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8/10/2019 10. AITS-10 (09-Nov-14)
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GENER L INSTRUCTIONS
1. In addition to this question paper, you are given a separate answer sheet.
2. Fill up all the entries carefully in the space provided on the OMR sheet ONLY IN BLOCK CAPITALS.
Incomplete/incorrect/carelessly filled information may disqualify your candidature.
3. A student has to write his/her answers in the OMR sheet by darkening the appropriate bubble with the
help of HB Pencil as the correct answer(s) of the question attempted.
4. Paper carries 80 questions each of 3 marks.
5. Any rough work should be done only on the blank space provided at the end of question paper.
6. Foreach correct answer gets 3 marks, each wrong answer gets a penalty of 1 mark.
7. Blank papers, clip boards, log tables, slide rule, calculators, mobiles or any other electronic gadgets in
any form is "NOT PERMISSIBLE".
Time : 2 Hr. Date : 09-11-2014 Max. Marks : 240
ALL INDIA IJSO(STAGE-I) TEST SERIES
OPEN TEST/MOCK TEST PAPER # 10
Resonance Eduventures Pvt Ltd
CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Rajasthan)-324005
Tel No : 0744-3192222, 3012222, 3022222| Fax 022-39167222, 0744-2427144
PCCP Head Office
J-2, Jawahar Nagar, Main Road, Kota (Rajasthan)-324005
Contact. No. :+91-0744-2434727, 8824078330
Website :www.pccp.resonance.ac.in E-mail :[email protected]
Toll Free : 1800 200 2244 |SMS RESO DLP at 56677 |dlpd@resoancne ac in www dlpd resonance ac.in
mailto:[email protected]:[email protected]://www.pccp.resonance.ac.in/8/10/2019 10. AITS-10 (09-Nov-14)
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IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 2
Space For Rough Work
IJSO TEST PAPER
1. A cylinder of radius 6 cm and height h cm is filled with ice cream. The ice cream is then distributedamong 10 children in identical cones having hemispherical tops. The radius of the base of the cone is3 cm and its height is 12cm. Then the height h of the cylinder must be :(A) 100/7 cm (B) 18 mc (C) 15 cm (D) 200 / 11 cm
2. If a, b and c are distinct real numbers such that a : b + c = b : c + a then :(A) a, b, c are all positive (B) a, b, c are all negative
(C) a + b + c = 0 (D) ab + bc + ca + 1 = 0
3. L, M and N are mid points of sides AB, BC and CA of triangle ABC. If area of triangle ABC is 48 sq. units,
then the area of triangle LMN will be :
(A) 6 sq.units (B) 8 sq.units (C) 12 sq.units (D) 24 sq.units
4. The angle of elevation of the top of a tower as observed from a point on the horizontal ground is x. If we move
a distance dtowards the foot of the tower, the angle of elevation increases to y, then the height of the tower
is :
(A)xtanytan
ytanxtand(B) d(tan y + tan x) (C) d(tan y tan x) (D)
xtanytan
ytanxtand
!
