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IJSO-STAGE-I
WORKSHOP
HINTS & SOLUTIONS
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 1
WORKSHOP FOR IJSO STAGE-I _DIGNOSTIC TEST
1.2010
x2009+ 1 +
x
1= 0
x2010
2010x2010x2009 2 = 0
2009x2
+ 2010x + 2010 = 0
+ =2009
2010
=2009
2010
1
+ 1
=
2009
20102009
2010
= 1
2. Req. volume = 78
12
+ )1(
8
1
3
12
64
7+
192
=
192
22=
96
11
3. 913913 ........... (100 digit)we see that 913 is repeated.913 comes 33 times so the last digit (100 th
digit) will be 9.
4.cos
x2
siny
= 3
sincos
cosysinx2= 3
2xsin ycos = 3cossin ...(i)xsin 2ycos = 0 ...(ii)(ii) 2, then (i) (ii)
y = sinput y = sin in (i)2ycos = xsin2sincos = xsinx = 2cosx2 + 4y2= 4cos2 + 4sin2= 4
5.A
B C D
2x 100x
y y
80
80 = x + y100 = 2x + y
20 =
x
x = 20y = 60sin y. tan y + sec y= sin 60 tan 60 + sec 60
=2
3 3 + 2
=2
3+ 2 =
2
7
6. R
B C
3
K3 3 = r 3 3 = r
area of equilateral triangle ABC
=4
3 236 =
4
3 36 3
= 27 3
s =2
363636 = 39
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. B A D D A B C A C B A A B D D D B B C D
Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. C C D A B B B D A C B A C C B C A A D D
Ques. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. A A A B B D C B C D D B C B C B C B C D
Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. C C B A C C D A D D A D B B A D B A D B
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 2
r =s
=39
327
= 3
7. A B = B.
8. We have TRQ = 30Since, ST is a diameter and angle in a semi-
circle is a right angle.
SRT = 90Now, TRQ + SRT + PRS = 180 30 + 90 + PRS = 180 PRS = 180 30 90 PRS = 180 120 PRS = 60.
9. F E
BD
C
A
AB AC = 28.80 .....(i)BE CF = 20 .....(ii)(i) (ii)
2
1AB CF
2
1AC BE =
4
128.80 20
2 = 144 = 12
=2
1AD BC = 12
AD BC = 24
10. ax2 + bx + 1 = 0For real rootsb2 4ac 0 b2 4a(1) 0 b2 4aFor a = 1, 4a = 4, b = 2, 3, 4a = 2, 4a = 8, b = 3, 4
a = 3, 4a = 12 b = 4a = 4, 4a = 16, b = 4Number of equations possible = 7
11.
12
7
HC
GA
7
12D
B
Let a be the side of square :In triangle CHB
cos =a
12
In triangle ABG
cos(90) =a
7
sin =a
7
sin2 + cos2 = 1
2a
144+ 2a
49= 1
a2 = 193
12. AB and CB are two two-digit numbers withthe same unit digit.Therefore, R.H.S. should also be amultiplication of two two-digit numbers with thesame unit digit.R.H.S. = DDD = D x 111 = D x 3 x 37.
Now 37 is a two-digit number with 7 as theunit digit. Therefore , 3D should also be a two-digit number with 7 as the unitdigit D = 9 and 3D = 27. Therefore, 27 x 37 =999. Hence, A = 2, B = 7, C = 3 and D =9 and
A + B + C + D = 2 + 7 + 3 + 9 = 21.
13. (12 22) + (32 42) + (52 62) + ....+ (9921002)= (1 + 2) (1 2) + (3 + 4)(3 4) + (5 + 6) (5 6) +.....+ (99 + 100)(99 100)= 3 7 11..........199a = 3, d = 4 and = 199
= a + (n
1) d 199 = 3 + (n 1)( 4) 199 + 3 = (n 1)( 4)
n 1 =4
196
= 49
n = 50
S50
=2
50[ 3 199]
= 25 [ 202]= 5050.
14. b
a
+ a10b
b10a
= 2
b
a+
b
a101b
10b
ab
= 2
Letb
a= x
x +x101
10x
= 2
x101
10xx10x 2
= 2
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 3
10x2 + 2x + 10 = 2 + 20x10x2 18x + 8 = 05x2 5x 4x + 4 = 05x(x 1) 4(x 1) = 0(5x 4)(x 1) = 0
x =5
4or 1 = 0.8 or 1.
15. Given xa xb xc = 1
xa + b + c = x a + b + c = 0Hence, a3 + b3 + c3 = 3abc
16.
