IOQ-JS 2021 Question Paper

Paper: Shift I

PART-A-1

Question 1. Six circles each of radius 3cm are inscribed in an equilateral triangle ABC such that they touch each other and also touch the sides of the triangle as shown in the adjacent figure height of triangle ABC is

a. 6(2√3+3)
b. 6(2√3+6)
c. 3(2√3+3)
d. 6(2+√3)

Answer: (c)

ABC is an equilateral triangle.

We join the centres of the circle and form an equilateral triangle DEF with side 12 cm.

Height DG = √ 3/2 X 12

= 6√3

AD ⇒ sin 30° = 3 / X

x = 6 cm

Height of triangle ABC = AD + DG + GH

= 6 + 6 √3 + 3 cm

= 9 + 6 √3

= 3 (3+2√3)

Question 2. Find the remainder when x⁵¹ is divided by x² – 3x + 2

a. X
b. (2⁵¹ – 2)x + 2 – 2⁵¹
c. (2⁵¹ – 1)x + 2 – 2⁵¹
d. 0

Answer: (c)

x⁵¹ = (x² – 3x + 2) g(x) + r(x)

x⁵¹ = (x – 2) (x – 1)g(x) + (ax + b)

degree of r(x) < degree of division

On putting x = 1,

1⁵¹ = a + b

a + b = 1 …..(1)

Now let us put x = 2,

2⁵¹ = 2a + b ……(2)

(2) – (1)

a = 2⁵¹ – 1

Putting the value of ‘a’ in equation (1),

2⁵¹ – 1 + b = 1

b = 2 – 2⁵¹

r (x) = ax +b

= (2⁵¹ – 1) x + (2– 2⁵¹)

Question 3. If

a. 2 < x < 5
b. X < 2 or 5 < x
c. x < -2 or x > 5
d. None of these

Answer: (b)

x∈ (–∞, 2) ∪ (5, ∞)

So, x < 2 or 5 < x

Question 4. If x² + ax + b = 0 and x² + bx + a = 0 have one common root, then

a. a + b = 0
b. a + b = 1
c. a + b = – 1
d. a² + b² = 1

Answer: (c)

x² + ax + b = 0 ……(i)

x² + bx + a = 0 ……(ii)

Both have a common root α.

From equation (ii) and (iii),

α =1

From (i) and (ii),

a + b = 1

Question 5. If 100²⁵ – 25 is written in decimal notations, then the sum of its digits is

a. 444
b. 442
c. 424
d. 422

Answer: (a)

100²⁵ – 25

(10²)²⁵ – 25

10⁵⁰ – 25

Digit sum = 9 × 48 + 7 + 5

= 432 + 12

= 444

Question 6. ABC is a triangle. The bisector of angle A meets BC in D. The relation between AD, AB and AC is

a. AD > √AB ⋅ AC
b. AD > AB.AC
c. AD = √AB ⋅ AC
d. AD < √AB ⋅ AC

Answer: (d)

Thus, BD < CD and D lies to the left of E.

In ΔABC, ∠B > ∠C

∠A + ∠B + ∠C = 180^o

∠A + 2∠C < 180

∠A / 2 < 90^{o .}– ∠C

In ΔAFD and ΔAFE,

AD < AE

ΔABE ≅ ΔGCE

Then, CG = AB and AG = 2AE

From ΔACG,

AB + AC > 2AE > 2AD ⇒ AB + AC / 2 > AD

Now, ar ΔABC = ½ AB. AC sinA

=

Thus, we have

By Simplification,

AB. AC cos A/2 > AD²

AB. AC > AD²

Question 7. If in a wheat mutant, the length of chromosome 1B was found to be 6.7μm, instead of 5.0 μm, approximately how many additional base pairs are incorporated in the mutant chromosome?

a. 0.5 X 10⁴ bp
b. 5 X 10⁴ bp
c. 1.7 X 10⁴ bp
d. 5 .78 X 10⁴ bp

Answer: (a)

Wheat Mutant

Given → Length of chromosome 1B

→ 6.7 μm = new length

→ 5.0 μm = old length

Change in length after mutation = 6.7 – 5.0 μm = 1.7 μm

→ 3.4 Å is the length between any two base pairs.

→ 3.4 × 10^–10 meter

→ 3.4 × 10^–10 × 10⁶μm = 3.4 × 10^–4μm

→ 3.4 × 10^–4μm → 1 Base pair

→ 1μm → 1 / 3.4 X 10^-4 Base Pair

→ 1.7 μm →

=

= 0.5 × 10⁴ Base Pair

→ So, the correct option is a.

Question 8. Considering the following characteristics, identify the correct inheritance pattern from the given options.

The most affected individuals are male.

Affected sons result from the female parents who are either affected or who are known to be carriers because they have affected brothers, fathers or maternal uncles.

Affected daughters are born to affected fathers and either affected or carrier mothers.

The sons of affected mothers should be themselves affected.

Approximately, half the sons of carrier mothers should be affected.

a. Autosomal Recessive Inheritance
b. Autosomal Dominant Inheritance
c. Sex-Linked Recessive Inheritance
d. Sex-Linked Dominant Inheritance

Answer: (c)

X-linked recessive inheritance is a mode of inheritance in which a mutation in a gene on the X chromosome causes the phenotype to be always expressed in males. A male carrying such a mutation will be affected, because he carries only one X chromosome. The following are some of the characteristics of X-linked recessive inheritance:

(1) Males with an X-linked recessive disorder always inherit the disease-associated allele from their mother.

(2) Females (who have two X chromosomes) must have a mutation on both X chromosomes in order to be affected with the condition. If only the father or the mother has the mutated X-linked gene, the daughters are usually not affected and are called carriers because one of their X chromosomes has the mutation but the other one is normal.

(3) Sons will be affected if they inherit the mutated X-linked gene from their mother. Fathers cannot pass X-linked recessive conditions to their sons.

(4)

From the chart, it is observed that half the sons of carrier mothers should be affected. While no sons will be affected if the father is affected.

Hence, the inheritance pattern is X-linked recessive inheritance.

Question 9. The transpiration pull is maximum under which of the following conditions?

a. Closed stomata, low light intensity, humid air
b. Open stomata, dry air, moist soil
c. Open stomata, dry air, dry soil
d. Open stomata, high humidity in the air, moist soil

Answer: (b)

Environmental factors that affect the rate of transpiration are as follows:

a) Stomata – When stomata are open, transpiration rates increase. When they are closed, transpiration rates decrease. Transpiration occurs through the stomatal apertures and can be thought of as a necessary “cost” associated with the opening of the stomata to allow the diffusion of carbon dioxide gas from the air for photosynthesis.

b) Humidity – The rate of diffusion of any substance increases as the difference in concentration of the substances in the two regions increases. When the surrounding air is dry, diffusion of water out of the leaf goes on more rapidly.

c) Soil – Plants cannot continue to transpire without wilting if the soil is very dry because the water in the xylem that moves out through the leaves is not being replaced by the soil water. This condition causes the leaf to lose turgor or firmness and the stomata to close.

Question 10. In a marine ecosystem of phytoplanktons and zooplanktons with a rich diversity of fauna, which of the following images would be a correct representation of a pyramid of biomass?

Answer: (d)

The pyramid of biomass in a marine ecosystem of phytoplanktons and zooplanktons with rich diversity of fauna will be inverted one because of the following reasons:

(i) Small standing crops of phytoplanktons support the large standing crop of zooplanktons.

(ii) A number of generations of phytoplanktons may thus be consumed by a single generation of zooplanktons.

(iii) Biomass of fish may be larger as fishes are large in size with a longer lifespan and a huge number of generations of phytoplanktons and zooplanktons can be consumed by fishes.

Question 11. Curcuma longa, Azadirachta indica, Basmati Rice and Indian Ginseng are all related to which of the following concepts?

a. Bioterrorism
b. Biomagnification
c. Biopiracy
d. Biodegradation

Answer: (c)

Biopiracy is the term used to refer to the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment. Curcuma longa, Azadirachta indica, Basmati Rice and Indian Ginseng are all related to biopiracy.

Most of the industrialised nations are rich financially but poor in biodiversity and traditional knowledge. In contrast, the developing and the underdeveloped world is rich in biodiversity and traditional knowledge related to bio-resources. Traditional knowledge related to bio-resources can be exploited to develop modern applications and can also be used to save time, effort and expenditure during their commercialisation.

In 1997, an American company got patent rights on Basmati rice through the US Patent and Trademark Office. This allowed the company to sell a ‘new’ variety of Basmati, in the US and abroad. This ‘new’ variety of Basmati had actually been derived from Indian farmer’s varieties. Several attempts have also been made to patent uses, products and processes based on Indian traditional herbal medicines, e.g., turmeric (Curcuma longa), neem (Azadirachta indica) and Indian Ginseng.

