Class: SZ2 JEE-ADV(2014-P1) MODEL Date: 12-07-20 Time ... · Narayana CO Schools 1 Class: SZ2...

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Class: SZ2 JEE-ADV(2014-P1) MODEL Date: 12-07-20

Time: 3hrs WAT-03 INITIAL KEY Max. Marks:180

INITIAL KEY

SZ2 Physics Initial Key Dt. 12-07-2020

Q.No. 1 2 3 4 5 6 7 8 9 10

Ans. AC AC AB ACD AD ABCD AD CD AC AB

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

Ans. 6 1 6 8 6 4 5 4 3 8

SZ2 Chemistry Initial Key Dt. 12-07-2020

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

Ans. ACD ACD ABC BC A ACD BCD BD ABC B

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

Ans. 7 5 9 2 2 2 9 5 8 9

SZ2 Mathematics Initial Key Dt. 12-07-2020

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

Ans. ABD ABC BD ABC BD BC ABC AC ABCD ABCD

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

Ans. 1 6 2 5 0 2 4 8 5 2

SZ2_JEE-ADV(2014-P1)_WAT-03_Key&Solutions_Exam.Dt.12-07-20

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01.

1 0

1

2 0

2

2 1 2 1

1 2

1 2

4

4

so p <p

4 1 14

0.004

Tp p

n

TP P

n

n n

p p p

TT

R n n

R cm

= +

= +

= −

= −

=

So smaller concave, larger convex

02. T

HRdg

=

Given that h, T, d and g are fixed hence R must be same As weight

of the liquid by force due to surface tension (T cos and R are fixed

for a given liquid, given material of capillary at a constant

temperature) hence, weight of liquid in both the capillaries must be

equal

So, option 1 and 2 are correct

03. ( )( ), ' 2a b QQ T F =

( )

.316 10' 0.08 8

2 2 0.1

FQQ m cm

T

− = = = =

4Qm cm =

2 25 4 3Om cm = − =

Required distance = (5+3) cm or (5-3) cm

i.e 8cm or 2cm


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04.

31023

4

1 2

2 1 1 2 7.5 10 1 1 12 7.5 10

10 3 6 3

Th

eg r r

−−

= − = − =

05. 1/3

3 1 121 5

5

T TV

rejs n reJs

= − = =

06. 2T

hreg

=

2

2 70

2 10 980h

−

=

7003.57 2

98= =

07. 2

/ 2

T TP

h h

=

2T

F Ah

=

( )

33

2

5

2 2 2 70 70 102 49 10

1 0.01

10 1

T m Tm

h h h

N

− =

=

08. App weight = ( )B TW F F= + + 73 20

8.2 15.4 8980

= − + =

09. 2 cos

( )

Th

re g a

=

11. 2 2

0 0

4 416 / 12 /A o A B

A B

T TP P P P N m P P N m

R R= + = + = = + =

3

6B B B

A A A

n P R

n P R

= =

12. 2

hrdgT =

13.

2

4

12

2 2

0.03 10

3 10

Lh cm

T SR

T SR

X

X N

−

−

= =

=

=

=

=


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14. When the ring is about to leave the water surface, surface tension

force on it is

( )2 2 2STF RT rT R T T = + = +

Spring force SF kx=

( )2kx R T T mg = + +

( ) ( )

2 41

3

0.9 4 10 8 10 100.080

2 2 3.14 50 20 10

kx mgT Nm

R r

− −−

−

− − = = =

+ +

15. 3 34 4

3 3R r K = (Conservation of volume)

R = radius of bigger drop

r = radius of smaller drop

3 3R r K=

( )

( )

( )

2

2

2 2 3

2 2 3

1/3 2 2 2

/3

/3

4

4

4 4 10

4 10

10

10 1 100

10 101

6

i

f

U S R

U KS r

KS r S R

S Kr R

K R R

−

−

−

=

=

− =

− =

− =

− =

=

16. 2.w T r= 3 4 412.5675 10 36 10 8.5 10 , 4

4y− − −= = =

17. 0 1cos 60 2R cm

R= =


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18.

( )3 38.2 10 9.8 80.36 10W mg N− −= = =

The water is in contact along the length,

( )20 10 0.2 20.4L cm= + =

( )( )2 2 420.4 10 7.3 10 148.92 10TF N− − − = =

6 310 0.2 1.5410 1000 9.8 15.092 10

2FB N− − = =

Apparent weight

( ) 380.36 14.892 15.092 10T FW F B −+ − = + − 380.16 10 N−=

19. Pressure inside smallest bubble is largest; Hence air will from both

bubble’s into bubble B

20. As the drop coalesce, the final surface area is decrease. Hence the

surface energy of final drop is less that total initial drops. Hence

energy is released


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42.

