Physics MCQs for NEET — Practice Questions with Answers

The value of gravitational acceleration (g) increases as we move from the equator to the poles. This occurs because the Earth is not a perfect sphere but an oblate spheroid, meaning it is slightly flattened at the poles and bulging at the equator. The radius of the Earth is smaller at the poles than at the equator, resulting in a stronger gravitational pull at the poles.

$ g = { GM \over R^2 } and M = { 4 \over 3 } \pi R^3 \rho $ $ \therefore g = { G \over R^2 } . {4 \over 3 } \pi R^3 \rho \Rightarrow \rho = { 3g \over 4 \pi RG } $

for condition of weight lessness at equator $ \omega = \sqrt { g / R } $ $ = \sqrt { 10 \over 6400 \times 10^3 } = { 1 \over 800} rad /sec $

Time peripd of simple pendylym $ T = 2 \pi \sqrt { l /g } $ In artificial satellite g ' = 0 $ \therefore T = \infty $

$ { g' \over g } = \left( { R \over (R+h) } \right)^2 = \left( {6400 \over 6400+64} \right) \Rightarrow g' = 960.400 ms^{-2} $

$ g' = g - \omega^2 R cos^2 \lambda $ Rotation of the earth results in the decreased weight apparently.this decrease in weight is not felt at the poles as the angle of latitude is 90'

using $ g = { GM \over R^2 } $we get $ g_m = g/5 $

$ g^1 = g ({ 1 - { d\over R }} ) \Rightarrow d = { (n-1) \over n } R $

$ g = { 4 \over 3 } \pi GR \rho \, and, g' = {4 \over 3 } \pi GR' \rho $ $ \therefore { g' \over g } = {R' \over R } = 0.2 \Rightarrow g' = 0.2 g $

mass does not vary from place to place