2) The acceleration due to gravity

when the lift is moving with an

upward acceleration becomes(g + a). Hence the pressure at a

depth. h below the free surface of

the liquid is ( )p h g a= + .

25. (2)

A

B C

P Q

When the diode X is removed, thegiven circuit would become asfollows. During positive half cycle ofthe input the diodes A, B and Cconduct. Consequently currentpasses through A, PQ, C and B.During the negative half cycle, thediode B is reverse biased andhence the negative half-cycle issuppressed. As a result the outputwaveform will be as shown in (2)

26. (4) A B A B A B = + = +

A B A B A B+ = =

( ) ( )A B A B A B A B AB AB = + = +

T T 0 0 0 (AND gate)+ = + =

27. (3) Let the mass of the gas in moles be

n when its pressure is 20 atm. Let n/2moles be the mass of the gas afterreleasing n/2 moles. Since thevolume is constant, Charles law is

applicable. Hence

20 n 300

n

11.7 T2

20 n 300 600

11.7 n/2 T T

= =

600 11.7T 30 11.7

20

= =

351K.=

+ =300K t C 351

= t C 51 C


9/17

9


28. (4) We have c sf f 10.009+ =

c sf f 9.991 =

=cf 10 MHz

=sf 0.009 MHz

we have

cmE 2mV2

= , when m is modulation

factor and cE is amplitude of

carrier wave.

m 10mV2mV

2

=

m 0.4 =

Also s

c

Em

E=

sE 0.4 10 4mV = =

29. (1)

Since the point of contact O slips to

the left, the frictional force acts

towards right. If a be the

translational acceleration a of

the centre of mass, then we have

F f 1 a+ = ...(1)

considering rotational motion about

the centre of mass, we have

FR fR =

Now 22 a

MR and5 R

= =

( ) 22 a

RC F f R5 R

=

i.e., ( )2

F f a5

= ...(2)

subtracting (2) from (1) we have

= 2

2f a a5

=3

a5

210a 3 10 ms3

= =

using a in (2) we get

2F 3 10 4

5 = =

F 7N =

30. (3) We have2

GMg

R=

= =

3

2

G 4 4

R GR3 3R

Hence g R

Then

=

A A A

B B B

g R

g R

If AR and B are doubled, the ratio

A

B

g

g

would remain the same.

Further by Keplers 3rd law, g will not

be independent of the radius and

hence of the period of rotation.


10/17

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31. (2) We have

( ) ( ) ( )2 2 2

a b b c c a 0 + +

2 2a b ab bc ca 0 +

2 2 2a b c1

ab bc ca

+ +

+ +

2 2 2

ab bc ca1

a b c

+ +

+ +

Also, 2 2 2a b c 2ab cos c 2ab+ = <

2 2 2b c a 2bc+ < and

2 2 2c a b 2ca+ <

( )2 2 2a b c 2 ab bc ca + + < + +

2 2 2

ab bc ca 1

2a b c

+ + >

+ +

Range1

,12

=

32. (3) 1, 2, 3 R and ( )f 1 0= , ( )f 2 0= ,

( )f 3 0=

( ) ( )f 1 f 2 = but 1 2 .

f is not one-to-one mapping.

From the graph, it is clear that thisfunction takes all values from to + .

It is onto.

1 2 3

33. (4) Let us add one more number, n 1a + ,

to the given sequence. The number

+n 1a is such that n 1 na a 1+ = +

squaring all the numbers, we have

21a 0=

2 22 1 1a a 2a 1= + +

2 23 2 2a a 2a 1= + +

2 2n n 1 n 1a a 2a 1 = + +

2 2n 1 n na a 2a 1+ = + +

Adding the above equalities, we

get

2 2 2 2n1 n 12a a .... a a ++ + + +

2 2 21 2 na a ... a= + + + +

( )1 2 n2 a a ... a n+ + + +

( )1 2 n2 a a ... a + + +

2n 1n a n+= +

1 2 na a .... a 1

n 2

+ + +

34. (1) Since , are the roots of

( )2

x px p c 0 + = , + = p and

( )p c = +

Now, ( ) ( )1 1 1 c + + =

2 2

2 2

2 1 2 1

2 c 2 c

+ + + + +

+ + + +

( )

( ) ( )

( )

( ) ( )

+ += +

+ +

2 2

2 2

1 1

1 1 c 1 1 c

( )

( ) ( ) ( )

2

2

1

1 1 1

+=

+ + +

( )

( ) ( ) ( )

2

2

1

1 1 1

++

+ + +

PART B : MATHEMATICS


11/17

11


( )

( ) ( )

( )

( ) ( )

2 21 1

1 1

+ +

= + + +

1 11

+ + = = =

.

