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10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 1
Time : 3 hrs
Marks : 30
(i) The magnitude of the gravitational force experienced by the mass at a depth
from the surface (see figure) will be
[0.5]
(A) (B)
(C) (D)
Correct answer: (A) [0.5]
g(r) = GM(r)/r2 M(r) = (4/3)r3 = M /(4/3)R3
g(r) = (GM/R2)(r/R) = gr/R F = [mg/R]r = mg(1-h/R)
(ii) In the grid provided on your answer sheet, plot a graph of , where
is the force on the mass at a distance from the centre of the Moon, as a
function of , as varies from 0 to .
[1.0]
(iii) If kg, what is the minimum time (in seconds) it will take, from the
moment the mass is dropped through the hole at the surface, for it to reach the
centre of the Moon?
1
r/R
F/mg
(0,0) 1 2
0.25
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 2
Time : 3 hrs
Marks : 30
[1.0]
Correct answer: T = 1619 s [1.0]
Substitute mg/R as k in the formula for the frequency, Time period = 6476
s, time take to reach the centre = 1619 s.
0 = 1/2(k/m)0.5
Time taken to reach the centre, T = 2(m/k)0.5
T = (R/g)0.5 T = 1619 s
(iv) The oxygen molecule O2 has a force constant, N m-1
. The equilibrium
bond length is m and the change in the bond length when it is
fully stretched is 6.0% of . Calculate the vibrational energy, that is the sum of
kinetic and potential energies per mole of oxygen (in kJ-mol-1
).
[1.5]
(Avogadro's number, NA = 6.0231023
)
Vibrational energy per mole = UN0 = 28.052 kJ/mole [1.5] Ls = 1.59 x 10-10 m [0.5]
Vibration energy is the work done in changing the separation from
equilibrium position to the fully stretched position [1.0].
Force, F = k(Ls-L) Work, W = 0.5k(Ls-L)2
U = 0.5k(Ls-L)2
Vibrational energy per mole = UN0 = 28.052 kJ/mole [0.5]
(v) The atomic weights of the halogen elements listed in the periodic table are:
F Cl Br I
19.0 35.5 79.9 126.9
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 3
Time : 3 hrs
Marks : 30
Two halogen elements, X and Y, form diatomic molecules X2 and Y2 with force
constants N-m-1
and N-m-1
respectively. The vibration
frequencies are measured to be Hz and Hz.
Identify the halogen elements X and Y by writing their symbols. Write your
answer in the form X = , Y = in the answer sheets.
[1.0]
Correct answer: X = Cl and Y = F [1.0]
The ratio of the vibration frequencies is 0.623.Only X = Cl and Y = F are
the elements such that the ratio of the square roots of the masses
(my/mx)0.5 ~ 0.64.
Question 2
(i) If sunlight is shone through a transparent container (with walls of negligible
thickness) filled with nitrogen gas, what will be the ratio of the scattered light
intensity for colours corresponding to wavelengths 400 nm and 650 nm
respectively? [1.0]
Ratio of the scattered intensities = 6.2 [1.0]
Correct ratio of the light intensity = 0.9 [0.5] or
Correct ratio of the scattered light intensity without taking into account
solar spectrum [0.5].
The ratio of the intensity of 400 nm and 650 nm light in the sunlight is 0.9.
s(400)/s(650) = (650/400)4*0.9 = 6.2 [0.5].
Visibility range for polluted air 60 x 103 m or 75 x 103 m [1.5]
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 4
Time : 3 hrs
Marks : 30
Scattering losses due to the pollutants will be proportional to 0.1*40 = 4
times larger compared to pure air. [0.5]
Pollutants act as additional scattering hence the visibility range reduces by
5 times (60 x 103 m)
(ii) Milk is a colloidal solution in which droplets of liquid
fat, of size around 100 nm, are suspended in water.
These droplets scatter light more strongly than the water
molecules, causing normal milk to appear white rather
than transparent.
Consider the following experiment. A few drops of milk
are added to a glass of water illuminated from above by
a beam of sunlight, as shown in the figure on the right.
