What should be the angular speed of earth, so that body lying on equator may appear weightlessness $ (g = 10\,m/{s^2},\,\,R = 6400\,km)$
$\frac{1}{{800}}rad/s$
$\frac{1}{{400}}rad/s$
$\frac{1}{{600}}rad/s$
$\frac{1}{{100}}rad/s$
Time period of simple pendulum increases by an amount $\sqrt 2 $ times at height $'h'$ from the surface of earth. Then the value of $h$ is
What should be the angular speed of the earth, so that a body lying on the equator may appear weightlessness $(g = 10\,m/s^2, R = 6400\,km)$
A particle of mass $M$ is at a distance $'a'$ from surface of a thin spherical shell of uniform equal mass and having radius $a$
If the gravitational acceleration at surface of Earth is $g$ , then increase in potential energy in lifting an object of mass $m$ to a height equal to half of radius of earth from surface will be
Two spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F.$ The space around the masses is now filled with a liquid of specific gravity $3.$ The gravitational force will now be