Physics MCQs for NEET — Practice Questions with Answers

The escape velocity ( u) from the Earth's gravitational field is given by the formula $v = \\sqrt{2gr}$. Here, g is the acceleration due to gravity and r is the radius of the Earth. This formula is derived from equating the kinetic energy required to escape Earth's gravity to the gravitational potential energy.

Because it does not depend on the mass of projectile

The escape velocity ( u) can be derived from the relationship between gravitational force and the required kinetic energy to escape. Given the mass of the Earth (M), its mean density ( ho), its radius (R), and the gravitational constant (G), the formula is $ v = R \\sqrt{\\frac{8 \\pi}{3} G \\rho} $. This formula is derived from the standard escape velocity formula and the relation between density, mass, and volume.

Gravitational potential of A at 0 = $ - { GM \over r/2 } = - { 2GM \over r } $

of B at 0 = $ - { GM \over r/2 } = - { 2GM \over r } $ Total potential at 0 = $- { 4GM \over r } $

$ Ve \alpha { 1\over \sqrt r } $ $ { V_1 \over V_2 } = \sqrt{ r_2 \over r_1 } = \sqrt { R+7R \over R } = 2 \sqrt 2 \Rightarrow V_2 = { V_1 \over 2 \sqrt 2 } $

$ v = \sqrt { GM \over R+ h } = {1 \over 2 } \sqrt { 2GM \over R } $ $ \therefore 4R = 2 (R+h )$ $ \therefore h = R = 6400 km$

$ { V_p \over V_e } = \sqrt {Mp \over Me} \times { Re \over Rp } = \sqrt { 6 \times {1 \over 2 } } = \sqrt 3 $ $ \therefore Ve = \sqrt 3 Ve $

$ \upsilon = \sqrt{2gR} \therefore {V1 \over V2 } = \sqrt {g1 \over g2 } .{R1\over R2} = \sqrt {gk} = (kg)^ {1/2} $

$ \therefore Ve = \sqrt {2GM \over R } \therefore Ve \alpha \sqrt { M \over R } $ If M becomes double and R becomes half then escape velocity; becomes two times

Due to three particles net intensity at the center $ \vec I = \vec I_A + \vec I_B + \vec I_C = 0 $ because out of these three intensities ARE equal in magnitude and between each other is 120’