35. (1) , , are the roots of

3x px q 0+ + =

0 + + =

3 3 3 3 0 + + =

( ) ( )

=

2 2

( )2+

3 3 33 0= = .

36. (3)

1

n

n

1 2 nLt 1 1 ..... 1 k

n n n

+ + + =

(say)

n

1 1log k lim log 1

n n

= +

2 nlog 1 ... log 1

n n

+ + + + +

n

n r 1

1 rlim log 1

n n =

= +

( )1

0

log 1 x dx= +

( )

1

1

0

0

xx log 1 x dx

1 x

= + +

1

0

x 1 1log 2 dx

1 x 1 x

+=

+ +

( )1

0

log 2 x log 1 x = +

( )log 2 1 log 2 2 log 2 1= =

4log 4 log e log

e

= =

4k

e =

37. (1) 2 3 4dy

x y x y cos xdx

=

3 2 4dyx x y y cos xdx

+ =

+ =

34

3

dy y cos xy

dx x x (1)

put 3y z, = Then 4dy dz

3ydx dx

=

4 dy 1 dzydx 3 dx

=

(1) becomes3

1 dz z cos x

3 dx x x+ =

3

dz 3 3 cos xz

dx x x + = . It is linear in z

I.F3 dx

3 log x 34e e x

= =

38. (4) Let ( )1 1x , y be any point on the

required locus.

( )1 1 1 1 1c x , y , r y = =

( )2 2c 0, 3 , r 2= =

If the circles touch, then ( )1 2r r d =

( )22

1 1 1y 2 x y 3 = +

2 2 21 1 1 1 1y 4y 4 x y 6y 9 + = + +

Locus of ( )1 1x , y is2x 10y 5 0, + =

or 2x 2y 5 0 + = , a parabola


12/17

12


39. (4) Equation of OA isx y z

r1 2 3

= = = (say)

Any point on OA is of the form

(r, 2r, 3r) P (say). Direction ratios

of OA are (1, 2, 3). Direction ratios of

BP are (r 4, 2r 5, 3r 6). OA BP.

Then 1(r 4) + 2(2r 5) + 3(3r 6) = 0

16r

7 =

P is the point16 32 48

, ,7 7 7

40. (4) Let D be the mid point of BC

= + =BD DC DB DC 0

A

B CD

O

S

Now, we know that AO 2 SD=

=2 SD AO

+ +OA OB OC

= + + + +

OA OD DB OD DC

= + + +

OA 2 OD DB DC

= + + +

OA 2 OS SD 0

= + + =OA 2 OS AO 2 OS

41. (4) Let A, B, C denote the events of

passing the tests I, II and III

respectively. Clearly, A, B, C are

independent events.

( ) ( )( ) ( )= = 1

P A B A C P A B2

( ) ( )+ P A C P A B C

( ) ( ) ( ) ( )= +P A P B P A P C

( ) ( ) ( ) P A P B P C

1 1pq p pq

2 2= +

( )1 pq p p q 1 1 = + + =

Now, p(q + 1) = 1 is satisfied by

infinite number of values of p and q

42. (1) From the given conditions, we have

2 sin sin sin = + (1)

2cos cos cos = (2)

squaring (1),

2 2 24 sin sin sin 2 sin sin = + +

using (2),

( ) 24 1 cos cos 1 cos =

21 cos 2 sin sin+ +

2 2cos cos 4 cos cos +

( )2 sin sin 1=

2 2cos cos 4 cos cos

1 sin sin

+ =

2=

43. (3)( )sin 2A B 5sin B 1

+=

By componendo and dividendo,

we get

( )tan A B 3tan A 2

+=

44. (4)

45. (1) ( )+ = + + + 3 3 2A A 3A 3A

i.e., ( )( )+ + + = 2A A 2A

( )

+ + = + 12A 2A A


13/17

13


46. (1) The number of ways of placing

1 2A and A in 10 places so that 1A

is always above 2A is10

2C . There

are 8! ways of arranging the eightother candidates in the remaining 8places. So the total number ofarrangements is

( )( )10 210! 10!