Let the two objects be C & DIn ABC
tan 60 =BC
AB=
x
50
x =3
50
tan 45 = BD
AB
1 =yx
50
x + y = 50
y = 50 3
50
y =3
50( 3 1)
Distance between two objects is
3
50 ( 3 1)
17.h
ah=
h2
2
a2
h
a
h2
a22
ha
1
ah=
2
a
2h 2a = a
2h = 3a
cubeofvolume
coneofvolume=
3
2
a
hr3
1
= 3
2
3
h2
h)h2(3
1
= 2.25 18. Assume the weight of alloy A is 100 kg
The weight of alloy B is 400 kg Gold Silver CopperA 40kg 60kg 0
B 140kg 160kg 100kg
total 180kg 220kg 100kg Ratio of Gold and Silver in new alloy
=500
180:
500
220
= 36% : 44%
19. From the given figure
X
12
RP Q
x x 2
We know thatPR RQ = RX2
(2x 2) (x 2) = 122
(x 1)(x 2) = 72 x2 3x 70 =0
x2
10x + 7x
70 = 0 x(x 10) + 7(x 10) = 0
(x + 7) (x 10) = 0 x = 7 not possible x = 10
20. x = 0.d25d25d25.....
x =999
25d=
3727
25d
=
27
n
n37
25d
d25 is multiple of 37.So, d25 = 37 25 n = 25 and d = 9.Now, n + d = 34.
25. A fully loaded elevator has a mass of 6000 kg.
mgT = ma
mg
a
T
T = m(ga)
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 4
= 6000 (102) = 4.8 104 N26. At point A body has only PE.
PE = mg (h + x)KE = 0At point C
C
B
A
h
x
KE = 0By applying work - energy theorem betweenpoint A & C.Work done by gravity + work done by
resistance= KE at pt A KE at pt CMg (h + x) Fx = 0 0Fx = Mg (h + x) [Here F is the resistanceoffered.]
F = Mg
x
h1
27. We know,
21 R
1
R
1)1(
f
1
The given lens is plano-convex,
R2
=
Given : = 1.5, R1
= 10 cm
1
10
1)15.1(
f
1
f
1= 0.05 f =
05.0
1= 20 cm
28. When the heaters are connected in series then
the equivalent resistance would be 2R.
Power, P1
=R2
V2
When the heaters are connected in parallelthen the equivalent resistance would be R/2
Power, P2
=2/R
V2=
R
V2 2
so,4
1
V2
R
R2
V
P
P2
2
2
1
P1
: P2
= 1 : 4
29. Output current, IS
= 4A
Output voltage, ES
= 20V
and 1
2
N
N
S
P
PP
S
S
P 4
1
2
N
N
II
I I
P= 2A
20
E
1
2
E
E
N
N P
S
P
S
P EP
= 40V
30. (C) The displacement over a quarter circle
comes out to be 2 m. The time required for
this is 2 second. This is 2 second which can
be found using kinematical equations and
hence the answer.
31. Initial momentum of ball = 0 (as the ball is
initially at rest)
Final momentum of ball,P2= mv
Here m = 0.25 kg and v = 10 m/s
P2
= 0.25 10 = 2.5 Ns
Impulse imparted to ball = chage in
momentum of the ball = p2 p
1= 2.5 Ns.
33. outout = 12 100
75=
4
312 = 9J
m = 1kg
mgh = 9
h =110
9
gm
9
=0.9 m
v2 = u2+2gh
v2 = 2 10 0.9
v2 = 18
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 5
34. P = 60cosPP2PP 2122
21
2
13023030 222
60
60
P2
P1
P = 230
F =t
P
=2.0
10330 10= 3150
35.2
1mv2 =
2
1kx2
2
1 0.5 (1.5)2 = 50 x2
x2 =500
)5.1(5.0 2
x =10
5.1= 0.15 m
36. Xcm=4321
44332211
Xcm
=10
16941 = 3 m
42. (i) Ferrous sulphate [FeSO4]
(ii) 2FeSO4(s) Fe2O3(s) + SO2(g) +
SO3(g)
Ferrous sulphate Ferric oxide
(green) (black)
43. In the reaction 2H2S + SO
2 2H
2O + 3S
H2S is losing hydrogen and removal of hydrogen
from any substance is the oxidation.
47. Only (Mg) and (Mn) metals give H2
gas with
dil. HNO3
48. Copper is more reactive than silver so (Ag)
can not displace copper from its salt solution.
