Solution All Paper

IJSO-STAGE-I WORKSHOP HINTS & SOLUTIONS

WORKSHOP FOR IJSO STAGE-I _DIGNOSTIC TEST Que s.

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B

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Que s. 21

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Que s. 41

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Que s. 61

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Ans.

Ans.

Ans.

1.

A

y = sin put y = sin in (i) 2ycos = xsin 2×sincos = xsin x = 2cos x2 + 4y2= 4cos2 + 4sin2 =4

2009 x 1 +1+ =0 2010 x

2009 x 2  2010 x  2010 =0 2010 x 2009x2 + 2010x + 2010 = 0 +=  =

–2010 2009

5.

2010 2009

y° y°

–2010 1     2009 1 + = 2010 = – 1    2009 2

2.

2x° 80° 100°

B

4.

2

1  1  1 Req. volume =    7 +    (1) 8 3   8

3 × 2

=

913913 ........... (100 digit) we see that 913 is repeated. 913 comes 33 times so the last digit (100th digit) will be 9.

D

=

3 +2

3 7 +2= 2 2 3

2x y – =3 cos  sin  2 x sin  – y cos  =3 cos  sin  2xsin – ycos = 3cossin xsin – 2ycos = 0 (ii) × 2, then (i) – (ii)

x°

C

80 = x + y 100 = 2x + y – – – – 20 = – x x = 20 y = 60 sin y. tan y + sec y = sin 60 × tan 60 + sec 60

7  22 11 + = = 64 192 192 96

3.

A

R

6.

C

B

K 33=r 33=r

...(i) ...(ii)

area of equilateral triangle ABC =

3 4

2

6 3 

=

3 × 36 × 3 4

= 27 3 s=

6 3 6 3 6 3 = 9 3 2 WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 1

= 7. 8.

12 a In triangle ABG

 s

r=

cos =

27 3 cos(90 – ) =

9 3

=3 A  B = B. We have TRQ = 30º Since, ST is a diameter and angle in a semicircle is a right angle.  SRT = 90º Now, TRQ + SRT + PRS = 180º  30º + 90º + PRS = 180º  PRS = 180º – 30º – 90º  PRS = 180º – 120º  PRS = 60º.

7 a sin2 + cos2 = 1

sin =

144 2

F B

13.

(12 – 22) + (32 – 42) + (52 – 62) + ....+ (992 – 1002) = (1 + 2) (1 – 2) + (3 + 4)(3 – 4) + (5 + 6) (5 – 6) +.....+ (99 + 100)(99 – 100) = – 3 – 7 – 11..........199 a = – 3, d = – 4 and  = – 199  = a + (n – 1) d – 199 = – 3 + (n – 1)(– 4) – 199 + 3 = (n – 1)(– 4)

.....(i) .....(ii)

1 1 1 AB × CF × AC × BE = 28.80 × 20 2 2 4

2 = 144  = 12 =

1 AD × BC = 12 2

AD × BC = 24

n–1= 10.

ax 2 + bx + 1 = 0 For real roots b2 – 4ac  0  b2 – 4a(1)  0  b2  4a For a = 1, 4a = 4,  b = 2, 3, 4 a = 2, 4a = 8,  b = 3, 4 a = 3, 4a = 12  b=4 a = 4, 4a = 16,  b=4 Number of equations possible = 7 C D

12

H 12 B

11.

7 A

Let a be the side of square : In triangle CHB

196 = 49 4

n = 50

50 [– 3– 199] 2 = 25 [– 202] = – 5050.

S50 =

14.

a a  10b + =2 b b  10a

a  b  10  b  a +  a = 2 b b1  10  b  Let

a =x b

x+

x  10 =2 1  10 x

7 G

=1

AB and CB are two two-digit numbers with the same unit digit. Therefore, R.H.S. should also be a multiplication of two two-digit numbers with the same unit digit. R.H.S. = DDD = D x 111 = D x 3 x 37. Now 37 is a two-digit number with 7 as the unit digit. Therefore , 3D should also be a twodigit number with 7 as the unit–digit  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.

C

AB × AC = 28.80 BE × CF = 20 (i) × (ii)

49

12.