The water turns cloudy, but some sunlight still passes through, since the
concentration of milk is small. The glass is now viewed (I) from below, and (II)
from the side, as shown in the figure.
When compared to the emerging light viewed from below (I), the emerging light
viewed from the side (II) will appear [0.5]
(A) bluish (B) orange (C) reddish (D) the same
Correct answer: (A) [0.5]
Shorter wavelengths are scattered much more than the longer
wavelengths, hence blue will be scattered more along the direction (II)
compared to direction (I)
II
I
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 5
Time : 3 hrs
Marks : 30
(iii) Which of the following atmospheric phenomena is mainly governed by Mie
scattering of light?
[0.5]
Images taken from:
(A) http://bostern.wordpress.com (B) http://www.kaneva.com
(A) red sunset (B) white clouds
(C) blue sky (D) rainbow
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 6
Time : 3 hrs
Marks : 30
(C) http://lisathatcher.wordpress.com (D) http://www.freefoto.com
Correct answer: (B) white clouds [0.5]
(a) and (c) are due to Rayleigh scattering (wavelength dependent), (d) is due to
wavelength dependent refraction and dispersion through water droplets.
(i) The green colour of leaves and shoots of plants is usually due to the presence of
chlorophyll, the compound mainly responsible for photosynthesis. Which of the
following graphs depicts the correct absorbance spectrum of chlorophyll?
[1.0]
Correct answer: (B) [1.0]
For green colour the absorption is minimum and hence is reflected or
scattered the most
(ii) Assuming that the rate of photosynthesis is proportional to the amount of light
absorbed (see above figure), what will be the wavelength (in nm) corresponding to
the maximum photosynthesis rate in green plants?
[0.5]
450 nm [0.5]
Maximum absorption in Fig. b is at 450 nm.
[1.0]
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 7
Time : 3 hrs
Marks : 30
Correct answer: (A) [1.0]
Larger absorbance where the solar spectrum has a maximum intensity.
Question 3
(a) The dissociation of in blood occurs in two steps. Write down balanced equations
for these two steps.
[0.5]
H2CO3 (aq.) H +(aq.) + HCO3- (aq.) [0.25]
HCO3- (aq.) H +(aq.) + CO3
2-(aq.) [0.25]
(i) Calculate the concentration of in a solution at 37 0C, and hence its pH value, if
and are present in equal concentrations in mol/l in that solution.
[0.5]
[H2CO3] = [HCO3-] K1 = [H
+]
[H+] = 2.2 ×10
-4 M [0.25]
pH = 3.65 [0.25]
(ii)
[1.0]
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 8
Time : 3 hrs
Marks : 30
pH = 7.4 [H+]=3.98 × 10 -8 [0.5]
[0.5]
(b) Usually in the human body, is in equilibrium with the dissolved in the blood.
At 370 C, 5.0 x 10
-3.
Calculate the total equilibrium constant, for the reaction
[0.5]
[0.5]
(c) Blood plasma contains a total carbonate buffer pool, which is a mixture of and
with a total concentration of M at 38 0C. At this temperature the value
of the equilibrium constant is . The concentration of the is
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 9
Time : 3 hrs
Marks : 30
negligible. Calculate the ratio of concentrations of and , and
their individual concentrations, in this blood sample at pH 7.4.
[1.5]
pH = 7.4 [H+]=3.98 × 10 -8
or [0.5]
[0.5]
[0.5]
Question 4
.
(a) Which of the above carry de-oxygenated blood?
[1.0]
1,4,5, RA and RV
(b)
[0.5]
What would be the individual’s heart rate (in beats/minute) as calculated from the table?
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 10
Time : 3 hrs
Marks : 30
Heart rate: 60/0.7 = 85.7 beats/min [0.5]
(c)
[1.5]
(d) Blood flows from the heart into the aorta during a cardiac cycle. If the diameter of the aorta
is approximately 2.4 cm, then, using the table in 4(b) calculate the average speed (in
cm s-1
) of blood flowing into the aorta in one full cardiac cycle.
