140

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(PART A-1)

1. As m = 200g = 8 g/cm3

V = 8200m

= 25 cm3

so volume displaced = 25 cm3

mass = V × = 25 × 1 = 25 gtotal reading = 1100 + 25 = 1125 g

2. Sum of all natural no less the 400S1 = 1+ 2 + 3 + 4 + .............. 399

= 2

400399 = 79800

Sum of all natual number divisible by six.S2 = 6 + 12 + 18 + ......... 396No of terms396 = 6 + (n – 1)6n = 66

Sum = 2

66 [2× 6 + (66 – 1)6]

S2 = 13266 S1 – S2 = 79800 – 13266 = 66534.

3. In ABC

1–r

A r C

rB

AB2 = BC2 + AC2

(1 – r)2 = r2 + r2

(1 – r)2 = 2r2

1 – r = 2 r

1 = r 12

r = 121

ANSWER KEY

HINTS & SOLUTIONS (YEAR-2009-10)_NSEJS (STAGE-I)

Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. B B D A D C B B D B D D B D D

Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. C D B A C A B C B B B A B D B

Ques. 31 32 33 34 35 36 37 38 39 40Ans. AB AD CD ACD ABC ABCD AC BD ABD AD

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4. N2 = 28g , NO2 = 46g, CO2 = 44g Ascending order = A, C, B

5. M1V1 + M2V2 = M3V3

1 × 3.7 + 0.3 × 5 = M3 × 8.7M3 = 0.5977

14. Half-life of element = 14 hoursTime period = 56 hoursNo. of half lives = 4

N = No n

21

N = No 4

21

= 16No {taking No as 1g}

= 0.0625 g The fraction of substance thatdisintegrates = 1 – 0.0625 = 0.9375 g

15. a + b = c

squaring both sids

a + b + ab2 = c ...(i) [For any triangle a + b > c]

[ ab2 is a positive quantity]

Which contradict equation (i)Hence, no triangle formed.

19.O2

O'A

1

B C

OO’A is a right angled triangle.OA = BC length of direct common tangent. OO’ = 2 + 3 = 5 52 = 12 + x2

x = 24 x = 62 .

142

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26. D = a2 + b2 + c2 and c = ab.Let a = x and b = x + 1.D = (x)2 + (x + 1)2 + [x (x + 1)]2

= x2 + x2 + 2x + 1 + x2 (x2 + 2x + 1) = 2x2 + 2x + 1 + x4 + 2x3 + x2

= x4 + 2x3 + 3x2 + 2x +1Which is always a square of odd numbers.So, D is always an odd number..

27. a2 + 2b = 7, b2 + 4c = – 7 and c2 + 6a = – 14 a2 + 2b + b2 + 4c + c2 + 6a = 7 – 7 – 14(a2 + 6a + 9) + (b2 + 2b + 1) + (c2 + 4c + 4) = – 14 + 9 + 1 + 4(a + 3)2 + (b + 1)2 + (c + 2)2 = 0It is possible only when a = – 3, b = – 1 and c = – 2.So, a2 + b2 + c2 = (– 3)2 + (– 1)2 + (– 2)2

= 9 + 1 + 4 = 14.

29. One mole of oxalic acid is neutralised by two moles of NaOH.

30. cot2

sin1

1–sec+ sec2

sec1

1–sin

=

2

2

sincos

)sin1[cos

cos1 +

2cos1

cos1

1sincos

2sin

cos

sin1cos1

+ cos

1

cos1

1sin

=

2sin

cos

sin1sin1

)sin1()cos1(

+ cos

1

cos1cos1

cos1)1(sin

=)sin1(sin

)sin1)(cos1(cos22

–

)cos1(cos)cos1)(sin1(

2

=

22 cossin

)sin1)(cos1(cos –

2sincos

)cos1)(sin1(

=

2sincos

)cos1)(sin1( –

2sincos

)cos1)(sin1( = 0.

143

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(PART A-2)

32. Sodium hydrogen carbonate and sodium carbonate both are salts of weak acid (H2CO3) and strong base(NaOH). Thus their aqueous solutions are alkaline with pH higher than 7.

33.• •

R1 R2 R3

AB

2

1

3

4•

A B1,2,3,4 R2

R3

R1

RAB = R1 + 32

32RR

RR

Here R2 and R3 are in parallel

35.

A B

CD 2x + 5y – 7

5x + 2y + 2

x + y + 43x + 2y – 11

AB = CD 5x + 2y + 2 = 2x + 5y – 73x – 3y + 9 = 0x – y + 3 = 0 ....... (i)AD = CB3x + 2y – 11 = x + y + 42x + y – 15 = 0 ........ (ii)Using (i) & (2) x = 4, y = 7So, length of rectangle = 5 (4) + 2 (7) + 2 = 36 cm.

Breadth of rectangle = 4 + 7 + 4 = 15 cm. Area of rectangle = 15 × 36 = 540 sq. cm.

Diagonal of rectangle = 22 3615 = 39 cm.Perimeter of rectangle = 2 (15 + 36) = 102 cm.

37. U = {X| X is a point on straight line AB},P = {M|M is a point on ray AB},Q = {N|N is a point on ray BA}R = {L|L is point on segement AB}.

Option A : PQ

: M is a point on ray AB N is a point on ray BA : Point on segment AB.

So, option (A) is true.

144

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Option B : P’Q’ = Q P

: (PQ)’ [By Demorgans law] : Not a point on segment AB

So, option (B) is false.Option C : [P’Q’]’ = R

: [(PQ)’]’ [By Demorgans law]

: PQ

: M is a point on ray AB N is a point on ray BA : Point on segment AB.

So, option (C) is true.

Option D : P Q’ : Point on ray AB and Not a point on ray BA

P’ Q : Not a point on ray AB and Point on ray BA

So, option (D) is false.

39. (i) Velocity at middle point will be

2

vu 22 = 2155 22 = 55 answer is(B)

(ii) v = u+ at

55 = 5 + at1

15 = 55 + at2

5555515

tt

1

2

= 15

53

1515

1553

=

215 answer is (A) also

(iii) as2

uv 22

as2

25225

as = 100 m2/s2 answer is (D) also

40. Na, Hg is not sonorous, since Na is very soft and Hg is liquid.