Let ;x y u x y u− = + =

constant

43.

Replace by

Replace by in (1)

Similarly,


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44. Replace x by 2,

Replace x by 1

Replace x by

,

Solve (1) & (3)

45.

given replace x by ,

domain of is [-1, 1] and range of is i.e.,

46.

Given .

Replace then

Now

and

Which give and


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Thus, from (1)

47.

Put x = y = 0 we get

Put we get

Put is equation (1)

Hence

48.

49. Given, f(a+x)=f(a-x)

a)f(2a-x)=f(a+(a-x))=f(a-(a-x))=f(x)

( ) ( )2f a x f x − = ………(i)

b) ( ) ( )( ) ( )( ) ( )2f a x f a a x f a a x f x− = + = − + = −

( ) ( )2f a x f x + = − …….(ii)

c) f(2b+x) = f(-x) from eq. (ii) ……..(iiii)

d) from eqs. (ii) and (iii), we get


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( )2 (2 )f a x f b x+ = +

Period is (2b-2a)

50.

(A) f(x) = sgn

for every real x

Every constant function is a periodic function.

(B) Let T > 0 be a rational number then f(x + T)

f(x) is periodic function

(C)

this is a periodic function.

(D)

Since {.} is a periodic function hence this function is periodic.

51. ( )

( ) ( )

2

2

3, 5

5 1

7 |1 | |1 | 1 1

6 1 5

3 5

x x

x x

x x x

f x x x

x x

− −

+ − −

− − + + − = + −


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( )

( )( )

2

2

3, 5 5

5 1 1 5

7 |1 | |1 | 1 1

6 1 1

3 5

x x x

x x x

x x x

f x x x

x x

− −

+ − − −

− − + + − − = − + − − − −

( ) ( ) ( )( )

0 5

6 5 1

2 7 |1 | |1 | 1 1

6 1 5

0 5

x

x x

f x f x x x x

x x

x

−

+ − −+ − − − − + + −

+

For f(x) to be an odd function

( ) ( ) 0f x f x x R+ − =

6x = − and ( )7 6 7 1x = + = − + =

53. Given ( ) ( ) ( ) ( ) ( ) ( )1 2 2 3 3 ......... 1f f f nf n n n f n+ + + + = +

On putting (n+1) in place on f, we get

( ) ( ) ( ) ( ) ( ) ( )( ) ( )1 2 2 ......... 1 1 1 2 1f f nf n n f n n n f n+ + + + + + = + + +

On subtracting eq. (i) from eq(ii), we get

(n+1)f(n+1)=(n+1)(n+2)f(n+1)-n(n+1)f(n)

( ) ( ) ( )1 1 0n f n nf n + + − =

( ) ( ) ( )1 1nf n n f n = + +

( ) ( ) ( )2 2 3 3 ........f f nf n = =

Eq(i)

( ) ( ) ( )1 ........ ( 1)f nf n nf n to n terms+ + + − =n(n+1)f(n)

( ) ( ) ( ) ( ) ( )1 1 1f n nf n n n f n + − = +

( ) ( ) ( )1 2 1 2f nf n nf n = =


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54. f

Put

Put x = 1

As

(ii)

55.

Let

Given

(i) put we get

(i) put

……A)

(i ) put


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For A & B

57. If ( ) ,ax b a

f x xcx a c

+=

−

Then, f(f(x))=x

( )( )f f x x = and 4 4

f fx x

=

( )( )4 4

4f f x f f xx x

+ = +

59.

60. ( ) 0 5 5f = = −

( )( )( ) 5 5f f f x = −

Since ( ) ( )2

5 5f x x= + −

( ) ( )2

5 5 5f x x= + −

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

5 5 5f f f f x + = −

( )( )2

5 5f f + =

( ) 15 54

f f + =

( ) 15 54

f f −

( )2 1/45 5 5 5f + − = −


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( )2 1/45 5f + =

1/45 5f + =

1/85 5f + =

Class: SZ2 JEE-ADV(2014-P1) MODEL Date: 12-07-20 Time ... · Narayana CO Schools 1 Class: SZ2...

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