Class: SZ2 JEE-ADV(2014-P2) MODEL Date: 26-12-20 Time: 3hrs … · 2020. 12. 27. · Narayana CO...

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Class: SZ2 JEE-ADV(2014-P2) MODEL Date: 26-12-20

Time: 3hrs WAT-14 Max. Marks: 180

Jee-Adv_Model_Initial Key

SZ2 Physics Initial Key Dt. 26-12-2020

Q.No. 01 02 03 04 05 06 07 08 09 10

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

Q.No. 11 12 13 14 15 16 17 18 19 20

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

SZ2 Chemistry Initial Key Dt. 26-12-2020

Q.No. 21 22 23 24 25 26 27 28 29 30

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

Q.No. 31 32 33 34 35 36 37 38 39 40

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

SZ2 Mathematics Initial Key Dt. 26-12-2020

Q.No. 41 42 43 44 45 46 47 48 49 50

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

Q.No. 51 52 53 54 55 56 57 58 59 60

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

SZ2_JEE-ADV(2014-P2)_WAT-14_Key&Solutions_Ex.Dt.26-12-20

Narayana CO Schools

2

1.

du=- 0.Gdm m

x

2

2

ld

dl

u du

+

−

=

2

0 2log2

e

Gm d lu

l d l

+ = −

−

2. For no acceleration

1 2mg mr=

2

3

GMr r

R=

0

4

3G =

3. From E=2

2Gm

R

sin2

2

12 .

Gm

R and L R=

2

2( )

GmE j

L

=

4. Let “ ” is density then

3 34. 83

M P R R = − 34. .7

3R =

For full sphere 1 34

. 83

M R =

1 8

7M M=


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3

Potential due to cavity is potential due to full sphere-potential due

to cavity, so

3 8 772

2

M

MV G

R

R

= − −

9

14

GM

R= −

5. Given A BE E= SO

1 2 1

2 2 20

4 4

GM GM GM

R R R+ = +

212 2

3.

4 4

GMGM

R R=

2 13M M=

3 3 32 14 4

(2 ) 3. .3 3

R R R − =

2 17 3 =

1

2

7

3

=


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4

6. due to removed sphereincavity sphereV V V= −

2 12

3

3(3 )

2 4 2( )

2

incavity

GM R GMV R

RR= − − +

11 3.

8 2 8

2

GM GM M

RR− +

113

8 8

GM GM

R R− +

GM

VR

= −

7. For r R2

2.

mv GMm

r r=

1v and

r

r R 2

3.

mv GMm r

r R=

8. 1 2(4 )

P

GME

a=

9. 1 2(4 )

P

GME

a=

2

2

GMm mv

r r=

And 2

GMmK

r= and


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5

2KrM

Gm=

(OR)

2 dv =

Kdr

Gm

2 2.4K

r dr drGM

=

22

Kand

r Gm

=

2 22

e K

m Gm r=

10. 2

1 1

2

2 2

2 2

3 1

E M

E M= =

1

2

1

6

M

M=

11. For r= 2

R,up to

3

R, 0 = and between

3

Rand

2

R, 0

2

=

3 3 3

00

4 4 4

3 3 3 2 3 3 2

R R Rm

= + −

=0.108 3 0R

Gravitational field at a distance r=R/2 is

( )

( )3 02 2

0.0184

/ 2g

G RGmE

RR

= = 0

35

81

GR=

12. For 05 R

, Up to ,6 3 8

Rr

= = ,between

R

3and

3R

4,

0 03R 5 ,2 4 6 8

Rand between and

= =

The mass of the sphere of radius 5r/6 is

0 03R 5 ,2 4 6 8

Rand between and

= =


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3 3 3

0 0

4 4 3 4/ 2

3 3 3 4 3 3

R R Rm

= + −

+

3 3

0

4 5 4 3/8

3 6 3 4

R R

−

=0.332 3 0R

Gravitational field at a distance, r=5

6

R is

2(5 / 6)g

GmE

R=

3

0

2

0.33236

25

G R

R

= 00.48 GR =

13. 2 2

1 1 11

4 16 64

GM GME

R r

= = − + +

2 114

GM

r

= −

2

4

3

GME

r=

14. Gm

VR

= −

1 1 11

2 4 8

GmV

r

= − + + +

2 3

1 1 11

2 2 2

GmV

r

= − + + +

1

11

2

GmV

r= −

−

2GmV

r=

15. 3

11 4 64 6413

MM M

M= = =

Gravitational potential at

‘p’ on the circle 2 2 26Y Z+ = is

1

26 2 10

GM GMV = − +

64

6 10

GM GMV = − +

10 32

10 3V GM

= −


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16. Potential at any point P on the circle 2 2 22Y Z+ = is