Bioterrorism- Intentional release of virus, bacteria or other germs. Biomagnification- (Bioamplification) Increase in concentration of toxin, successively higher level in the food chain.

Biodegradation – Naturally occurring breakdown of material by microorganisms such as bacteria and fungi.

Question 12. Read the following criteria carefully.

Slow evolutionary change relative to similar entities.

Gross similarity to an ancestral fossil.

Very low taxonomic richness at present as compared to the past.

Phylogenetic inference of specific characters as plesiomorphic.

Phylogenetic inference of a genealogical divergence between other groups that diverged in the distant past.

Known in the fossil record before being discovered alive.

These criteria can be used to categorize a group of organisms most probably into

a. Connecting links
b. Living fossils
c. Endangered species
d. Extinct species

Answer: (b)

Living fossils have the following characteristics:

Living organisms are members of a taxon that has remained recognizable in the fossil record over an unusually long-time span.

They show little morphological divergence, whether from early members of the lineage, or among extant species (slow evolutionary change relative to the similar entities).

They tend to have little taxonomic diversity (very low taxonomic richness at present as compared to the past).

Exceptionally, little change throughout a long fossil record, gives the impression that the taxon had remained identical through the entire fossil and modern period (known in the fossil record before being discovered live).

Living fossil being a surviving representative of an archaic lineage which retain some primitive features of its ancestral lineage (plesiomorphies)

So, the correct answer is living fossils.

Question 13. Gravitational collapse is the contraction of an astronomical object under its own gravity. This draws the matter inward towards the centre of gravity. A neutron star is an example of the collapsed core of a giant star. A certain neutron star of radius 10 km is of mass 1.5 times mass of sun. The acceleration due to gravity on the surface of the neutron star is nearly

a. 2.0 x 10⁸ m / s²
b. 2.0 x 10¹² m / s²
c. 2.6 x 10¹⁶ m / s²
d. 2.6 x 10²⁰ m / s²

Answer: (b)

Given:

Radius of neutron star, R = 10 km = 10000 m

Mass of neutron star = 1.5 times mass of sun = 1.5 x 1.99 x 10³⁰ kg

To find:

Acceleration due to gravity on the surface of the neutron star, g=?

We know that the gravitational force between two objects is given by,

Where,

m = mass of object

M_n = mass of neutron star

G = universal gravitational constant

R = radius of neutron star

Also, according to Newton’s second law of motion,

F = mg ………….(2)

Here,

g = acceleration due to gravity

From (1) and (2)

⇒ g = 2 × 10¹² m / s²

Question 14. The tympanic membrane (eardrum) is a very delicate component of the human ear. Typically, its diameter is 1 cm. The maximum force that the ear can withstand is 2.5 N. In case, a diver has to enter the seawater of density 1.05 X 10³ kg / m³ without any protective gear, the maximum safe depth for the diver to go into the water is about

a. 12m
b. 9m
c. 3m
d. 1.5 m

Answer: (c)

Given:

Diameter of eardrum, D = 1 cm = 0.01 m

Maximum force which it can withstand, F= 2.5N

Density of seawater, ρ = 1.05 X 10³ kg / m³

To find,

Maximum safe depth for the diver to go into the water, h=?

We know that pressure at depth h inside any static fluid is given by,

P = ρgh

Where ρ is the density of the fluid, g is the acceleration due to gravity.

⇒P = 1.05 X 10³ X 10 X h …………(1)

Also, from formula of pressure,

P = F / A, where F is force and A is area of the surface

From (1) and (2)

Question 15. A nuclear reactor is working at 30 % efficiency (i.e. conversion of nuclear energy to electrical energy). In this reactor,

a. 1.04 X 10¹⁷
b. 6.5 X 10¹²
c. 3.125 X 10¹²
d. 3.25 X 10³²

Answer: (a)

Given: Efficiency, η = 30%

Nuclear energy released per atom, E = 200 MeV

= 20 × 10⁶× 1.6 × 10^–19 J

Electrical power, P = 1000 kW = 10⁶ W

Time, t=1 s

To find: Number of atoms disintegrated (undergone fission) per second in the reactor, n=?

We know that amount of nuclear energy converted to electrical energy per atom is given by,

E_o = efficiency X nuclear energy

E_o = 0.3 X 200 X 10⁶ X 1.6 X 10^-19 = 9.6 X 10^-12 J

So, total energy for n atoms = n X 9.6 X 10^-12 J

From formula of power,

Power = Total energy / time

Question 16. Two illuminated point objects O₁ and O₂ are placed at a distance 24 cm from each other along the principal axis of a thin convex lens of focal length 9 cm such that the images of both the objects are formed at the same position. Then, the respective distances of the lens from O₁ and O₂ (in cm) are

a. 12 and 12
b. 18 and 6
c. 14 and 10
d. 16 and 8

Answer: (b)

Given:

Distance between the objects, d=24 cm

Focal length of a convex lens, f = 9 cm

To find: Distances of the lens from O₁ and O₂ = ?

Let us suppose:

The first object be at a distance x from the convex lens then the second object will be at 24-x distance from the convex lens.

By using the lens formula i.e.,

For first object,

Now for second object,

Given, Image is formed at the same distance from the lens for both the objects.

∴ From equation (1) and (2),

x(24–x) = 24 X 9 / 2= 108

x² – 24x + 108 = 0

From Sridharacharya rule,

x = 18, 6

Therefore, the first object will be at a distance 18 cm or 6 cm from the convex lens and the second object will be at 24 – x = 24 – 18 = 6 cm or 24-x = 24 – 6 = 18 cm distance from the convex lens.

Question 17. Two blocks A and B are in contact with each other and are placed on a frictionless horizontal surface. A force of 90 N is applied horizontally on block A (situation I) and the same force is applied horizontally on block B (situation II). Mass of A is 20 kg and B is 10 kg. Then, the correct statement is

a. Magnitude of force by block A on B is 90 N (situation I) and the magnitude of force by block B on A is also 90 N (situation II).
b. Magnitude of force by block A on B is 30 N (situation I) and the magnitude of force by block B on A is 60 N (situation II).
c. Magnitude of force by block A on B is 60 N (situation I) and the magnitude of force by block B on A is 30 N (situation II).
d. Magnitude of force by block A on B is 45 N (situation I) and the magnitude of force by block B on A is also 45 N (situation II).

Answer: (b)

Given:

Magnitude of force in both situations, F = 90 N

Mass of block A, m_A=20 kg

Mass of block B, m_B=10 kg

Situation I:

On applying ∑ F = ma for the system.

For block B, on applying ∑ F = ma

F_BA = m_Ba

F_BA = 10 X 3 = 30 N

Here, F_BA is a force on block B by block A.

Situation II:

On applying ∑ F = ma for the system.

For block A, on applying ∑ F = ma

F_AB = m_Aa

F_AB = 20 X 3 = 60 N

Here, F_AB is a force on block A by block B.

Question 18. In the adjoining circuit, R = 5 Ω and V_X = 6. The value of R_X will be

a. 4Ω
b. 12Ω
c. 16Ω
d. 20Ω

Answer: (a)

Given:

Value of R = 5Ω

Value of potential across R_x, V_X = 6

To find: Value of R_x = ?

According to the question,

On reducing the circuit,

⇒

⇒

⇒

⇒

So, R_eq = 8R/5 + R_x=8+R_x

According to the Ohm’s law,

V = IR

Now, V_x = I × R_x

⇒48 +6R_x = 18r_x

⇒ 12R_x = 48

⇒ R_x= 4Ω

Question 19. In a process of waterproofing, a fabric is exposed to (CH₃)₂ SiCl₂ vapors. It reacts with the hydroxyl groups on the surface of the fabric or with traces of water to form a waterproofing film [(CH₃)₂ SiO]_n, through the following reaction;

n(CH₃)₂ SiCl₂ + 2nOH^– → 2Ncl^– + nH₂O + [(CH₃)₂ SiO]_n

Where, n stands for a large integer.

The waterproofing film is deposited on the fabric, layer upon layer. Each layer is 6Å thick [the thickness of the [(CH₃)₂SiO] group. How much (CH₃)₂ SiCl₂ is needed to waterproof one side of a piece of fabric, 1m by 2m, with a film 300 layers thick? The density of the film is 1.0 g/cm³.

a. 0.63g
b. 0.36g
c. 6.3g
d. 3.6g

Answer: (b)

Given:

Thickness of each layer of film, t = 6Å = 6 X 10^-10 m

Surface dimensions of fabric to be waterproofed = 1m by 2m

Number of film layers, n = 300

Density of film, ρ = 1.0 g / cm³ = 1000 kg / m³

To find: Amount of (CH₃)₂ SiCl₂needed for waterproofing, m = ?