[1.0]
Speed = Volume/(cycle time*Area) [0.5]
Speed = 42/4.52 = 60/4.52
Speed = 13.26 cm/s (for cycle time of 0.7 s) or 23.23. cm/s (for cycle time of 0.4 s)
[0.5]
(e) Blood flows from the aorta and its major arteries into arterioles and fine-walled capillaries.
If all the major arteries in the body have a total cross-sectional area of about 7.0 cm2
calculate the average speed (in cm-s-1
) in the major arteries which have the same volume
of blood as the aorta flowing through them. [0.5]
V2 = 8.56 cm/s (13.26 cm/s), V2 = 14.99 cm/s (23.23 cm/s)
A1V1 = A2V2
FULL [0.5] Credits to be given for the correct value.
HALF [0.25] Credits to be given for the formula even if the value is wrong.
Time Mitral valve Aortic valve
0.2 s C O
0.6 s O C
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 11
Time : 3 hrs
Marks : 30
[1.0]
Question 5
(i) Calculate the average acceleration of this cheetah required to reach its maximum
speed.
[0.5]
Average acceleration, [0.25]
[0.25]
(ii) Calculate the distance travelled during the first 3.0 s, assuming that the
acceleration is uniform.
[0.5]
Distance travelled during the acceleration phase , [0.25]
[0.25]
(iii) The cheetah has to do work against friction, mostly due to air. Assume that this
frictional force is always 100 N. Calculate the total mechanical work done by the
Aorta Renal
vein
Alveolar
space in
lungs
Pulmonary
artery
Y X Y X
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 12
Time : 3 hrs
Marks : 30
cheetah during the first 23.0 s of its motion.
[1.0]
Work done in
Change in kinetic energy, [0.25]
[0.25]
Overcoming friction due to air, [0.25]
Total work done
[0.25]
(a) During the first 23.0 s, the body temperature of the cheetah rises from 38.5 0C to 40.0
0C. Take the specific heat of the body of the cheetah to be 4.2 kJ kg
-1 K
-1.
(vi) If the rise in body temperature is linear during this time, calculate the total heat
generated by the cheetah’s metabolism. Neglect any heat loss to the surroundings.
[1.0]
Heat generated in the body of the Cheetah, [0.5]
[0.5]
(vii) Assume that some of the energy generated by the cheetah’s body increases its
temperature and the rest corresponds to the mechanical work done. Calculate the
fraction of the total generated energy that is converted to kinetic energy.
[1.0]
Total energy generated in the body of Cheetah, [0.5]
Fraction = = 0.06 [0.5]
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 13
Time : 3 hrs
Marks : 30
(viii) Write down the balanced chemical reaction for aerobic respiration.
[1.0]
The balanced reaction:
(ix) If the cheetah requires 400 kJ of energy, calculate the volume of oxygen required
if all this energy is to be obtained by aerobic respiration. Take the molar volume
of oxygen gas to be 24.5 litres.
[1.0]
Number of glucose molecules required to produce 400kJ of energy is
[0.25]
Number of mol of is
[0.25]
Volume of required, [0.5]
(x) The cheetah extracts oxygen from the air while breathing. The inhaled air (about
500 ml per breath) contains 20.0 % oxygen (by volume), while the exhaled air is
assumed to contain 15.0 % oxygen (by volume). Calculate the volume of oxygen
that the cheetah can use during the 23.0 s of its run, at a breathing rate of 150
breaths per minute.
[1.0]
Amount of absorbed by the lungs per breath is
[0.5]
Total number of breaths during its entire motion,
10 th International Junior Science Olympiad,
Pune, India
Theory Questions
Page 14
Time : 3 hrs
Marks : 30
[0.25]
Thus the amount of intake of , [0.25]
(xi) Anaerobic respiration converts the energy from glucose into ATP. If glucose were
to be completely burnt up, one mole would release 2872 kJ of energy. What is the
efficiency of anaerobic respiration compared to complete combustion of glucose?
[1.0]
Energy generated in production of one mol of ATP due to oxidation of glucose is
[0.5]
Efficiency of the anaerobic respiration, [0.25]
[0.25]
(xii) If all the 400 kJ required by the cheetah for its run were to be produced by
anaerobic respiration, calculate the total amount of glucose (in kg) that would be
required.