E D

+

a a2 2 a = 193

A

9.

7 a

x  10 x 2  x  10 =2 1  10 x WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 2

10x 2 + 2x + 10 = 2 + 20x 10x 2 – 18x + 8 = 0 5x 2 – 5x – 4x + 4 = 0 5x(x – 1) – 4(x – 1) = 0 (5x – 4)(x – 1) = 0 x= 15.

1 1 2 ( 2 h)2 h r h volume of cone 3 3 = = 3 volume of cube a3  2h     3 

4 or 1 = 0.8 or 1. 5

18.

Given xa × xb × xc = 1  xa + b + c = xº a+b+c=0 Hence, a3 + b3 + c3 = 3abc

= 2.25  Assume the weight of alloy A is 100 kg  The weight of alloy B is 400 kg  Gold Silver Copper A 40kg 60kg 0 B 140kg 160kg 100kg total  180kg 220kg 100kg  Ratio of Gold and Silver in new alloy =

180 220 : 500 500

= 36% : 44% 16. 19.

From the given figure

X

Let the two objects be C & D In ABC tan 60º =

12

P

3

AB BD

tan 45º =

50 1= xy x + y = 50

50 y = 50 –

20.

x=

y=

( 3 – 1) 3 Distance between two objects is 50 ( 3 – 1) 3

17.

2a 2 2h

h–a a = 1 2

n d25 d25 = = 999 27  37 27

d25 n 37 d25 is multiple of 37. So, d25 = 37 × 25  n = 25 and d = 9. Now, n + d = 34. 25.

h–a

A fully loaded elevator has a mass of 6000 kg.

2a 2

h

T

a 2 h

R

x = 0.d25d25d25.....

3

50

h–a = h

Q

We know that PR × 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

50 x=

x– 2

x

AB 50 = BC x

a

mg–T = ma

2h – 2a = a 2h = 3a

mg T = m(g–a) WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 3

P1 : P2 = 1 : 4

 26.

= 6000 (10–2) = 4.8 × 104 N At point A body has only PE.

29.

Output current, IS = 4A Output voltage, ES = 20V

PE = mg (h + x) KE = 0 A At point C

NP 2 and N  1 S

h B



NP IS 2 4    NS IP 1 IP

 IP = 2A



NP EP 2 E    P NS ES 1 20

 EP = 40V

x C KE = 0 By applying work - energy theorem between point A & C. Work done by gravity + work done by resistance = KE at pt A – KE at pt C Mg (h + x) – Fx = 0 – 0 Fx = Mg (h + x) [Here F is the resistance offered.] h  F = Mg 1   x 

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

27.

1 1 1  We know, f  (  1) R  R  2  1

Here m = 0.25 kg and v = 10 m/s

The given lens is plano-convex,

Impulse imparted to ball = chage in

R2 = 

momentum of the ball = p2 – p1 = 2.5 Ns.

P2 = 0.25 × 10 = 2.5 Ns

Given :  = 1.5, R1 = 10 cm 1  1 1   (1.5  1)    f 10  

1 = 0.05 f 28.

f=

1 = 20 cm 0.05

33.

outout = 12 × m = 1kg mgh = 9

When the heaters are connected in series then the equivalent resistance would be 2R. 2

V 2R When the heaters are connected in parallel then the equivalent resistance would be R/2

h=

Power, P1 =

Power, P2 =

so,

V2 2V 2 = R/2 R

75 3 = 12  = 9J 100 4

9 9  =0.9 m gm 10  1

v2 = u2+2gh v2 = 2 × 10× 0.9 v2 =

18

P1 V 2 R 1    P2 2R 2V 2 4

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 4

48. 34.

P12  P22  2P1P2 cos 60

P =

Copper is more reactive than silver so (Ag) can not displace copper from its salt solution.

P2 60°

30 2  30 2  2  30 2 

51.

1 2

Density of water = 1 g/mL weight 1 mL water = 1g

60° P1

weight of 1000 mL water = 1 × 1000 = 1000 g moles of water in one litre =

P = 30 2

F=

Weight 1 litre water 1000 Molecular weight = 18 = 55.56

P 30 3  10 10 = = 150 3 t 0.2

1 moles of H 2O contain = 6.023 × 10 23 molecule

35.