Volume of a single film,

V = L X B X H

⇒ V = 1 X 2 X 6 X 10^-10= 12 X 10^-10 m³

So, volume of 300 layers,

Total volume of film, V_t = 36 x 10^-8 m³

We know that density is given by,

Density = total mass / total volume

⇒ total mass, m = ρ X V_t

⇒ M = 1000 X 36 X 10^-8

⇒ M = 0.36 g

Question 20. Given that at a certain temperature, in a 1.5L vessel, 5.0 mole of A, 7.0 mole of B and 0.1 mole of C are present. Then, the value of equilibrium constant for the reaction A + B ⇌ 2C + Heat is about

a. 7.22 x 10^-5
b. 2.31×10^–4
c. 7.22×10^–4
d. 6.11 x 10^-5

Answer: (b)

1.5 L

n_A = 5 mol

n_B = 7 mole

n_C = 0.1 mol

A + B ⇌ 2C

K_C = 2.86 × 10^–4

Choose the closest option.

Hence, option (b) is correct i.e., 2.31 × 10^–4.

Question 21. An alcohol (A) on dehydration with conc. H₂SO₄ at a high temperature yields compound (B). On ozonolysis, every molecule of compound (B) yields two molecules of acetaldehyde. Which of the following is the starting alcohol (A)?

a. 1 – butanol
b. 2 – butanol
c. Propanol
d. 2 -propanol

Answer: (b)

Question 22. In an experiment with 100 mL 0.1 M solution of Copper Chloride, by mistake
5 gms of a mixture containing equal weights of Tin, Silver, Lead and Calcium, was added. Finally, after some time, the solution gets completely decolorized. This is mainly due to:

a. Silver reacts with Copper Chloride.
b. Calcium reacts with Copper Chloride.
c. All the metals react with Copper Chloride.
d. Only Lead reacts with Copper Chloride Forming white precipitate of lead chloride.

Answer: (b)

Given: CuCl₂ → 100 ml 0.1 M

5g mixture of equal weight of Sn, Ag, Ca and Pb.

Wt of Ca in mixture = 5 / 4 = 1.25 g

Moles of Ca = 1.25 / 40 = 0.03125 mol

According to the electrochemical series, reactivity order of metals is

Ca > Sn > Pb > Cu > Ag

We know that M =

Moles of CuCl₂ = 0.01 mol

Here, Ca has the highest reactivity. So, it reacts firstly with CuCl₂ and forms CaCl₂ compound as follows;

Hence, due to the formation of CaCl₂ colour of solution decolorized.

Question 23. Suppose that A and B forms compound B₂A₃ and B₂A. If 0.05 mole of B₂A₃ weighs 12 g and 0.1 mole of B₂A weighs 10 g, what are the atomic weights of A and B respectively?

a. 70 and 25
b. 50 and 20
c. 40 and 30
d. 30 and 40

Answer: Bonus

Given:

We know, mole = weight / molecular weight

Molecular weight = weight / mole

2B + 3A = 240 ..…(i)

2B + A = 100 …(ii)

Subtracting equation (ii) from equation (i),

Question 24. Triclosan (C₁₂H₇Cl₃O₂) is an antibacterial and antifungal agent. It is a polychlorophenoxy phenol. It is widely used as a preservative and antimicrobial agent in personal care products such as soaps, skin crasms, and deodorants etc. A label on a 200 mL hand sanitizer bottle claims that it contains Triclosan 0.2% w/v. What will be the number of molecules of Triclosan present in the bottle? (N_A is Avogadro’s Number)

(A) 1.4 × 10²⁵ N_A

(B) 1.4 × 10²⁴ N_A

(C) 1.4 × 10²³ N_A

(D) 1.4 × 10²² N_A

Sol.

Bonus

Given:

200 mL → 0.2 % w/V

In 100 mL weight of triclosan = 0.2 g

So, in 200 mL → 0.4g

Weight of Triclosan (C₁₂H₇Cl₃O₂) = 0.4 g

Mole = Weight / Molecular weight = 0.4 / 289.5

Mole = 0.001382 = 0.0014 = 1.4 ×10^–3

Mole = Number of molecules / N_A

No. of molecules = 1.4×10^–3 × N_A

Part-A-2 Q.25 to Q.32 (MCQ)

Question 26. Given (a – b)² + (a – c)² = (b – c)², then which of the following statements are true?

a. Equation is valid when b = c and a ≠ c
b. Equation is valid when a = b
c. Equation is valid when a = c
d. Given equation is not valid when a, b and c are distinct.

Answer: (b,c,d)

(a – b)² + (a – c)² = (b – c)²

(A) a b = c, a ≠ c

(a – c)² + (a –c)² = 0

2(a –c) = 0

a = c (Not Follow)

(B) a = b

(a –a)² + (a – c)² = (a – c)²

a = c

(C) a = c

(a – b)² + (a –a)² = (b –a)²

(a –b)² = (b – a)²

a – b = b – a

2a = 2b

a = b

(D) We can say from option (B) and (C) that option (D) is also correct.

Question 27. Choose the correct statement(s) from the following options.

a. A robust adaptive immune response is initiated using a weakened form of the bacterium known as live attenuated vaccines.
b. Administration of a killed or chemically inactivated virus can trigger a weaker adaptive immune response, but can be strengthened with booster doses.
c. A conjugate or multivalent component always reduces immunogenicity of the vaccine.
d. Inclusion of alum, cytokines and/or lipids always reduces the immune response to a vaccine.

Answer: (a and b)

Live vaccines use a weakened (or attenuated) form of the germ that causes a disease. Because these vaccines are so similar to the natural infection that they create a strong and long-lasting immune response. Just 1 or 2 doses of most live vaccines can give you a lifetime of protection against a germ and the disease it causes. Live vaccines are used to protect against Measles, mumps, rubella (MMR combined vaccine); Rotavirus; Yellow fever. That makes the option “a” the correct statement.

A killed or chemically inactivated vaccine uses the killed version of the germ that causes a disease. Inactivated vaccines usually don’t provide immunity (protection) that’s as strong as live vaccines. So, you may need several doses over time (booster shots) in order to get ongoing immunity against diseases. Inactivated vaccines are used to protect against; Hepatitis A; Rabies. So, the statement given in option b is correct.

A conjugate or multivalent vaccine use specific pieces of the germ (conjugate) — like its protein, sugar, or capsid (a casing around the germ). Because these vaccines use only specific pieces of the germ, they give a very strong immune response, i.e., a high immunogenicity, that’s targeted to key parts of the germ. So, a multivalent component is added to achieve a strong immune system response because it increases the immunogenicity of vaccines. So, option c is an incorrect statement.

Inclusion of alum, cytokines or lipids does not always reduce the immune response to a vaccine. In fact, alum, cytokines and lipid are adjuvants (a preparation that enhances the immunogenicity of an antigen) most often used in vaccines in humans. Hence, all of them have an effect on immune response to a vaccine (weakened/ attenuated or killed version of the germ/ antigen that causes a disease), it may be enhanced or suppressed but not always reduce it that makes the statement of option d incorrect.

Question 28. The minimum energy required to exist that is the energy required to perform chemical reactions even when a person is at rest is called the basal metabolic rate (BMR), which accounts for about 50 to 70% of the daily energy expenditure in most sedentary individuals. It is influenced by many factors. Some statements are made about these factors.

Choose the correct statements from the following options.

a. The Thyroid hormone decreases the metabolic rate.
b. The growth hormone increases the metabolic rate.
c. Fever decreases the metabolic rate.
d. Malnutrition decreases the metabolic rate.

Answer: (b and d)

Option (b) – correct.

Pituitary growth hormone (GH) is a peptide hormone predominantly secreted by the anterior pituitary. GH plays critical roles in regulating somatic growth and the metabolism of carbohydrates, lipids, and protein. This results in the increase of metabolic rate, making option b the correct statement.

Prolonged periods of abnormal nutrition cause an adaptive change in BMR. In prolonged malnutrition, the BMR declines, while in prolonged overnutrition, the BMR is increased. So, option d is correct that malnutrition decreases the metabolic rates.

Thyroid hormones have a marked effect on BMR, since thyroid hormones regulate the rate of cellular metabolism. Hyperthyroidism in which there is an increase in the production of thyroid hormones leads to a high BMR, while hypothyroidism in which thyroid hormones are depleted causing a low BMR. So, thyroid hormone increases BMR, making option “a” an incorrect statement.

Fever is a common pathological cause for a high BMR, since a rise in body temperature increases the rate of cellular metabolic reactions that makes option c incorrect.