[1.5]
Amount of glucose of required,
[0.5]
Approximate molecular mass of glucose,
[0.5]
[0.5]
10 th
International Junior Science Olympiad, Pune, India
Experimental Tasks
Page 1
Time : 3 hrs
Marks :40 A B C + +
Task: This task is divided into three parts. Total marks: 14
A1: To determine the centre of gravity of a triangular plate, A.
A2: To record the time period of oscillation for different
suspension points for the plate.
A3: To analyze the above data and results.
A.Q1 Determination of CG: [1.0 mark]
Mark “X” on Sheet 1 at the appropriate position to denote the CG (large sized sheet).
CG within 5 mm [1.0]
CG within 10 mm [0.5]
A.Q2 Table A.1: Oscillation measurements: [4.0 marks]
h
(m)
h2
(m2)
Time taken for 50 oscillations (s) T = T1/50
(s)
T2
(s2)
hT2
(ms2)
1st
(t1)
2nd
(t2)
3rd
(t3)
Mean (T1)
(t1+t2+t3)/3
H1 0.243 53.65
H2 0.155 47.77
H3 0.179 49.08
H4 0.098 45.48
For each value of h within ±5 mm [0.25 x 4.0= 1.0]
For each value of T1 within ± 1 s [0.5 x 4.0= 2.0]
For each value of T1 within ± 2 s [0.25 x 4.0= 1.0]
For calculating h2 and hT
2 [0.25 x 4.0= 1.0]
A
10 th
International Junior Science Olympiad, Pune, India
Experimental Tasks
Page 2
Time : 3 hrs
Marks :40 A B C + +
A.Q3Results of the data analysis
(a) Grid1: hT
2 (y-axis) versus h
2 (x-axis) [2.0 marks]
0.000.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
hT
2 (
m s
2)
h2 (m
2)
Grid 1
Plotting 4 points correctly [0.25 x 4.0= 1.0]
For best fit straight line [1.0]
10 th
International Junior Science Olympiad, Pune, India
Experimental Tasks
Page 3
Time : 3 hrs
Marks :40 A B C + +
A.Q4Table A.2: Calculations from Grid 1 [3.0 marks]
Quantity Numerical
value
FULL
CREDITS
HALF
CREDI
TS
Slope of the graph (s) 4.06 ± 0.2 [0.5] 4.06 ± 0.4 [0.25]
y-intercept of the graph (c) 0.040 ± 0.005 [0.5] 0.040 ± 0.01 [0.25]
Acceleration due to gravity (g) 9.8 ± 0.5 [1.0] 9.8 ± 1.0 [0.5]
Radius of gyration (K) 0.10 ± 0.01 [1.0] 0.10 ± 0.02 [0.5]
Values outside the above mentioned range = Zero
A.Q5(a) Table A.3:
[3.0 marks]
Holes h (m) hˈ (m)
H1 0.243 0.045
H4 0.098 0.11
Each correct value of hˈwithin ± 10 mm [0.5x 2 = 1.0] Values outside the above mentioned range = Zero
(b) Sheet 1: Mark the positions of points of oscillation J1 and J4 on Sheet 1.Label them as J1
and J4 clearly.
[1 x 2 = 2.0]
A.Q6Table A.4: Lengths of equivalent simple pendulums [1.0 mark]
Holes h (m) L (m)
H1 0.243 0.288
H4 0.098 0.209
For each correct calculation within ± 0.015 m [0.5]
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 1
Time : 3 hrs
Marks : 40 A B C
Task : In this set of experiments we will investigate, Total Marks: 20
A1: The buffering capacity of milk
A2: Enzymatic digestion of milk proteins
A3: Determining the calcium content of milk
B1 The buffering capacity of milk
B.Q1.A pH of water = [0.25 Mark]
[6.0 – 7.0: 0.25, Other values = Zero]
B.Q1.B pH of sodium carbonate = [0.25 Mark]
[>10: 0.25, Other values = Zero]
B.Q1.C pH of acetic acid = [0.25 Mark]
[2 – 4: 0.25, Other values = Zero]
B.Q1.D pH of milk = [0.25 Mark]
[6.0 – 8.0: 0.25, Other values = Zero]
B
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 2
Time : 3 hrs
Marks : 40 A B C
B.Q2 Observation Table B.1 [2.0 Marks]
Marks will be given based on the volume of the solution required to get the desired pH in
each case.
