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Day 383: JEE Mains Physics

5 questions. Try each one first, then open its answer.

  1. Question 1

    Atoms. Asked once: 31 Jan 2023 (same type asked 6 times) See all 6 of this type

    117
    If the binding energy of ground state electron in a hydrogen atom is 13.6eV13.6eV, then, the energy required to remove the electron from the second excited state of Li2+Li^{2 +} will be: x×10−1eVx \times10^{- 1}eV. The value of xx is___

  2. Question 2

    Thermodynamics. Asked once: 25 Jan 2023 (same type asked 6 times) See all 6 of this type

    106
    Match List I with List II
    List IList II
    A. Isothermal ProcessI. Work done by the gas decreases internal energy
    B. Adiabatic ProcessII. No change in internal energy
    C. Isochoric ProcessIII. The heat absorbed goes partly to increase internal energy and partly to do work
    D. Isobaric ProcessIV. No work is done on or by the gas
    Choose the correct answer from the options given below:

    Tap an option to check your answer.

  3. Question 3

    Kinetic Theory. Asked once: 10 Apr 2023 (same type asked 6 times) See all 6 of this type

    93
    A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be:

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  4. Question 4

    Moving Charges and Magnetism. Asked once: 31 Jan 2023 (same type asked 6 times) See all 6 of this type

    97
    The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%50\%. The percentage change in voltage sensitivity of the galvanometer will be:

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  5. Question 5

    Atoms. Asked once: 11 Apr 2023 (same type asked 6 times) See all 6 of this type

    21
    A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x×1015 Hzx \times10^{15}\text{ }Hz. The value of xx is ____. (Given  h=4.25×10−15eVs)\left. \ h = 4.25 \times10^{- 15}eVs \right)