Question 29. Infinitely long conductor when carrying current I produces a magnetic field B around it. The magnitude of B at a distance r is given by the relation

Following figure shows such an infinitely long conductor placed along the X-axis carrying current I and magnetic field B at point S is 2 × 10^-4 T, directed into the plane of the paper. Given r = 1 cm. Then, the correct statement(s) is/are

a. I = 10 A
b. Number of electrons transported across the cross section of the conductor during 1 s is 6.25 × 10¹⁹
c. Direction of current is from x₂ to x₁
d. Electrons are flowing from x₂ to x₁

Answer: (a,b,c)

Solution

Given:

Magnitude of magnetic field at point S, B=2 × 10^-4 T

Distance of point S from the x-axis is, r= 1 cm=10^-2 m

Value constant,

Also, it is given that,

Substituting the values, we get

2 × 10^–4 = 10⁷ x 21 / 10^-2

⇒ I = 10 A

∴ Option (A) is correct.

From the definition of current, I = Charge (q) / time (t)

⇒ q = I × t = 10 × 1 = 10 C

From quantization of charge, q = ne, where n is number of electrons and e is the charge on one electron.

⇒ n = q / e

⇒ n = 10 / 1.6 x 10^-19 = 6.25 × 10¹⁹

Option (b) is correct.

From the right-hand thumb rule, as the magnetic field is directed into the plane of paper, therefore, the direction of current is from x₂ to x₁.

∴ Option (c) is correct.

We know that electrons flow opposite to the direction of the current, therefore the direction of electrons flow is from x₁ to x₂.

∴ Option (d) is incorrect.

Option (a, b, c) are correct.

Question 30. The ratio of the charge of an ion or subatomic particle to its mass (q/m) is called a specific charge. Then, the correct option(s) is/are

a. SI unit of specific charge can be written as A-s / kg
b. If all the isotopes of hydrogen are ionized then tritium will have the least specific charge among them
c. Specific charge of an α-particle will be greater than that of an electron
d. Specific charge of an electron is 1.75 × 10¹¹ C/ kg

Answer: (a,b,d)

Given,

Specific charge = charge/mass

We know that, current=charge/time. As a unit of current is A and of time is s, so, the unit of charge is A-s. Also, the unit of mass is kg, therefore the unit of specific charge is A-s/kg.

Option (A) is correct.

Ionized isotopes of hydrogen: (₁H¹)⁺, (₁H²)⁺, (₁H³)⁺,

The charge is the same for all the ionized isotopes and mass is highest for tritium. Thus, the specific charge of tritium will be lowest as specific charge is inversely proportional to mass.

Option (b) is correct.

Since the mass of an electron is very less than the mass of an alpha particle, its specific charge will be greater than alpha particle as a specific charge is inversely proportional to mass.

Option (c) is incorrect.

Charge of an electron 1.6 x 10^-19 C and mass is 9.1 x 10^-31 kg, hence specific charge q/m is 1.75 x 10^-11 C/kg

Option (d) is correct.

Question 31. Acetylene torches and burners used by glassblowers produce intense ultraviolet light. Glassblowers wear special glasses that contain which of the following elements to absorb the UV?

a. Neodymium
b. Praseodymium
c. Cerium
d. Didymium

Answer: (d)

At almost 1600°C, our furnaces and these torches give off harmful UV rays that could permanently damage any part of the human eye.

Didymium works by filtering yellowish light that is closely related to UV rays. This is because didymium does not absorb all visible light.

Question 32. Equal lengths of magnesium ribbons are taken in four test tubes A, B, C and D. In test tube A, 1M acetic acid is added; in test tube B, 1M HCl is added; in test tube C, 1M HNO₃ is added; and in test tube D, 1M NaOH is added. The observed results will be:

a. The fizzing occurs more vigorously in A.
b. The fizzing occurs more vigorously in B.
c. The fizzing occurs more vigorously in C.
d. The fizzing occurs more vigorously in D.

Answer: (b,c)

Both HCl and HNO₃ are strong acids. So, they vigorously react with Mg ribbon and produce H_{2 (}↑). Hence, options (b) and (c) are correct.

IOQ-JS 2021 Shift II Question Paper

SHIFT – II

Section A

Question 1. The autonomous nervous system regulates involuntary functions of the body and can be subdivided into the sympathetic and the parasympathetic nervous system. Both of these systems control the same group of body functions but have opposite effects on the functions they regulate. The sympathetic nervous system prepares the body for intense physical activity like the fight-or-flight response. The parasympathetic nervous system has the opposite effect and relaxes the body and inhibits or slows many high energy functions. Which of the following involuntary effects in the body are brought about by the sympathetic nervous system during a fight-or-flight situation?

i. Increased salivation

ii. Increased digestion

iii. Loss of bowel and bladder control

iv. Body shivering

v. Crying

vi. Pupil dilation

a. i, ii and vi
b. i, iv and v
c. iii, iv and vi
d. iii and v

Answer: (c)

Humans protect themselves whenever they sense a threat. Their basic impulse to protect themselves is automatic and unconscious. Amygdala (a part of the limbic system; plays an important role in emotion and behaviour) is responsible for detecting fear and preparing our body for an emergency response. It triggers an alarm by sending signals to the hypothalamus. After which hypothalamus activates the sympathetic nervous system by sending signals via the autonomic nerves to the adrenal glands. These glands release stress hormones that are adrenaline and cortisol in the blood. These hormones immediately prepare us for fight or flight and allow the mind to be as open as possible.

Some immediate changes occur and these sensations are not exactly pleasant as they are not meant for relaxation. They are designed to move us to action.

Some of these changes are: –

1. Salivation– Increased total protein concentration after short-term acute stress changes the chemical properties of saliva, such as the adhesion or lubrication of oral surfaces and salivary production can be decreased

2. Digestion – Activity in the digestive system also decreases because food is not as important as fighting off danger or fleeing to safety. The energy needed to digest food is therefore used for more immediate survival purposes like providing glucose to the brain.

3. Loss of bowel and bladder control – Urination occurs when our bladders become full. The brain can stimulate the desire to pee by sending an inhibitory signal to the brainstem (control bladder). When we become stressed or anxious, electrical signals from the limbic system (signals of fear) become so intense that the brainstem has trouble following the brain’s commands to hold urine. That’s why many people urinate more frequently during stress so the body loses bladder control.

4. Body shivering- Strong emotions can cause a person to shake or shiver. Nerve impulses are sent by the hypothalamus to the skeletal muscles to bring about rapid contractions that generate heat. 

5. Crying – Crying is a response to the stress experienced by the sympathetic nervous system.

6. Pupil dilation- In an emergency situation, the brain wants complete information of surrounding to get out of the situation so pupils open up, or dilate, to let in more light.

By reading the above lines, iii, iv and vi statements are correct.

Question 2. When a person starts exercising, many body parameters change from the original state of rest. The trends in two such parameters are shown in the graph during the initial phase of exercise.

P and Q most likely represent:

a. P: carbon dioxide level in vein Q: oxygen level in artery
b. P: breathing rate Q: carbon dioxide level in artery
c. P: oxygen level in artery Q: carbon dioxide level in vein
d. P: oxygen level in artery Q: oxygen level in vein

Answer: (d)

‘P’ is the oxygen level in the artery – Oxygen saturation in the blood is 96–98% regardless, breathing is fast or slow or shallow. Our atmosphere has 21% oxygen but we breathe only around 5% of it. The cell only uses around one-fourth of the oxygen we breathe. That means we use only 1.25% of the oxygen available in the air we breathe. Even in the feeling of breathlessness, there is more than enough oxygen.

‘Q’ oxygen level in vein – More oxygen is used in respiration in body tissue during exercise. Hence, the level of oxygen will reduce in the vein.

Question 3. Descriptions of four biological samples (I – IV) are given below.

I: Can be viewed using a light microscope with a total magnification of 1000X; possesses cell wall and does not possess mitochondria

II: Can be seen using a light microscope with a total magnification 100X; possesses a cell wall and has a nucleus.

III: Needs electron microscope for viewing; can be found attached to the membrane system in the cytoplasm.

IV: Needs electron microscope for viewing; cannot replicate on its own, needs other specific cells for replication.

I, II, III and IV respectively represent:

a. virus; plant cell; ribosome; bacteria
b. plant cell; bacteria; vacuole; virus
c. bacteria; plant cell; ribosome; virus
d. bacteria; protist; plant cell vacuole; mitochondria

Answer: (c)

I) Bacteria – At 1000x magnification we will be able to see 0.180mm, or 180 microns. Bacteria are too small to see without the aid of a microscope. While some eukaryotes, such as protozoa, algae and yeast, can be seen at magnifications of 200X-400X, most bacteria can only be seen with 1000X magnification.

Bacteria cells have ribosomes and a cell wall, but they don’t have organelles such as nuclei, mitochondria or chloroplasts. Bacteria cells do have a cytoplasm and cell membrane. One of the key structures of a bacteria cell is the plasmid.

II) Plant cell – Using a light microscope of 100X magnification power, we can view cell walls, vacuoles, cytoplasm, chloroplasts, nucleus and cell membrane of plant cell.