Range of volume: 0.3 - 0.5 ml for each solutions [1 x 2 = 2.0]
0.2 - 0.3 ml or 0.5-0.6 ml for each solutions [0.5 x 2 = 1.0]
Other values = zero.
Stepwise addition to 40 ml water
Sodium carbonate solution Acetic acid solution
Stepwise
volume
added in
ml
pH value Stepwise
volume
added in
ml
pH value
1 0
0
2 0.1
0.1
3 0.1
0.1
4 0.1
0.1
5 0.1
0.1
6 0.1
0.1
7 0.1
0.1
Total …… Volume of Na2CO3 solution
added to reach pH 10.0
…… Volume of CH3COOH solution
added to reach pH 4.0
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 3
Time : 3 hrs
Marks : 40 A B C
B.Q3 Observation Table B.2 [2.0 Marks]
Stepwise addition to 40 ml Milk
Sodium carbonate solution Acetic acid solution
Stepwise
volume
added in
ml
pH value Stepwise
volume
added in
ml
pH value
1 0
0
2 0.5
0.5
3 0.5
0.5
4 0.5
0.5
5 0.5
0.5
6 0.5
0.5
7 0.5
0.5
Total …… Volume of Na2CO3
solution added to reach
pH 10.0
…… Volume of CH3COOH
solution added to reach pH
4.0
Marks will be given based on the volume of the solution required to get the desired pH in
each case.
Range of volume: 1.5 - 2.5 ml for each solutions [1 x 2 = 2.0]
1.0 - 1.5 ml or 2.5-3.0 ml for each solutions [0.5 x 2 = 1.0]
Other values = zero
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 4
Time : 3 hrs
Marks : 40 A B C
Questions:
B.Q4 [1.0 Mark]
Comparing the observations in Table B.1 and B.2 which of the following statements describe
the role played by milk?
a) You require more acetic acid solution to lower the pH of milk to 4 than to lower
the pH of water to 4.
True (T) False (F)
b) Less sodium carbonate solution is required to raise the pH of milk to 10 than to
raise the ph of water to 10
True (T) False (F)
B.Q5 [1.0 Mark] As compared to water, milk resists change in pH of the resulting solution when acetic acid is
added. This is because components of milk:
a) lead to increase in concentration of the OH- ions in the resulting solution
b) prevent increase in concentration of the free H+ ions in the resulting solution
c) lead to decrease in concentration of CH3COO- ions in the resulting solution
Write the correct option in the appropriate box
X
X
b)
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 5
Time : 3 hrs
Marks : 40 A B C
B2 Enzymatic digestion of Milk protein
B.Q6.A Iw = [0.5 Mark]
Iw > 0.6 mA [0.5]
Iw = 0.2-0.6 mA [0.25]
B.Q6.B IO = [0.5 Mark]
Anything below 0.2mA or the value based on the curve [0.5]
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 6
Time : 3 hrs
Marks : 40 A B C
B.Q7 Observation Table B.3 [2.0 Marks]
Time
(in s)
Current
(in mA)
Time
(in s)
Current
(in mA)
1.
16.
2.
17.
3.
18.
4.
19.
5.
20.
6.
21.
7.
22.
8.
23.
9.
24.
10.
25.
11.
26.
12.
27.
13.
28.
14.
29.
15.
30.
20 Readings [0.1 x 20 = 2.0]
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 7
Time : 3 hrs
Marks : 40 A B C
B.Q8 Graph plotting: [3.5 Marks]
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 8
Time : 3 hrs
Marks : 40 A B C
Plot a graph of current versus time.
