KVPY Class-XII (SB/SX)

Result

KVPY Class-XII (SB/SX) - 06-11-2016

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Weekly Quiz Competition

Question 1
1 / -0

Suppose, we have to cover xy-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. We can choose tiles of shapes like equilateral triangle, square, regular pentagon and regular hexagon. Then, the tiling can be done with tiles of

A
all four shapes

B
exactly three of the four shapes

C
exactly two of the four shapes

D
exactly one of the four shapes

Solution

We can cover the plane using squares and equilateral triangles. Also, regular hexagon is made of equilateral triangles. But, pentagon cannot cover the plane because of its shape.
Question 2
1 / -0

Physical processes are sometimes described visually by lines. Which one of following can intersect each other?

A
Streamlines in fluid flow

B
Lines of forces in electrostatics

C
Rays in geometrical optics

D
Lines of force in magnetism

Solution

A ⇒ If stream lines intersect, then there will be two directions of fluid flow at a point, which is absurd.

B ⇒ Lines of forces in electrostatic never intersect.

D ⇒ Line of force in magnetism never intersect each other.
Question 3
1 / -0

A nurse measures the blood pressure of a seated patient to be 190 mm of Hg.

Choose the correct option.

A
The blood pressure at the patient's feet is less than 190 mm of Hg.

B
The actual pressure is about 0.25 times the atmospheric pressure.

C
The blood pressure at the patient's neck is more than 190 mm of Hg.

D
The actual pressure is about 1.25 times the atmospheric pressure.

Solution

Blood pressure = gauge pressure = 190 mm Hg
Atmospheric pressure = 760 mm Hg
Actual pressure = 190 + 760 mm Hg = 950 mm Hg = 1.25 × 760 mm Hg
Question 4
1 / -0

An electron with kinetic energy E collides with a hydrogen atom in the ground state. The collision will be elastic

A
for all values of E

B
for E < 10.2 eV

C
for 10.2 eV < E < 13.6 eV only

D
for 0 < E < 3.4 eV only

Solution

When an e^- collides with an atom which is massive in comparison to e^-; if E is less than minimum excitation energy, then the collision is elastic.
∴ E < 10.2 eV (Minimum excitation energy)
Question 5
1 / -0

The continuous part of X-ray spectrum is a result of:

A
Photoelectric effect

B
Raman effect

C
Compton effect

D
Inverse photoelectric effect

Solution

X-ray emission is the inverse of the photoelectric effect. In the photoelectric effect, the energy of a photon is transformed into the kinetic energy of a photoelectron, while during X-ray emission, the kinetic energy of a photon is converted into X-rays.
Question 6
1 / -0

Thermal expansion of a solid is due to the

A
symmetric characteristic of the inter-atomic potential energy curve of the solid

B
asymmetric characteristic of the inter-atomic potential energy curve of the solid

C
double well nature of the inter-atomic potential energy curve of the solid

D
rotational motion of the atoms of the solid

Solution

Thermal expansion of a solid is due to asymmetric characteristic of inter-atomic potential energy curve of the solid.
Question 7
1 / -0

The packing efficiency of the face centred cubic (fcc), body centred cubic (bcc) and simple/primitive cubic (pc) lattices follows the order:

A
fcc > bcc > pc

B
bcc > fcc > pc

C
pc > bcc > fccq

D
bcc > pc > fcc

Solution

fcc bcc sc

η 74% 68% 52.4%

Order: fcc > bcc > pc
Question 8
1 / -0

The most stable Lewis acid-base adduct among the following is:

A
H₂O → BCI₃

B
H₂S → BCI₃

C
H₃N → BCI₃

D
H₃P → BCI₃

Solution

The stability of the Lewis acid-base adduct depends on the strength of the Lewis base (ability of the to donate the L.P of electrons) as well as the tendency of the Lewis acid to accept the electrons.
In the given options the Lewis acid (BCl₃) is common.
NH₃ is the strongest base.
Due to smaller atomic size the density of lone pair electrons on N in NH₃ is larger than that of P in PH₃. So, NH₃ is a stronger Lewis base than that of PH₃.
Oxygen has a greater electronegativity than nitrogen so the non-bonded electron pair on a nitrogen atom is more available for sharing than a non-bonded electron pair on an oxygen atom.
Question 9
1 / -0

A unit cell of calcium fluoride has four calcium ions. The number of fluoride ions in the unit cell is

A
2

B
4

C
6

D
8

Solution

The number of F^-ions will be equal to 8 since for one Ca^+2, there should be two F^- ions.
Question 10
1 / -0

Which of the following molecules is a primary acceptor of CO₂ in photosynthesis?

A
Pyruvate

B
3-phosphoglycerate

C
Phosphoenolpyruvate

D
Oxaloacetate

Solution

In C₄plants, the primary acceptor of CO₂ is phosphoenolpyruvate which is a 3C compound, such as in maize, sugarcane etc.
Question 11
1 / -0

Which of the following pairs of molecules does not form a hydrogen bond?