Plant cells have a cytoplasm, cell membrane, mitochondria and nucleus which all perform the same functions as animal cells.

III) Ribosome – A ribosome is a cell organelle which can be seen only through electron microscope and cannot be seen with a light microscope. It functions as a micro-machine for making proteins. Ribosomes are found ‘free’ in the cytoplasm or bound to the membrane system i.e., the endoplasmic reticulum (ER) to form rough ER.

IV) Virus – Viruses are very small and most of them can be seen only by electron microscopy. A virus is a small collection of genetic code, either DNA or RNA, surrounded by a protein coat. A virus cannot replicate alone. Viruses must infect cells and use components of the host cell to make copies of themselves. Often, they kill the host cell in the process and cause damage to the host organism.

Hence, option (c) is correct.

Question 4. Raja’s mother collects all the kitchen wastes every day and puts it in a pot. She then adds a few cut pieces of old papers, a spoonful of sour buttermilk and some soil. She covers the pots and keeps it aside with intermittent mixing. After several days, it turns into a nutrient-rich compost to grow plants. In the context of decomposition in this composting process, the most appropriate statement among the following is:

a. Paper acts as a good source of carbon while buttermilk gives the correct acidity to the mixture
b. Soil acts as a good source of inorganic nitrogen while buttermilk is a good source of proteins
c. Paper is a good source of carbon while buttermilk is a good source of starter bacteria
d. Paper is a good source of fibre while buttermilk is a good source of fat

Answer: (c)

Solid wastes are discarded solid materials which are produced due to various human activities.

(1) Kitchen waste – Vegetables and fruit peelings are the number one food remnants of kitchen containing some living tissues of plants and nutrients of plants and animals.

(2) Cut pieces of old papers – Paper is a valuable material for composting because it’s a great source of carbon. Paper is made mostly out of organic compounds: that is carbon, hydrogen and oxygen (C, H and O).

(3) Spoonful of sour buttermilk – Buttermilk consists mostly of water, the milk sugar lactose, and the milk protein casein and many bacteria. These bacteria are a good source of starter bacteria that make compost.

(4) Soil – Soil is a major source of nutrients needed by plants for growth. The three main nutrients are nitrogen (N), phosphorus (P) and potassium (K). Together they make up the trio known as NPK.

By reading the above lines, it can be said that statement (c) is correct.

Question 5. A girl G walks into a room along the path shown by the dashed line (see figure). She tries to observe images of small toys numbered 1, 2, and 3 in the plane mirror on the wall.

The order in which she will see images of the toys is:

a. 3, 2, 1
b. 3, 2
c. 1, 2, 3
d. 2, 3

Answer: (d)

The image of the 1st toy will never be visible to the girl. According to the concept of the field of view, it is clear that the image of the 2^nd toy will appear earlier than the 3^rd toy.

So, option (d) is correct

Question 6. A heating element in the form of a wire with a uniform circular cross-sectional area has a resistance of 310 Ω and can bear a maximum current of 5A. The wire can be cut into pieces of equal length. The number of pieces, arranged suitably, so as to draw maximum power when connected to a constant voltage of 220 V, is

a. 7
b. 8
c. 44
d. 62

Answer: (a)

Data given:

Resistance of wire, R = 310Ω

Maximum current, I = 5A

Voltage, V = 220 V

To find: Number of pieces in which wire can be cut and arranged suitably for maximum power n=?

According to Ohm’s law,

V = I_eq

⇒ 220 = 5R_eq

⇒ R = 44Ω

It is given that the wire can bear a maximum current of 5A. So, the equivalent resistance of arrangement of pieces of wire can’t be less than R_eq= 44Ω.

We know that for maximum power, all pieces should be connected in parallel. The equivalent resistance of the parallel combination of an identical resistor having resistance R is given by,

R_eq = R / n

⇒ 44 = 310 / n

⇒ n = 7

So, option (a) is correct.

Question 7. Consider the following two statements:

S1: If you put 100g ice at 0 °C and 100g water at 0 °C into a freezer, which is maintained at –10 °C, the ice will eventually lose a larger amount of heat.

S2: At 0 °C, water is denser than ice.

Choose the correct statement among the following.

a. Both S1 and S2 are true and S2 is the correct explanation of S1.
b. Both S1 and S2 are true but S2 is not the correct explanation of S1.
c. S1 is true but S2 is false.
d. S1 is false but S2 is true.

Answer: (d)

Given:

Mass of ice, m_i= 100g

Initial temperature of ice, T_i = 0 °C

Mass of water, m_w= 100g

Initial temperature of water, T_w = 0 °C

Temperature of freezer, T= –10 °C

The process of conversion of state of water is

Water (0 ^oC) → ice (0 ^oC) + Q₁ → ice (-10 ^oC) + Q₂

Total heat lost by water to form ice at –10 °C

Q_water = Q₂ + Q₁

Q_water = m_wS∆T + m_wL ,

where, S = specific heat capacity of ice = ½ cal g^{-1 o}C^-1

L = latent heat capacity of ice = 80 cal g^-1

m = mass

Q_water = 100 x ½ x 10 + 100 x 80

Q_water = 8500 cal

Similarly, for ice,

ice (0 ^oC) → ice (-10 ^oC) + Q_ice

Total heat lost by ice to decrease its temperature to –10 °C from 0 °C

Q_ice = mS ∆T

Q_water = 100 x ½ x 10 x 100 = 500 cal

As Q_ice < Q_water , hence statement 1 is false.

At 0 °C, ice is less dense than water because the orientation of hydrogen bonds causes molecules to push further apart, which lowers the density of ice.

So, statement 2 is true.

Question 8. Consider the paths of (1) Halley’s comet near the sun and (2) an alpha particle scattered by a nucleus. In the figures given below, the dots represent the Sun/Nuclei, and the curves with arrows mark the paths of the comet/alpha particles schematically.

The correct statement about the trajectories is:

a. I represent the trajectory for Halley’s Comet and II for the scattering of the alpha particles.
b. III represents the trajectory for Halley’s Comet and II for the scattering of the alpha particles.
c. II represents the trajectory for Halley’s Comet and I for the scattering of alpha particles.
d. II represents the trajectory for Halley’s Comet and III for the scattering of alpha particles.

Answer: (d)

The path of the comet will be elliptical. So, the trajectory will be II according to the figure.

The alpha particle will be repelled by the nucleus since both are of the same nature (positive). So, its trajectory will be III according to the figure. Therefore, option (d) is correct.

Question 9. When water changes phase from liquid to vapour, some bonds are broken. The correct statement relating to this change is:

a. New bonds are formed between nearby H / H and O / O while H – O bonds break.
b. Hydrogen bonds between H₂O molecules are broken.
c. Covalent bonds existing within the H₂O molecules are broken.
d. Ionic bonds existing between H⁺ ions and OH^– ions are broken.

Answer: (b)

Liquid water contains H-bond, when we heat the water (i.e. change it from liquid to vapour), H-bonds break between H₂O molecules and it gets vaporized as H₂O(g).

Question 10. Jyoti was asked by her mother to add a pinch of potassium permanganate to water in a container to disinfect it. As she added the crystals and observed the changes in water, the phenomena of diffusion came to her mind. She wrote the following statements. Identify the statement made by Jyoti that is incorrect.

a. When the entire liquid is of uniform color no further diffusion can be observed.
b. The diffusion gets completed almost instantaneously.
c. Diffusion will take place slower if the water is colder.
d. Maximum color in liquid originates from the bottom of the flask.

Answer: (b)

Diffusion will be time consuming and cannot be completed almost instantaneously because dissolution of crystals will take time.

Question 11. Ramen collected rain water and measured its electrical conductivity. He boiled the water for a few minutes. Then he covered the container and allowed the water to cool to room temperature. Electrical conductivity of water now measured was lower than that measured before boiling. The reason for this most likely is:

a. Precipitation of CaCO₃ from the water during boiling.
b. Removal of dissolved oxygen from the water.
c. Removal of dissolved carbon dioxide from the water.
d. Reaction of cationic species in the water with atmospheric oxygen.

Answer: (c)

Rain water contains oxides of carbon, sulphur like CO₂ and SO₂ dissolved in it. On boiling, these gases escape out. So, the electric conductivity of water gets lowered after boiling.

Question 12. Consider a setup in which two graphite rods are immersed in a 2M NaCl(aq.) solution. The rods are connected to two terminals of a 9V battery with a bulb in series as shown in the figure. Of the following, the change that will NOT be observed when the circuit is closed for a few minutes is:

a. The bulb will glow.
b. The pH of solution near the cathode will increase.
c. Oxygen gas would be generated near the +ve electrode which will oxidize the graphite electrode.
d. Total mass of liquid in the beaker will decrease.