Labels on each axis [0.25]
Scale on each axis [0.25]
Marks for occupying more than 60 % of graph paper [0.25]
Drawing smooth curve [0.25]
Based on the plot, within Iw and I0 ranges and the nature of the curve [0.1 x 20 = 2.0]
B.Q9 [1.0 Mark]
Mark a point K on the graph paper where the casein concentration is maximum, a point L where
casein concentration is minimum and a point M where the casein concentration is half-way
between the maximum and minimum values.
Based on the plot, all 3 correct [1.0]
2 correct [0.5]
1 correct [0.25]
B.Q10 [1.0 Mark]
If the increase in current is proportional to the amount of digested casein and maximum current
represents complete digestion of casein, deduce from the graph the time taken for digestion of
50% casein.
Time taken based on the value of M in the above question [1.0]
Other values = Zero
10 th
International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 9
Time : 3 hrs
Marks : 40 A B C
B3 Estimation of calcium content in milk
B.Q11 Observation Table B.4 [3.5 Marks]
Sr. No. Titration I Titration II Titration III
1 Initial burette reading
ml
2 Final burette reading ml
3 Difference in burette
reading ml
Average difference in burette reading (m1) = (6.8 – 7.2) [2.0 marks]
Average difference in burette reading (m1) = (6.6 – 7.4) [1.5 marks]
Average difference in burette reading (m1) = (6.2 – 7.8) [1.0 marks]
Average difference in burette reading (m1) = (6.0 – 8.0) [0.5 marks]
Any other Average difference in burette reading: [0.0 marks]
An average value deduced using anomalous reading will result in deduction of 0.25 mark.
B.Q12 [1.0 Mark]
Deduce the amount in milligrams of Ca2+
per 10 ml of the diluted solution (the atomic weight
of Ca is 40).
0.756 mg based on 7.0 ml in the previous question or for correctly calculated value based on the
previous question [1.0]
Wrong calculations [0.5]
Average burette reading: (A)…………………………………ml of 0.001 M Na2EDTA
10 th International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 1
Time : 3 hrs
Marks : 40 A B C
Task : Extraction of lycopene from tomato Total Marks: 6.0
Absorbance of extract:
C.Q1 Observation Table C.1 [3.5 Marks]
Current
in mA
Blue
LED
White
LED
1 Is 0.8 – 1.0
mA
0.6 mA and
above
2 Il 0.0 – 0.08
mA
~ 0.4 mA
Percentage of light
transmitted
0 to 8%
[0.25]
30 – 70%
[0.25]
Is (Blue LED) 0.8 – 1.0 mA [1.5]
Is (Blue LED) 0.6 – 0.8 mA and 1.0 – 1.2 mA [0.75]
Is (White LED) 0.6 mA and above [1.5]
Is (White LED) 0.3-0.6 mA [0.75]
Percentage transmitted in each case [2 x 0.25 = 0.5]
Questions
C.Q2 [1.0 Mark]
If the test tube Ab (containing the solvent) was removed from between the photodiode and the
white LED,
a) The current measured would be less than Is
b) The current measured would be more than Is
c) The current measured would be equal to Is
Write the correct option in the box below.
C
a)
10 th International Junior Science Olympiad,
Pune, India
Experimental Tasks + +
Page 2
Time : 3 hrs
Marks : 40 A B C
C.Q3 [1.5 Marks]
Which of the following can you deduce from your observations in the experiments on transmitted
light. Indicate your answers as YES (Y) or NO (N) in the table below.
a) Lycopene absorbs more blue light relative to other parts of the visible spectrum.
b) Lycopene preferentially absorbs light in the red and yellow parts of the spectrum.
c) Lycopene is an antioxidant.
d) Red and yellow parts of the spectrum are absorbed relatively less compared to
blue parts of the spectrum.
e) Blue light passes through the solution better compared to red light.
f) Lycopene absorbs light equally across the spectrum.
a) Y
b) N
c) N
d) Y
e) N
f) N
Each correct option [0.25]
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 1
1 ) Answer : A Solution : a: 27m3 / 22.4 lt = 10^3 moles = 10^26 molecules
b: 1000/18 = 50 moles = 10^25 atoms
c: 5 seconds for 1 breath => 10^8 breaths
d: 14 billion years => 10^17 seconds
2 ) Answer : A Solution: The sun is just below the horizon. The Moon is just above the
horizon. This implies crescent moon. Since the illumination is from below, answer is A.
