Gravitation Test

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Gravitation Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The variation of g with height or depth (r) is shown correctly by the graph in the figure (where R $$=$$ radius of the earth ),

A

B

C

D

Solution

for $$r\ge R;\quad g=\cfrac { GM }{ { r }^{ 2 } } $$ - eq. 1
For r<R; g=$$\cfrac { GMr }{ { R }^{ 3 } } $$ - eq.2
According to relations 1 and 2
the g vs r curve will be like
Question 2
1 / -0

An object is dropped at the surface of the earth from the height of $$3600$$ km. Calculate the ratio of the weight of the body at that height and on the surface of the earth.

A
$$1.34$$

B
$$2.44$$

C
$$6.25$$

D
$$12.32$$

Solution

$$g\alpha \dfrac{1}{(R+h)^2}$$
$$\therefore \dfrac{mg_{surface}}{mg_{height}}=\left(\dfrac{R+H}{R} \right)=\left(\dfrac{10000}{6400}\right)$$
$$\implies 2.44$$
Question 3
1 / -0

Where will g be greatest when one goes from the centre of earth to an altitude equal to radius of earth?

A
at the surface of earth

B
at the centre of earth

C
at the highest point

D
none of the above

Solution

Let at any height $$h$$ gravity due to earth will be $$g$$.
$$\Rightarrow mg=\dfrac{GMm}{(R+ h)^2}$$ $$\Rightarrow g=\dfrac{GM}{(R+h)^2}$$
$$\Rightarrow g \propto \dfrac{1}{(R+h)^2}$$
$$\therefore $$ As $$h$$ increases $$g$$ decreases.
$$\Rightarrow $$ At $$h=0$$ , $$g$$ will be maximum, $$g= \dfrac{GM}{R^2}$$
So , $$g$$ will be maximum at the surface of the earth.
Question 4
1 / -0

SI unit of G is $$Nm^{ 2 }{ kg }^{ -2 }$$. Which of the following can also be used as the SI unit of G?

A
$$ m^{ 3 }{ kg }^{ -1 }{ s }^{ -2 }$$

B
$$ m^{ 2 }{ kg }^{ -2 }{ s }^{ -1 }$$

C
$$ m { kg }^{ -3 }{ s }^{ -1 }$$

D
$$ m^{ 2 }{ kg }^{ -3 }{ s }^{ -2 }$$

Solution

Unit of $$G$$ is $$Nm^2kg^{-2}$$ as $$1N=kgms^{-2}$$.
So $$G$$ can be written as
$$kgms^{-2}\times m^2kg^{-2}=m^3kg^{-1}s^{-2}$$
Question 5
1 / -0

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

A
$$\dfrac { 4\pi G }{ 3gR } $$

B
$$\dfrac { 3\pi R }{ 4gG } $$

C
$$\dfrac { 3 G}{ 4\pi RG } $$

D
$$\dfrac { pi g }{ 12RG } $$

Solution
Question 6
1 / -0

What will be acceleration due to gravity on the surface of moon if its radius is $$\frac { 1 }{ 4 } $$th the radius of the earth and its mass is $$\frac { 1 }{ 80 }$$th the mass of the earth?

A
$$\frac { g }{ 2 } $$

B
$$\frac { g }{ 3 } $$

C
$$\frac { g }{ 7 } $$

D
$$\frac { g }{ 5 } $$

Solution

We know, $$g=\frac { { GM }_{ e } }{ { R }_{ e }2 } \\ and\quad g\quad '=\frac { { GM }_{ m } }{ { R }_{ m }2 } \\ \frac { g\quad ' }{ g } =\frac { M_{ m } }{ M_{ e } } \left( \frac { { R }_{ e } }{ { R }_{ m } } \right) ^{ 2 }\\ =\frac { 1 }{ 80 } \times \left( \frac { 4 }{ 1 } \right) ^{ 2 }=\frac { 1 }{ 5 } \\ \Rightarrow g' = \frac { g }{ 5 } $$
Question 7
1 / -0

SI unit of G is.