Answer: (c)

Due to the flow of electrons, current is generated so that the bulb will glow.

Cathode – 2H₂O(ℓ) + 2e^– → H₂(g) + 2OH^– (aq)

Anode – 2Cl^–(aq) → Cl₂(g) +2e^–

Overall reaction – 2H₂O + 2Cl^–(aq) → H₂ + Cl₂ + 2OH^–(aq)

Na⁺ + OH^– → NaOH, pH > 7

At room temperature, graphite electrodes cannot be oxidized by O₂.

Question 14. Iodine, an essential element for humans, is naturally present in some marine fishes, plants and ecosystems at large. Solubility of elemental iodine in water is negligible but is high in non-polar organic solvents. The most common form of iodine used in the diet of humans and animals is potassium iodide (KI), a white solid powder at room temperature, which is highly soluble in water. (12 marks)

14.1. In a chemistry laboratory period, 36 students of a class had to perform the following tests.

i. 0.5 gram KI is dissolved in about 5cm³ distilled water. A drop of this solution is put on a moist pH paper.

ii. 0.5 gram KI is dissolved in about 5cm³ distilled water. Part of this solution is mixed with lead (II) nitrate solution. The colour changes in the mixture are observed.

iii. 0.5 gram KI is put in a test tube containing about 5cm³ distilled water. Then they are to observe whether the test tube becomes hot or cool on mixing.

In test ii, a yellow precipitate is observed In test iii, the test tube becomes colder as KI dissolves.

(a) Identify the colour imparted on pH paper in test i.

(b) Being very expensive, KI should be economically used. What is the minimum amount of KI (in grams) required for the complete class for carrying out the above three tests procedures?

Write the necessary calculations/reasoning needed to arrive at your answer.

14.2 An aqueous solution of KI treated with an acidified solution of hydrogen peroxide (in sulphuric acid) gives a precipitate of Iodine crystals.

(a) Write the balanced molecular equation for the reaction.

(b) Identify the reducing agent in the reaction.

(c) The most appropriate option to separate iodine from the above mixture is:

(i) Filtration

(ii) Distillation

(iii) Steam distillation

(iv) Chromatography

(v) Using a magnet

14.3 When solid KI is heated in an open dry test tube, a gas is liberated from the test tube.

(a) What is the colour of the gas?

(b) After the gas evolution stops, what remains in the test tube? Write its chemical symbol/formula (if mixture, write formulae of components) and its state (solid/liquid).

(c) The reaction can be classified as (identify the correct option(s)):

(i) thermal combination

(ii) thermal decomposition

(iii) double displacement

(iv) displacement reaction

14.4 Tincture iodine is an antiseptic, also effective in inactivating the novel coronavirus. It is prepared by dissolving 20 g of Iodine and 25 g of KI in 500mL alcohol and then adding distilled water to make the volume 100mL. In this process, iodine combines with I^– to produce I^–3 species. Sumit and Rekha were separately preparing tincture iodine using the above procedure. Sumit was working hurriedly, as he wanted to join a birthday party. By mistake, he added carbon tetrachloride in the flask instead of alcohol.

At the end of the procedure, two immiscible liquid layers appeared in his flask. Sumit shook the flask vigorously and kept it for some time. The two layers remained separate. He observed that the lower layer was strongly coloured, while the upper layer had a faint colour different from the lower layer. Rekha followed the protocol perfectly and got a homogenous mixture.

Identify the compositions of the top and the bottom layers in Sumit’s flask.

Answer: (14.1 – 14.4)

14.1 Solutions

(a)

Hence, pH of the solution = 7 and colour of the pH paper will be green.

(b) Since KI is expensive, on preparing the solution, test III can be performed initially followed by test I and then test II, so the minimum amount of KI per student = 0.5 g.

Total number of students = 36

∴ Total amount of KI = 36 × 0.5 = 18g

14.2 Solutions

(a) 2KI + H₂O₂ + H₂SO₄ → I₂ + K₂SO₄ + 2H₂O

(b)

Hence, KI is a reducing agent.

(c) Filtration

14.3 Solutions

(b) Potassium remains in test tube

Symbol → K(s)

(c) ∵ Here, first we heat then liberation of I₂ occurs. Hence, it is a thermal decomposition reaction.

14.4 Solutions

Reaction of KI with I₂ follow as

KI + I₂ → I₃^O + K⁺ (I₂ → Non-polar)

Here, I₃^O & K⁺ are ionic or polar in nature.

So, KI dissolved only in water.

⇒ After addition of CCl₄, two layers are formed due to density difference.

But when we are vigorously stirring the solution, I₂ (non-polar) gets dissolved in CCl₄ (non-polar) and I₃^O (polar) in H₂O (polar) due to this we get two layers of different colour as follows –

Question 15. Flame is a hot bright stream of burning gases, Flames have different structures and properties depending on fuel and burning conditions. The attached figure (drawn approximately to scale) shows a candle flame burning in open air in which three regions are distinctly visible surrounding a dark zone: an innermost zone that is pale yellow in colour, surrounded by a red zone, with a bluish envelope at the outside. Point 1–6 represents different locations in the inside and surrounding region of the flame. Consider wax to have chemical formula C₂₄H₅₀.

15. 1. Among points 1-6, identify

(a) the hottest point.

(b) the coldest point.

(c) the point where water vapour concentration is the highest.

15.2. From the following list, identify two substances that are present at point 3 but not at point 6. Also, write chemical equations for the reactions causing removal of these substances.

List: Oxygen, Nitrogen, Carbon, Wax, Carbon dioxide, Carbon monoxide, Water.

15.3. The space at point 2 prominently has (identify the correct option):

a. only air.

b. air with freshly evaporating wax vapour.

c. air with extra carbon dioxide released from combustion.

d. oxygen rich air (as oxygen concentration has locally increased due to diffusion).

15.4 Another flame used in laboratories is produced from Bunsen burner. It is used for heating, combustion, sterilisation process etc. By adjusting the ratio of gas (fuel) and air in Bunsen burner, it is possible to get a stable blue flame, which is largely non-luminous. Shlok was given two different organic compounds: naphthalene (C₁₀H₈) and citric acid (C₆H₈O₇). He burned 1.0g of each compound separately in a porcelain piece in a blue Bunsen burner flame.

For which of the two compounds, the flame would emit more yellow light? Write the reason for your answer, along with the necessary supporting calculations/arguments.

Answer: (15.1 to 15.4)

15.1. (a) The hottest point–5, because of the complete combustion in the outermost layer.

(b) The coldest point-2 because of no combustion in the dark zone.

(c) At point-6

15.2. Carbon and carbon monoxide

15.3. Ans (d). Oxygen rich air (as oxygen concentration has locally increased due to diffusion)

15.4. Naphthalene (C₁₀H₈) will give more yellow light because it is an unsaturated hydrocarbon and due to the presence of more percentage of carbon, it will produce sooty yellow coloured flame.

Question 16. A famous experiment performed by Tolman and Honzik (in 1930) in which they studied the behaviour of rats in a complex maze for a period of 17 days as shown in the figure. The rats had to find their way around the maze once every day. All rats were healthy and were given regular meals throughout the experiment. The rats were divided into 3 groups, which were treated as follows on reaching the end of the maze. (8 marks)

Group I: Day 1–17: Every time the rats reached the end, they were given additional food

Group 2: Day 1–10: Every time the rats reached the end, they were removed from the maze.

Day 11–17: Every time the rats reached the end, they were given additional food

Group 3: Day 1–17: Every time the rats reached the end, they were removed from the maze.

The average number of errors (any deviation from the shortest correct path to reach the end) observed for each group of rats is shown in the graph below.

16.1 A few statements are listed below. Based on the results of the experiment, identify each of the statements as True or False.

(a) Rats need good nutritional status to perform well in the maze.

(b) Result shows characteristic stimulus (maze) – response (reaching the end) behaviour which is genetically determined and hence not changeable.

(c) The find the end of the maze is by trial and error method and not due to learning.

(d) Rewarding the rats has improved the end results.

(e) There was active learning happening in rats in group 2 even before day 11.

16.2 What response can be expected if the rats in the group I were kept hungry before the experiment? Assume that all other conditions in the above experiment setup remain the same. Choose the most appropriate option from the choices below and justify your choice based on the experimental observations presented above (only). Also, give reasons for rejecting the other three options.

(A) Overall rise of line 1 above line 3.

(B) Increase in errors as the experiment proceeds.

(C) Steeper decreases in line 1 in less time.

(D) Same response as line 3 in the graph.

Answer: (16 to 16.2)

Answer 16: Simplified version of given scenario: In this question, the whole experiment is about latent learning. It is a type of learning which is not apparent in the learner’s behaviour at the time of learning, but which manifests later when a suitable motivation and circumstances appear. This shows that learning can occur without any reinforcement of behaviour.