12 2

3(3 4 2 ) 2

2 4 2 2

GM GMV

Y= − − +

4(44)

2 2

GM MV = − +

222

2V GM

= −

17. A) F = Eq =

M=2kg a=g=5: N/kg ( 2 23 4+ )

F=2*5=10N

= 053

B) centripenton GravitationF F=

2 2 2 2 0cos60mrw F F F= + +

2 3mrw F=

22

2. 3

3

a Gmm w

a

=

3

3Gmw

a=

C) Decreases

D) 1 221

r r F R Rr

= =


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2

1 1

2 2

F r

F r 2

R

R

= =

12

F

4F =

18. A) 2g

kgmh

−=

B) 1X

gR

−

=

21

hg

R

−

2X h=

C) 2

GMmF

R=

D) 2

Gmg

R=

2 2

(0.9 )0.625

(1.2 )

G m Gma

R R

= =

62.5%= of earth’s acceleration due to gravity

= 5

8 time that on the surface of the earth

19. ( )A q→

( )B p→

( )C p→

( )D q→

3

2 2

4

3G R

GmE

R R

= = E R

At surface,

34

3G R

GmV

R R

= − = − 2V R

3

2centre

GMV

R= −

At centre of a solid sphere field strength is zero


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20. A) Resultant 12 sindE dE = 2 2 2 2

( )2

( ) ( )

G dm d

d x d x=

+ +

2 2 2 22

( ) ( )

GM ddx

L d x d x

=

+ +

Total gravitational field

3

2

2

2 20

2 dx

( )

L

GmdE

L d x

=

+

Integrating the above equation it can be found that,

2 2

2

4

GmE

d L d=

+

B) 2

22CB

GmF j

R

=

2

24CD

GmF j

R

= −

2 2

2 2cos 45 sin 45

4 4CA

Gm GmF i j

R R

= − +

C CA CB CDF F F F= + +

2 2

2 2

1 12 2

4 2 4 2

Gm Gmi j

R R

= − + + +

C) The value at a depth h is

1h

g gR

= −

-2 5.0(9.8 ms ) 16400

km

km

= −

29.79ms−=

D) ( )

2

24

h

g gRg

R h= =

+

21

4

R

R h

+

1

2

R

R h =

+

h=R


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41.

The equation of line PQ is

( )h

y k x hk

− = − −

Or 2 2hx ky h k+ = +

2 2 2 2

,0 0,h k h k

Q and Ph k

+ +

Also, 2 21 12a x y= +

Or 2 2 2

1 1 4x y a+ =

Eliminating 1x and 1y , we have

( )2

2 2 2

2 2

1 14x y a

x y

+ + =

42. the given circle is ( ) ( )2 2

1 2 9x y+ = + = , which has radius 3.

The points on the circle which are nearest and farthest to the point

P(a,b) are Q and R, respectively.

Thus, the circle centered at Q having radius PQ will be the smallest

circle while the circle centered at R having radius PR will be the

largest required circle


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Hence, the difference between their radii is PR-PQ=QR=6.

43.

Since 075B C = = , we have

030BAC =

060or BOC =

( )0 03

cos30 , sin 30 ,2 2

a aB a a

− −

And 3

,2 2

a aC

−

44.

( )( ) ( )( )1 2 1 2 0x x x x y x y x− − + − + =

( ) ( )2 2 1 2 2 1 0x y x x x x x y+ − + + − =

( )2 2 2 4 0x y ax a b y+ + − =

45.

