Narayana CO Schools

1

Class: SZ2 JEE-MAIN MODEL Date: 11-04-20

Time: 3hrs WTM-10 INITIAL KEY Max. Marks: 300

Jee-Main_Model_Initial Key

SZ2 Physics Initial Key Dt. 11-04-2020

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

Ans. 3 1 3 1 4 3 2 1 1 2

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

Ans. 1 3 4 1 1 1 2 3 1 2

Q.No. 21 22 23 24 25

Ans. 7.00 8.00 6.00 1.00 11.00

SZ2 Chemistry Initial Key Dt. 11-04-2020

Q.No. 26 27 28 29 30 31 32 33 34 35

Ans. 4 2 4 3 1 4 4 1 2 3

Q.No. 36 37 38 39 40 41 42 43 44 45

Ans. 4 1 2 3 4 2 2 3 4 2

Q.No. 46 47 48 49 50

Ans. 5.00 9.00 6.00 5.00 4.00

SZ2 Mathematics Initial Key Dt. 11-04-2020

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

Ans. 3 1 2 1 1 1 1 4 2 2

Q.No. 61 62 63 64 65 66 67 68 69 70

Ans. 3 4 2 1 2 4 2 4 2 2

Q.No. 71 72 73 74 75

Ans. 2.00 4.00 2.00 50.00 9.00

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

2

01. ( )( ) ( )( )540 1 100 80 1x x t y y t c + − = + −

640 80x xt y yt− = +

( ) ( )80 80x y t x t− = +

( )80 8x y

xx y

−=

+

t=100,

100x+100y=640x-80y

180y=540x

1

3

x

y=

02. 1 2 1 2;L L C C

03. More energy is required to separate the molecules of lower pressure.

04. ( )2 1

1 2

mc T TdQ mcPower

dt t t

−= = =

05. 50 gr of ice requires ( )50 80 cal of heat to completely melt into water

at 00 .C . 30 gr of water at 080 C loses ( )30 80 cm of heat to cool down

to water at 00 C

Hence final temperature is 00 C

30 gr of ice melts to water

20 gr ice remains

06. 50 100 200 1 100 20 80 20 100 540m + + + =

53m=

07. Because low specific heat temperature increased by a greater

amount on absorption of small heat energy

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

3

08. Temperature increases at uniform rate at all points except during

melting and boiling

09. 80 40 1 60m = 30m g=

Ice to water ratio=50 30 20

2 : 730 40 70

−= =

+

10. ( ) ( ) ( )100 35 20 35 10ms m s− = − −

50 / 9m kg=

11. Tension in the wire is 4.0 3.1 .F N=

The area of cross section is ( )2

2 32.0 10A r m−= =

6 24.0 10 .m−=

Thus , the tensile stress developed

2

6

4.0 3.1

4.0 10

FNm

A

−

−

= =

6 23.1 10 .Nm−=

12. Nature length 1 2 2 1

2 1

l T l T

T T

−=

−

13.

2

1 1

2 2 2

500 12000

4

F rN

F r F

= = = =

14. Fl

yAe

=

y F Same= =

1 2

2 2

1 1 2 2

l l

r f r f=

2

2 2

1 2

4 2 16l l

r l r l

=

+

1

2

1

8

l

l=

15. The longitudinal stress( )( )

( )

2

23

4.0 3.1

2.0 10

kg ms

m

−

−

=

6 23.1 10 Nm−=

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

4

The longitudinal strain 30.31 10

2.0

m

m

−=

30.0155 10−=

Thus 6 2

11 2

3

3.1 102.0 10

0.0155 10

NmY Nm

−−

−

= =

16. The change in pressure

6 2 5 210 10P Nm Nm− −= = −

5 29 10 Nm−=

Compressibility 1 /

mod

V V

Bulk ulus P

= =

Or ( ) ( )

11 2 1

3 3 5 250 10

10 9 10

Vm N

m Nm

− −

− −

= −

Or 11 3 5 350 10 10 9 10V m− − = −

7 3 34.5 10 0.45m cm−= − =

Thus the decrease in volume is 30.45 .cm

17. As the cross sections of the wires are equal and same tension exists

in both, the stresses developed are equal. Let the original lengths

of the steel wire and the copper wire be sL and cL respectively and

the elongation in each wire be .l

( )11 2...........