The idea of latent learning was developed by Tolman; that humans engage in this type of learning every day as we drive or walk the same route daily and learn the locations of various buildings and objects. Only when we need to find a building or object does learning become obvious. Tolman conducted experiments with rats and mazes to examine the role that reinforcement helps rats learn their way through complex mazes. These experiments eventually led to the theory of latent learning.

In their study, 3 groups of rats (healthy and fed regularly) had to find their way around a complex maze. At the end of the maze, there was a food box. Some groups of rats got to eat the food, some did not, and for some rats, the food was only available after 10 days.

Group 1: Rewarded

Day 1 – 17: Every time they got to end, they were given food (i.e., reinforced).

Group 2: Delayed Reward

Day 1 – 10: Every time they got to the end, they were taken out.

Day 11 -17: Every time they got to end, they were given food (i.e., reinforced).

Group 3: No reward

Day 1 – 17: Every time they got to the end, they were taken out.

Following is the simplified version of the graph given in the question

Spatial Learning in Rats Tolman & Honzik, 1930

For the delayed reward group (Group 2):

According to the observations made in this experiment, group 2 rats learned the route on days 1 to 10 and formed a cognitive map of the maze. They took longer to reach the end of the maze because there was no motivation/ circumstance for them to perform. From the day 11 onwards, they had a motivation to perform (i.e., food) and reached the end before the reward group.

Case 1: When the food reward was not introduced in the maze, the error scores and time scores of the rewarded rats showed a large increase.

Case 2: When the reward was introduced into the maze, the rats showed a large decrease in time and error scores.

The drop in the error curve for the group of rats that were rewarded on the eleventh day brought the curve significantly below the curve of a control group of rats that had been rewarded from the first. This shows that between stimulus (the maze) and response (reaching the end of the maze) a mediational process was occurring; the rats were actively processing information in their brains mentally by using their cognitive map (which they had latently learned).

So, with this knowledge now let us start solving the questions.

Answer: 16.1 Statement (a) – TRUE: From the perspective of neuropsychology, adequate nutrition is essential for healthy brain functioning and optimal learning. The consumption of healthy foods is closely related to adequate brain function and which is a prerequisite for efficient cognition and the performance of organized behaviour. Hence, statement (a) is TRUE that rats need good nutritional status to perform well in the maze.

Statement (b) – FALSE: This statement is false because the stimulus-response behaviour is never genetic. It is a learnt behaviour and hence, changeable also. This is proved by the drop in the error curve for the group of rats that were rewarded on the eleventh day brought the curve significantly below the curve of a control group of rats that had been rewarded from the first. This shows that between stimulus (the maze) and response (reaching the end of the maze) a mediational process was occurring; the rats were actively processing information in their brains by using their cognitive map (which they had latently learned).

Statement (c) – FALSE: This statement which states that the find of the end of the maze is by trial and error method and not by learning is wrong. If we talk about group-2; The rats learned the route on days 1 to 10 and formed a cognitive map of the maze. They took longer to reach the end of the maze because there was no motivation/ circumstance for them to perform earlier. From day 11 onwards they had a motivation to perform (i.e., food) and reached the end before the reward group (group-1). It means they learned it, not merely a trial and error method.

Statement (d) – TRUE: Yes, rewarding the rats has improved the end results. This can be seen by comparing the performances of Group 2 with Group 1 and 3. Group-2 rats from day 11 onwards had a motivation to perform (i.e., food) and reached the end before the reward group (group-1). While group 3 performance improved only slightly as getting out of the maze was just a smaller reward.

Statement (e) – TRUE: Yes, there was active learning in the Group-2 rats even before day 11. They learned the route on days 1 to 10 and formed a cognitive map of the maze. Although they took longer to reach the end of the maze because there was no motivation for them to perform. But from day 11 onwards, they had a motivation to perform (i.e., food) and hence reached the end of the maze even before the Group 1.

Answer 16.2 (C)

It is well known that the consumption of healthy foods and particular nutrients is closely related to adequate brain functions; a prerequisite for efficient cognition and the performance of organized behaviour. Keeping in mind the importance of a healthy diet to achieve a cognitively organized behaviour and also the importance of the reward in a learning process; here option C is the most appropriate answer.

In the beginning of the experiment when group 1 rats are kept hungry, the average error at the beginning would be higher than other groups. But as all the other conditions are kept the same, means they will be fed (rewarded) at finding the end of the maze; the end reward would be more motivational hence, the most appropriate event happening with respect to their performance would be a steeper decrease in line 1.

Not-well fed rats (Group 1-here) will show a large decrease in time and error scores. The drop in the error curve would be significantly below the curve of the control group (earlier well-fed Group 1) hence a steeper line 1 in less time.

Question 17. In the early nineteenth century, two scientists Payen and Persoz ground barley seed in water to prepare a crude extract (A). The scientists then carried out a series of treatments on the extract (A). At every step, iodine tests were carried out as follows. (7 marks)

Iodine test: Mixture (Starch + sample) → wait for 10 mins →Add iodine → Check for colour changes.

The different steps of treatment and the result recorded are shown in the flowchart below.

17.1 Blue colour indicates (identify the correct option)

a. That starch is a polymer of glucose units

b. That starch is digested into small units of glucose

c. Glucose units released from starch have formed a complex with iodine

d. Iodine is trapped in the intact polymer of starch

17.2 Based on the observations, identify each of the following statements as True or False.

a. Barley seeds contain a substance that converts glucose to starch.

b. Barley seed coat contains a substance that can convert search to glucose but it gets destroyed by heat. The substance present in barley seeds in water-soluble and breaks starch into small units.

d. The process of heating up to 70°C enhances the chemical activity of the barley filtrate but heating above 70°C inactivates it.

17.3 Which of the preparations (A to I) indicate the presence of the “active substance” being analyzed in barley?

Answer: (17 to 17.3)

Answer 17: Previous knowledge required to solve the questions:

The iodine–starch test was first described by J. J. Colin and H. F. Gaultier de Claubry, and independently by F. Stromeyer. The triiodide anion instantly produces an intense blue-black colour upon contact with starch.

It is observed that the helix (coil or spring) structure of the glucose chain is the key to this test. Further, the resulting colour depends on the length of the glucose chains. The triiodide and pentaiodide ions formed are linear and slip inside the helix structure.

The intensity of the colour decreases with the increase in temperature and the presence of water-miscible organic compounds like ethanol.

On heating, the blue colour amylase-iodine complex dissociates but is formed again on cooling because the helical structure is disrupted; thereby amylose loses its iodine binding capacity and the blue colour. The blue colour reappears on cooling due to the recovery of iodine binding capacity due to regaining of the helical structure.

Answer 17.1 (D)

If a solution is giving a positive iodine test i.e., the appearance of blue colour is there, then it indicates that the iodine is trapped in the intact polymer of starch. Because the helix (coil or spring) structure of the glucose chain/polymer of starch is the key to this test; the triiodide and pentaiodide ions formed are linear and slip inside the helical structure of the polymer.

Answer: 17.2 Statement (a) – FALSE: Barley seeds contain a substance (enzyme- diastase) that converts starch to maltose; not glucose to starch. Hence, this statement is false.

Statement (b) – FALSE: No, barley seed coat does not contain a substance (enzyme) that converts starch to glucose. Rather barley seeds contain a substance (enzyme- diastase) that converts starch to maltose but it gets destroyed by heat approximately above 100 degrees.

Statement (c) – TRUE: The substance (enzyme) in barley seeds is water-soluble, and it breaks starch into small units.

Statement (d) – TRUE: The process of heating up to 70 degrees enhances the chemical activity of barley filtrate, evidenced by step number 3 in the question; whereby even after heating; enzymes were still present in filtrate (E) and also performed by dissolving the starch, hence giving no colour in iodine test. And heating above 70 degree inactivates the active substance in barley filtrate evidenced by step 4; where heating above 100 degrees must have deactivated the active agent (enzyme) and hence, starch is very much present in filtrate as well as a precipitate, thereby giving positive iodine test.

Answer: 17.3 Preparations A, C, E and H indicate the presence of an active substance (an enzyme – diastase) because they all show negative results in the iodine test.

Question 18. Different types of respiratory organs in animals occupying different habitats are represented in the figure (W–Z) given below. (7 Marks)

18. 1. The organs most likely belong to (choose from the options) cockroach, prawn, tadpole and rabbit.

Fick’s law of diffusion shows how various factors influence the rate of diffusion and is represented as:

Q = DA (P1 -P2) / L

Where Q = rate at which a gas such as O₂ diffuses between two locations, D = diffusion coefficient, which is characteristic of the diffusing substance (eq., a gas), the medium and the temperature.

A = cross-sectional area over which the gas is diffusing

P1 and P2 are the partial pressure of the gas at the two locations.

L = path length or distance between the two locations.