51. Density of water = 1 g/mL
weight 1 mL water = 1g
weight of 1000 mL water = 1 1000 = 1000 g
moles of water in one litre =
weightMolecular
waterlitre1Weight=
18
1000= 55.56
1 moles of H2O contain = 6.023 1023
molecule
55.56 moles of H2O contain = 55.5 6.023
1023 molecule
52. Molecular weight of NaCl = 23 + 35.5 = 58.5 g
58.5 g of NaCl contain = 23 g of sodium
11.7 of NaCl will contain =5.58
23 11.7 g
= 4.6 g
53. Ratio of moles of H and O atoms in the sample
of (NH4)
3PO
4= 12 : 4 or 3 : 1
If moles of H atoms are 3.18
then, moles of O atoms are =3
18.3= 1.06
54. Milli equivalent of a base = Milli equivalent of
acid
1
1
E
1000w =
2
2
E
1000w
40
10002 =
2E
10003
E2
= 60
55. H3PO
4is tribasic acid
its basicity (n) = 3
nmolarity)M(
Normality)N(
1
N= 3
N = 3
therefore H3PO
4has the highest normality.
56. Nitride ion is N3,it is formed when nitrogen
atom gains three electrons. Thus it will contain
7 protons and 10 electrons.
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 6
57.
O H
OH
Non-polar
Polar
58. Molecular mass = 2 vapour density
= 2 30
= 60
Empirical formula of the compound = CH2O
Its empirical formula mass = 12 + 1 2 + 16= 30
massformulaEmpirical
massMolecularn
=30
60= 2
Molecular formula of the compound =
(empirical formula)2
(CH2O)
2
C2H
4O
2
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 7
WORKSHOP FOR IJSO STAGE-I _TEST PAPER-1
1. Lel the possible number be N then it can beexpressed asN = 9k + 6and N = 21l + 12
9k + 6 = 21l + 12 9k 21l = 6or 3 (3k7l) = 6
or 3k = 7l + 2 or k =3
2l7
So put the min. possible value of l such thatthe value of k is an integer or in other wordsnumerator (i.e., 7l +2) will be divisibnle by 3.Thus at l =1 , we get k = 3 (an integer). so theleast possible number N = 9 3 + 6 = 21 1+ 12 = 33.Now the higher possible values can beobtained by adding 33 in the multiples of LCMof 9 and 21. i.e., the general form of the number
is 63m + 33. So the other number in the givenrange including 33 are 96, 159, 222, 285,348,...1104. Hence there are total 18 numberswhich satisfy the given condition.
2. The required HCF = (2)HCF of (315. 25)1= 251 = 31
Hence (c) is the correct option.
3. r
2
r2
2
r11r4
r
2
r
1
2
r1
2
1r2
3
3
3
33
= r 2r2
2
2r3
3
= r r23 = rr2
3= 32
Hence (b) is the correct option.
4. The unit digit of the whole expression will be
the equal to the unit digit of the sum of the
unit digits of the expression.
Now adding the unit digits of 12 + 22 + 32 +
.............. + 102
We get 1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0= 45
Hence the unit digit of 12 + 22 + 32 +.......+ 102
is 5
now since there are 10 similar columns of
numbers which will yield the same unit digit
5. hence the sum of unit digits of all the 10
columns is 50 (= 5 + 5 + ........ + 5)
hence, the unit digit of the given expression is
0 (zero).
5. Consider some appropriate values :As p = 3.99 , q = 4.99, r = 6.99A = [p +q +r] = [3.99 + 4.99 + 6.99] = [15.97]= 15B = [p] + [q] + [r] = [3.99] + [4.99] +[6.99]= 3 + 4 + 6 = 13Hence A B = 2
6. (a+ 1) (b 1) = 625But 625 = 1 625 = (a + 1) (b 1) a = 0,
b = 6265 125 = (a +1) (b1) a = 4, b = 12625 25 = (a + 1) (b 1) a = 24, b = 26125 5 = (a + 1) (b 1) a = 124, b = 6
625 1 = (a + 1) (b 1) a = 624, b = 2Thus (a+b) is always equal to or greater than50. SInce the min (a+b) = 50 = (24 + 26)Alternatively : (a +1) (b1) = 625ab + b a = 626b ( a + 1) a = 626b(a+1) = a + 626
b=)1a(
625
)1a(
625)1a(
)1a(
)626a(
+1
Let us consider a = 4 then b = 126, (a = 4,b = 126), (a = 24, b = 26), (a = 124, b = 6),(a = 624, b = 2)
7. Let there be x bangles each side, then thetotal number of bangles he had = x2 + 38If he increases the size of the square by oneunit each side , then the total number ofbangles= (x + 1)2 25Thus x2 + 38 = (x +1)2 25
x2
+ 38 = x2
+ 1 + 2x
25 2x = 62 x = 31
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. C C B A B B B C C A D A A B A B A C A C
Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. B D C D D B A C A D A C A B C A B B D A
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 8
Thus total number of bangles = x2 + 38= 961 + 38 = 999
Alternatively : Go through options, consider asuitable value from the options and check thatin each case it must produce a perfect squarenumber.As 999- 38 = 961 is a perfect squareand 999 + 25 = 1024 is also a perfect square.
