Acceleration Due to Gravity

The density of a planet is three times that of earth and radius $2.5$ times that of earth. If $g_p$ and $g_e$ represents acceleration due to gravity on planet and earth respectively then ratio $g_p$ to $g_e$ is

Time period of a simple pendulum is $T_1$ when on the earth's surface and $T_2$ when taken to a height '2R' above the earth's surface, where 'R' is the radius of the earth. The ratio $T_1 : T_2$ is

The time taken by simple pendulum for one oscillation is T on earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. The value of x is

The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $\frac{g}{9}$, where $g$ is acceleration due to gravity on earth's surface, is

______.

Select the correct statement out of the following

Given below are two statements:

Statement I: Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.

Statement II: Acceleration due to earth's gravity is same at a height '$h$' and depths '$d$' from earth's surface, if $h = d$.

In the light of above statements, choose the most appropriate answer from the options given below.

The percentage decrease in the weight of a body when taken to a height of $48$ km above the surface of the earth is (Radius of the earth is $6400$ km)

The lengths of seconds pendulums on the surface of the earth and at an altitude '$h$' from the surface of the earth are $l_s$ and $l_h$ respectively. The radius of the earth is

If a capillary tube is immersed in water, rise of water in capillary is $h_1$. When the whole arrangement is taken to a depth $d$ in a mine, the water level rises to $h_2$. The ratio $\frac{h_2}{h_1}$ is (R=radius of earth)

The acceleration due to gravity at a height 'h' above the surface of earth is '$g_h$'. At the depth 90 km below the earth's surface the acceleration due to gravity is also '$g_h$'. The value of 'h' is

When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface, the change in g is approximately $\alpha\%$. The value of $\alpha$ is ______.

(Take radius of the earth $= 6400\text{ km}$.)

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$th as much as at present is ($g$ = gravitational acceleration, $R$ = equatorial radius of the earth.)

What should be the angular velocity of earth due to rotation about its own axis so that the weight at $th$

equator becomes $\left(\frac{3}{5}\right)$ of initial value? ($g =$ acceleration due to gravity, R = radius of earth)

Two planets A and B have densities ' $\rho_1$ ', ' $\rho_2$ ' and have radii ' $r_1$ ', ' $r_2$ ', respectively. The ratio of acceleration due to gravity on $A$ to that of $B$ is

Acceleration due to gravity on the surface of earth is ' $g$ '. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ________ g.

A body weighs 45 N on the surface of the earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be

Time period of a simple pendulum on earth's surface is ' T '. It time period becomes ' xT ' when taken to a height ' $2R$ ' above earth's surface. The value of $x$ will be $(R =$ radius of earth)

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

The depth at which the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is

( $R =$ radius of the earth)

The gravitational pull of the moon is $\left(\frac{1}{6}\right)^{th}$ of the earth and mass of moon is $\left(\frac{1}{8}\right)$th of the earth. This implies that the

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is ( $R =$ radius of earth, $g =$ acceleration due to gravity) ( $n =$ integer)

Water rises up to height ' $x$ ' in a capillary tube immersed vertically in water. When the whole arrangement is taken to a depth ' $d$ ' in a mine, the water level rises up to height Y. If R is the radius of the earth then the ratio $Y : x$ is

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth, then radius of the planet would be

The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

The mass of a planet is $\frac{1}{10}$th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

The acceleration due to gravity on the surface of earth is $g$. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:

If ' $R$ ' is the radius of earth \& ' $g$ ' is acceleration due to gravity on earth's surface, then mean density of earth is

A boy weighs 72 N on the surface of earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be

A simple pendulum has a periodic time ' $T_1$ ' when it is on the surface of earth of radius ' $R$ '. Its periodic time is ' $T_2$ ' when it is taken to a height ' $R$ ' above the earth's surface. The value of $\frac{T_2}{T_1}$ is

The height ' $h$ ' above the earth's surface at which the value of acceleration due to gravity $(g)$ becomes $\left(\frac{g}{3}\right)$ is ( $R =$ radius of the earth)

The height at which the weight of the body becomes $\left(\frac{1}{16}\right)^{\text{th}}$ of its weight on the surface of the earth of radius ' $R$ ' is

The height above the earth's surface at which the acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface is ( $R =$ radius of earth)

A $90\text{ kg}$ body placed at $2R$ distance from surface of earth experiences gravitational pull of: ($R =$ Radius of earth, $g = 10\text{ m s}^{-2}$)

The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R =$ radius of the earth)

Assuming the earth to be a sphere of uniform mass density, a body weighed $300\text{ N}$ on the surface of earth. How much it would weigh at R/4 depth under surface of earth?