A
Water and water

B
Water and glucose

C
Water and ethanol

D
Water and octane

Solution

Hydrogen bond is a weak bond between two molecules resulting from an electrostatic attraction between a proton in one molecule and an electronegative atom in another molecule. This is not possible in case of water and octane.
Question 12
1 / -0

Which of the following statements is incorrect regarding biological membrane?

A
Biological membrane is composed of lipids and proteins.

B
Peripheral proteins are loosely associated with the membrane.

C
Integral proteins span the lipid bilayer.

D
Lipids and membrane proteins do not provide structural and functional asymmetry.

Solution

Proteins provide asymmetry to plasma membrane. They are of peripheral and integral types.
Question 13
1 / -0

The percentage of sunlight captured by plants is

A
2 - 10%

B
10 - 20%

C
60 - 80%

D
100%

Solution

Plants capture only 2 - 10% of the photosynthetically active radiation (PAR) and this small amount of energy sustains the entire living world.
Question 14
1 / -0

The hard outer layer of pollens, named exine, is made of

A
cellulose

B
tapetum

C
sporopollenin

D
pectin

Solution

Sporopollenin forms the exine of pollen grain which is resistant to acids, high temperature and radiations.
Question 15
1 / -0

Insectivorous plants, such as Venus flytrap catch and digest insects in order to supplement the deficiency of

A
sulphur

B
nitrogen

C
potassium

D
phosphorus

Solution

Insectivorous plants grow in nitrogen deficient soils. In order to compensate the deficiency of nitrogen, they catch and digest insects and obtain N from chitin (NAG).
Question 16
1 / -0

Which of the following statements about nucleosomes is true?

A
They consist of only DNA.

B
They are nucleus-like structures found in prokaryotes.

C
They consist of DNA and proteins.

D
They consist of only histone proteins.

Solution

Nucleosome is the smallest unit of DNA packaging containing 200 nitrogen bases and four types of histone proteins (H₂A, H₂B, H₃ and H4). H₁ type of histone is used in plugging.
Question 17
1 / -0

Epithelial cells in animals are held by specialised junctions, one of them being "gap junctions". Which of the following is the function of "gap junctions"?

A
They facilitate cell-cell communication by rapid transfer of small molecules.

B
They cement the neighbouring cells.

C
They stop substances from leaking.

D
They provide gaps in the tissue to facilitate uptake of small molecules across tissues.

Solution

Gap junctions are cytoplasmic communications between two epithelial cells for rapid transfer of some ions and small molecules.
Question 18
1 / -0

Which of the following statements is true about glandular epithelium in salivary glands?

A
It consists of isolated single cells.

B
It consists of multicellular clusters of cells.

C
Its secretions are endocrine.

D
It consists of squamous epithelial cells.

Solution

Salivary glands are multicellular exocrine glands, made up of cuboidal epithelium.
Question 19
1 / -0

Which of the following ion pairs is involved in nerve impulses?

A
Na⁺, K⁺

B
Na⁺, Cl^-

C
K⁺, Cl^-

D
K⁺, Ca2⁺

Solution

Na⁺ - K⁺ pump is essential for impulse conduction.
Question 20
1 / -0

Which of the following hormones controls blood pressure and is secreted by the human heart?

A
Erythropoietin

B
Atrial natriuretic factor

C
ACTH

D
Glucocorticoid

Solution

ANF [Anti-Natriuretic Factor] is antagonist of renin and acts as a vasodilator to reduce blood pressure. ANF is secreted by the human heart.
Question 21
1 / -0

Oxytocin and vasopressin are synthesised in

A
hypothalamus

B
adrenal gland

C
pituitary gland

D
ovary

Solution

Oxytocin and vasopressin are synthesised in hypothalamus. Then, they come into neurohypophysis of pituitary gland to be released in blood.
Question 22
1 / -0

The path of passage of stimulus, when you accidentally touch a hotplate, is

A
Receptor → Brain → Muscles

B
Muscles → Spinal cord →Receptor

C
Muscles → Brain → Receptor

D
Receptor → Spinal cord → Muscles

Solution

Reflex arch
Receptor → Sensory neuron → Spinal cord → Motor neuron → Muscle [Effector]
Question 23
1 / -0

In the presence of glucose and lactose, Escherichia coli utilises glucose. However, lactose also enters the cells because

A
low level of permease is always present in the cell

B
it uses the same transporter as glucose

C
if diffuses through the bacterial cell membrane

D
it is transported through porins

Solution

The preferred molecule is glucose. Then, lactose is used by E.coli because lactose is also broken down into galactose and glucose. Permease is a carrier protein which helps in facilitated diffusion. Lac operon is always operational at low levels.
Question 24
1 / -0

Passive immunisation is achieved by administering

A
heat killed vaccines

B
toxoids

C
live attenuated vaccines

D
antibodies

Solution

Administration of already prepared antibodies is called passive immunisation.
There are two ways to acquire passive immunity.
(1) Naturally acquired passive immunity
(2) Artificially acquired passive immunity
Question 25
1 / -0

If ¹⁴CO₂ is added to a suspension of photosynthesising chloroplasts, then which of the following will be the first compound to be radioactive?

