The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth,is (radius of earth is $6400 \ km$) (in $km$)

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then:

The mass of a spherical planet is $4$ times the mass of the earth, but its radius $(R)$ is the same as that of the earth. How much work is done in lifting a body of mass $5 \,kg$ through a distance of $2 \,m$ on the planet (in $\,J$)? (Take $g = 10 \,ms^{-2}$ for Earth)

The highest temperature, density, and pressure on Earth are found:

The time period of a simple pendulum increases by a factor of $\sqrt{2}$ at a height $h$ from the surface of the Earth. Then the value of $h$ is: