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Day 232: JEE Mains Physics

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

  1. Question 1

    Thermal Properties of Matter. Asked once: 27 Jun 2022 (same type asked 2 times) See all 2 of this type

    99
    A lead bullet penetrates into a solid object and melts. Assuming that 40%40\% of its kinetic energy is used to heat it, the initial speed of bullet is: (Given, initial temperature of the bullet =127∘C=127^{\circ}C, Melting point of the bullet =327∘C=327^{\circ}C, Latent heat of fusion of lead =2.5×104 JKg−1= 2.5 \times10^{4}\text{ }JKg^{- 1}, Specific heat capacity of lead =125 J/kgK= 125\text{ }J/kgK )

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  2. Question 2

    Work, Energy and Power. Asked once: 14 Jun 2022 (same type asked 2 times) See all 2 of this type

    9
    A particle experiences a variable force F→=(4xi^+3y2j^)\overrightarrow{F}=\left( 4x\widehat{i}+ 3y^{2}\widehat{j} \right) in a horizontal x−yx - y plane. Assume distance in meters and force is newton. If the particle moves from point (1,2)(1,2) to point (2,3)(2,3) in the x−yx - y plane, the Kinetic Energy changes by

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  3. Question 3

    Current Electricity. Asked once: 29 Jul 2022 (same type asked 2 times) See all 2 of this type

    94
    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils. Reason R: Constantan and manganin have very small value of temperature coefficient of resistance. In the light of the above statements, choose the correct answer from the options given below.

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  4. Question 4

    Electrostatics. Asked once: 5 Apr 2024 (same type asked 2 times) See all 2 of this type

    117
    The electric field at point pp due to an electric dipole is EE. The electric field at point RR on equatorial line will be Ex\frac{E}{x}. The value of xx :

  5. Question 5

    Gravitation. Asked once: 2021 Paper 3 (same type asked 2 times) See all 2 of this type

    2
    A solid sphere of radius RR gravitationally attracts a particle placed at 3R3R from its centre with a force F1F_{1}. Now a spherical cavity of radius (R2)\left( \frac{R}{2} \right) is made in the sphere (as shown in figure) and the force becomes F2F_{2}. The value of F1:F2F_{1}:F_{2} is:

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