C 8! 8!2! 8! 2

= =

47. (4) ( )1 1 1cos cos x cos x sin x + +

1cos cos x2

= +

( )1sin cos x=

1 2 2sin sin 1 x 1 x = =

1 2 61

25 5

= =

48. (1) We have

= + +

x a b y b c z c a

Taking dot products with

a, b ,c ,

we get

= =

a y a b c y 8 a

Similarly,

=

z 8 a b and

=

x 8 c

+ + = + +

x y z 8 a b c

49. (2) x y 1+ = represents the sides of a

square formed by the linesx y 1, x y 1,+ = = x y 1 + = and

=x y 1. 2 2 2x y a+ = represents

the circle of radius a and centre

(0, 0). Since1

a 1,2

< < each side of

square intersects the circle in twopoints. The number of solutions = 8

50. (3) 50th group contains 50 consecutive

positive integers of which the last

number is50 51

50 12752

= =

The sum of the 50th group

= 1226 + 1227 + + 1275

501226 1275 62, 525

2= + =

51. (3) f(x) is continuous in 0, and

derivable in ( )0,

Also, ( )f x 2 cos x 2 cos 2x = +

By Lagranges MVT, there exists

( )c 0, such that

( ) ( ) ( )f f 0

f c0

=

2 cos c 2 cos 2c 0+ =

2cos c 2 cos c 1 0 + =

( ) ( )cos c 1 2 cos c 1 0 + =

cos c 1 or 1/ 2 =

c /3 =

52. (4)

( )

=

n

on

o

x dx

G.E

x x dx

( )

( )

+ + + + =

1

o

0 1 2 ... n 1

n x x dx

( )1

o

n 1 n/2

n xdx

=

( )n 1 n/2n 1

n/ 2

= =


14/17

14


53. (2) Take 1 2 2 1n n 100 n 100 n+ = =

1x = average marks of boys = 52,

2x = average marks of girls = 42

x = average marks of boys and

girls = 50

combined mean 1 1 2 2

1 2

n x n xx

n n

+=

+

( )1 1n 52 100 n 4250100

+ =

1n 80 =

54. (4) Integer: 1 2 3 2n

Probability: log1 log2 log3 log 2n

A integer chosen is even, B chosen

integer is 2. Then ( )A B implies, the

chosen integer is 2. Required

probability ( ) ( )

( )

p A Bp B/A

p A= =

log 2

log 2 log 4 ... log 2n= + + +

( )nlog 2

log 2 4 ..... 2=

( )nlog 2

log 2 1. 2. 3. ... n=

( )

log 2

n log 2 log n!=

+

55. (3)6 61 1 0 = =

( ) ( )2 4 21 1 0 + + =

4 2 1 0 + + = ( )2 1

Now,

( )4 + + + + + + = + +

25 3

2 2

1 11

1 1

2

2 4 2

1

1

=

+ +

22

1

= =

56. (2) =

zw

1z i

3

=

zw 1

z i3

2

22

x y1

1x y

3

2+ =

+

221y y

3

=

1 2y 0

9 3 =

2y 3 =

3y ,

2 = a straight line

57. (2) ( ) ( )+ + = + +6

6 63 30 11 2 3 C C 2 3

( ) ( )+ + + +2 6

6 63 32 6C 2 3 ... C 2 3

In ( )+6

32 3 ,

( ) ( )

+ = r6 r6 3

r 1 rT C 2 3

It is rational if r = 0, 6. Similarly in

other cases.

The total number of rational

terms 1 0 1 1 1 1 2 7= + + + + + + =


15/17

15


58. (4) Let

2

2

3x 9x 17

y 3x 9x 7

+ +

= + +

( ) ( ) + 23y 3 x 9y 9 x

( )+ =7y 17 0

x is real 0

( ) ( ) ( )2

9y 9 4 3y 3 7y 17 0

23y 126y 123 0 +

2y 42y 41 0 +

1 y 41 The maximum value is 41

59. (2) 3G A B C= + + ( )3x,3y

( )a cos t b sin t 1, a sin t b cos t= + +

3x a cos t b sin t 1, = + +

3y a sin t b cos t=

3x 1 a cos t b sin t, = +

3y a sin t b cos t=

( ) ( )