8. (10a + b) (10c + d) (10b + a) (10d + c)= (100 a.c + 10 b. c + 10a d + b.d) (100b.d + 10b.c + 10a.d + a.c)
= 99 (a.c b.d)Now, inorder to the difference be maximum soa c will be maximum and b.d will be minimum,thus a c = 9 8 = 72 and b d = 1 2 = 2Hence 99(ac bd) = 99(722) = 99 70= 6930
9. Since all the numbers u,v, w, x are negative,
but when uv + vw +wx = 0, then uv it meansthere must be some terms whose value willbe positive and thus it adding up with negativevalue makes the expression zero.e.g., (k) + (k) = 0Hence we can say that there must be someeven integers which converts a negative numberinto a positive number. Further we know thatan even number when multiplied with any other(even or odd) number it finally makes an evennumber.Explanation : uv + vw + wx = 0
( u)v + (v)w+ (w)x = 0 { u, v,w, x
}Again ifk + l = m or k + m = l or l + m = ki.e., half of the numerical value will be positiveand half of the numerical value will be negativesince there are every non zero integer.Now, if (k)even positive valueSo, if there exists some even integer n, then k l m n even (though k, l, m can be odd).
10. Unit digit of (1!)1! is 1Unit digit of (2!)2! is 4Unit digit of (3!)3! is 6UNit digit of (4!)4! is 6Unit digit of (5!)5! and there after is 0 so theunit digit of the sum of the whole expressionis 7.To know the remainder , when any number isdivided by 5, we just need to know the unitdigit of the dividend . further from the previousequcation, we know that the unit digit of thesum of the whole expression is 7.So divide 7 by 5 and get 2 as the remainder.
11. Given circuit is equivalent to following figure
RAB
= 4.1
12.
[balanced wheat-stone bridge] then
equivalent resistance, =3
2
R= 1.5 R
13.
A R
R RR
R B
R
C
RAB
=9
R5.
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 9
14.
Req
= 3/2
i =
2/3
30= 20 Amp.
From figure current through B D branch= 5 Amp.
20A
2 2
2
3
So current comingfrom this branch= 15 Amp.
5A
20A
15. The equivalent circuit of the given combina-
tion is shown in figure. Let i be the current
flowing through the circuit then the terminal
voltage across ends a and b will be :
Vad
= (Va V
h) + (V
h V
c) +
(Vc V
d)
Vad
= 1
= iR1
+ 2
a d6.0V9.0V
b c
The potential drop across resistance R is :
R = 4V
ad= (V
d V
a) = iR
2
Therefore, 1
= iR1
+ 2
= iR2
or i (R1
+ R2) =
1
2
or i =21
21
RR
=0.6
0.3
42
0.60.9
= 0.5 AA
16. Applying KVL along ABCDA
12 = ix + i 500 ................(1)
Applying KVL along CDEFC
ix = 2 ....................(2)
i =500
10=
50
1Amp. ;
x =i2 = 2 50 = 100
17. After removing charge from P, net force on
central charge will be :
F =12
10510109
r
qKq 559
221
F = 4.5 N
m = 0.5 kg
so, acceleration,
a = 5.0
5.4
M
F
= 9 m/s2
upwards
18. Resistance of wire R = A
A = volume of wire = constant
If becomes n then A bewilln
A
hence the resistance of wire becomes
R = n2 R
The resistance of each
5
1th part is
x =5
R=
5
Rn2
This is a balanced wheat stone bridge
Equivalent resistance across AB is
Req = x = 5
Rn2
19. 2 V
20. 6 : 3 : 2
21. ascending positions
22. H.G.J. Moseley
23. 6th period
24. After losing one electron electronic
configuration of C, N, O and F is as follows -
C = 1s2
, 2s2
, 2p1
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 10
N = 1s2, 2s2, 2p2
O = 1s2, 2s2, 2p3
F = 1s2, 2s2, 2p4
C, N, O and F belong to same period. In a
period on moving left to right, I.E. increases.
In case of second I.E., oxygen is an exception.
It has stable half filled orbital configuration,
hence it is most stable among four. Thereforeorder of second I.E. is - O > F > N > C.
26. I.E.2
27. When one electron is added to oxygen atom
it becomes O ion. It has high charge density.