The height ' h ' from the surface of the earth at which the value of ' $g$ ' will be reduced by $64\%$ than the value at surface of the earth is ( $R =$ radius of the earth)

An object of mass $100\text{ kg}$ falls from point $A$ to $B$ as shown in figure. The change in its weight, corrected to the nearest integer is ($R_E$ is the radius of the earth)

If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = \pi^2\text{ m/s}^2$, then the length of the second's pendulum at a height $h = 2R$ from the surface of earth will be:

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4\text{ m}$, then the time period of small oscillations will be __________ s. [take $g = \pi^2\text{ m s}^{-2}$]

A pendulum is oscillating with frequency ' $n$ ' on the surface of earth. If it is taken to a depth $\frac{R}{4}$ below the surface of earth, new frequency of oscillation of depth $\frac{R}{4}$ is ( $R =$ radius of earth)

The weights of an object are measured in a coal mine of depth ' $h_1$ ', then at sea level of height ' $h_2$ ' and lastly at the top of a mountain of height ' $h_3$ ' as $W_1, W_2$ and $W_3$ respectively. Which one of the following relations is correct? [$h_1 \ll R, h_3 \gg h_2 = R, R =$ radius of the earth ]

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as $R$.)

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ( $g =$ gravitational acceleration, $R =$ equatorial radius of the earth)

If two planets have their radii in the ratio $x : y$ and densities in the ratio $m : n$, then the acceleration due to gravity on them are in the ratio

Consider a planet whose density is same as that of the earth but whose radius is three times the radius '$R$' of the earth. The acceleration due to gravity '$g_n$' on the surface of planet is $g_n = x$. $g$ where $g$ is acceleration due to gravity on surface of earth. The value of '$x$' is

Two planets A and B of radii $R$ and 1.5 R have densities $\rho$ and $\rho/2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to $A$ is:

The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are $\rho$ and $\rho/3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_A : g_B)$ will be:

Considering earth to be a sphere of radius '$R$' having uniform density '$\rho$', then value of acceleration due to gravity '$g$' in terms of $R, \rho$ and $G$ is

If $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be:

Time period of simple pendulum on earth's surface is '$T$'. Its time period becomes '$xT$' when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then the value of '$x$' will be

The height at which the weight of the body becomes $\left(\frac{1}{9}\right)$th its weight on the surface of earth is $(R =$ radius of earth)

The weight of a body on the surface of the earth is $100\text{ N}$. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

A body weight $W$, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: A pendulum clock when taken to Mount Everest becomes fast.

Reason R: The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.

In the light of the above statements, choose the most appropriate answer from the options given below

The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth $R_e = 6400\text{ km}$):

A mine is located at depth $R/3$ below earth's surface. The acceleration due to gravity at that depth in mine is ($R =$ radius of earth, $g =$ acceleration due to gravity)

The depth at which acceleration due to gravity becomes $\frac{g}{2n}$ is $(R =$ radius of earth, $g =$ acceleration due to gravity on earth's surface, $n$ is integer)

A body weighs $300\text{ N}$ on the surface of the earth. How much will it weigh at a distance $\frac{R}{2}$ below the surface of earth? ( $R \rightarrow$ Radius of earth)

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d = \frac{R}{2}$ from the surface of earth, if its weight on the surface of earth is 200 N, will be:

(Given R = radius of earth)

The weight of a body on the earth is $400\text{ N}$. Then weight of the body when taken to a depth half of the radius of the earth will be:

Given below are two statements:

Statement I: Acceleration due to gravity is different at different places on the surface of earth. Statement II: Acceleration due to gravity increases as we go down below the earth's surface. In the light of the above statements, choose the correct answer from the options given below

Given below are two statements:

Statement I: Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.

Statement II: Acceleration due to earth's gravity is same at a height 'h' and depth 'd' from earth's surface, if h = d.