2 2 2 23x 1 3y a b + = +

60. (1) The points on the minor axis, at a

distance of =2 2a b ae from the

centre are ( )0, ae . Length of the

perpendiculars from ( )0, ae to

the tangent 2 2 2y mx a m b= + +

are

+ +

+

2 2 2

2

ae a m b

1 mand

+ +

+

2 2 2

2

ae a m b

1 m

sum of their squares is

( )+ ++

2 2 2 2 2

2

2 a e a m b

1 m

( )(

)+ + =

+

2 2 2 2 2 2

2

2 a e a m a 1 e

1 m

22a=

A

B

C

P

Q

R

S

a

a

Aliter: Consider the tangent AB and

two points P and Q on the minor

axis. Clearly PR = QS = a

2 2 2PR QS 2a + =


16/17

16


61. (4)

62. (4)

63. (3) Adsorption is a spontaneous

change taking place on an

exothermic reaction, the gas phase

is retained by the adsorbent.

64. (4) 1 alkyl halide cannot form stable

carbocation. The CX bond

breaking and addition of X to the

carbon site taking place at the

same time. The rate of reaction is

directly proportional to the

concentration of the alkyl halideand nucleophile. It is an NS 2

reaction.

65. (3) 2CaH is ionic hydride giving rise

more H ions than 2BeH a covalent

species

66. (2) Number of electrons in +CN , CO and NO are 14. Bond

order is 3. Molecular orbital

configuration is as that of 2N .

67. (4)

= =

2 1m

, 400 Scm eq

68. (3)

= = = 2m400 2

8 Scm eq100

= n1 1000

cell constantR C

=

m

cellconstant 1000R

C

= =

0.4 1000500

8 0.1

69. (2) =% S

4Mass ofBaSO formed32

233 Mass of organiccompound

= =

32 0.699 100100 20 %

233 0.480

70. (4)

71. (3) In cubical close packing structure

(ccp), anions occupy primitive of

the cube, while cations occupy

voids. In ccp, there are 2

tetrahedral voids for every

octahedral void per anion. For one

oxygen atom there are two

tetrahedral voids and 1 octahedral

void cations ( )+2 . Number ofdivalent cation in tetrahedral void

= 2 1/5 since 1/2 octahedral voids

are occupied by trivalent cation+ 3Y . Number of trivalent cation

= 1 1/ 2, so formula of the

compound is

( ) = =1 2 1 1 4 5 1022 5 2

1X Y X Y O X Y O

5

72. (4) G H T S, =

At equilibrium, G 0 =

H T S =

and

= = H 35300

S JT 350

1 1100 J K mol

73. (4) As osmotic pressure of two solutions

are same molarity of 2 solutions are

equal.

Molar mass of glucose = 180

Number of gram of glucose in molar

solution 180 0.05 9= =

=9 6

180 M

= =

180 6M 120

9

Empirical

formula = 2CH O 12 2 16 30= + + =

( )n

Empirical formula = Molar mass

PART C : CHEMISTRY


17/17

17


( )n30 120=

4 8 4n 4 C H O=

74. (4) 2 2 2 2 2z1s , 1s , 2s , 2s , 2p ,

1 2

2 2 2 2x y x y2p , 2p 2p 2p3 4

75. (3)

76. (4) F has highest oxidation potential.

None of the oxidising agent can

reduce 0F to F

77. (3) The constituent of nucleic acids are,nitrogenous base, sugar and

phosphoric acid. The sugar present

in DNA is D () 2 deoxyribose, and

the sugar present in RNA is D ()

ribose. Due to D sugar com-

ponent, DNA and RNA molecules

are chiral

78. (2)

79. (3)

80. (2)

3CH

3CH

3H C

3H C

non superposable mirror image

81. (2) -hydroxyl carbonyl compounds

readily undergo dehydration to

form , -unsaturated aldehydes or

ketones

82. (4) Molar mass of compound

= = =

f 2

1

K W 1000 5.12 2 1000148.4

W T 100 0.69=i van'tHoff factor

=Theoretical Molar Mass

Experimental Molar mass

94i 0.6334148.4

= =

( )6 6C H

6 5 6 5 22C H OH C H OH

1

2

Total number of moles in the solution

= + = 1 1

2 2

=1

2i1

= 1 i2

( ) = 2 1 i

( )= 2 1 0.6334

= =0.3666 2 73.32% or 0.73.32

83. (2) Configuration varying at the

carbon atom is called anomer

84. (3)

85. (2)

86. (4)

87. (2)

88. (2)

2CH CH=

Cl

2CH CH=

Cl

( )2 nCH CHCl=

polyvinyl ch oride

89. (3)

90. (4)

b.mat full syllabus test vii main solns

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