46. By observation, directly equation must be

( ) ( )2 24 4 2 2 1 0x xy y x y− + + − + =


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Or ( )2 24 4 2 2 1 0x xy y x y− + − − + =

( ) ( )2 2

2 1 0 2 1 0x y or x y− + = − − =

i.e 2 24 4 4 2 1 0x xy y x y− + + − + =

or 2 24 4 4 2 1 0x xy y x y− + − + + =

1 2 1 2 2 2 1 4 2a b c a b c + + + + + = + −

+1-4+2=2

48. PAB is the required triangle

Area of 1

2 2 22

PAB = =

51 & 52

Let ( ),D be an interior point of a circle 2 2 2x y a+ = . Then

2 2 2a +

Now, the equation of BC is

cos sin

x y

− −=


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Let DB=r. Then the coordinates of B are ( )cos , sin .r r + + It lies

on 2 2 2x y a+ = . Therefore,

( ) ( )2 2 2cos sinr r a + + + =

Or ( )2 2 2 22 cos sin 0r r + + + + − =

This is quadratic in r, which gives the values of r, one each for

points B and C.

Let 1DB r= and 2DC r= .

Then 1r and 2r are the roots of (ii). Therefore,

( )1 2 2 cos sinr r + = − +

Also 2 2 21 2r r a = + −

The harmonic mean of 1r and 2r is

( )

( )

2 2 2

1 2

1 2

22

2 cos sin

ar r

r r

+ −= −

+ +

( )

( )

2 2 2

cos sin

a

+ −= −

+

53. 1 9 9 1r = + − =

54. Diameter =2 (reading) ( )2 5 10= = 55 &56

As the lines are perpendicular,

Coefficient of 2x +Coefficient of 2 0y =

2 0a − =

Or a=2

Also, 2 2 22 0abc fgh af bg ch+ − − − =

3c =−

Hence, the given pair of lines is

2 22 3 2 5 5 3 0x xy y x y+ − − + − =

Factorizing, we get lines

X+2y-3=0 and 2x-y+1=0


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The point of intersection of the lines is C(1/5, 7/5)

The points of intersection of the lines with the x-axis are A(3,0) and

B(-1/2,0)

57.

( ), lies on the line y=x

From the diagram, we get a-s; b-r; c-q; d-p

59. ( ) ( )2 2 2 2 2 21 2 2 3 3 1 1 2 3 1 2 3 1 2 2 3 3 12 2 2 2 2 2 0x x x x x x x x x y y y y y y y y y+ + + + + + + + + + + =

( ) ( )2 2

1 2 3 1 2 3 0x x x y y y + + + + + =

1 2 3x x x+ + =0 and 1 2 3y y y+ + =0

circumcenter and centroid of ABC coincide

ABC is equilateral

( ) ( ) ( )2 2 2

PA PB PC+ +

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2

1 1 2 2 3 3x x y y x x y y x x y y= − + − + − + − + − + −

( )2 23 1 6x y= + + =

B) , 33

OBQ k

= =


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C) Max. distance of R from S, 2

1 2 12

d = + = +

2 3 2 2d = +

a+b=3+2=5

D) I and g coincide with origin IA=IB=IC=GA=GB=GC=1

IA+IB+IC+GA+GB+GC=6

60.

A. ( )( )3 3 2 20 0x y x y x xy y+ = + − + =

( ) 0x y + =

the equation represents one real line

B. ( )22 2 32 0 . 0x y xy y x y y+ + + = + =

0x y + = or y=0

the equation represents two distinct line

C. ( ) ( )2 2

2 2 2 21 4 0 1 0 4 0x y x and y− + − = − = − =

1 2x and y= =

the equation represents four points (1,2), (-1,2), (1,-2) and (-1,-

2)

D. ( )( ) ( ) ( )22 2 21 1 1 1 0x x x y y x− − + − + =

( ) ( ) ( )2 22 21 1 1 0x x x y y + − + − =

( ) ( )2 22 21 0 1 1 0x or x x y y + = − + − =

( ) ( )1 0 1 0 1 0x or x x and y y + = − = − =

The equation represents a line x+1=0 and four points

(0,0),(0,1),(1,0) and (1,1)

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Class: SZ2 JEE-ADV(2014-P2) MODEL Date: 26-12-20 Time: 3hrs … · 2020. 12. 27. · Narayana CO...

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