3 ) Answer: B Solution :
m =sin i
sin r
sin r =sin i
m=R / 1+ (3 / 4)2R
4
3
=3
5'
cot r =4
3=
R
¢R - R
¢R =7
4R
Area of shadow = 49
16pR2
4 ) Answer B. Solution: The magnetic field above the wire is out of the plane. This flux is
decreasing and
should be compensated by the current in the loop F and so the current in loop F will be
anti- clockwise. For loop G the situation is opposite
5 ) Answer A: Solution : (a) corresponds to p = 0. Then solve for T and put V = 1, a = α, b
= β
6 ) Answer C
7 ) Answer B
8 ) Answer B: Solution : We assume the sign convention where distance measured in the
direction of light propagation are positive, and in opposite direction, negative. For the morning
experiment, since the image of the outside world (object distance u ≈ inf) is real, sharp and
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 2
inverted, the lens must be converging one, and the image distance (v) is equal to the focal length
(f) of the lens. Thus f = 0.120m. for the evening experiment, the hole in the card serves as the
object of size 0.005m. The image size on the wall is 0.020m.
Distance between the lens and the card u = - uc (uc > 0)
Distance between the lens and the wall v = + vc (vc > 0)
Distance between the lens and wall (question) d = uc + vc
Magnification = v c /−u c = 0.020 / -0.005 = -4 (real images by converging lenses
always have negative magnification, i., inverted image).
Thus, applying the lens formula
1
v−
1
u=
1
f
1
v c+
1
uc=
1
f
1
4uc+
1
uc=
1
f
5
4uc=
1
f
uc =5f / 4 = 5 x 0.120m/4 = 0.150m
vc = 4 uc – 4 x 0.150m = 0.600m
Therefore, the distance between the lens and the wall is
d = uc + vc = (0.15 0 + 0.600)m = 0.750m
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 3
9 ) Answer D :Solution:
V 1= 300 V t1= 5 min V 2= 200 V t2= 20 min
V 3= 100 V =t3 ? min
Rate of heat supplied by heating container = V
2
R
For first and second container (resistance R is same)
t 1
t 2
=1
4≠
V 1
2
V 2
2=
4
9
Second container should take 2.25 times that of container one but it is taking four times
that of container one. There is some heat loss ( Ploss ) to the environment which is at
lower temperature than ice temperature ( 00
C).
Same amount of heat is being melt in both containers, that means amount of heat supplied
is same.
(V 12
R− P loss)t1=(V 2
2
R− Ploss)t2
which gives Ploss R=5
3×10
4= V loss
2>V 3
2= 10
4
.
Container will loose all heat given by 100 V power supply and ice will cool down instead
of melting.
Additional:
Say outside temp is T 0 and final temperature of ice is T f .
k (0− T 0)= 104×
5
3
k (T f− T0)= 104×1
which gives T f=2
3T 0
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 4
10 ) Answer C:
Force at the bottom : Fb
Weight of water : W
Downward vertical force of curved surface on water is : Fg
Then Fb = W + Fg
=> Fg = Fb - W
=> Pb.A -Vg
Fg =ghA – Vg
Fg =g (hA – V)
= 103 x 10 x [ 0.25 x 3.14 x (0.20)
2/4 – 2.5 x 10
-3] N
= 104 x [7.85 -2.50] x 10
-3 N
= 53.5 N
By 3rd law, Fg = vertical force of water on glass
11 ) Answer B : Fig 2 violates Pauli’s exclusion principle and Hund’s rule.
12 ) Answer C: Solution : Grahams law of diffusion
13 ) Answer A : Frendlich isotherms: (lower the temp higher is the adsorption)
14 ) Answer C: Solution : The conductivity that is measured in an electrolyte solution depends
on the type and concentration of the ions. As long as the reaction is taking its course the
conductivity drops, when the standard solution is in surplus the conductivity rises again
15 ) Answer D: Statements (i) and (iii) are correct
16 ) Answer B:
17 ) Answer B: Iron acts as the cathode and oxygen is reduced.
