1. The answer is 17 (Seventeen). Explanation: It's given that - P is a date in the month of October Q is a date in the month of November. P and Q both are perfect squares. Perfect squares are numbers whose square root is a natural number, including 0. The available dates in October that are perfect numRead more

    The answer is 17 (Seventeen).

    Explanation:
    It’s given that –
    P is a date in the month of October
    Q is a date in the month of November.

    P and Q both are perfect squares.

    Perfect squares are numbers whose square root is a natural number, including 0.

    The available dates in October that are perfect numbers are:

    1*1 = 1st October
    2*2 = 4th October
    3*3 = 9th October
    4*4 = 16th October, and
    5*5 = 25th October

    It’s mentioned that Pth October and Qth November are only 10 days apart.

    As all the perfect square dates before 25th October do not occur 10 days before any date of November, the only choice we’re left with is 25th October.

    10 days after 25th October comes 4th November.

    here, both 25 and 4 are perfect squares.

    thus, P = 25 and Q = 4.

    Now, P – 2Q = 25 – 2(4) = 25 – 8 = 17 (Required Answer).

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  2. This answer was edited.

    The Indian scientists’ name have been correctly matched with their contributions/discoveries below: (i) J.C. Bose → (d) Ultrashort radio waves (ii) C.V. Raman → (f) Inelastic scattering of light by molecules (iii) M.N. Saha → (a) Thermal ionization (iv) S.N. Bose → (c) Quantum statistics (v) H.J. BhRead more

    The Indian scientists’ name have been correctly matched with their contributions/discoveries below:
    (i) J.C. Bose → (d) Ultrashort radio waves
    (ii) C.V. Raman → (f) Inelastic scattering of light by molecules
    (iii) M.N. Saha → (a) Thermal ionization
    (iv) S.N. Bose → (c) Quantum statistics
    (v) H.J. Bhabha → (b) Cosmic radation and nuclear weapons
    (vi) S. Chandrasekhar → (e) Structure and evolution of stars
    (i ) – (d); (ii) – (f); (iii) – (a); (iv) – (c); (v) – (b); (vi) – (e).

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  4. The scientists' name have been correctly matched with their contributions/discoveries below: (i) Faraday → (g) Law of electromagnetic induction (ii) Rutherford → (j) Nuclear model of atom (iii) Chadwick → (i) Neutron (iv) Bohr → (b) Quantum model of hydrogen atom (v) Newton → (a) Law of gravitation,Read more

    The scientists’ name have been correctly matched with their contributions/discoveries below:

    (i) Faraday → (g) Law of electromagnetic induction
    (ii) Rutherford → (j) Nuclear model of atom
    (iii) Chadwick → (i) Neutron
    (iv) Bohr → (b) Quantum model of hydrogen atom
    (v) Newton → (a) Law of gravitation, Laws of motion
    (vi) Maxwell → (c) Unification of light and electromagnetism
    (vii) Salam → (f) Unification of weak and electromagnetic interactions
    (viii) Einstein → (h) Theory of relativity, Explanation of photoelectric effect
    (ix) Roentgen → (e) X-rays
    (x) Madam Curie → (n) Discovery of radium
    (xi) J. J. Thomson → (m) Electron
    (xii) de Broglie → (k) Wave nature of matter
    (xiii) John Bardeen → (l) Transistor
    (xiv) Hubble → (h) Expansion of the universe

    (i ) – (g); (ii) – (j); (iii) – (i); (iv) – (b); (v) – (a); (vi) – (c); (vii) – (f); (viii) – (d); (ix) – (e); (x) – (n); (xi) – (m); (xii) – (k); (xiii) – (l); (xiv) – (h).

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  5. In any nuclear reaction, the number of nucleons remains unchanged. The radioactive decay of uranium nucleus is: 92U238 → 90Th234 + 2He4 The total number of nucleons is 238 before as well as after the decay.

    In any nuclear reaction, the number of nucleons remains unchanged. The radioactive decay of uranium nucleus is:

    92U23890Th234 + 2He4

    The total number of nucleons is 238 before as well as after the decay.

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  7. In CuSO4.5H2O, we have: 1 Copper (Cu) atom, 1 Sulphur (S) atom, 4 Oxygen (O) atoms, 5*2 = 10 Hydrogen (H) atoms, and 5*1 = 5 Oxygen (O) atoms. Thus, in total, the number of atoms present in the compound is: 1+1+4+10+5 = 21.

    In CuSO4.5H2O, we have:
    1 Copper (Cu) atom,
    1 Sulphur (S) atom,
    4 Oxygen (O) atoms,
    5*2 = 10 Hydrogen (H) atoms,
    and 5*1 = 5 Oxygen (O) atoms.

    Thus, in total, the number of atoms present in the compound is: 1+1+4+10+5 = 21.

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