Now another electron to be added to make
O2 ion, would feel repulsion. Hence this pro-
cess would be endothermic.
28. Na+ and O2 have two shells only, while K+
has three shells. Hence size of K+ is more
than those of Na+ and O2.
29. Inert gases have zero valency , these gases
are He, Ne, Ar, Kr, Xe & Rn
30. It is more difficult to remove an electron from
M2+ rather than other species, as in it the
remaining electrons are bound by high nuclear
charge.
31. Inside the lysosomes for the digestion ofingested particles
32. 3 4 2 1
33. Mg2+
34. Glycolipids or glycoproteins
35. Collecting tubule
36. Active transport
37. stem and root tips, vascular cambium, cork
cambium
38. calcium and magnesium
39. Collenchyma
40. Spindle-shaped, unbranched, unstriated,
uninucleate and involuntary
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 11
WORKSHOP FOR IJSO STAGE-I _TEST PAPER-2
1. A = /2A + B + C = 180B + C = 180 A = 180 /2 = 90cos2A + cos2B + cos2C= cos2 (/2) + cos2B + cos (90 B)= 0 + cos2B + sin2 B= 0 + 1 = 1
2. sin + .......sinsinsin = sec4
sin + 4sec = sec4
sin + sec2 = sec4 sin = sec4 sec2= sec2 (sec2 1)= sec2 tan2
3. SQB
sin =a.3
a=
3
1
4.
AC = 22 34 = 5
area ABC =2
1 4 3 =
2
1 5 BD
BD =5
12= 2.4
In BDC
DC = 22 )4.2(3 = 76.59 = 24.3
= 1.8
tan BDDC 4.2
8.1 43
5. tan4 + cot4 = ALet tan2 = x
cot2 =x
1
x2 + 2x
1= AA
(x x
1)2 + 2 = AA
as minimum value of (x x
1)2 is zero. AA
2
6. Let a = 2, b = 3
from option (A) cos = a +a
1= 2 +
2
1=
2.5and we know that max. value of cos is 1so cos = 2.5 is not possible
from option (B) sec = 22 ba
ab2
= 22 32
)3)(2(2
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. D C B B B D C A D D B A A C A C D B D C
Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. D C C C C B D A C A C D D B B D C A D A
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 12
=13
12i.e less then 1
and we know the value of sec is not layingbetween 2 and 2
so sec =13
12is not possible
from option (i)
cosec2 = 2)ba(
ab4
= 2)32(
)3)(2(4
=
25
24i.e
less than 1
and we know that value of cosec2 4
cosec2 =25
24is not possible so none of
the above option are correct
so option (D) is correct.
7.
In ABC
sin 30 =20
x
x = 20 sin 30 = 20 21 = 10 m.
Length of remaining part = 10 m.
8. A + B + C + D = 360C + D = 180 (A + B)sin (A + B) + sin (C + D)= sin (A + B) + sin (180 (A + B))= sin (A + B) sin (A + B)
= 0
9. y = 7 sin x + 3 cos x .... (1)y = 7 cos x + 3 sin x .... (2)from (1) & (2)7 sin x + 3 cos x = 7 cos x + 3 sin x4 sin x = 4 cos xtan x = 1x = 45
y = 7 sin x + 3 cos x= 7 sin 45 + 3 cos 45
= 7 2
1+ 3
2
1
=
2
10= 5 2
10. Option (A)30tan1
30tan12
2
= 2
2
3
11
3
11
=
3
11
3
11
=
3
43
2
=2
1(rational)
option (B) 4 cos3 30 3 cos 30 = cos 3 30= cos 90 = 0 (rational)option (C) 3 sin 30 4 sin3 30 = sin 3 30 =sin 90 = 1 (rational)
option (D)130cot
30cot22
=
1)3(
)3(22
=2
32
= 3 (irrational)
11.
T1
NT2
2g T1= 0
T1= 2g
T1
+ 2g sin 30 = T2
T2
= 2g + 2g 2
1= 3g
mg = T2
= 3gm = 3kg
12. v2 = u2 + 2as0 = 722 + 2 0.9 a
a = 01.724 then v2 722 = 2
01.
724 .08
v = 24 m/s
13. Let T be the tension in the string .T = ma (equation for mass A )Let a is acceleration of mass B.ma = FTma = F ma
a =
a
m
F
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 13
14. For first case tension in spring will beT
s= 2mg just after 'A' is released.
