KVPY Class-XI (SA) - 2015

n + 2n + 3n + … + 99n = n(1 + 2 + 3 + … + 99)
= n(90×100/2) = 9 × 25 × 22 × n is a perfect square when n = 22
Hence, number of digits in n² = 3.

For option (iii), if x = z = 1 and y = 2, then option (iii) is right. So, by option (iii) we can't say x = y = z. Remaining options imply x = y = z.

If n = 3m, then n¹⁹ is a multiple of 3.
If n = 3m + 1 or 3m + 2, then n³⁸ – 1 is a multiple of 3 by binomial expansion.

12! + 13! + 14!
= 12! (1 + 13 + 14.13)
= 12! x 196; which is only divided by possible distinct primes 2, 3, 5, 7 and 11.

Methane is a greenhouse gas because it absorbs longer wavelengths of the electromagnetic spectrum while transmitting shorter wavelengths.

0² = 2² - 2a 1/4
a = 8m/s²

a^T = 10 + 8 = 18 m/s²

f = ma = 75 × 18 = 1350 N

The beta particles of a radioactive metal originate from the nucleus of the metal atoms.

According to Mendeleev's periodic law, the properties of elements are a periodic function of the atomic mass of the elements. 

Baking soda is sodium bicarbonate (NaHCO₃) and vinegar is acetic acid (HCH₃COO). Both reacts with each other due to acid-base neutralisation reaction to produce salt (sodium acetate), carbon dioxide and water.

NaHCO₃(s) + CH₃COOH(l) → CO₂(g) + H₂O(l) + CH₃COO^-(aq) + Na⁺(aq)

The element having electron configuration 1s²2s²2p⁶3s¹ has the lowest first ionisation energy due to larger size and less nuclear charge as compared to the given elements. Therefore, according to the given electronic configurations, the element having electronic configuration 1s²2s²2p⁶3s¹ will form ionic bond easily.

'Cu' is present below H₂ in activity series. So, it cannot displace hydrogen from acid. Thus, it cannot produce H₂ gas on reaction with HCl.

Salt provides a hypertonic medium to the cell. Due to hypertonic solution outside the bacterial cell, bacteria will die by plasmolysis (Exosmosis).

Restriction endonucleases cut DNA from specific base sequence (Palindromic sequence). They are essential tools for recombinant DNA technology. The enzyme "scans" a DNA molecule, looking for a particular sequence, usually of four to six nucleotides.

Trypsin is a proteolytic enzyme. It is present in pancreatic juice and is essential for efficient protein digestion.

A person with blood group AB has both A and B antigens in membrane of his RBC, but lacks antibodies (a, b) in his plasma.

In glycolysis, one molecule of glucose (C₆H₁₂O₆) forms two molecules of pyruvic acid (CH₃COCOOH).

The electron transport chain is a process in which the NADH and [FADH2] produced during glycolysis, β-oxidation, and other catabolic processes are oxidised, thus releasing energy in the form of ATP. The electron transport chain in the mitochondrion is the site of oxidative phosphorylation in eukaryotes.

The innate immune system includes: Physical Barriers - such as skin, the gastrointestinal tract, the respiratory tract, the nasopharynx, cilia, eyelashes and other body hair; Defense Mechanisms - such as secretions, mucous, bile, gastric acid, saliva, tears, and sweat.

Vitamin K is useful in synthesis of prothrombin and fibrinogen in the liver, which are necessary for blood clotting.

Pseudomonas is a denitrifying bacterium. It changes nitrates back to free nitrogen. This lowers the nitrogen content in the soil. Pseudomonas aeruginosa can, under anaerobic conditions (as in swampy or water-logged soils), reduce the amount of fixed nitrogen (as fertiliser) by up to 50 percent.

Annelids show metameric segmentation. Metameric segmentation is the repetition of organs and tissues at intervals along the body of an animal, thus dividing the body into a linear series of similar parts or segments (metameres). It is most strikingly seen in Annelida.

Typhoid is caused by Salmonella typhi, which is diagnosed by Widal test.

Vulture feeds on the dead tiger and hence incorporates the chemicals into its body. (It is due to biomagnification).

1 Base = 500 Da, ds DNA having 10 BP or 20 bases, thus 20 x 500 =10 kDa

Cellulose is polymer of β, D-glucose and glycogen of α, D-glucose. Glucose, fructose and ribose are monosaccharides.

The bulb of an onion contains a modified stem in the centre, but most of what we eat are layers that are modified underground leaves.

Let 'a' mole XeF₂ form and 'b' mole XeF₆ form:
Xe + F₂ → XeF₂ + XeF₆
(262/131) 152/38 'a' mole 'b' mole
Now, by applying P.O.A.C. on Xe:
a + b = 2 …………(1)
By applying P.O.A.C. on F:
2a + 6b = 8 ………..(2)
After solving equations (1) and (2):
a = 1 mole and b = 1 mole
Therefore, The molar ratio XeF₂ : XeF₆ is: 1 : 1.

The formula is 4ⁿ, where n = length of base pair cut by enzyme. Alu I is a restriction endonuclease, which is a four base pair cutter and its frequency is 1/256 BP, while the frequency of 6 cutter Bam HI, ECORI is 1/4096.

At sea level, water boils at 212°F, but at an altitude of 7,500 feet, it boils at about 198°F. Foods that are prepared by boiling or simmering will cook at a lower temperature, and it will take longer to cook them.

Glycogen is insoluble, thus, storing it as glycogen will not upset the osmotic pressure rather than glucose which is soluble in water, and if it is stored as glucose, it will disturb the osmotic pressure (hypertonic) that will cause the cell to lyse.

Starting from the exterior of the cell, there are 3 things that the line will cross before it reaches the centre of nucleus – 1) Cell membrane, 2) Mitochondrion and 3) Nuclear membrane.
Cell membrane is made of 1 phospholipid bilayer. Mitochondrion has 2 membranes - outer and inner (each one is a phospholipid bilayer) and the line will cross mitochondrion across and touch both the membranes on both sides. This works out to 4 bilayers. Finally, the line will cross nuclear membrane which again has 2 membranes (inner and outer - both are lipid bilayers). So the line will cross 2 bilayers here. Hence, the total number of bilayers that the line will cross comes to 1 + 2 + 2 + 2 = 7. But 7 is not given in the options. So, the nearest option is 6.