18.2 If the temperatures of the habitats, in which the four animals having the organs of type W–Z live, are the same, then based on the medium used for gas exchange, the value of D would be higher for animals possessing respiratory organs of the types (a) ________as compared to animals with organs of types (b) ______(choose from W–Z).

18.3 Two features of respiratory organs in animals are listed in Column I in the given table. Fill in.

– column II with the appropriate factor from Fick’s law equation that will be affected by the feature mentioned in the column I,

– column III with the effect that the feature will have on the factor mentioned in Column II, and

– column IV with the corresponding effect on the rate of diffusion (Q).

(Marks will be given only for the completely correct row.)

Answer: (18.1 to 18.3):

Answer 18 Given Information in this question

Fick’s law of diffusion: Q= D A (P1-P2)/L;

This equation tells us that for a given molecule diffusion rate is:

(a) Directly proportional to the area of the absorptive surface (A). Greater the area for diffusion, greater will be the rate of diffusion.

(b) Directly proportional to the concentration gradient (P1 – P2). The steeper the gradient, or in other words, greater the difference in concentration between two areas, greater the rate of diffusion will be.

(c) Indirectly proportional to the distance of travel (L). The longer the distance of travel the slower the rate of diffusion. So, there must be a short diffusion distance.

Answer 18.1: Here,

figure W → respiratory organ of the tadpole → external gills → aquatic habitat

figure X → respiratory organ of the prawn → internal gills → aquatic habitat

figure Y → respiratory organ of the rabbit → lungs → terrestrial habitat

figure Z → respiratory organ of the cockroach → spiracles → terrestrial habitat

Answer 18.2 Hereby we kept the temperature of habitats of all given animals the same.

Only based on the type of medium used for gas exchange, we have to find the value of D. Diffusion coefficient, D is inversely proportional to the viscosity of the medium. Higher the viscocity, less is the diffusion coefficient. Water is more viscous than air which means D of oxygen in water is less than that of air. That means D in terrestrial habitat will be higher than D in aquatic habitat. Hence, for gas exchange, the value of D would be higher for animals possessing respiratory organs of the types Y and Z as compared to animals with organs of types W and X.

Answer 18.3:

Question 19. Four identical beakers, as shown below, contain the same amount of water, Beaker ‘a’ contains only water. A steel ball (mass 0.800 kg) is held submerged in the beaker ‘b’ by a string from above. A same-sized plastic TT ball (mass 0.020 kg) is held submerged in beaker ‘c’ by a string attached to a stand from outside, as shown in the figure. Beaker ‘d’ contains a same sized TT ball held submerged from a string attached to the bottom of the beaker. The volume of each ball is 10^–4 m³. These beakers (without stands) are placed on weighing pans and register readings W_a, W_b, W_c, and W_d for a, b, c, and d, respectively. (7 Marks)

If W_a = 1 kg, then obtain the relation between W_a, W_b, W_c, and W_d. Show the main steps of your calculations. For calculation purposes, ignore the part of the stand and the thread submerged in water.

Answer: W_b = W_c > W_d > W_a

Solution

Given:

Mass of the steel ball, m_s= 0.800 kg

Mass of TT ball, m = 0.020 kg

The volume of each ball, V = 10^–4 m³

Reading of pan a, the mass of water, W_a= 1 kg

We know that the weighing pan reads the normal reaction N.

For beaker b, let’s suppose the Normal force applied by the weighing pan on the water is N_b.

Water will apply buoyant force on the steel ball upward and according to Newton’s third law, the steel ball will apply the same amount of force on the water in the downward direction.

From equilibrium of forces on water,

N_b = Mg + ρ gV

⇒ N_b = 1 X 10 + 1000 X 10 X 10^-4

So, reading of pan for b (W_b) =

Similarly, for beaker c, W_c=1.1 kg

For beaker d, let’s suppose the Normal force applied by weighing pan on the water is N.

From equilibrium of forces on TT ball,

T + mg = ρgV

⇒ T = ρgV – mg ………(1)

From equilibrium of forces on water,

N = W _ag + ρgV ………(2)

From equilibrium of forces on weighing pan,

N_d + T = N

⇒ N_d = N – T

From (1) and (2),

N_d = W _ag + ρgV – ρgV + mg

⇒ N_d = W _ag + mg

⇒ N_d = 1 x 10 + 0.02 x 10

⇒ N_d = 10.2 N = W _dg

⇒ W_d = 1.02 kg

So, we can say that

W_b = W_c > W_d > W_a

Question 20. Smartphones can be used to perform simple experiments related to sound. There are various apps that record the intensity of an audio signal. An app (WaveEditor here) displays the audio signal in the form of a wave, whose amplitude is proportional to the loudness of the audio signal. (6 marks)

Two students Fatima (F) and Bharat (B) conduct a simple experiment using smartphones. In an open field, both place their smartphones at a distance d from each other as shown in the figure. They stand next to their smartphones and clap one after another. The audio signals from the claps are digitally recorded by WaveEditor and the output produced on the screens of their smartphones are shown next to their sketches. Note that the figure is not to scale. The time mentioned above the screen image is the time of the peak amplitude for each clap’s audio signal received in their phones, respectively. They determine the speed of sound from this experiment to be 363 m/s.

Calculate the distance d (in m). Show the main steps of your calculation.

Answer: d = 1315.512 m

From the figure, we can say that when Fatima claps the intensity of the sound will be maximum and is recorded by the app in its smartphone at time 5.099 s.

This sound wave will propagate and travel towards Bharat and recorded at the time 8.723 s.

Now, when Bharat claps the intensity of the sound will be maximum and is recorded by the app in its smartphone at time 6.615 s. This sound wave will propagate and travel towards Fatima and recorded at the time > 6.615 s whose reading has not been shown in Fatima’s smartphones.

So, only reading of Fatima’s clap is fully available,i.e. time (t) = 5.099 s, 8.723 s

Therefore, distance between Fatima and Bharat is

d = vt = 363 × (8.723–5.099)

d = 1315.512 m

Question 21. With about half of its surface always having a day, Earth constantly receives heat from the Sun and maintains an average temperature of 288 K. From this heat, an average power of 4.3×10¹⁶ W goes into the evaporation of water. The water evaporated from the Earth finally precipitates over its surface. Suppose one collects this water for one year and the thickness of this water shell is over the surface of the Earth; this value in the meter is the well-known average annual rainfall on the globe. For the following two questions, make suitable assumptions wherever needed. (6Marks)

21.1 Estimate h.

21.2 The freshwater requirement is about 6800 liters/day per head, which includes domestic water usage and water used for irrigation and industry. Estimate the ratio of water requirement for the population of the world and the total water received through the rain over the land annually.

d = 1315.512 m

Answer (21.1 to 21.2)

Answer 21.1 (A)

Given:

Average power goes into the evaporation of water, P = 4.3×10¹⁶ W

Temperature of earth, T=288 K =15 °C

Freshwater requirement = 6800 liters/day per head

So, for 1 s energy can be calculated as,

P = E/ t

⇒ E = Pt = 4.3 X 10¹⁶ X 1 = 4.3 X 10¹⁶ J ……..(1)

Which is used in the form of heat.

Now, for water when it goes from 15°C to 100° C and then converts to vapour at 100° C, the heat required is

Q = mS∆T + mL = m(S∆T+L), ………(2)

Where,

S = specific heat capacity of water

L = latent heat capacity of water

m = mass

This is provided by the heat energy from the sun, so, from (1) and (2)

⇒ 4.3 × 10¹⁶ = m(SΔT+L)

⇒ 4.3 × 10¹⁶ = m [4200 × 85 + 2268 × 10³]

⇒ m=1.64 × 10¹⁰ kg

It is the mass evaporated in 1 s.

Now, mass evaporated in 1 year can be calculated as,

M = 1.64 × 10¹⁰ × 365 × 24 × 60 × 60 = 5.17 × 10¹⁷ kg

Let there be a shell formed by water of thickness h near the surface of the earth.

So, volume, V = Surface area x height = 4 𝜋R²h where R is the radius of the earth.

From the density of water,

ρ = M / V

⇒ ρV = M

⇒ 1000 X 4 X 3.14 X (6400 X 0³)² X h = 5.1 X 10¹⁷

⇒ h = 1m

Answer 21.2 (B)

Ratio = Water required annually for earth / annual rainfall

Water required = 6800 × 800 × 10⁷ × 365

Annual water received from rainfall, = 4 x 22 / 7 x (6400 x 10³)² x 1 x 10³

∴ Ratio = 0.0385

IOQ-JS 2021 was successfully conducted on Jan 17th 2021, despite the disruption caused by the COVID-19 pandemic. Since the exam is over, we have got our hands on the question paper and our subject experts have further solved them. Students can view and download this free question paper for IOQ-JS from this page.

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