Gravitation Test

Result

Gravitation Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The value of 'g' at a certain height above the surface of the earth is 27% of its value on the earth's surface.The height is : (R=6300 km)

A
2100 km

B
13031 km

C
14031 km

D
14700 km
Question 2
1 / -0

If the gravitational potential on the surface of earth is $$V_0$$ then potential at a point at height half of the radius of earth is

A
$$\dfrac{V_0}{2}$$

B
$$\dfrac{2}{3}V_0$$

C
$$\dfrac{V_0}{3}$$

D
$$\dfrac{3V_0}{2}$$

Solution

$${u_r} = \frac{{ - GMm}}{r}\left( {Gravitational\,\,potential} \right)$$
$$ = {v_0}$$ at surface
at height $$h \sim \frac{r}{2}$$
$${u_{{r^2}}} \sim \frac{{ - GMm}}{{\frac{{3r}}{2}}} = \frac{2}{3}{V_0}$$
Hence,
option $$(B)$$ is correct answer.
Question 3
1 / -0

The amount of work done in lifting a body of mass 'm' from the surface of the earth to a height equal to twice the radius of the earth is

A
$$\dfrac{2GMm}{3R}$$

B
$$\dfrac{3GMm}{2R}$$

C
$$\dfrac{5GMm}{3R}$$

D
$$\dfrac{3GMm}{5R}$$

Solution
Question 4
1 / -0

At what temperature will the root mean square velocity of oxygen molecule be sufficient so as to escape from the earth? Escape velocity from the earth is $$11.0\ km/s$$ and the mass of the one molecule of oxygen $$5.34\times 10^{-25}\ kg$$ (Boltzmann constant $$k=1.38\times 10^{-23}\ J/K$$):

A
$$3\times 10^{5}\ K$$

B
$$3.5\times 10^{5}\ K$$

C
$$1.56\times 10^{5}\ K$$

D
none of these

Solution
Question 5
1 / -0

The loss in wight of a body taken from earth's surface to a height $$h$$ is 1%. The charge in weight taken into a mine of depth $$h$$ will be

A
$$1$$% loss

B
$$1$$% gain

C
$$0.5$$% gain

D
$$0.5$$% loss

Solution
Question 6
1 / -0

The weights of an object in a coal mine, at the surface and at the top of a mountain at the same piece on the earth are $$W_1, W_2, W_3$$ respectively then

A
$$W_1 > W_2 > W_3$$

B
$$W_3 > W_2 > W_1$$

C
$$W_1 = W_2 = W_3$$

D
$$W_1 < W_2 > W_3$$

Solution
Question 7
1 / -0

Rockets are lunched in Eastward direction to take advantage of:

A
The clear sky on Eastersn side.

B
The thinner atmosphere on this side.

C
both A and B

D
non of the above

Solution

Rockets are lunched in Eastward direction because, earth’s rotational energy is maximum at equator. When a spacecraft is launched, it should end up spinning around the earth quickly enough by the earth’s gravity. so, in east direction, initial velocity is equal to the earth surface i.e. $$1670\,Km/h$$
Question 8
1 / -0

The decrease in trhe value of g on going to a height above the earth's surface will be :-

A
$$g/2$$

B
$$\dfrac{5g}{9}$$

C
$$\dfrac{4g}{9}$$

D
$$\dfrac{g}{3}$$

Solution

$$g=\dfrac { GM }{ { R }^{ 2 } } $$
$$g$$ at height $$h$$
$${ g }_{ h }=\dfrac { GM }{ { \left( R+h \right) }^{ 2 } } \quad \quad \left( h=R/2 \right) $$
$${ g }_{ h }=\dfrac { GM }{ { \left( R+\dfrac { R }{ 2 } \right) }^{ 2 } } =\dfrac { 4GM }{ 9{ R }^{ 2 } } =\dfrac { 4g }{ 9 } $$
Decrease value of $$g$$
$$g-\dfrac { 4g }{ 9 } =\dfrac { 5g }{ 9 } $$
Question 9
1 / -0

The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

A
$$\dfrac{R_e}{4}$$

B
$$\dfrac{R_e}{2}$$

C
$$\dfrac{3R_e}{4}$$

D
$$R_e$$

Solution
Question 10
1 / -0

A particle of mass $$m_1$$ lies inside a spherical shell of mass $$m_2$$ and radius R at a distance r from the centre. The gravitational potential energy of the system is

A
$$-\dfrac{G(m_1+m_2)}{r}$$

B
$$\dfrac{Gm_1m_2}{R^2}$$

C
$$+\dfrac{Gm_1m_2}{r^2}$$

D
$$+\dfrac{Gm_1m_2}{R}$$