55.56 moles of H2O contain = 55.5 × 6.023 ×

1 1 mv2 = kx2 2 2

1023 molecule 52.

Molecular weight of NaCl = 23 + 35.5 = 58.5 g 58.5 g of NaCl contain = 23 g of sodium

1 × 0.5 (1.5)2 = 50 × x2 2

 11.7 of NaCl will contain = x2 =

0.5  (1.5)2 500

23 × 11.7 g 58.5

= 4.6 g 53.

Ratio of moles of H and O atoms in the sample of (NH4)3 PO4 = 12 : 4 or 3 : 1

1.5 x= = 0.15 m 10

If moles of H atoms are 3.18 3.18 = 1.06 3 Milli equivalent of a base = Milli equivalent of

then, moles of O atoms are = 36.

Xcm = 1 1  2  2  3  3  4  4 1 2  3  4

54.

acid

w 1  1000 w 2  1000 = E1 E2

1  4  9  16 Xcm = =3m 10 42.

2  1000 3  1000 = E 40 2

(i) Ferrous sulphate [FeSO4]  (ii) 2FeSO4(s) 

Fe2O3(s) + SO2(g) +

E2 = 60

SO3(g) Ferrous sulphate (green)

43.

Ferric oxide

55.

H3PO4 is tribasic acid its basicity (n) = 3

(black)

(N) Normality n (M) molarity

In the reaction 2H2S + SO2  2H2O + 3S

N =3 1 N=3

H2S is losing hydrogen and removal of hydrogen from any substance is the oxidation.

therefore H3PO4 has the highest normality. 47.

Only (Mg) and (Mn) metals give H2 gas with dil. HNO3

56.

Nitride ion is N3–,it is formed when nitrogen atom gains three electrons. Thus it will contain 7 protons and 10 electrons.

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 5

57.

H

O Non-polar

O H 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

Molecular mass Empirical formula mass 60 = =2 30  Molecular formula of the compound = n

(empirical formula)2 (CH2O)2 C2H4O2

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 6

WORKSHOP FOR IJSO STAGE-I _TEST PAPER-1 Que s.

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C

Que s. 21

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Ans.

1.

Lel the possible number be N then it can be expressed as N = 9k + 6 and N = 21l + 12  9k + 6 = 21l + 12  9k – 21l = 6 or 3 (3k–7l) = 6

or

3k = 7l + 2 or k

= 7l  2 3

So put the min. possible value of l such that the value of k is an integer or in other words numerator (i.e., 7l +2) will be divisibnle by 3. Thus at l =1 , we get k = 3 (an integer). so the least possible number N = 9 × 3 + 6 = 21 × 1 + 12 = 33. Now the higher possible values can be obtained by adding 33 in the multiples of LCM of 9 and 21. i.e., the general form of the number is 63m + 33. So the other number in the given range including 33 are 96, 159, 222, 285, 348,...1104. Hence there are total 18 numbers which satisfy the given condition. 2.

3.

5.

Consider some appropriate values : As p = 3.99 , q = 4.99, r = 6.99 A = [p +q +r] = [3.99 + 4.99 + 6.99] = [15.97] = 15 B = [p] + [q] + [r] = [3.99] + [4.99] +[6.99] = 3 + 4 + 6 = 13 Hence A–B=2

6.

(a+ 1) (b – 1) = 625 But 625 = 1 × 625 = (a + 1) × (b – 1) a = 0, b = 626 5 × 125 = (a +1) × (b–1) a = 4, b = 126 25 × 25 = (a + 1) × (b –1) a = 24, b = 26 125 × 5 = (a + 1) × (b –1) a = 124, b = 6 625 ×1 = (a + 1) × (b –1) a = 624, b = 2 Thus (a+b) is always equal to or greater than 50. SInce the min (a+b) = 50 = (24 + 26) Alternatively : (a +1) (b–1) = 625 ab + b – a = 626 b ( a + 1) – a = 626 b(a+1) = a + 626

The required HCF = (2)HCF of (315. 25)–1 = 25 –1 = 31 Hence (c) is the correct option.

r

3

2r 

1  1r  2 3 2  1

3 =

r

r 2

3 2r =

r

2r 3r

4r 11r 3 2 2 r 3 2

=

r

 3r  2   2  r     2   2 

b=

3

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

(a  626) (a  1)  625 625   +1 (a  1) (a  1) (a  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)

= 32

Hence (b) is the correct option. 4.