2.0 10s

l stressi

L Nm−=

And 11 21.1 10c

l stress

L Nm−=

( )......... ii

Dividing (ii) by (i)

2.1

20 :111.1

s

c

L

L= =

18. F

yA=

Fl

yAl

= l e=

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

5

19. Fl

nA

=

2

4 2

0.25 5 10

25 10 0.05 10n

−

− −

=

2769.2 /n N m=

20. ( )Ay R r

Fr

−=

21. Heat lost = Heat gained

1 1 1 2 2 2

1 1 2 2

m S m S

m S m S

+=

+

22. When A and B are mixed, ( ) ( )5 10A BmS mS=

2A BS S =

When B and C are mixed ( ) ( )5 10B CmS mS=

2B CS S = so 4A CS S== when A and C are mixed

( ) ( )10 40A CmS mS− = −

23. Q

H

=

24. PV

kV

−=

5 3

6

10 10 10

10k

−

−

=

6 6 310 10 10−= 910=

4 21 10 /k N m=

25. 3 2

11 2

5

10 1010 /

10 0.1

Fly N m

Ae −

= = =

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

6

26. Side chain (–CH2OH) is the parent chain

30.

31.

32

1)

2)

3)

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

7

4)

41.

5-methylcyclopent-2-en-1-ol

47.

49.

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

8

51. cot tan2 2

b c A B C

b c

− − =

+ and 0150B C+ =

1 3 2 3 3 2 4 2 3 2 2 2

b c b c b c+ −= =

+ + + − + −

2 1

2 3 4 2 3 2 2 2

b c b c b c

b c

− − − −= =

+ + + −

045B C− = , 0

0 1150 52

2B C C+ = =

52. ( )( )cos sin cos B sin 2A A B+ + =

( ) ( )cos sin 2A B A B− + + =

( ) ( )cos 1,sin 1A B A B− = + =

,2 2

A B A B C

= + = =

2sin

2sin

a b A

c C

+= =

53. 1 2 9 3

sin 182 3 2

bc bc= =

( ) ( )22 2 23 3 36 27b c b c− = + − =

2 2 63b c + =

( )2 2 2 2 cos 63 2 18 812

a b c b A

= + − = − − =

9a =

54. ( ) 2cos cos cos

4

bca B C a B C

l

− = +

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

9

2

3cos cos cos cos cos cos

4

abca B C b C A c A B

R= + + +

2

3cos . cos cos

4

abcC c c A B

R= + +

( )( )2

3cos cos cos

4

abcc A B A B

R= + − + +

2 2 2 2

3 3.sin sin

4 4 4

abc abc abc abcc A B

R R R R= + = + =

55. ( )cos cos 2 1 cosA B C+ = −

22cos cos 2.2sin2 2 2

A B A B C+ − =

cos 2sin2 2

A B C− =

2cos .cos 4sin .cos2 2 2 2

C A B C C− =

2sin cos 2sin2 2

A B A BC

+ − =

sin sin sin , , .A B C a c b are in AP+ =

56. tan . tan2 2

A C S b

S

−=

5 2

. 3 3 2 36 5

s bs b s s b

s

−= − = =

3 2 , ,a b c b a c b a b c are in AP+ + = + =

57. 2 22 cos cos 2 cosbc A ca B ab C a b+ + = +

2 2 2

2 2 2 2 2 2 2 2

2

c a bb c a a b c a b

+ −+ − + + + − = +

2 2 2

2 2 222

c a bb a b

+ −+ = +

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

10

2 2 2 2 2 2 2 2 03 2 2 90b c a a b c b a A+ + = + + = =

58. A,B,C are in AP 02 60B A C B = + =

sin sin

sin 2 sin 2 .2sin cos .2sin cossin sin

a c A CC A C C A A

c a C A+ = +

( )3

2sin 2sin 2. 32

A C B= + = = =

59. ( ) ( ) ( )21 0c ab a c a b b c− − − + − =

2 2 2 0a b c ab bc ca+ + − − − =

060a b c A B C= = = = =

2 0 2 0 2 0 3 3 3 9sin 60 sin 60 sin 60

4 4 4 4+ + = + + =

60. 6 1

sin sin ,sin sin2 4

A B A B+ = =

Let ( )A B A B

2 2 1 6sin sin 2.