A
$$N^2 -m^2/kg$$

B
$$N -m^2/kg$$

C
$$N -m/kg$$

D
$$N -m^2/kg^2$$

Solution

Hint: using the formula of gravitional constant
Step1:We know that
$$\mathrm{F}=\mathrm{G} \times \dfrac{M_{1} M_{2}}{r^{2}} \\$$
$$\mathrm{~F} \times \mathrm{r}^{2}=\mathrm{G} \times \mathrm{M}_{1} \mathrm{M}_{2} \\$$
$$\mathrm{~F} \times \dfrac{r^{2}}{M_{1} M_{2}}=\mathrm{G} \\$$
$$\mathrm{G}=\mathrm{F} \times \dfrac{r^{2}}{M_{1} M_{2}}$$
SI Unit of
$$\text { Force }=\mathrm{F}=\text { Newton } \\$$
$$\text { Distance }=\mathrm{R}=\mathrm{m} \\$$
$$\text { Mass }=\mathrm{M}=\mathrm{kg}$$
Step2: Now,
$$\mathrm{G}=\mathrm{F} \times \dfrac{r^{2}}{M_{1} M_{2}}$$
$$\mathrm{G}=\dfrac{\text { Newton } \times \text { Meter }^{2}}{\text { Kilogram } \times \text { Kilogram }}$$
$$\mathrm{G}=\dfrac{\text { Newton Meter }^{2}}{\text { Kilogram }^{2}}$$
$$\mathrm{G}=\mathrm{Nm}^{2} / \mathrm{kg}^{2}$$
So, SI Unit of $$\mathrm{G}$$ is $$\mathrm{Nm}^{2} / \mathrm{kg}^{2}$$

$$\textbf{Thus, option (D) is correct.}$$
Question 8
1 / -0

At what height, is the value of $$g$$ half that on the surface of earth? ($$R =$$ radius of the earth)

A
$$0.414\ R$$

B
$$R$$

C
$$2\ R$$

D
$$3.5\ R$$

Solution

$$g\quad '=\frac { { gR }^{ 2 } }{ { r }^{ 2 } } =\frac { g }{ 2 } \\ \Rightarrow \quad { r }^{ 2 }={ 2R }^{ 2 }\\ \Rightarrow r=R\sqrt { 2 } =R+h\\ \Rightarrow h=R(\sqrt { 2 } -1)\\ =R(1.414-1)=0.414R$$
Question 9
1 / -0

The value of g near the earth's surface is.

A
$$8.9\ ms^{-2}$$

B
$$8.9\ ms^{-1}$$

C
$$9.8\ ms^{-2}$$

D
$$9.8\ ms^{-1}$$

Solution

Its value is $$9.8\ ms^{-2}$$ on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is $$9.8\ ms^{-2}$$. When discussing the acceleration of gravity, it was mentioned that the value of $$g$$ is dependent upon location.
Question 10
1 / -0

The magnitude of acceleration due to gravity at an altitude 'h' from the earth is equal to its magnitude at a depth 'd'. Find the relation between 'h'and 'd'. If the 'h'and 'd' both increases by $$50$$ %, are the magnitudes of acceleration due to gravity at the new altitude and the new depth equal.

A
$$d \ =\ h$$

B
$$d\ =\ 2h$$

C
$$\displaystyle h = \frac { d }{ 3 } $$

D
$$d\ =\ 4h$$

Solution

$${ g }_{ n }=g\left( 1-\cfrac { 2h }{ R } \right) \\ { g }_{ d }=g\left( 1-\cfrac { d }{ R } \right) $$
If $${ g }_{ n }={ g }_{ d }\\ g\left( 1-\cfrac { 2h }{ R } \right) =g\left( 1-\cfrac { d }{ R } \right) \\ \Rightarrow \cfrac { 2h }{ R } =\cfrac { d }{ R } \\ \therefore 2h=d$$