A
ATP

B
NADPH

C
NADH

D
3-phosphoglycerate

Solution

During Calvin cycle, ¹⁴CO₂ is incorporated with a 5C compound ribulose 1-5 biphosphate and forms 2 molecules of 3C compound 3-phosphoglycerate.
Question 26
1 / -0

Which of the following species makes the largest true flower in the world?

A
Amorphophallus titanium

B
Rafflesia arnoldii

C
Nelumbo nucifera

D
Helianthus annuus

Solution

Rafflesia arnoldii is a total root parasite and forms the largest flower in angiosperms while Amorphophallus titanium is the largest inflorescence in the angiosperms.
Question 27
1 / -0

Match the following organelIes in Group I with the structures in Group II.

Group - I Group - II

P. Mitochondrion i. Cisternae

Q. Golgi ii. Cristae

R. Chloroplast iii. Thylakoids

S. Centrosome iv. Radial spokes

A
P - ii, Q - i, R - iii, S - iv

B
P - i, Q - i, R - ii, S - iv

C
P - iv, Q - i, R - ii, S - iii

D
P - iv, Q - ii, R - i, S - iii

Solution

Cristae are invaginations of inner mitochondrial membrane. Cisternae is a unit of Golgi body, where glycosidation and glycosylation take place. Grana of chioroplast are composed of thylakoids. Centrosome contains centrioles with radial spokes.

Question 28

1 / -0

Match the human disorders shown in Group I with the biochemical processes in Group II.

Group - I	Group - II
P. Phenylketonuria	i. Melanin synthesis
Q. Albinism	ii. Conversion of phenylalanine to tyrosine
R. Homocystinuria	iii. Tyrosine degradation
S. Argininemia	iv. Methionine metabolism
	v. Urea synthesis

P - ii, Q - i, R - iv, S - v

P - i, Q - iv, R - ii, S - v

P - ii,Q - i, R - v, S - iii

P - v, Q - iii, R - i, S - ii

Solution

Phenylketonuria is due to non-conversion of phenylalanine into tyrosine. Albinism is non-synthesis of melanin pigment. Homocystinuria is associated with methionine metabolism. Argininemia is associated with urea synthesis.

Question 29
1 / -0

If the initial number of template DNA molecules in a PCR reaction is 1000, then the number of product DNA molecules at the end of 20 cycles will be closest to

A
10³

B
10⁶

C
10⁹

D
10¹²

Solution

During PCR, the number of DNA molecules increases by 2ⁿ, where 'n' is the number of divisions.
2 to the power of 20 = 1,048,576 copies for a double stranded DNA.
Therefore, for 1,000, it is 1000 × 1,048,576 = 1,048,576,000 copies = 10⁹

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Re-Attempt Weekly Quiz Competition

Group - I	Group - II
P. Mitochondrion	i. Cisternae
Q. Golgi	ii. Cristae
R. Chloroplast	iii. Thylakoids
S. Centrosome	iv. Radial spokes

	fcc	bcc	sc
η	74%	68%	52.4%

KVPY Class-XII (SB/SX) - 06-11-2016

Result

KVPY Class-XII (SB/SX) - 06-11-2016

Score

-

Rank

-

SHARING IS CARING

Question 1

all four shapes

exactly three of the four shapes

exactly two of the four shapes

exactly one of the four shapes

Solution

Question 2

Streamlines in fluid flow

Lines of forces in electrostatics

Rays in geometrical optics

Lines of force in magnetism

Solution

Question 3

The blood pressure at the patient's feet is less than 190 mm of Hg.

The actual pressure is about 0.25 times the atmospheric pressure.

The blood pressure at the patient's neck is more than 190 mm of Hg.

The actual pressure is about 1.25 times the atmospheric pressure.

Solution

Question 4

for all values of E

for E < 10.2 eV

for 10.2 eV < E < 13.6 eV only

for 0 < E < 3.4 eV only

Solution

Question 5

Photoelectric effect

Raman effect

Compton effect

Inverse photoelectric effect

Solution

Question 6

symmetric characteristic of the inter-atomic potential energy curve of the solid

asymmetric characteristic of the inter-atomic potential energy curve of the solid

double well nature of the inter-atomic potential energy curve of the solid

rotational motion of the atoms of the solid

Solution

Question 7

fcc > bcc > pc

bcc > fcc > pc

pc > bcc > fccq

bcc > pc > fcc

Solution

Question 8

H2O → BCI3

H2S → BCI3

H3N → BCI3

H3P → BCI3

Solution

Question 9

2

4

6

8

Solution

Question 10

Pyruvate

3-phosphoglycerate

Phosphoenolpyruvate

Oxaloacetate

Solution

Question 11

Water and water

Water and glucose

Water and ethanol

Water and octane

Solution

Question 12

Biological membrane is composed of lipids and proteins.

Peripheral proteins are loosely associated with the membrane.

Integral proteins span the lipid bilayer.

Lipids and membrane proteins do not provide structural and functional asymmetry.

Solution

H₂O → BCI₃

H₂S → BCI₃

H₃N → BCI₃

H₃P → BCI₃

Na⁺, K⁺

Na⁺, Cl^-

K⁺, Cl^-

K⁺, Ca2⁺