Self Studies

KVPY Class-XI (SA) - 2018

Result Self Studies

KVPY Class-XI (SA) - 2018
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    The number of real roots of the polynomial equation x4 - x2 + 2x - 1 = 0 is

    Solution

    x4 - x2 + 2x - 1 = 0
    (x2)2 - (x -1)2 = 0
    ⇒ (x2 + x - 1) (x2 - x + 1) = 0
    (x2 - x + 1) = 0 (This has imaginary roots.)
    x2 + x - 1 = 0 (This has two real roots.)
    Hence, the number of real roots is 2.

     

  • Question 2
    1 / -0

    Suppose the sum of the first m terms of an arithmetic progression is n and the sum of its first n terms is m, where m ≠ n. Then, the sum of the first (m + n) terms of the arithmetic progression is

    Solution

    Sm = m/2 [2a + (m - 1)d] = n ... (1)
    Sn = n/2 [2a + (n - 1)d] = m ... (2)
    By (1) and (2),
    (m - n)a + (m - n){m + n - 1} d/2 = -(m - n)
    ⇒ 2a + (m + n - 1)d = -2 (m ≠ n)
    ⇒ Sm + n = m+n/2 [2a + (m + n - 1)d] = -(m + n)

     

  • Question 3
    1 / -0

    A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show

    Solution

    Since both box and water are in a state of free fall (g = 0), weight is zero. So, water will not come out of the hole.

     

  • Question 4
    1 / -0

    One can define an alpha-Volt (αV) to be the energy acquired by a particle when it is accelerated by a potential of 1 Volt. For this problem you may take a proton to be 2000 times heavier than an electron. Then,

    Solution

    Using conservation of energy,
    Increase in the kinetic energy of the alpha particle = Decrease in the electrostatic potential energy
    ΔK = q (ΔV)
    As we have charge on alpha particle = 2 times, the charge on electron,
    ( e = charge on electron)
    qα = 2qe = 2e
    Δ Ke = change of kinetic energy on accelerating an electron by 1 V
    ΔKα = 2ΔKe
    1αV = 2eV

     

  • Question 5
    1 / -0

    Which of the following is NOT true about the total lunar eclipse?

    Solution

    A lunar eclipse can occur on a new moon and full moon day.

     

  • Question 6
    1 / -0

    Among the following, the correct statement is:

    Solution

    Based on the principle of common ion effect, the dissociation of a weak electrolyte decreases in the presence of a common ion from a strong electrolyte.
    dil. aq. soln of NH3 is NH4OH(aq)
    NH4OH ⇌ NH4+ + OH
    NH4Cl → NH4+ + Cl
    Due to common ion effect, equilibrium of NH4OH shifts in backward direction, so [OH] ↓.
    Decrease in the concentration of [OH-] ions causes the [H+] to increase, and therefore, the pH decreases.

     

  • Question 7
    1 / -0

    Among the following, the INCORRECT statement is:

    Solution

    The maximum number of e in the shell with principal quantum number n is equal to 2n2 (not n2 + 2).

     

  • Question 8
    1 / -0

    The correct order of energy of 2s orbitals in H, Li, Na and K, is

    Solution

    The energy of the electron in the 2s orbital decreases with the increase in the nuclear charge.

     

  • Question 9
    1 / -0

    A filter paper soaked in salt X turns brown when exposed to HNO3 vapour. Salt X is

    Solution

    A filter paper soaked in KI turns brown when exposed to HNO3 vapour due to liberation of iodine. The reaction is as follows:
    6KI + 8HNO3 → 6KNO3 + 4H2O + 2NO + 3I2
    Hence, X is KI.

     

  • Question 10
    1 / -0

    The role of haemoglobin is to

    Solution

    Since oxygen has very little solubility in water, it does not dissolve in the blood plasma and needs a carrier molecule to be transported through the body.
    Haemoglobin is the red pigment in the RBCs which combines with oxygen to form oxyhaemoglobin, and helps transport it to different parts of the body.

     

  • Question 11
    1 / -0

    Which of the following molecules is a secondary metabolite?

