NEET Questions

Assertion: The time period of a pendulum, on a satellite orbiting the earth is infinity.Reason: Time period of a pendulum is inversly proportional to $\sqrt{g}$

A body is projected from earth's surface to become its satellite, its time period of revolution will not depend upon

Assuming the radius of the earth to be $6.4 imes 10^6 \mathrm{~m}$. What is the time period $T$ and speed of satellite for equatorial orbit at $1.4 imes 10^3 \mathrm{~km}$ above the surface of the earth?

Pertaining to two planets, the ratio of escape velocities from respective surfaces is $1: 2$, the ratio of the time period of the same simple pendulum at their respective surfaces is $2: 1$ (in same order). Then the ratio of their average densities is

I. The geo-stationary satellite always has circular orbit.II. The time period of satellite depends on mass of the satellite, mass of the earth and height of satellite from the surface of earth.Which of the following statement(s) is/are incorrect?

If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is $(g=$ acceleration due to gravity near the poles and $R$ is the radius of earth) (Ignore equatorial bulge)

The ratio of product of velocity and time period of electron orbiting in $2^{ext {nd}}$ and $3^{ext {rd}}$ stable orbits.

Two planets ${A}$ and ${B}$ have radius in the ratio 1: 6 . A satellite revolving near the surface of planets have same time period of revolution. The ratio of density of planets $A$ and $B$ is:

If a simple pendulum based clock is taken to deep inside earth surface, then it

A hole is bored along the diameter of the earth and a particle is dropped into it. If $R$ is the radius of the earth and $g$ is the acceleration due to gravity at the surface of the earth, then the time period of oscillation of the particle is