5. When the curves y = log10
x and y = x-1are drawn in the x-y plane many thies do they intersect for values
x "1 ?(A) Never (B Once (C) Twice (D) More than twice
6. The number of isossceles triangles with integer sides such that no side is greater than 4 units is :
(A) 8 (B) 9 (C) 16 (D) 12
7. Find the sum of (2.5)2 + 52+ (7.5)2+ 102+ 12.52+ -------- + (50)2 .
(A) 35175 (B) 17587 (C) 17937.5 (D) None of these
8. If the difference of (1025 7) and (1024 + x) is divisible by 3 then x is equal to
(A) 3 (B) 2 (C) 6 (D) 1
9. The sum of the reciprocals of the roots of the equation,2010
2009x + 1 +
x
1= 0, is
(A) 2009
2010
(B)
1 (C) 2010
2009
(D) 1
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IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 3
Space For Rough Work
10. If the perimeter of a rectangle is p and its diagonal is d, then the difference between the length and width
of the rectangle is
(A)2
pd8 22 #(B)
2
pd8 22 !(C)
2
pd6 22 #(D)
4
pd8 22 #
11. If a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then (a + b + c + d) is equal to :
(A) 5 (B) 10/3 (C) 7/3 (D) 5/3
12. Which one of the following pairs of lines are consistent having unique solution
(A) x + y = 7 and 2x + 2y = 14 (B) x y = 5 and 2x 2y = 15
(C) x y 1 = 0 and 4x 4y 15 = 0 (D) 2x + y 6 = 0 and 4x 2y 4 = 0
13. If sin x + sin2x = 1, then :
(A) cos x + cos2x = 1 (B) cos x cos2x = 1 (C) cos2x + cos3x =1 (D) cos2x + cos4x = 1
14. If the altitudes of a triangle are in the ratio 2 : 3 : 4, then the lengths of the corresponding sides are in theratio :
(A) 2 : 3 : 4 (B) 6 : 4 : 3 (C) 3 : 2 : 4 (D) 3 : 2 : 1
15. AD, BE and CF are the medians of $ABC. The sum of lengths of segments BE and CF is :
(A)
3
5BC (C) >
2
3BC (D) t2(B) t1= t2(C t1< t2(D) t1can be greater or smaller depending upon the initial velocity of the body
41. What volume of a 0.8 M solution contains 100 millimoles of the solute?
(A) 100 mL (B) 125 mL (C) 500 mL (D) 62.5 mL
42. 75 ml of H2SO4(specific gravity = 1.18) containing 49% H2SO4by mass is diluted to 590 ml. Calculatemolarity of the diluted solution. [S = 32]
(A) 0.7 M (B) 7.5 M (C) 0.75 M (D) 0.25 M
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IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 9
Space For Rough Work
43. For 100 ml of 0.3 M CaCl2solution + 400 ml of 0.1 M HCl solution, correct datas are :
(A) Total concentration of cation(s) = 0.14 M (B) Total concentration of cation(s) = 0.07 M
(C) [Cl] = 0.1 M (D [Cl] = 0.2 M
44. The atomic weight of an element is 'a'. If this element occurs in nature as a triatomic gas, then the correct
formula for the number of moles of gas in its 'w' g is :
(A) a
w3
(B) a3
w
(C) 3wa (D) w3
a
45. Atomic radius is of the order of 108cm and nuclear radius is of the order of 1013cm. The fraction ofatom that is occupied by nucleus is :(A) 105 (B) 105 (C) 1015 (D) None of these
46. Which of the following is/are isotones :
(A) 21 H,31 H (B)
157 N,
168 O (C)
4018 Ar,
4020 Ca (D)
31 H, He
42
47. The ratio of the energy of a photon of wavelength 3000 to that of a photon of wavelength 6000 respectively
is:
(A) 1 : 2 (B) 2 : 1 (C) 3 : 1 (D) 1 : 3
48. Find out the number of photons emitted by a 60 watt bulb in one minute, if wavelength of an emittedphoton is 620 nm.(A) 1.125 1022 (B) 1.225 1022 (C) 1.5 1022 (D) 1.125 1012
49. Bombardment by 2-particle leads to artificial disintegration in three ways, (I), (II) and (III) as shown.Products X, Yand Zrespectively are :
(ii)9
Be412
C + Y6
(i)
13
N + X7(iii)16
O + Z8
(A) 3-particle, proton, positron (B) positron, neutron, proton(C) 3-particle, neutron, proton (D) positron, proton, neutron
50. A positron is emitted from Na2311 . The ratio of the mass number and atomic number of the resulting
nuclide is:(A) 22/10 (B) 22/11 (C) 23/10 (D) 23/12
Hint : During the emission of positron from the nucleus atomic number changes.
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IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 10
Space For Rough Work
51. The correct representation of Charles law is given by :
(A) (B) (C ) (D)
52. A flask is of capacity one litre. What volume of air will escape from the open flask, if it is heated from 27C to37C ?Assume pressure to be constant.(A) 33.3mL (B) 23.3 mL (C) 43.3 mL (D) 13.3 mL
53. If an ideal gas at 1 atmosphperic pressure, is spreading from 20 cm3to 50 cm3at constant temperature,then find the final pressure :(A) 0.4 atm (B) 2.5 atm (C) 5 atm (D) None of these.