    Solution

    Molecules which cannot be directly used and synthesised by animal body, but may change primary process of a living system, are known as secondary metabolite.
    Penicillin is an antibiotic obtained from Penicillium notatum, which is used for treatment of bacterial infection and provides artificial passive acquired immunity.

     

  • Question 12
    1 / -0

    Lecithin is a

    Solution

    Lecithin is also called phospholipid molecule. It consists of two fatty acids, one glycerol, one phosphate and one choline molecule. It is also called phosphatidyl choline, which is an amphipathic molecule. It is a major membrane lipid constituting plasma membrane of the cell.

     

  • Question 13
    1 / -0

    The water potential (ψw) of pure water, at standard temperature and atmospheric pressure, is

    Solution

    ψw of pure water is zero at standard temperature & atmospheric pressure because there is no solute in pure water. So, water potential (ψw) is zero. When some solute is dissolved in water, the water potential of pure water decreases.

     

  • Question 14
    1 / -0

    In neurons, action potential is generated by a rapid influx of

    Solution

    An action potential is a rapid rise and subsequent fall in voltage or membrane potential across a cellular membrane with a characteristic pattern. Sufficient current is required to initiate a voltage response in a cell membrane. If the current is insufficient to depolarise the membrane to the threshold level, an action potential will not fire. Na+ channels open at the beginning of the action potential, causing depolarisation that stimulates neural conduction.

     

  • Question 15
    1 / -0

    Erythropoietin is produced by

    Solution

    Juxtaglomerular cells of kidney produce erythropoietin hormone during oxygen deficiency. This hormone promotes red bone marrow to form RBCs (Erythropoiesis). Hemoglobin is the protein in red blood cells that helps blood carry oxygen throughout the body.

     

  • Question 16
    1 / -0

    Tendrils are modifications of

    Solution

    In plants, tendril is a modification of stem, leaves or petiole with a thread-like coiled shape. Functions of tendril are providing support, helping in climbing, providing attachment or cellular invasion by parasitic plant.
    Examples:
    1. Leaves modified into tendrils for climbing in pea
    2. Stem tendrils, which develop from axillary bud in cucumber, pumpkins, watermelon and grapevines for climbing

     

  • Question 17
    1 / -0

    Which of the following combinations of biomolecules is present in ribosomes?

    Solution

    The ribosome is a complex molecule made of ribosomal RNA molecules and proteins that form a factory for protein synthesis in cells. Ribosomes read the nucleotide sequence of a messenger RNA (mRNA) into a protein sequence, using the genetic code. They use transfer RNAs (tRNAs) to mediate this process of translation from the nucleotide language of RNA and DNA into the amino acid language of proteins.

     

  • Question 18
    1 / -0

    Which of the following proteins does not play a role in skeletal muscle contraction?

    Solution

    Actin, myosin (contractile protein), troponin,and tropomyosin (regulatory protein) are essential for muscular contraction. Calcium is required by two proteins, troponin and tropomyosin, that regulate muscle contraction by blocking the binding of myosin to filamentous actin.

     

  • Question 19
    1 / -0

    Which of the following statements is true about trypsinogen?

    Solution

    Trypsinogen is an inactive enzyme produced from pancreatic juice, which gets activated by the action of the enterokinase produced from intestinal gland in small intestine, into its active form trypsin.

     

  • Question 20
    1 / -0

    Which of the following organisms respires through the skin?

    Solution

    Salamanders have soft and moist skin, covering their entire body and tail. These secretive and small creatures are principally active at night, and remain hidden under fallen logs and within moist leaf litter during the day. Specific species, such as the lungless salamanders, lack the primitive lungs that other amphibians have and breathe exclusively through their skin.

     

  • Question 21
    1 / -0

    Which of the following human cells lacks a nucleus?

    Solution

    Prokaryotic cell, mature erythrocyte (animal), and mature sieve tube cell of phloem are examples of nucleus-less cell. In humans, mature red blood cells are flexible, oval, biconcave disks. They lack a cell nucleus and most organelles, in order to accommodate maximum space for hemoglobin. The biconcave shape allows RBCs to bend and flow smoothly through the body's capillaries. It also facilitates oxygen transport.

