ANSWERS...= 0.2006 + 0.0295 = 0.2301 V 14. Answer (4) According to Bent’s rule, the halogen atom...

ANTHE-2020 (Engineering) (For Class XII Studying moving to XII Passed) - (Sample Paper)_Answers

1

ANSWERS

Physics Chemistry Mathematics

1. Answer (2)

2. Answer (1)

3. Answer (3)

4. Answer (2)

5. Answer (2)

6. Answer (1, 2, 4)

7. Answer (2, 3)

8. Answer (1)

9. Answer (4)

10. Answer (1)

11. A R

B S

C P

D Q

12. Answer (3)

13. Answer (1)

14. Answer (4)

15. Answer (2)

16. Answer (3)

17. Answer (1, 3, 4)

18. Answer (2, 4)

19. Answer (2)

20. Answer (3)

21. Answer (2)

22. A R

B P

C P, R, S

D Q

23. Answer (2)

24. Answer (2)

25. Answer (3)

26. Answer (1)

27. Answer (4)

28. Answer (2)

29. Answer (1, 3, 4)

30. Answer (2, 3)

31. Answer (1, 4)

32. Answer (3)

33. Answer (3)

34. Answer (1)

35. A R

B Q

C P

D S

(For Class XII Studying moving to XII Passed) - (Sample Paper)_Solutions ANTHE-2020 (Engineering)

2

ANSWERS & SOLUTIONS

PHYSICS

1. Answer (2)

0 2sin60º4

IB

d

0total

3 3

4

IB

d

2

tan30ºd

a

0 0total

3 3 93

3 2

I IB

a a

2. Answer (1)

Maximum potential is at x = 6

v – 5 = 30

v = 35

3. Answer (3)

0 1t

i iT

220

0

1

Tt

H i R dtT

3

20

0

1

3

T

t

Ti RT

20

3

i RT

d

3

ANTHE-2020 (Engineering) (For Class XII Studying moving to XII Passed) - (Sample Paper)_Solutions

4. Answer (2)

At steady state current through capacitor is zero.

0 23

C

VV R

R

5. Answer (2)

tan45ºcos

v

H

B

B

tan cos

1tan (cos )

6. Answer (1, 2, 4)

01

2

VI

R

02

2

3

VI

R

1

2

3

4

I

I

7. Answer (2, 3)

Theoretical

8. Answer (1)

9. Answer (4)

Solution of Q. Nos. 8 and 9

0

4

VI

R

2

0 002

4 2

V VP V

R R

00 0

32 0

4

C

VV V V

0

4C

VV

10. Answer (1)

Sensitivity length of wire.

11. Answer A(R); B(S); C(P); D(Q)

| x |

tan L Cx

R

In LR circuit, current lags voltage.

In RC circuit, current leads voltage.


4

CHEMISTRY

12. Answer (3)

50%

ln2T

k

93.75% 50%

1 100 4ln2t ln 4t

k 6.25 k

13. Answer (1)

2 2Pt | H (10 atm) | HCl(pH 4.7) || HCl(pH 1.3) | H (1 atm) | Pt

The overall reaction is

2 a c a 2 c(H ) 2(H ) 2(H ) (H )

2a 2 c

cell 2c 2 a

[H ] (pH )0.059E 0 log

2 [H ] (pH )

a c

0.0590.059[(pH) (pH) ] log10

2

= 0.2006 + 0.0295 = 0.2301 V

14. Answer (4)

According to Bent’s rule, the halogen atom being more electronegative, will extract higher p-character of sp3

hybridised P-atom in the given species. In P2H4, there is no halogen atom bonded to P-atom. So relatively

P – P bond will have higher p-character and hence higher bond length.

15. Answer (2)

Cl2O6 in the solid state exists as 2 4ClO ClO

. Oxidation states of Cl in the cationic and anionic parts are +5

and +7 respectively.

16. Answer (3)

2 4 2 4

(A)

Na SO BaCl 2NaCl BaSO

4(B)

BaSO 4C BaS 4CO

2 2

(C)

BaS 2HCl BaCl H S

17. Answer (1, 3, 4)

The lobes of xyd , 2 2x yd

and 2z

d will favour the pentagonal bipyramidal geometry.

18. Answer (2, 4)

Protective colloids must be lyophilic in nature.

5


19. Answer (2)

20. Answer (3)


2 2Cu(s) Cl CuCl (s)

In CuCl2 (s), each Cu2+

ion is surrounded by 4 Cl– ions in its layer and 2 Cl

– ions of the adjoining layers.

22 4

Yellow

CuCl 2HCl 2H [CuCl ]

2 24 2 2 4

Yellow Blue

[CuCl ] 4H O [Cu(H O) ] 4Cl

21. Answer (2)

According to Henry’s law, the solubility of a gas in a solvent (expressed as mole fraction Xg of the gas in

solution) is directly proportional to partial pressure of gas in contact with solution

g H gP K (X )

where KH is Henry’s law constant and Pg is partial pressure of gas. As temperature increases Xg decreases.

So KH increases because Pg is kept constant.

22. Answer A(R); B(P); C(P, R, S); D(Q)

RbBr has rock salt structure. Rb+ and Br

– have co-ordination number 6 each. BeO has zinc blende type

structure. Be2+

and O2–

have co-ordination number 4 each. Mn3O4 has a spinel structure. Cs+ and Cl

– have

co-ordination number 8 each.

MATHEMATICS

23. Answer (2)

2( ) 1 3 0f x x

and ( ( )) (1 5 ) f f x f x

31 1 5 x x x

3 4 0 x x

( 2)( 2) 0 x x x

( , 2) (0, 2) x

24. Answer (2)

2

0

2 [sin cos ] 2( )

I x x dx

0 2–2

+ +––


6

25. Answer (3)

( ) 2(1 sin4 )f x x

2(1 0) 24

f

26. Answer (1)

1 1 1 1( ) ( ) I P Q P Q P Q P P Q P

1 1 1( ) ( ( ) )Q Q P P P I P Q P

1 1( ) Q Q P Q P I

27. Answer (4)

1/

2 4

0

lim cot 4

x

x

x e k e

28. Answer (2)

cos3

x

29. Answer (1, 3, 4)

/22

2

0

sin

m

mI x dx

2 1 2 3 1.......

2 2 2 2 2

m m

m m

and 2 1

2 2 2 2.......2 1 2 1 3

m

m mI

m m

and 2 1

2 2 2.......2 1 3

m

mI

m

30. Answer (2, 3)

( ) 0f x at x = 1

(1) ( 1) 0 f f

( 2, 2) a

31. Answer (1, 4)

2 4 12 2 3 5 30 8 4 5 4 8 0x x x x

2 4 14 15 66 52 16 0x x x x

2 x 35 0x

0, 35x

7


32. Answer (3)

33. Answer (3)


2

1 18 ( ) 6 1f x f

x x

1

(1)2

f

and 2 (1) (1) dy

f fdx

6 1

7 2 =

19

14

34. Answer (1)

2

2

(tan )sec

2( )

d xxdx

xd x

dx

35. Answer A(R); B(Q); C(P); D(S)

(A) 2

12

(B) ( ) | ln| ||f x x is continuous and non-differentiable at 1 x

(C) max( ) 4f x

(D) Area =

13

1 2

0

0

42 3 2 3

3 32

xx dx

Edition: 2020-21

ANSWERS...= 0.2006 + 0.0295 = 0.2301 V 14. Answer (4) According to Bent’s rule, the halogen atom...

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