54. 5 L of a sample of a gas at 27C and 1 bar pressure is compressed to a volume of 1000 mL keeping thetemperature constant. The percentage increase in pressure is :
(A) 100 % (B) 400 % (C) 500% (D) 80%
55. For a fixed amount of ideal gas at constant temperature, which of the following plots is correct :
(A) (B) (C) (D)
56. Starting with one mole of nitrogen and 3 moles of hydrogen, at equilibrium 50% of each had reacted
according to the reaction : N2(g) + 3H
2(g) 2NH
3(g)
If the equilibrium pressure is P, the partial pressure of hydrogen at equilibrium would be :
(A) P/2 (B) P/3 (C) P/4 (D) P/6
57. For a fixed amount of ideal gas at constant temperature, which of the following plots is/are not correct :
(A) (B) (C) (D)
58. The standard molar enthalpies of formation of ethane, CO2& liquid water are #21.1, #94.1 and #68.3
kCal respectively. Calculate the standard molar enthalpy of combustion of ethane.
(A) 372.0 kCal. (B) 352.0 kCal. (C) 322.0 kCal. (D) 362.0 kCal.
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IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 11
Space For Rough Work
59. For which temperature the pOH of pure water can be greater than 7.(A 20 C (B) 30 C (C) 40 C (D) 50 C
60. The self ionization constant for pure formic acid , K = [ HCOOH2+ ] [HCOO#] has been estimated as 10#6
at room temperature .The density of formic acid is 1.15 g/cm3. The percentage of formic acid converted toformate ion are(A) 0.002 % (B) 0.004 % (C) 0.006 % (D) 0.008 %
61. The plasma membrane is(A) permeable. (B) impermeable. (C) selectively permeable (D) impervious
62. Biogeochemical cycles are(A) dynamic but stable systems of constant interaction between the biotic and abiotic components ofthe biosphere.(B) transferring of matter and energy between the abiotic components of the biosphere.(C) essential in the maintenance of balance between the biotic components of the biosphere.(D) maintainance of water cycle between biotic components of ecosystem.
63. On mitotic division of a cell with 20 chromosomes(A) two daughter cells will be created, each having 20 chromosomes.(B) two daughter cells will be created, each having 40 chromosomes.
(C) four daughter cells will be created, each having 10 chromosomes.(D) two daughter cells will be created, each having 10 chromosomes
64. Match List I with List II and select the correct answer using the codes given below the lists.List List(Plant tissue) (Function)A. Sclerenchyma 1. Conduction of waterB. Xylem 2. Transport of foodC. Phloem 3. Mechanical strengthD. Meristem 4. Cell Division
A B C D(A) 4 2 1 3(B) 1 4 3 2
(C) 3 1 2 4(D) 2 3 4 1
65. Synctial epidermis is found in(A) pheretima. (B) ascaris. (C) taenia. (D) hydra.
66. Plants having seeds but lacking flowers are(A) Pteridophytes. (B) Gymnosperms. (C) Angiosperms. (D) Bryophytes.
67. Phylloclade is a feature of(A) vallisneria (an aquatic plant) (B) maize (a terrestrial plant)(C) cactus (a xerophytic plant) (D) mango (a tall tree)
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IJSO STAGE-I _OPEN TEST/MOCK TEST PAPER-5_PAGE # 12
Space For Rough Work
68. At which stage of meiosis I, the homologous chromosomes separate but are held by chiasmata ?(A) Diakinesis (B) Diplotene (C) Pachytene (D) Zygotene
69. Ptyalin enzyme present in which fluid of the body ?(A) Bile (B) Saliva (C) Pancreatic Juice (D) None of the above
70. False statment amonst the following is(A) ozone is a non-poisonous gas.
(B) UV radiations split oxygen molecule into atomic oxygen.(C) oxygen molecule combines with oxygenatom to form ozone.(D) ozone is found in thestratosphere layer of the atmosphere
71. Process of accumulation of pesticides in the human body is called(A) biomagnification. (B) bioaccumulation. (C) bioconcentration. (D) biodegradation
72. One of the factor required for the maturation of erythrocytes is(A) Vit D (B) Vit A (C) VitB
12(D) Vit C
73. Syphilis is caused by(A) Treponema pallidum (B) HIV(C) Neisseria (D) Trichomonas
74. If a plant is heterozygous for tallness, the F2
generation has both tall and dwarf plants. This proves theprinciple of :(A) Dominance (B) Segregation(C) Independent assortment (D) Incomplete Dominence
75. DNA differs from RNA :(A) In the nature of sugar alone (B) In the nature of purines alone(C) In the nature of sugar and pyrimidines (D) None of the above
76. The first organism was(A) aerobic (B) anaerobic (C) catabolic (D) both (A) and (B)
77. Total number of canines in permanent dental set of human is(A) 4 (B) 6 (C) 2 (D) 12
78. The exchange of gases in alveoli of lungs takes place by-(A) passive transport (B) simple diffusion (C) osmosis (D) active transport
79. The functional unit of contractile system in striated muscle is(A) myosin. (B) sarcomere. (C) I-band. (D) actin
80. Microbial diseases like Syphilis and AIDS are transmitted through(A) contaminated water. (B) sneezing. (C) sexual contact. (D) unhygenic food.
