solution all paper

8/3/2019 Solution All Paper

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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


3/21


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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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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= 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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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.


7/21


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


8/21


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


9/21


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.


10/21


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


11/21


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


12/21



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


13/21


=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


14/21


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.


15/21


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.


16/21



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


17/21


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


18/21


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


19/21



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


20/21


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.


21/21


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

solution all paper

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