In the light of above statements, choose the most appropriate answer from the options given below

Given below are two statements:

Statement I: Rotation of the earth shows effect on the value of acceleration due to gravity (g) Statement II: The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.

In the light of the above statements, choose the correct answer from the options given below

There is a second's pendulum on the surface of earth. It is taken to the surface of planet whose mass and radius are twice that of earth. The period of oscillation of second's pendulum on the planet will be

A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing $W$ on earth will weigh on that planet:

The acceleration due to gravity at height $h$ above the earth if $h \ll R$ (Radius of earth) is given by

A simple pendulum is oscillating with frequency '$F$' on the surface of the earth. It is taken to a depth $\frac{R}{3}$ below the surface of earth. ( $R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

The value of acceleration due to gravity at a depth '$d$' from the surface of earth and at an altitude '$h$' from the surface of earth are in the ratio

At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3R$ above earth surface. Where $R$ is Radius of earth (Take $R = 6400\text{ km}$ ). The depth $d$ is equal to

Earth is assumed to be a sphere of radius R. If '$g_\phi$' is value of effective acceleration due to gravity at latitude $30^\circ$ and '$g$' is the value at equator, then the value of $|g - g_\phi|$ is ($\omega$ is angular velocity of rotation of earth, $\cos 30^\circ$ $= \frac{\sqrt{3}}{2}$)

The radius of earth is $6400\text{ km}$ and acceleration due to gravity $g = 10\text{ m s}^{-2}$. For the weight of body of mass $5\text{ kg}$ to be zero on equator, rotational velocity of the earth must be (in $\text{rad/s}$)

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)

(Take g = 10 m s$-2$, radius of earth = 6400 km)

If the radius of earth shrinks by $2\%$ while its mass remains same. The acceleration due to gravity on the earth's surface will approximately:

The radii of two planets A and B are in the ratio 2: 3. Their densities are 3$\rho$ and 5$\rho$ respectively. The ratio of their acceleration due to gravity is:

The percentage decrease in the weight of a rocket, when taken to a height of $32\text{ km}$ above the surface of earth will, be:

$($ Radius of earth $= 6400\text{ km})$

The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be: (Given g = acceleration due to gravity at the surface of earth).

A body weighs $500\text{ N}$ on the surface of the earth. At what distance below the surface of the earth it weighs $250\text{ N}$? (Radius of earth, $R = 6400\text{ km}$)

The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by:

(Given R = radius of earth)

The length of a seconds pendulum at a height h = 2R from earth surface will be: (Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $\pi$2 ms${}^{-2}$)

An object is taken to a height above the surface of earth at a distance $\frac{5}{4}$ R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be:

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is:

[Radius of earth R = 6400 km and $\sqrt{3}$ = 1.732]

If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth $\alpha h\ (h \ll R_e)$ from the earth surface. The value of $\alpha$ will be _________.

(use $R_e = 6400\text{ km})$

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude. Reason R: At equator, the direction of acceleration due to the gravity is towards the center of earth. In the light of above statements, choose the correct answer from the options given below:

The average density of the earth is [g is acceleration due to gravity]

At a height 'R' above the earth's surface the gravitational acceleration is (R = radius of earth, g = acceleration due to gravity on earth's surface)

When the value of acceleration due to gravity '$g$' becomes $\frac{g}{3}$ above surface of height '$h$' then relation between '$h$' and '$R$' is ( $R =$ radius of earth)

The depth from the surface of the earth of radius $R$, at which acceleration due to gravity will be $60\%$ of the value on the earth surface is

The depth at which acceleration due to gravity becomes $\frac{g}{n}$ is [ $R$ = radius of earth, $g =$ acceleration due to gravity, $n =$ integer $]$

The depth 'd' below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is $(R =$ radius of the earth)

If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is: (Given: r < RE)

The length of the seconds pendulum is lm on earth. If the mass and diameter of the planet is 1.5 times that of the earth, the length of the seconds pendulum on the planet will be nearly

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

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be:

For a body of mass '$m$', the acceleration due to gravity at a distance '$R$' from the surface of the earth is $\left(\frac{g}{4}\right)$. Its value at a distance $\left(\frac{R}{2}\right)$ from the surface of the earth is ( $R =$ radius of the earth, $g =$ acceleration due to gravity)