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).

7.

Let there be x bangles each side, then the total number of bangles he had = x2 + 38 If he increases the size of the square by one unit each side , then the total number of bangles = (x + 1)2 – 25 Thus x2 + 38 = (x +1)2 – 25 x2 + 38 = x2 + 1 + 2x – 25  2x = 62  x = 31 WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 7

Thus total number of bangles = x2 + 38 = 961 + 38 = 999 Alternatively : Go through options, consider a suitable value from the options and check that in each case it must produce a perfect square number. As 999- 38 = 961 is a perfect square and 999 + 25 = 1024 is also a perfect square. 8.

9.

(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 so a ×c will be maximum and b.d will be minimum, thus a × c = 9 × 8 = 72 and b ×d = 1 × 2 = 2 Hence 99(ac – bd) = 99(72–2) = 99 × 70 = 6930

Given circuit is equivalent to following figure RAB = 4.1 

12.

Since all the numbers u,v, w, x are negative, but when uv + vw +wx = 0, then uv it means there must be some terms whose value will be positive and thus it adding up with negative value makes the expression zero. e.g., (k) + (–k) = 0 Hence we can say that there must be some even integers which converts a negative number into a positive number. Further we know that an even number when multiplied with any other (even or odd) number it finally makes an even number. Explanation : uv + vw + wx = 0 (– u)–v + (–v)–w + (–w)–x = 0 { u, v,w, x

  } Again if k + l = m or k + m = l or l + m = k i.e., half of the numerical value will be positive and half of the numerical value will be negative since there are every non zero integer. Now, if (–k)even positive value So, if there exists some even integer n, then k × l ×m ×n  even (though k, l, m can be odd). 10.

11.

Unit digit of (1!)1! is 1 Unit digit of (2!)2! is 4 Unit digit of (3!)3! is 6 UNit digit of (4!)4! is 6 Unit digit of (5!)5! and there after is 0 so the unit digit of the sum of the whole expression is 7. To know the remainder , when any number is divided by 5, we just need to know the unit digit of the dividend . further from the previous equcation, we know that the unit digit of the sum of the whole expression is 7. So divide 7 by 5 and get 2 as the remainder.

[balanc ed wheat- s tone br idge] then equivalent resistance, = A

R R

13.

RAB =

C R

3R = 1.5 R 2

R R

B R

5R . 9

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 8

16.

Applying KVL along ABCDA 12 = ix + i 500 ................(1) Applying KVL along CDEFC ix = 2 ....................(2)

14.

10 1 = Amp. 500 50

i=

;

2 = 2 × 50 = 100  i After removing charge from P, net force on central charge will be : x=

17.

F=

30 = 20 Amp. 3/2

a=

From figure current through B  D branch = 5 Amp.

2

2 2

3 20A

15.

9  10 9  10 5  5  10 5 12

5A 20A

becomes will be  If     n then A  

b 9.0V

Resistance of wire R = 

A n hence the resistance of wire becomes R = n 2 R

So current coming from this branch = 15 Amp.

a

F 4.5  = 9 m/s2 upwards M 0.5

 A A = volume of wire = constant

18.

The resistance of each

The equivalent circuit of the given combination 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– Vh) + (Vh – Vc) + (Vc – Vd) Vad =  1 = iR1 +  2

x =

Therefore, –  1 = iR1 +  2 = – iR2 or i (R1 + R2) =  1 –  2

1   2 9.0  6.0 3.0  i = R R = = 0.5 A 24 6.0 1 2

1 th part is 5

R n 2R = 5 5

This is a balanced wheat stone bridge  Equivalent resistance across AB is

c d 6.0V

The potential drop across resistance R is : R = 4 Vad = (Vd – Va) = –iR2

or

r2



F = 4.5 N m = 0.5 kg so, acceleration,

 Req = 3/2 i=

Kq1q2

R eq = x =

n 2R 5

19.

