Physics MCQs for NEET — Practice Questions with Answers

The intensity of the gravitational field inside a spherical shell is zero. This can be understood by applying the shell theorem, which states that a uniform spherical shell of mass exerts no net gravitational force on a particle located inside it. Hence, the gravitational field intensity at the center of the spherical shell is zero.

$ P.E U = - { GMm \over r } = { GMm \over R+ h } $ $ U _ { initial } = - { GMm \over 3R } and U_{ final } = -{ GMm \over 2r } $ $ loss of P.E = gain in K.E = { GMm \over 2 R } - { GMm \over 3R } = { GMm \over 6R } $

$ \triangle U = U _2 - U_1 = { mgh \over 1+h/Re } = { mgRe \over 1 + { Re \over Re} } = { mgRe \over 2 } $ $ \therefore U_2 - ( - mgRe ) = { mgRe \over 2 } $ $ \therefore U_2 = - { 1 \over 2 } mgRe $

If body is projected with velocity Ï… (Ï… Ve) then height up to which it will rise but $ u = { ve \over 2 } $ $ \therefore h = { R \over \left( { Ve \over Ve/2 } \right)^2 -1 } $ $ { R \over 4-1} = R/3 $

change in potential energy in displacing a body from $r_1$ and $r_2$ is given by $ \triangle U = GMm \left[ { 1 \over r_1 } - { 1 \over r_2 } \right] = GMm \left[ { 1 \over 2R } - { 1 \over 3R } \right] = { GMm \over 6R} $

$ K.E = { GMm \over 2R } \therefore K.E \alpha { 1 \over R} $

this should be equal to escape velocity i.e = $ \sqrt { 2gR } $

Escape velocity does not depend on the mass of the projectile

$ { V_p \over V_e } = \sqrt { {g_p \over g_e } . { R_p \over R_e} } = \sqrt { 2 \times 2 } = 2 $ $ \Rightarrow V_e = 2 V_e = 2 \times 11.2 = 22.4 km^{-1} $

$Ve = \sqrt { 2GM \over R} = R \sqrt { 8/3 \pi Gp } $ $ \therefore Ve \alpha R if \rho constant $ since the planet is having double radius incomparision to earth therefore the escape velocity becomes twice i.e. $ 22 kms^{-1}$