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SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 1
NSWER KEY
HINTS & SOLUTIONS
ALL INDIA IJSO(STAGE-I) TEST SERIES
OPEN TEST/MOCK TEST PAPER # 10DATE : 09-11-2014
1. Volume of icecream in cylinder = volume oficecream in 12 cones
!(6)2 h = 10 "#$%&' !()!32 )3(
3212)3(
31
36 h = 10 [36 + 18]
h = 15 cm.
2.cb
a
(=
ac
b
(
[a, b, c *distinct & real and b + c +c + a
[!a +b]
a (c + a) = b (b + c)
ac + a2
= b2
+ bcac bc = b2a2
c(a b) = (b a)(b + a)
c = a + b [,a +b]
Hence, a + b + c = 0.
3. LN =2
1BC [by mid-point theorem]
A
L N
BM
C
-ABC ~ -MNL [By AA similarity]
MNLArea
ABCArea
-
-
=
2
LN
BC
./
0
12
3= (2)
2
= 4
area -LMN =4
ABCarea-=
4
48= 12 sq. unit.
4.
Let the height of the tower be h.
, In -ABC
tan y =b
h
b =ytan
h
In -ABD
tan x =db
h(
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. C C C A B D C B B D B D D B C C A B A B
Ques. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. D B B C C C A A C A A B B C B A D C D C
Ques. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. B C A B C B B A C C C A A B A A D A A B
Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans. C A A C B B C B B B A C A B C B A B B C
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SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 2
tan x =d
ytan
h
h
(
xtan
h=
ytan
h+ d
d = h ..
/
0
112
34
ytan
1
xtan
1
d = h ../
0112
3 4ytanxtan
xtanytan
h =xtanytan
ytanxtand
4 .
5. For the curves to intersect, log10
x = x-1
Thus, log10x = x
1or xx= 10
This is possible for only one value of
x (2 < x < 3).
6. Isosceles triangles
Maximum side = 4 unitsA
CB
c b
a
a < b + c
b < a + c
c < a + b
Possible combinations :
4, 4, 1 3, 3, 4 2, 2, 3
4, 4, 2 3, 3, 2 2, 2, 1
4, 4, 3 3, 3, 1 2, 2, 2
4, 4, 4 3, 3, 3 1, 1, 1
12 such combinations are possible.
7. (2.5)2[12+ 22+ 32+ 42+ ------ + 202]
= (2.5)26
)140()120(20 ()()
= 6.25 6
412120 ))
=6100
412120625
))))
=65
4121625
)))
=2
417125 ))=
4
35875= 17937.5
8. Given (1025 7 ) (1024+ x) is divisible by 3
10257 1024x
1024(10 1) 7 x
9 510247 x
If divisible by 3 value of x = 2
9.2010
x2009+ 1 +
x
1= 0
x2010
2010x2010x2009 2 ((= 0
2009x2+ 2010x + 2010 = 0
6+ 7=2009
2010and 67=
2009
2010
61
+ 7
1= 67
7(6
2009
201020092010
8= 1
10. Since perimeter of a rectangle p = 2(l + b) and
diagonal of the rectangle d = 22 bl (
9 (l + b) =2
por (l + b)2=
4
p2
l2+ b2+ 2lb =4
p2
or d2+ 2lb =4
p2
or 2lb =4
p2d2
,(l b)2= l2+ b22lb = d24
p2+ d2
=4
pd8 22 4
or (l b) =4
pd8 22 4
11. a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d +
5 = k(a + 1) + (b + 2) + (c+ 3) + (d + 4) = 4k
a + b + c + d + 10 = 4K
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SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 3
a + b + c + d + 5 + 5 = 4K
K + 5 = 4K
3K = 5
K =3
5
a + b + c + d + 5 = K
a + b + c + d = K 5 =3
55 =
3
10
12. Option (D) is correct because
2
1
4
2+ ..