2V

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

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 9

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. Therefore order 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 process 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 of ingested particles

32.

3

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

4

2

1

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 10

WORKSHOP FOR IJSO STAGE-I _TEST PAPER-2 Que s.

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D

C

B

B

B

D

C

A

D

D

B

A

A

C

A

C

D

B

D

C

Que s. 21

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35

36

37

38

39

40

D

C

C

C

C

B

D

A

C

A

C

D

D

B

B

D

C

A

D

A

Ans.

1.

A = /2 A + B + C = 180 B + C = 180 – A = 180 – /2 = 90° cos2A + cos2B + cos2C = cos2 (/2) + cos2B + cos (90° – B) = 0 + cos2B + sin2 B =0+1=1

2.

sin +

4.

AC =

4 sin   sin   sin .......  = sec 

4 sec 4  = sec  sin  + sec2  = sec4  sin  = sec4  – sec2  = sec2  (sec2  – 1) = sec2  tan2 

area ABC =

SQB

DC =

sin  +

3.

=5

42  32

1 1 43 =  5  BD 2 2

12 = 2.4 5 In BDC BD =

3 2  ( 2 . 4 )2 =

9  5.76 =

3.24

= 1.8 tan  5.

DC BD



1 .8 2.4



3 4

tan4 + cot4 = A Let tan2  = x  cot2  = x2 + (x –

1 x2

1 x

=A

1 2 ) +2 =A x

as minimum value of (x –

a sin  =

3.a

1 =

3

1 2 ) is zero.  A x

 2 6.

Let a = 2, b = 3 from option (A) cos = a +

1 1 =2+ = a 2

2.5 and we know that max. value of cos  is 1 so cos = 2.5 is not possible from option (B) sec =

2ab 2

a b

2

=

2(2)(3) 22  3 2

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 11

12 i.e less then 1 13 and we know the value of sec is not laying between 2 and – 2 =

10.

12 is not possible 13 from option (i)

Option (A)

1  tan 2 30 1  tan 2 30

 1   1    3 =

 1   1    3

so sec =

cosec2  =

4ab

( 2  3 )2

2

=

1 2 3 3 1 (rational) 1 = 4 = 1 2 3 3 1

4(2)(3) =

( a  b )2

2

=

24 i.e 25

less than 1 and we know that value of cosec2  4

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)

24 is not possible so none of 25 the above option are correct so option (D) is correct. cosec2  =

option (D) =

3

2 cot 30  2

cot 30  1

2( 3 ) =

( 3 )2  1

=

2 3 2

(irrational)

T2

N

7.

T1

In ABC

11.

x sin 30º = 20

1 = 10 m. 2  Length of remaining part = 10 m. x = 20 sin 30º = 20 ×

8.

2g – T1 = 0 T1 = 2g T1 + 2g sin 30° = T2

A + B + C + D = 360° C + D = 180° – (A + B)

 T2 = 2g + 2g ×

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 x 4 sin x = 4 cos x tan x = 1 x = 45°  y = 7 sin x + 3 cos x = 7 sin 45° + 3 cos 45°

1 =7

2

1 +3

10 =

2

=5 2

2

1 = 3g 2

mg = T2 = 3g m = 3kg 12.

v2 = u2 + 2as 0 = 722 + 2 × 0.9 × a

4  72 .01 4  72 × .08 .01

a=–

then v2 – 722 = – 2 ×

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’ = F–T ma’ = F – ma F  a =  – a  m  

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 12

14.

For first case tension in spring will be Ts = 2mg just after 'A' is released.

19.

F cos  .....(i) 10

Comman acceleration, a =

a For block 2kg 2mg – mg = ma  a = g

30N

30– T 1 = 2a

T

............(ii)

a

In second case Ts = mg

T1

For block 1kg :

1 kg

T1 = a ............(iii) from equation (ii) 30 – a = 2a  a = 10 m/s 2

2mg – mg = 2mb b = g/2 a/b = 2

from equation (i) 10 =

F cos 60 º 10

F = 200N

T2

2

2.2 m/s

15. 20.

