The acceleration due to gravity at a height $1\, km$ above the earth is the same as at a depth $d$ below the surface of earth. Then $d\,=$ ......... $km$
$\frac{4}{3}$
$\;\frac{3}{2}$
$\;\frac{2}{3}$
$2$
At a given place where acceleration due to gravity is $‘g’$ $m/{\sec ^2}$, a sphere of lead of density $‘d’$ $kg/{m^3}$ is gently released in a column of liquid of density $'\rho '\;kg/{m^3}$. If $d > \rho $, the sphere will
If mass of a body is $M$ on the earth surface, then the mass of the same body on the moon surface is
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V.$ Due to the rotation of planet about its axis the acceleration due to gravity $g$ at equator is $1/2$ of $g$ at poles. The escape velocity of a particle on the planet in terms of $V.$
$Assertion$ : The length of the day is slowly increasing.
$Reason$ : The dominant effect causing a slowdown in the rotation of the earth is the gravitational pull of other planets in the solar system.
The weight of an object in the coal mine, sea level, at the top of the mountain are ${W_1},\;{W_2}$ and ${W_3}$ respectively, then