A mass falls from a helght $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of earth it is found that $t =2 T$. The entre setup is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as $t'$ and $T'$.
A mass falls from a helght $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of earth it is found that $t =2 T$. The entre setup is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as $t'$ and $T'$.
- [NEET 2019]
- A
$\mathrm{t}^{\prime}=\sqrt{2} \mathrm{T}^{\prime}$
- B
$\mathrm{t}^{\prime} > 2 \mathrm{T}^{\prime}$
- C
$\mathrm{t}^{\prime} < 2 \mathrm{T}^{\prime}$
- D
$\mathrm{t}^{\prime}=2 \mathrm{T}^{\prime}$
Similar Questions
The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth ?
The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth ?
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
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
- [AIIMS 2013]
Find the gravitational field at a distance of $2000\, km$ from centre of earth. (in $m / s ^{2}$)
(Given $\left.R_{\text {earth }}=6400 km , r =2000 km , M _{\text {earth }}=6 \times 10^{24} kg \right)$
Find the gravitational field at a distance of $2000\, km$ from centre of earth. (in $m / s ^{2}$)
(Given $\left.R_{\text {earth }}=6400 km , r =2000 km , M _{\text {earth }}=6 \times 10^{24} kg \right)$
- [AIIMS 2019]
The mass of the moon is $\frac{1}{{81}}$ of the earth but the gravitational pull is $\frac{1}{6}$ of the earth. It is due to the fact that
The mass of the moon is $\frac{1}{{81}}$ of the earth but the gravitational pull is $\frac{1}{6}$ of the earth. It is due to the fact that
The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $'R'$ is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)
The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $'R'$ is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)
- [JEE MAIN 2019]