/
0112
3+
2
1
2
1
b
b
a
a.e.i
13. sin x + sin2x = 1
sin x = cos2x
Now sin2x + cos2x = 1
(cos2x)2+ cos2x = 1
cos4x + cos2x = 1
14.
Let AD = h1BE = h
2
CF = h3
Area of -ABC =2
1ah
1=
2
1bh
2=
2
1ch
3
9 ah1= bh2= ch3
Given : h1: h
2: h
3= 2 : 3 : 4
,h1 = 2k, h2= 3k and h3= 4kSo, a(2k) = b(3k) = c(4k)
2a = 3b = 4c
Divide by 12
6
a=
4
b=
3
c.
,Corresponding sides are in the ratio 6 : 4 : 3.
15.
EF
A
CB D
G
By mid point theorem
EF =2
1BC
In -BGC
BG + GC > BC ....... (i)
In -GFE
FG + GE > FE ....... (ii)
Add (i) and (ii) we get
BE + CF > BC + FE
BE + CF > BC +
2
BC
"#
$%&
'8
2
BCFE
BE + CF >2
BC3
16. (1 + 2x)20= a0+ a
1x + a
2x2+ .... + a
20x20
Let x = 1
320= a0+ a
1+ a
2........ + a
20....(i)
Let x = 1
120= a0a
1+ a
2........ a
19+ a
20.....(ii)
Multiply equation (i) by 5 and add with equation (ii)
5.320+ 1= 6a0+ 4a
1+ 6a
2+ ..... + 4a
19+ 6a
20
2
13.5 20 (= 3a
0+ 2a
1+ 3a
2........ + 2a
19+ 3a
20.
17. D = 0 9(k + 1)28k = 0
9 a = 3 2 2
18.
x
2x
2y
y
7/2 7/2
3
3
A
B C
E
D
4(4
49= 4y2 + x2 )
9 = 4x2+ y2
49 = 4x2
+ 16y2
- * -
40 = 15 y2
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SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 4
y2 =3
8
15
408
9 = 4x2 +3
8
9 -3
8= 4x2
x2=1219
4x2 + 4y2 = AB2
3
84
43
194 )(
))
= AB2
3
32
3
19(
= AB2
AB2 = 15/3
AB = 17
19. Draw the angle bisector BE of -ABC to meet
AC in E.Join ED.Since ;B = 2;C, ;EBC
= ;ECBNow BA = CD, BE = CE
;EBA = ;ECD in -BEA and CED-BEA ? = y = ;IDE. , ;>?= 8 ;DIE )So, ;DEI = ;ABI = ;DBI-BDE is isosceles and BD = DE = EA
ED || AB , BC = AC(as BD = AE , ;A = 2;C. So 5;C = 180);C = 30. Also ;CED = ;EAD + ;EDA =2y = ;A = 72
20. Let TR = y.
Since OT is perpendicular bisector of PQ.
, PR = QR = 4cmIn right triangle ORP, we have
OP2 = OR2 + PR2
OR2 = OP2 PR2 = 52 42= 9 OR = 3 cm.In right triangles PRT and OPT, we haveTP2 = TR2 + PR2 and, OT2 = TP2 + OP2
OT2 = (TR2 + PR2) + OP2
(y + 3)2 = y2 + 16 + 25
6 y = 32
y =3
16
TP2 =
2
3
16./
012
3+ 42 =
9
256+ 16
=9
400
TP= 3
20
cm.
22.
Req
= 7 + 4 + 9 = 20@
V = IReq= 1 20 = 20 V
25. From energy conservation
PEi + KE
i = PE
f + KE
f
./
012
3(
2/d
GM
2/d
GM 21m +
2
1mv2 = 0 + 0
V2 =d
G4 (m
1 +m
2)
V =d
)mm(G4 21(
26.
(4 ?) R = ?RV= 20 (4 ?) R = 20
4
?is less than 4So that, R is greater than 5@
27. Apply Newtons law for system along the string
mBg = A(mA+ mC) g
9 mC= ABm
mA
=2.0
910
= 15 kg
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SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 5
28. x =3
4/
1
1
24
=
./
012
3
4
3
24=
3
424 )= 32 cm
29.
