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Day 176: JEE Mains Physics

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

  1. Question 1

    Thermodynamics. Asked once: 21 Jan 2026 S1 (same type asked 4 times) See all 4 of this type

    24
    10 mole of oxygen is heated at constant volume from 30∘C30^{\circ}C to 40∘C40^{\circ}C. The change in the internal energy of the gas is ____\_\_\_\_ cal. (The molecular specific heat of oxygen at constant pressure, Cp=7cal./mol∘CC_{p} = 7cal./{mol}^{\circ}C and R=2cal./mol∘CR = 2cal./{mol}^{\circ}C.)

  2. Question 2

    Electromagnetic Waves. Asked once: 2021 Paper 23 (same type asked 4 times) See all 4 of this type

    12
    The magnetic field vector of an electromagnetic wave is given by B=Boi^+j^2cos⁡(kz−ωt)B=B_{o}\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\cos(kz-\omega t); where i^,j^\widehat{i},\widehat{j} represents unit vector along xx and yy-axis respectively. At t=0 st = 0\text{ }s, two electric charges q1q_{1} of 4π4\pi coulomb and q2q_{2} of 2π2\pi coulomb located at (0,0,πk)\left( 0,0,\frac{\pi}{k} \right) and (0,0,3πk)\left( 0,0,\frac{3\pi}{k} \right), respectively, have the same velocity of 0.5ci^0.5c\widehat{i}, (where cc is the velocity of light). The ratio of the force acting on charge q1q_{1} to q2q_{2} is:

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  3. Question 3

    Electrostatics. Asked once: 8 Apr 2024 (same type asked 4 times) See all 4 of this type

    15
    Two charged conducting spheres of radii a and bb are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:

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  4. Question 4

    Mechanical Properties of Fluids. Asked once: 24 Jan 2026 S2 (same type asked 4 times) See all 4 of this type

    12
    A cubical block of density ρb=600 kg/m3\rho_{b}= 600\text{ }kg/m^{3} floats in a liquid of density ρe=900 kg/m3\rho_{e}= 900\text{ }kg/m^{3}. If the height of block is H=8.0 cmH = 8.0\text{ }cm then height of the submerged part is ____\_\_\_\_ cm.

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  5. Question 5

    Units and Measurements. Asked once: 8 Apr 2024 (same type asked 4 times) See all 4 of this type

    10
    Young's modulus is determined by the equation given by Y=49000Mldynecm2Y = 49000\frac{M}{\mathcal{l}}\frac{dyne}{cm^{2}} where MM is the mass and l\mathcal{l} is the extension of wire used in the experiment. Now error in Young's modulus (Y) is estimated by taking data from M−lM -\mathcal{l} plot in graph paper. The smallest scale divisions are 5 g5\text{ }g and 0.02 cmcm along load axis and extension axis respectively. If the value of MM and l\mathcal{l} are 500 g500\text{ }g and 2 cm2\text{ }cm respectively then percentage error of YY is:

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