For block 8kg

8g T 2 – 8g = 8× 2.2 T 2 = 96 N

T1

mg + F – T = ma .. (i) F + T – mg = ma ... (ii) (i) + (ii) 2F = 2ma  a = F/m

For block 12 kg

on putting the value of a in eqn (i)

T 1 – 12g – T 2 = 12 × 2.2 T 1 = 240 N

T2 12g

mg + F – T = m × F/m  T = mg 16.

21.

Any element shows radioactivity due to its unstable nucleus. It has been found that the nuclei of those atoms are unstable whose ratio of the neutrons to the protons is greater than1.5

22.

The activity of an element is not affected by the state of chemical combination. So radium sulphate is as radioactive as the radium content.

Net pulling force, f = M2gsin – M1gsin acceleration of M2

F M2 sin   M1 sin  = M M  M1  M2 1 2 18.

3T = (50 +25) g T = 250 N

a

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 13

23.

24.

A radioactive disintegration differs from a chemical change in being a nuclear process.

Total time (t) No. of half lives = Half - life period (t ) ½ =

6000 =4 1500

2nd half - life 1st half - life 1 g       0.5 g  3rd half - life  0.25g     

4th half - life  0.0625g 0.125g     

25.

Uncontrolled nuclear chain reaction is the basis of atom bomb.

26.

235 92 U

93 1 + 10 n  140 56 Ba + 36 Kr + 3 0 n Above reaction is an example of nuclear fission because in this reaction a heavier atom (U–235) is splited into fragments (barium and krypton) and neutrons by bombarding of neutron.

30.

CO2 and N2O have same number of atoms and same number of electrons.

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 14

WORKSHOP FOR IJSO STAGE-I _TEST PAPER-3 Que s.

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

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

A

B

D

B

A

A

D

A

A

B

C

C

B

C

C

D

C

C

D

Que s. 21 C

Ans.

3. B

D F

1.

A

1 1 reflexAOC = × 310 = 155º 2 2 ABD is a line ABC + CBD = 180º CBD = 180º – 155 = 25º ABC =

105º

C xº 25º E

From E, draw EF || AB || CD. Now, EF || CD and CE is the transversal.  DCE + CEF = 180º [Co-interior angles]  xº + CEF = 180º  CEF = (180º – xº). Again, EF || AB and AE is the transversal.  BAE + AEF = 180º [Co-interior angles]  105º + AEC + CEF = 180º  105º + 25º + (180º – xº) = 180º  xº = 130 Hence, x = 130º.

4.

O1

C

Let radius of larger circle and smaller circle be s 1 and s 2 respectively Then, A.T.Q. r12 – r22 = r22  r12 – (2)2 = (2)2  r12 = 2(2)2  r1 = 2 2 In right angle triangle OMP PM2 = OP2 – OM2   

A

2.

AOC = 500 AOC + reflex AOC = 360º So reflex AOC = 310º

O2

PM2 = (2 2 )2 – (2)2 PM2 = 4 PM = 2 R

B

Given, O1 = 20 cm, O2 A = 37 AC = CB [ A line from centre to chord bisects common chord  O, C  AB and O2 C  AB]

1 1 AC = CB = AB = × 24 = 12 cm 2 2 In O,AC (O,A)2 = (AC)2 + (O,C)2 (20)2 = (12)2 + (O1C)2 400 – 144 = (O1C)2

S O

P

5.

Q U T

Let PR = r1 = 4r QS = r2 = 3r PR || SQ So, by BPT OQ SQ = OP PR

O 1C =

256 = 16 cm In O2 AC (O2A)2 = (AC)2 + (O2C)2 (37)2 = (12)2 + (O2C)2 1369 – 144 = (O2C)2 O 2C =

1225 O2C = 35 cm O 1O 2 = O 1C + O 2C = 16 + 35 = 51 cm 

OQ 3 = 28 4 OQ = 21 cm PQ = OP – OQ = 7cm PQ = r1 + r2 = 7 4r + 3r = 7 7r = 7 r = 1. r2 = 3r = 3 cm. WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 15

A

6.

