IJSO-2009-Solution
-
Upload
nilesh-gupta -
Category
Documents
-
view
26 -
download
4
description
Transcript of IJSO-2009-Solution
140
CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892
(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
141
CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892
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
CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892
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
CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892
(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
CAREER POINT, 128, Shakti Nagar, Kota-324009 (Raj.), Ph. 0744-2503892
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.