4 4A B+ + =

0 0 0sin cos 90 90 90A B B A A B C= = − + = =

1 1

sin .sin cos sin4 4

A B B B= =

01sin 2 2 30

2B B= = or 0150

015 75B or =

0 015 75B and A= =

( ) 0 1cos cos60

2A B− = =

61. : : 4 :1:1,A B C =

0 0 0120 , 30 , 30A B C= = =

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

11

sin 3 / 2 3

sin sin sin 3 1 1 2 3

2 2 2

a A

a b c A B C= = =

+ + + + ++ +

62. ( ) ( )0 0

/ 2

sin sin sin 90 sin 90

DC AC a b

DAC ADC A B= =

− +

2cos sin 2cos

cos sin cos

a A A A

b B B B= − = −

2 0TanA TanB+ =

63. 4 4 4 2 2 2 2 2 22 2a b c a b a c b c+ + − − =

( )2 2 2 2 23b c a b c+ − =

3

2 cos 3 cos2

bc A bc A= =

2 1 1sin sin

4 2A A= =

64. 11 , 12 , 13b c k c a k a b k+ = + = + =

18 , 7 , 6 , 5a b c k a k b k c k+ + = = = =

7

cos35

A = , 19 25

cos ,cos25 35

B C= =

cos :cos :cos 7:19:25A B C =

65. ( ) ( )3 2cos 2 sin cosa B C R a A B C− = −

( ) ( ) ( )2 22 sin cos sin 2 sin 2R a B C B C R a A C= + − = +

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

12

( ) ( )2 2 2 2sin 2 cosR b c A a b c A= + = +

( )cos cos 3ab b A a B abc abc= + = =

66. ( )

( )

( )

22

2 2

2 2cos sin 1

s a sa a

b c b c b c

− = = − =

+ + +

( )

( ) ( )2

2 2

44 .cos

2

s s a bc bc A

bc b c b c

− = =

+ +

sin 2

cos2

bc

A b c

=

+

67. If cot : cot : cot : : ,2 2 2

A B Cl m n= then

( ) ( ) ( ): : : : 16 :12 :10 8: 6 :5a b c m n n l l m= + + + = =

68. Given expression ( )

( )

2 2 2

2 2 2

1

2 22

a b cabc

R Ra b c

abc

+ +

= = =

+ +

69. ( ), 2 2 sin 2 sin3

C a c R A C

= + = +1 3 3 1

2 . 2. 2 222 2

R R b +

= + = =

70. 3 1,x = let 3 1, 3 1b c= + = −

( ) ( )2 2

2 2 2 2 cos 3 1 3 1 2.2.cos3

a b c bc A

= + − = + + − −

( )1

2 3 1 4. 6 62

a= + − = =

Perimeter 3 1 3 1 6 2 3 6a b c= + + = + + − + = +

71. 3 4 2 2a b c a s s a= + = =

2

cot .cot 22 2 2

B C s a

s a a a= = =

− −

SZ2_JEE-MAIN_WTM-10_Key&Solutions_Exam.Dt.11-04-20

Narayana CO Schools

13

72. 2 2 2 2 cosa b c bc A= + −

2 235 4 2.2. . 2 3 1 0

2c c c c= + − − − =

1 2 1 22 3, 1c c c c+ = = −

( )2 2

1 2 1 2 1 24 12 4 16c c c c c c− = + − = + = 1 2 4c c− =

73. ( ) ( )

cot .cot .2 2

s s a s s bA BC c

− −= =

1s c

cs c s c s

= =− −

1 1 1

2a b c

s a s b s c s s s+ + = + + =

− − −

74. 2 12 6s a b c s= + + = =

( )( ) ( )

sin cos2 2

s a s c s s bB B

ac ac

− − −+ = +

3.2 6.1 1 12

3.4 3.4 2 2= + = + =

75. ( )

2

2

22

1 tan 1 cot4 2 2

cos5

1 tan 1 cot2 2

A B a b C

a bA B

A B a b C

a b

− − − −

+ = − = =

− − + + +

2

2 2

2

11 cot / 2

5 49 5 cot / 2 4 cot1 9 9 2

1 cot / 29

CC

C

C

−

= − = +

+

2 0cot 1 45 902 2

C CC= = =

Area 1 1

sin .6.3.1 92 2

ab c= = =