A D

E

x 3

F

E H

x

G

C

8. From the figure, BC

18 2  24 2 = 30

B

Since all the three smaller triangle are similar to the bigger one and their bases are

1 rd that of the bigger one, their areas 3

1 are th that of the triangle ABC 9

 Area of the hexagon

  1  1 = (18 × 24) 1  3 9  = 144 cm 2.   2  7.

As  PAC ~  QBC [AA similarity] [AA le: irk]



x AC y = BC 1 BC 1 = . y AC x

... (1)

z AC y = BC 1 AB 1 = . y AC z

... (2)

1  BC AB  1 1  + = y    AC AC  x z

=

1  AC  = y  AC 

C

In ABC, by angle bisector theorem AC AE = BC BE 7 x = 5 3x

21 – 7x = 5x 21 = 12x x=

21 7 = 12 4

In ABC, by angle bisector theorem AB BD = AC DC

3y = 35 – 7y 10y = 35 y=

35 7 = . 10 2

AC AP = CD PD

(1) o (2) d kst ksM +usij

=

y

In ADC, by angle bisector theorem

on adding (1) & (2)

1  AB  BC  y  AC 

5–y D 5

5y 3 = y 7

As  RCA ~  QBA



7 P

3–

B

7 AP y = PD 7 AP ×2= 7 PD 2=

AP PD

AP = 2PD AD = AP + PD AD = 3PD

1 y

1 PD = 3 AD

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 16

D

C

21.

Q

9. A

P

D

b

mH2SO 4 =

2x

18 36/1000

= 500

22.

Milli mole of oxalic acid = 100 ×

23.

Weights are independent of temperature.

24.

Percent loss of H2O in one mole of Na2SO4. nH2O 18n  100 = = 55.9 (142  18n)  n  10

25.

2NaOH + H2SO4  Na2SO4 + 2H2O 1 mole of H2SO4 required 2 mole of NaOH to be neutralized.

26.

H2O

a A

Volume of solution in litre

0.02

C

2x

10.

Moles of solute

i.e., 18 mole H2SO4 or 18 × 98 g H2SO4 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

B

In QAB, QP is median So, ar QAB = 2 area APQ = 2 × 1 = 2 cm2 Area ABC = 2 area ABQ = 2 × 2 = 4 cm2 Area of rectangle ABCD = 2 ar. ABC = 2 × 4 = 8 cm2

x

E

B

Draw CE || AD  AECD is ||gm EC = AD = a, AE = DC = b AEC = ADC = 2x AEC = EBC + BCE  BCE = 2x – x = x.  BE = EC = a  AB = AE + EB = a + b 16.

M = 18 =

Magnetic field due to straight wire PQ at O B1 = 0 Magnetic field due to straight wire RS at O

2 =1 1000 milli mole = 6.023 × 1023 molecules  1 milli mole = 6.02 × 1020 molecules



H2

+

1 O 2 2 0.5 mole

1 mole 1 mole weight 1 mole of H2 = 1 × 2 = 2 g weight 0.5 mole of O2 = 0.5 × 32 = 16 g

0 I (sin 1  sin 2 ) 4r Here 1 = 90º and 2 = 0º B1 =

0I . O 4r Magnetic field due semicircle, at O B1 =

0 I . O 4r Net magnetic field at O. B3 =

B = B1 +B2 = =

0 I  0 I .  O 4r 4r

 0  4 0  4 .    O = 0  0 2 2 4  2 4  2

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 =

30.

22.4 = 5.6 litre 4

According to mole concept 16g of oxygen element contains no. of atom NA.  1g of oxygen element will contain = = or NA = 16x 27 g of Al contains no. of atoms = NA 1g of Al contains no. of atoms = =

NA 27

16 x 27 WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 17

NA 16

WORKSHOP FOR IJSO STAGE-I _TEST PAPER-4 Que s.

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

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

D

C

B

A

A

C

B

B

D

B

D

C

D

A

B

C

B

C

C

Que s. 21 Ans.

1.

2.

3.

4.

5.

6.

7.