18 ) Answer A: A) I and III are correct
19 ) Answer D: Solution:
[Pb+2
] = 1 x 10-3
cma
aor
m
m
r
r
NH
HCl
NH
HCl 12.815.36
17
20033
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 5
[Br-] = 5 x 10
-4
Ionic product = [Pb+2
] [2Br-]
2
Ionic product = [1 x 10-3
] [2x5 x 10-4
]2
10-3+2-8
10-9
As Ksp > ionic product, no precipitate will form.
20 ) Answer D: each of the molecules is planar and triangular.
21 )Answer A: Solution: T is present only in DNA, while U is present only in RNA. If A≠T or
U it is single stranded. If A = T it is most likely to be double stranded. Sample 1 could possibly
be single stranded. However, it is not one of the choices in the correct answer (A).
22 ) Answer B: B) (i), (iv) and (v)
Solution:
If trait is dominant then daughter would show the trait as it would inherit the father’s X-
chromosome. Since son shows the trait, X chromosome inherited from the mother carried
the recessive trait (carrier).
As the daughter would inherit the father’s X-chromosome the daughter is bound to be a
carrier.
As the trait is RARE, marriage outside the family ensures that the mother is likely to not
carry the trait. Hence the son will have ‘0’ probability of inheriting the trait as the X-
chromosome will come from the mother.
23 ) Answer B: (i) and (z); (ii) and (y); (iii) and (x)
Solution: It based on analysis. No prior knowledge is needed.
In case of ‘x’ level of T3 and T4 will be low. As T3 and T4 has a negative effect on TRH
levels, there will be an increase in the levels of TR . Thus ‘x’ matches with ‘iii’
In case of ‘y’ T3 and T4 is low, so TRH levels are high (as explained above). As TRH
positively regulates TSH, TSH levels will increase. Thus ‘y’ matches with ‘ii’
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 6
In case of ‘z’ TRH levels are low, thus TSH levels will be low and finally T3 and T4
levels will be low. Thus ‘z’ matches with ‘i’.
24 ) Answer C: C) 1/16
Solution: 14
N: 15
N (start); after 1 cycle two molecules of DNA, 14
N: 15
N and 14
N: 14
N; after 2 cycle
four molecules of DNA, one 14
N: 15
N and three 14
N: 14
N; after 3rd
cycle we will have 8
molecules of DNA, one 14
N: 15
N and rest 7 14
N: 14
N. Finally after the 4th
cycle there will be 16
molecules of DNA of which only one will be hybrid 14
N: 15
N in nature.
25 ) Answer B:
B) Cacti close their stomata during the day, so availability of CO2 for RuBP carboxylase
activity is low during the day.
Solution: To conserve water during the hot days, cacti close stomata during the day. This
reduces the CO2 concentration in the leaf, which cannot support RuBPcase activity.
26 ) Answer D: D) i and iv
Solution: ‘i’ has equal illumination so the seedling does not bend. In ‘iv’ the seedling bends
towards the higher illumination.
27 ) Answer C: C) These organelles have their own genetic material.
Solution: A and C are correct but only C supports the concept of endosymbiosis. B is possible
only for a short duration.
28 ) Answer C: C) T-cells, antibodies and HIV
29 )Answer A: Solution: Population increase under these ideal conditions is called exponential
population growth, giving rise to what is called a J-shaped growth curve. Curve presented in D is
based on logistic model of population growth. B and C are just created as wrong answers.
Comment: Typical growth-curves for populations is a topic mentioned in the syllabus
under Systems – Ecology. See Chapter on Population Ecology in Biology 7th
Edition,
Campbell and Reece for clarification on growth curves.
30 )Answer B: B) Ammonia in tadpole and urea in frog.
Time : 3 hrs Marks
: 30
10 th
International Junior Science Olympiad, Pune, India
Multiple Choice Questions
Page 7
Solution: As ammonia has maximum solubility in water, thus tadpole has evolved the
mechanism to generate ammonia as the waste. On the other hand a land animal would generate
urea or uric acid. Frog makes urea.