2mg mg = ma a = g
In second case Ts= mg
2mg mg = 2mbb = g/2a/b = 2
15.
mg + F T = ma .. (i)F + T mg = ma ... (ii)(i) + (ii)2F = 2ma a = F/mon putting the value of a in eqn (i)mg + F T = m F/m
T = mg
16.Net pulling force, f = M
2gsin M
1gsin
acceleration of M2
=21
12
21 MM
sinMsinM
MM
F
18. 3T = (50 +25) gT = 250 N
19. Comman acceleration, a =10
cosF .....(i)
For block 2kg
a
30N T
30 T1
= 2a ............(ii)
For block 1kg :
a
T1 1 kg
T1
= a ............(iii)from equation (ii)30 a = 2a a = 10 m/s2
from equation (i) 10 =
10
60cosF
F = 200N
20. For block 8kg
T2
8g
2.2 m/s2
T2
8g = 8 2.2T
2= 96 N
For block 12 kg
T1
12g
a
T2
T1 12g T
2= 12 2.2
T1
= 240 N
21. Any element shows radioactivity due to itsunstable nucleus. It has been found that thenuclei of those atoms are unstable whose ratioof the neutrons to the protons is greater than1.5
22. The activity of an element is not affected bythe state of chemical combination. So radiumsulphate is as radioactive as the radiumcontent.
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 14
23. A radioactive disintegration differs from achemical change in being a nuclear process.
24. No. of half lives = )(tperiodlife-Half
(t)timeTotal
=1500
6000= 4
1 g life-half1st
0.5 g life-half2nd
0.25g life-half3rd
0.125g life-half4th
0.0625g
25. Uncontrolled nuclear chain reaction is thebasis of atom bomb.
26. U23592 + n10 Ba
14056 + Kr
9336 + n3
10
Above reaction is an example of nuclear fis-sion because in this reaction a heavier atom(U235) is splited into fragments (barium andkrypton) and neutrons by bombarding of neu-tron.
30. CO2and N
2O have same number of atoms and
same number of electrons.
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 15
WORKSHOP FOR IJSO STAGE-I _TEST PAPER-3
1.
B
A105
C x
E
25
F
D
From E, draw EF || AB || CD.
Now, EF || CD and CE is the transversal.
DCE + CEF = 180 [Co-interior angles] x + CEF = 180 CEF = (180 x).Again, EF || AB and AE is the transversal.
BAE + AEF = 180 [Co-interior angles] 105 + AEC + CEF = 180 105 + 25 + (180 x) = 180 x = 130Hence, x = 130.
2.O1 O2
A
B
C
Given, O1= 20 cm, O
2A = 37
AC = CB [ A line from centre to chord
bisects common chord
O, C AB and O2
C AB]
AC = CB =2
1AB =
2
1 24 = 12 cm
In O,AC(O,A)2 = (AC)2 + (O,C)2
(20)2 = (12)2 + (O1C)2
400 144 = (O1C)2
O1C = 256 = 16 cm
In O2AC
(O2A)2 = (AC)2 + (O
2C)2
(37)2 = (12)2 + (O2C)2
1369 144 = (O2C)2
O2C = 1225
O2C = 35 cm
O1O
2= O
1C + O
2C
= 16 + 35 = 51 cm
3. AOC = 500
AOC + reflex AOC = 360So reflex AOC = 310
ABC =2
1reflexAOC =
2
1 310 = 155
ABD is a line
ABC + CBD = 180CBD = 180 155= 25
4. Let radius of larger circle and smaller circle
be s1 and s2 respectively
Then, A.T.Q.
r12r2
2 = r22
r12(2)2 = (2)2
r12 = 2(2)2
r1 = 2 2In right angle triangle OMP
PM2 = OP2 OM2
PM2 = (2 2 )2 (2)2
PM2 = 4 PM = 2
5.
R
S
QP
T
O
U
Let PR = r1 = 4r
QS = r2 = 3r
PR || SQSo, by BPT
OP
OQ=
PR
SQ
28
OQ=
4
3
OQ = 21 cm
PQ = OP OQ = 7cm
PQ = r1 + r2 = 7
4r + 3r = 7
7r = 7
r = 1. r2 = 3r = 3 cm.
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. A B A B B B C B C B C D C D C C A D C B
Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. C A B D B A A D A A B C C B C C D C C D
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 16
6. A
B C
D E
F
G H
From the figure, BC 22 2418 = 30
Since all the three smaller triangle are
similar to the bigger one and their bases
are3
1rd that of the bigger one, their areas
are9
1th that of the triangle ABC
Area of the hexagon
=2
1(18 24)
9
131 = 144 cm2.