B

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 ATQ 2m – 2n = 56 from option (B) 6,3 satisfying the above condition i.e. 26 – 23 = 56 64 – 8 = 56 56 = 56 n(U) = 800 = n(C  H  B) = n(C) + n(H) + n(B) – n(C  H) – n(H  B) – n(B  C) + n(H  B  C) = 224 + 240 + 336 – 64 – 80 – 40 + 24 = 824 – 184 = 640 No of student who did not play any game = n(  ) – n(H  B  C) = 800 – 640 = 160 3x2 – 12x = 0 3x (x – 4) = 0 x = 0, 4 so all the option are correct so option D is correct. A = {1, 2 {3, 4}, {5}} Option D is correct Which is {5} A. 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

xy xy =a

xy = ax + ay x(y – a) = ay

ay x = y a from (iii) yz = cy + cz y= 

cz zc

from (ii) xz = bx + bz z=

bx x b

c 

y=

=



bx x b

bx c x b

bxc bx  cx  bc

ay x = y a bxc bx  cx  bc x= bxc a bx  cx  bc a

x=

abxc bxc  abx  acx  abc

...(i)

bxc – abx + acx – abc = abc x (bc – ab + ac) = 2abc

xz =b xz

...(ii)



yz yz = c

...(iii)

8.

x=

2abc bc  ab  ac

.Ans.

p = 22/3 + 21/3 (given)

from (i) WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 18

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.

10.

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,

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. Sn S1n





1 1 1   v u f

16.

or u = – 5 cm Hence the farthest distance at which the man can read the book is 5 cm. sin–1 (8 / 9)

17.

Cw > Cg

18.

Yellow, orange, red

19.

0º, 0º

21.

CO2

22.

Acetic acid is a weak acid. So , in its aqueous solution it dissociates incompletely.

n 1 = 10 2

23.

HClO4

n = 21

24.

concentration of OH– ions per unit volume decrease.

25.

[H3O+] =

n [2a1  (n  1)d1 ] 7n  1 2 = = n 4n  27 [2a1  (n  1)d2 ] 2 2a1  (n  1)d1 7n  1 2a 2  (n  1)d2 = 4n  27 (n  1)d1 a1  2 7n  1 (n  1)d2 = 4n  27 a2  2 th

Ratio of 11 term means, Put



 11. 12. 13. 14. 15.

1 1 1 1 1 1      u v f  5 5

or

a1  10d1 7  21  1 a 2  10d2 = 4  21  27 a1  10d1 4 148 a 2  10d2 = 111 = 3

negative for a real image and positive for a virtual image. secondary colour violet colour As the speed of sound is greater in water, it bends away from normal. For the closest distance : Here, v = – 25 cm f = + 5 cm Using lens formula,

1 1 1 1 1 6      u v f 25 5 25

or

u=–

25 cm = – 4.2 cm 6

10 – 6

= 10–8 M [Neglecting ioniza-

tion of water] Consider ionization of water. [H3O+] = y [OH–] = ( y + 10–6) [H3O+][OH–] = Kw = 10–14 y[y + 10–6] = 10–14 y2 + 10–6 y – 10–14 = 0 on solving for y. y = 9.9 × 10–9 % error =

10 –8 – 9.9  10 –9 9.9  10 – 9

×100 = 1%

1 26.

pH = log

27.

Sodium reacts with cold water, and burns with golden yellow flame.

1 1 1   v u f or

10 –14

[H ] Given that [H+] = 10-6 M Using the above formula : pH = - log 10-6 = 6 log 10  pH = 6 (as log 10 = 1)

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 19

2Na + 2H2O  2NaOH + H2 Sodium Cold water Sodium Hydrogen hydroxide Calcium also react with cold water but does not burn. Ca + 2H2O  Ca(OH)2 + H2 Calcium Water Calcium Hydrogen hydroxide Magnesium reacts mildly with cold water but reacts vigorously with boiling water. Mg

+

2H2O  Mg(OH)2 + H2

Magnesium boiling Magnesium water hydroxide Red hot iron reacts with steam. 3 Fe + 4 H2O Iron

Steam

heat

Fe3O4

+

4H2

Ferro-ferric oxide 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

WORK SHOP_IJSO_ STAGE-I _SOL._PAGE # 20