The radii of two planets are respectively ${R_1}$ and ${R_2}$ and their densities are respectively ${\rho _1}$ and ${\rho _2}$. The ratio of the accelerations due to gravity at their surfaces is
${g_1}:{g_2} = \frac{{{\rho _1}}}{{R_1^2}}:\frac{{{\rho _2}}}{{R_2^2}}$
${g_1}:{g_2} = {R_1}{R_2}:{\rho _1}{\rho _2}$
${g_1}:{g_2} = {R_1}{\rho _2}:{R_2}{\rho _1}$
${g_1}:{g_2} = {R_1}{\rho _1}:{R_2}{\rho _2}$
A simple pendulum doing small oscillations at a place $\mathrm{R}$ height above earth surface has time period of $T_1=4 \mathrm{~s}$. $T_2$ would be it's time period if it is brought to a point which is at a height $2 R$ from earth surface. Choose the correct relation $[R=$ radius of Earth]:
The mass of a planet is $\frac{1}{10}^{\text {th }}$ that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
The value of $‘g’$ at a particular point is $9.8\,m/{s^2}$. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of $‘g’$ at the same point (assuming that the distance of the point from the centre of earth does not shrink) will now be ......... $m/{\sec ^2}$.
The diameters of two planets are in ratio $4:1$ . Their mean densities have ratio $1:2$ . The ratio of gravitational acceleration on the surface of planets will be
In both figures shown below a hole along the diameter of earth. In first, a particle is released from $A$ and it oscillated with time period $T_1$. In second figure, same particle is released from point $B$ and it oscillates with time period $T_2$ then [$O$ is centre of earth]