Gravitation Test

Result

Gravitation Test - 64

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$g_{1}$$ is the acceleration due to gravity on the surface of earth, $$g_{2}$$ that at a height $$h$$ ($$h <<$$ radius of earth) and $$g_{3}$$ that at a depth $$h$$, then their ascending order is

A
$$g_{3}<g_{2}<g_{1}$$

B
$$g_{3}<g_{1}<g_{2}$$

C
$$g_{1}<g_{2}<g_{3}$$

D
$$g_{2}<g_{3}<g_{1}$$

Solution
Question 2
1 / -0

The value of universal gravitational constant on earth for a particle of mass 5 kgs is

A
$$6.67\times { 10 }^{ -11 }$$

B
$$6.67\times { 10 }^{ -7 }$$

C
$$5\times 6.67\times { 10 }^{ -11 }$$

D
$$6.67\times { 10 }^{ -23 }$$

Solution

A universal gravitational constant is a constant number that does not depends on the masses of any planets or objects. And it is equal to $$G=6.67\times 10^{-11}$$ $$Nm^2kg^{-2}$$
Option A
Question 3
1 / -0

The value of universal gravitational constant depend upon:

A
Nature of material of two bodies

B
Heat constant of two bodies

C
Acceleration of two bodies

D
None of these

Solution

Every object in the universe attracts every another object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects.

$$F\propto \dfrac{m_1m_2}{r^2}$$

$$F=G \dfrac{m_1m_2}{r^2}$$
where $$G$$ is the constant of proportionality and is called the universal gravitation constant. $$G$$ does not depend on anything.
Question 4
1 / -0

Acceleration due to gravity at surface of a planet is equal to that at surface of the earth and density is $$1.5$$ times that of earth. if radius of earth is $$R$$, radius of planet is

A
$$ \dfrac {R }{ 1.5 } $$

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

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

D
$$ \dfrac {4 }{ 9 } R $$

Solution

Gravity at surface of a planet $$=$$ gravity at surface of the earth.
$${ g }_{ p }={ g }_{ e }$$
$${ P }_{ 1 }={ 1.5P }_{ e }$$
Radius of earth $$=R$$
$$g=\dfrac { G\left( P\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) }{ { R }^{ 2 } } $$
It's given, $${ P }_{ P }=1.5{ P }_{ E }$$
$$\dfrac { G\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 }\times { P }_{ e } }{ { R }^{ 2 } } =\dfrac { G\times \dfrac { 4 }{ 3 } \pi \times { \left( { R }^{ 1 } \right) }^{ 3 }\times { P }_{ P } }{ { \left( { R }^{ 1 } \right) }^{ 2 } } $$
or, $$R.{ P }_{ e }={ R }^{ 1 }{ P }_{ P }$$
or, $${ R }^{ 1 }=R\times \dfrac { { P }_{ P } }{ { P }_{ P } } =\dfrac { R }{ 1.5 } =\dfrac { 2R }{ 3 } $$.
Question 5
1 / -0

Which of the following graphs shows the variation of acceleration due to gravity $$g$$ at height $$h$$ from the surface of earth?
Assume ($$h$$ <<<<$$Re$$)

A
$$a$$

B
$$b$$

C
$$c$$

D
$$d$$

Solution
Question 6
1 / -0

In the figure it is shown that the velocity of lift is $$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ while string ins winding on the motor shaft with velocity $$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ and shaft A is moving downward with velocity $$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ with respect lift, then find out the velocity of block B

A
$$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \uparrow $$

B
$$2\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \downarrow $$

C
$$4\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \uparrow $$

D
None of these

Solution

Lift $$=2$$ m/s
velocity $$=2$$ m/s
down ward velocity $$=2$$ m/s
$$\therefore$$ velocity of Block $$B=\uparrow 2+2\uparrow -2\downarrow $$
$$=4-2$$
$$=4-2$$
$$=2m/s\uparrow $$
Question 7
1 / -0

The value of $$'g'$$ reduces to half of its value at surface of earth at a height $$'h'$$, then

A
$$h=R$$

B
$$h=2R$$

C
$$h=(\sqrt{2}+1)R$$

D
$$h=(\sqrt{2}-1)R$$

Solution
Question 8
1 / -0

If the earth were to suddenly contract to half its present size, without any change in its mass, the duration of the new day be

A
$$18\ hours$$

B
$$30\ hours$$

C
$$6\ hours$$

D
$$112\ hours$$

Solution

Given,
$$r_2=\dfrac{1}{2}r_1$$
$$m_1=m_2=m(say)$$
$$T_1=24hr$$
By the conservation of angular momentum,
$$I_1\omega _1=I_2\omega _2$$. . . . . . . . .(1)
Where, $$\omega =\dfrac{2\pi}{T}$$ angular velocity of the earth
$$I=\dfrac{2}{5}mr^2$$ moment of inertia of earth.
From equation (1),
$$\dfrac{2}{5}mr_1^2.\dfrac{2\pi}{T_1}=\dfrac{2}{5}mr_2^2.\dfrac{2\pi}{T_2}$$
$$r_1^2.\dfrac{1}{T_1}=r_2^2.\dfrac{1}{T_2}$$
$$\dfrac{T_2}{T_1}=(\dfrac{r_2}{r_1})^2=(\dfrac{r_1/2}{r_1})^2=\dfrac{1}{4}$$
$$T_2=T_1.\dfrac{1}{4}$$
$$T_2=\dfrac{24}{4}hr=6hr$$
The correct option is C.
Question 9
1 / -0

The acceleration due to gravity on a planet is $$1.96\,m/{s^2}$$. If it is safe jump from a height of $$3$$m on the earth, the corresponding height on the planet will be

A
3 m

B
6 m

C
9 m

D
1.5 m

Solution
Question 10
1 / -0

If the radius of the earth is suddenly contracts to half of its present value, then duration of day will be of

A
6 hours

B
12 hours

C
18 hours

D
24 hours

Solution