     

  • Question 22
    1 / -0

    The first enzyme that the food encounters in the human digestive system is

    Solution

    Salivary amylase is an enzyme in the mouth that initiates the digestion of carbohydrates in the form of starches by catalysing the hydrolysis of polysaccharides into disaccharides. However, the main site for carbohydrate digestion is the small intestine.

     

  • Question 23
    1 / -0

    Glycoproteins are formed in which of the following organelles?

    Solution

    Eukaryotic cells contain a unique cluster of membrane vesicles, known as Golgi apparatus. It principally performs the function of packaging materials. Golgi apparatus is the important site of formation of glycoproteins and glycolipids. The Golgi apparatus receives proteins and lipids (fats) from the rough endoplasmic reticulum. It modifies some of them and sorts, concentrates and packs them into sealed droplets called vesicles.

     

  • Question 24
    1 / -0

    An example of nastic movement:

    Solution

    Nastic movements are non-directional movements, which depend on external stimulus in plants.
    These movements can be due to changes in turgor or changes in growth.
    Examples:
    (1) Folding of leaf of Mimosa pudica by touching
    (2) Diurnal movement of leaves
    (3) Response of insectivorous plant, such as Venus fly trap to their prey

     

  • Question 25
    1 / -0

    Among the following, the species with identical bond order are

    Solution

    O22– contains 18 electrons that are the same as F2.
    So,
    Bond order = 1
    B2 → σ1s2 , σ *1s2, σ 2s2, σ *2s2, π2px1 = π2py1
    Bond order = 1/2 (Nb – Na) = 1/2 (6 – 4) = 1
    Thus, O22– and B2 have identical bond order.

     

  • Question 26
    1 / -0

    The lowest stability of ehthyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical, respectively, are due to

    Solution

    Ethyl anions is less stable than the methyl anion because the methyl group exerts a +I effect, which increases e density on carbanion. So, the stability decreases.
    Ethyl-free radical is more stable than the methyl-free radical because of the σ –p orbital conjugation (hyperconjugation) in the ethyl radical.

     

  • Question 27
    1 / -0

    A cross was carried out between two individuals heterozygous for two pairs genes. Assuming segregation and independent assortment, the number of different genotypes and phenotypes respectively obtained would be

    Solution

    Genotype = 3n
    n = Number of heterozygous pairs
    = 32 = 3 × 3 = 9
    Phenotype = 2n
    n = Number of heterozygous pairs
    = 22 = 2 × 2 = 4

     

  • Question 28
    1 / -0

    If the H+ concentration of aqueous solutions is 0.001 M, then the pOH of the solution would be

    Solution

    [H+] = 10–3 M
    pH = –log10 [H+]
    = –log10 [10–3]
    pH = 3
    pH + pOH = 14
    pOH = 14 – 3 = 11

     

  • Question 29
    1 / -0

    Consider the following vision defects listed in Column-I & II and the corrective measures in Column III.

    Choose the correct combination.

    I. Column II. Column III. Column
    P. Hypermetropia i. Nearsightedness a. Convex lens
    Q. Myopia ii. Farsightedness b. Concave lens
    Solution

    Myopia or shortsightedness is corrected using a concave (curved inwards) lens, which is placed in front of a myopic eye, moving the image back to the retina and making it clearer.
    Hyermetropia or longsightedness is corrected using a convex (outward facing) lens. This is placed in front of a hypermetropic eye, moving the image forward and focusing it correctly on the retina.

     

  • Question 30
    1 / -0

    Which of the following properties causes the plant tendrils to coil around a bamboo stick?

    Solution

    The part of the tendril in contact with the support does not grow as fast as the other parts of the plant. It coils around the support with the help of the plant hormone called auxin. This causes the tendril to circle around the object and thus, cling to it.
    The unequal growth is due to more secretion of auxin in the other end and less secretion in the end in contact with the support.

     

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now