Distance between lens is = f1+ f
2,
30. For series combination : total resistance
= R1\ + R
2
,Total resistance = 10@ + 20@ + 30@Accordintg to ohms law : V = IR
, I =R
V=
30
220=
3
22AA
,Heat produced = I2RT
= 6030303
22
3
22 ))))
(!t = 30 min.)
= 2904 KJ
31. mi= 10g,m
w= 10g
ti= 10C, t
w= 85C
Heat lost by water
Qw= 10 (85 0) 1
= 850 cal
Heat required for ice to come at 0Cand to change into water
Qi= 10[0(10)] 5 + 10 80 cal
10 10 5 + 800 = 850 cal
Qw= Q
i
So net temperature = 0C
So ice left =0g.
33. Consider an observer moving with speed v
with point A in the same direction.
//////// /////////// /////////// //////// /////////// ///////
v
Am
observer
v
In the frame of observer, block will have initial
velocity v towards left.
///////// ////////// /////////// //////// /////////// ///////
Am
v
During maximum extension, the block will
come to rest with respect to the observer.
Now, by energy conservation,
21 mv2=
21 k 2maxx
, xmax
=k
mv 2
34. Efficiency of machine, B= 75%Energy used = 12 J
So, energy used to lift the mass
= 12 100
75
= 9 J
!2
1mv2= 9 (Here m = 1kg)
v2= 18 9 v = 18 m/s
35. At point A body has only PE.
PE = mg (h + x)
KE = 0
At point C
C
B
A
h
x
KE = 0
By applying work - energy theorem betweenpoint A & C.
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SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 6
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.]
F = Mg ./01
23 (
xh1
37. Resistance of each samicircle, R1= R
2
= !r 0.5= 0.5!@Resistance of diameter (AB1, R
3= 2r 0.5
= 1@
ABR
1=
!5.01
+!5.0
1+ 1
R>= 8 4(!!
41. M =solution
solute
V
n
1000
8.0=
solutionof.vol
10100 34)
vol. of solution = 125 ml
(Here nsolute= mole of solute, Vsolution= vol. of solution).
42. M1V1= M2V2
98
49)1.18 )10 ) 75 = M2 590
43. Molarity of cation =21
2211
VV
VMVM
((
=500
4001.01002.0 )()=
5
6.0= 0.12 M
Molarity of Cl=500
4001.0100)2.0(3 )(
=5
4.06.0 (= 0.2 M
44. Molecular wt. of gas = 3a
no. of moles of gas =a3
w.
45. Fraction =
atomof.vol
nucleusof.vol= 38
313
)10(3
4
)10(3
4
4
4
!
!
= 1015.
46. Isotones have same number of neutrons.
47.1
2
2
1
E
E
CC
8 =3000
6000= 2.
48. Use E = Cnhc
60 60 =9
834
10620
1031064.6n4
4
)
))))
n = 1.125 1022
50. Na2311 *D Ne2310 + e
01( ; So ratio of
atomic mass and atomic number =10
23.
52. Suppose at T = 27C = 300 K
T1= 37C = 310 K
V = 1 litre
V1= ?
at constant pressureT
V=
1
1
T
V
300
1=
310
V1
V1=
300
310= 1.0333 litre
Since, capacity of flask is 1 litre.
,Volume of air escaped out = 1.0333 1= 0.0333 litre = 33.3 mL Ans.
53. P1V1= P2V2
, P2
= atm50
201)= 0.4 atm Ans.
8/10/2019 10. AITS-10 (09-Nov-14)
19/19
SOL. IJSO STAGE-I _OPEN TEST/ MOCK TEST-5_PAGE # 7
54. P2 = 5.0
5.21)= 5 atm
9, % increase = 1001
15)
4= 400%
56. N2 + 3H2 2NH31 3 0
0.5 1.5 1
PH2
= P3
5.1= P/2
59. For H2O K
wET.
, [H+] ET.
, pH ET
1.
60. K = [HCOOH2
+] [HCOO] = 106.
[HCOO] = 103mol/L.
1 liter solution of HCOOH has = 1150 g mass.
moles of (HCOOH) in 1 litre solution =46
1150
= 25 mol.
out of 25 mol HCOOH 103mol are ionised into
HCOOions.
,% dissociation =25
10 3 100
= 0.004%.
! ! ! ! !