NSEJS-Solution-2008-09

August 27, 2017 | Author: Nilesh Gupta | Category: Euclidean Geometry, Physics & Mathematics, Physics, Physical Sciences, Science
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NSEJS-Solution-2008-09...

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HINTS & SOLUTIONS (YEAR-2008-09)_NSEJS (STAGE-I) ANSWER KEY Que s.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

C 17

D 18

D 19

D 20

C 21

B 22

C 23

B 24

D 25

A

D

A

B

AD

Que s.

C 16

26

27

28

29

30

Ans.

C

B

C

D

B

B

D

A

C

B

A

C

B

B

Que s.

31

32

33

34

35

36

37

38

39

40

AD

ABCD

BC

AD

ABC

CD

AC

B

AD

Ans. ABD

A

(PART A-1) 1.

Least count = value of 1 division of main scale/Total division of vernier scale 0.1 =

2.

LN =

x 20

⇒ x = 2mm

1 BC 2

∆ABC ~ ∆MNL

[By AA similarity]

Area ∆ABC  BC  2   Area ∆MNL =  LN  = (2) = 4 2

area ∆LMN = 3.

A

[by mid-point theorem]

L B

N M

C

area ∆ABC 48 = = 12 sq. unit . 4 4

(1 – tan A + sec A) (1 – cot A + cosec A) 

sin A

1



cos A

1 

=  1 − cos A + cos A   1 − sin A + sin A      cos A − sin A + 1   sin A − cos A + 1    cos A sin A   

= 

 1 − (sin A − cos A )   1 + (sin A − cos A )    cos A sin A   

= 

5.

=

12 − (sin A − cos A )2 cos A sin A

=

1 − (sin2 A + cos 2 A − 2 sin A cos A ) cos A sin A

=

2 sin A cos A = 2. sin A cos A

Neopentane contain only one type of H i.e. only (1º H) 1º

CH3 1º



H3C – C – CH3 1º

CH3 136

CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892

10.

Work done = Change in kinetic energy

1 1 × 2 × 25 – × 2 × 36 = 189 J 2 2

= 11.

Let the speed of the scooter = v m/min. ∴ Distance BC = 18 v and Distance CD = t v. In ∆ACD tan 60º =

h=

=

3

h =

B

3tv

tan 30º = 1

h

AD CD

h = tv

3

A

30º 18 min

C

60º t min

D

..... (i) AD BC + CD

h 18 v + vt

v(18 + t ) 3

..... (ii)

From (i) & (ii) 3tv=

∴ 13.

v(18 + t ) 3

3t = 18 + t t = 9 min.

2AgNO3(aq.)+H2S(g) → Ag2S(s) + 2HNO3 (aq.) CuSO4 (aq.)+ H2S (g) → CuS (s) + H2SO4(aq.) According to the equations, ratio of the amounts of H2S needed for precipitation is 1:2.

14.

M=

=

no. of moles of solute × 1000 V(ml)

6.02 × 10 20 × 1000 6.02 × 10 23 × 100

= 0.01M

15.

P, Q → prime number PR is divisible by Q. R is a multiple of Q. Option A → Q divides R is correct. Option B → P is not a multiple of R.∴ wrong Option C → P is prime & R is not multiple of P ∴ wrong Option D → P divides PQ is correct.

18.

Volume of icecream in cylinder = volume of icecream in 12 cones 1

2



2 3 π(6)2 × h = 10  3 π(3) × 12 + 3 π(3)   

36 h = 10 [36 + 18] h = 15 cm.

137

CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892

19.

Isosceles triangles Maximum side = 4 units A c B

b a

C

a
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