7. As PAC ~ QBC [AA similarity] [AAle:irk]
y
x=
BC
AC
x1 =
ACBC .
y1 ... (1)
As RCA ~ QBA
y
z=
BC
AC
z
1=
AC
AB.
y
1... (2)
on adding (1) & (2)(1) o (2) dks tksM+us ij
x
1+
z
1=
y
1
AC
AB
AC
BC
=
AC
BCAB
y
1
=
AC
AC
y
1
= y
1
8.
A
E
CDB y5
y
x
3x
3
5
7
P
In ABC, by angle bisector theorem
BC
AC=
BE
AE
5
7=
x3
x
21 7x = 5x
21 = 12x
x =12
21=
4
7
In ABC, by angle bisector theorem
AC
AB=
DC
BD
7
3=
y
y5
3y = 35
7y10y = 35
y =10
35=
2
7.
In ADC, by angle bisector theorem
CD
AC=
PD
AP
y
7=
PD
AP
7
7 2 =
PD
AP
2 =PD
AP
AP = 2PD
AD = AP + PD
AD = 3PD
AD
PD=
3
1
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 17
9.
D C
Q
BPA
In QAB, QP is medianSo, ar QAB = 2 area APQ
= 2 1 = 2 cm2Area ABC = 2 area ABQ
= 2 2 = 4 cm2
Area of rectangle ABCD = 2 ar. ABC= 2 4 = 8 cm2
10.
A
DC
BE
2x x
2xa
b
Draw CE || AD
AECD is ||gmEC = AD = a, AE = DC = b
AEC = ADC = 2xAEC = EBC + BCE BCE = 2x x
= x.
BE = EC = a AB = AE + EB = a + b
16. Magnetic field due to straight wire PQ at O
B1
= 0
Magnetic field due to straight wire RS at O
B1
= )sin(sinr4
210
I
Here 1
= 90 and 2= 0
B1
= .Or4
0
I
Magnetic field due semicircle, at O
B3
= .Or40I
Net magnetic field at O.
B = B1
+B2
= .Or4r400 II
= .O2424
00
44
=
22
00
21. M = 18 =litreinsolutionofVolume
soluteofMoles
i.e., 18 mole H2SO
4or 18 98 g H
2SO
4
are present in 1000 mL solution.
Since density of solution = 1.8 g/mL
weight of solution = 1.8 1000 = 1800 g weight
of water = 1800
18 98= 1800 1764 = 36 g
42SOHm =
36/1000
18= 500
22. Milli mole of oxalic acid = 100 2
0.02= 1
1000 milli mole = 6.023 1023 molecules
1 milli mole = 6.02 1020 molecules
23. Weights are independent of temperature.
24. Percent loss of H2O in one mole of Na
2SO
4.
nH2O
=18n)(142
10018n
= 55.9
n 10
25. 2NaOH + H2SO
4 Na
2SO
4+ 2H
2O
1 mole of H2SO
4required 2 mole of NaOH to
be neutralized.
26. H2O H
2+
2
1O
2
1 mole 1 mole 0.5 mole
weight 1 mole of H2
= 1 2 = 2 gweight 0.5 mole of O
2= 0.5 32 = 16 g
27. One mole of an element contains number of
atoms equal to Avogadro number (6.023 1023)
atoms.
23 g of Na contains = 6.023 1023 atoms
0.023 g of Na contains= 6.023 1020 atoms
29. GMM of He = 4 g
Volume 4 g He at NTP = 22.4 litre
Volume 1 g of He at NTP = 4
4.22
= 5.6 litre
30. According to mole concept -
16g of oxygen element contains no. of atom
NA.
1g of oxygen element will contain =16
NA
= or NA
= 16x
27 g of Al contains no. of atoms = NA
1g of Al contains no. of atoms =27
NA
=27
x16
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 18
WORKSHOP FOR IJSO STAGE-I _TEST PAPER-4
1. No. of non empty subset of a set containing n
element = 2n 1
so for set {1,2,3,4} no. of non empty subset =
24 1 = 15
2. ATQ 2m 2n = 56
from option (B) 6,3 satisfying the above condi-
tion i.e. 26 23 = 56
64
8 = 5656 = 56
3. n(U) = 800
= n(CHB) = n(C) + n(H) + n(B) n(CH)
n(HB) n(B C) + n(HBC)
= 224 + 240 + 336 64 80 40 + 24
= 824 184
= 640
No of student who did not play any game
= n( ) n(H BC)
= 800 640 = 160
4.3x
2
12x = 0
3x (x 4) = 0
x = 0, 4
so all the option are correct so option D is
correct.
5. A = {1, 2 {3, 4}, {5}}
Option D is correctWhich is {5} A.
6. a2 + b2 = 13 a = 2, b = 3
x3 + y3 = 65 x = 4, y = 1
{(ax + by) + (ay + bx)}
{a(x + y) + b(x + y)}
{(a + b)(x + y)}
{(5)(5)}
25
7. yx
xy
= a ...(i)
zx
xz
= b ...(ii)
zy
yz
= c ...(iii)
from (i)
xy = ax + ay
x(y a) = ay
x = ay
ay
from (iii)
yz = cy + cz
y =cz
cz
from (ii)
xz = bx + bz
z =bx
bx
y =c
bx
bxbx
bxc
=bccxbx
cxb
x = ay
ay
x =a
bccxbx
bxcbccxbx
cxba
x =abcacxabxbxc
abxc
bxc abx + acx abc = abc
x (bc ab + ac) = 2abc
x =acabbc
abc2
.Ans.
8. p = 22/3 + 21/3 (given)
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. A B A D D C D C A C A B D D A C A A D A
Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. B D C B A A C B B D B D C D A B C B C C
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 19
p3 = (22/3 + 21/3)
= (22/3)3 + (21/3)3 + 3 22/3 21/3(22/3 + 21/3)
= 4 + 2 + 3 2(p)
p3 6p 6 = 0.
9. p(x) = 2x4 x3 7x2 + ax + b
p(x) is divisible by x2 2x 3 = (x 3)(x + 1)
So, p(3) = 0, we get 3a + b = 72
and by p( 1) = 0, we get a + b = 4
Solving both equation
a = 19, b = 15
So, a + b = 34.
10. 1n
n
S
S=
]d)1n(a2[2
n
]d)1n(a2[2
n
21
11
=27n4
1n7
22
11
d)1n(a2
d)1n(a2
= 27n4
1n7
2
d)1n(a
2
d)1n(a
22
11
=27n4
1n7
Ratio of 11th term means,
Put2
1n = 10
n = 21
22
11
d10a
d10a
=27214
1217
22
11
d10a
d10a
=111
148=
3
4
11. negative for a real image and positive for a
virtual image.
12. secondary colour
13. violet colour
14. As the speed of sound is greater in water, it
bends away from normal.
15. For the closest distance :
Here, v = 25 cm
f = + 5 cm
Using lens formula,
f
1
u
1
v
1
or25
6
5
1
25
1
f
1
v
1
u
1
or u =
6
25 cm =
4.2 cm
Hence the closest distance at which the man
can read the book is 4.2 cm.
For the farthest distance :
Hear, v =
Using lens formula,
f
1
u
1
v
1
or5
1
5
11
f
1
v
1
u
1
or u = 5 cm
Hence the farthest distance at which the man
can read the book is 5 cm.
16. sin1 (8 / 9)
17. Cw
> Cg
18. Yellow, orange, red
19. 0, 0
21. CO2
22. Acetic acid is a weak acid. So , in its aqueous
solution it dissociates incompletely.
23. HClO4
24. concentration of OH ions per unit volume de-
crease.
25. [H3O+] = 6
14
10
10= 108 M [Neglecting ioniza-
tion of water]
Consider ionization of water.
[H3O+] = y [OH] = ( y + 106)
[H3O+][OH] = K
w= 1014
y[y + 106] = 1014
y2 + 106 y 1014 = 0
on solving for y. y = 9.9 109
% error = 9
98
109.9
109.910
100 = 1%
26. pH = log]H[
1
Given that [H+] = 10-6 M
Using the above formula :
pH = - log 10-6 = 6 log 10
pH = 6 (as log 10 = 1)
27. Sodium reacts with cold water, and burns withgolden yellow flame.
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WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 20
2Na + 2H2O 2NaOH + H
2
Sodium Cold water Sodium Hydrogen
hydroxide
Calcium also react with cold water but does
not burn.
Ca + 2H2O Ca(OH)
2+ H
2
Calcium Water Calcium
Hydrogen hydroxide
Magnesium reacts mildly with cold water but
reacts vigorously with boiling water.
Mg + 2H2O Mg(OH)
2+ H
2
Magnesium boiling Magnesium
water hydroxide
Red hot iron reacts with steam.
3 Fe + 4 H2O heat Fe
3O
4+ 4H
2
Iron Steam Ferro-ferricoxide
or iron (II, III) oxide
28. Froth floatation process
29. CuSO4
30. metallurgy
31. All dominant
32. 9 : 3 : 3 : 1
33. Organism with dominant phenotype is
heterozygous
34. Golden algae
35. Mammals with a pouch
36. Ray fish
37. cnidaria
38. Euplectella
39. Herdmania
40. UV radiations and lighting