Gravitation Test

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Gravitation Test - 38

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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 figure (where $$R$$ is radius of the earth).

A

B

C

D

Solution

For depth: $$g \propto (R-d)$$
For height: $$g \propto \cfrac{1}{R^2}$$
Hence, the graph will be linear for depth from centre of earth $$(R=0)$$ till the surface of earth $$(R)$$ and then parabolic with increase in height from earth.
Question 2
1 / -0

Above the earth's surface, the variation of g w.r.t. the height (r) is correctly represented by which of the given proportionalities

A
$$g \propto \dfrac {1}{r^2}$$

B
$$g \propto r$$

C
$$g \propto \ r^2$$

D
$$g \propto r^o$$

Solution

The variation of g with the height(r) $$=g_r=\dfrac{GM}{(R+r)^2} $$
$$\Rightarrow g_r $$ $$\alpha $$ $$\dfrac{1}{r^2}$$

Hence correct answer is option $$A $$
Question 3
1 / -0

The SI unit of G is.

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

B
$$Nm^2/kg$$

C
$$N \ ml \ kg$$

D
$$Nm^2/kg^2$$

Solution

$$\textbf{Correct option: Option (D).}$$

$$\textbf{Explanation:}$$

$$\bullet$$Gravitational force is the universal force of attraction experienced by all matters.

$$\bullet$$It is given by, $$F=\dfrac{GMm}{{{R}^{2}}}$$ (Where G is the universal gravitational constant whose value remains constant throughout the space, and M and m are the masses of the two objects and R be the distance between the bodies.)

Then, $$G=\dfrac{F{{R}^{2}}}{Mm}$$

For a force of 1 Newton between bodies of mass 1 kg and 1 meter apart, G will be
$$\bullet$$ S.I unit of G

$$\Rightarrow G=\dfrac{N\times {{m}^{2}}}{kg\times kg}$$

$$\Rightarrow G=N{{m}^{2}}/k{{g}^{2}}$$
$$\textbf{Thus, option (D) is correct.}$$
Question 4
1 / -0

Where will 'g' be greatest when one goes from the centre of earth to an altitude equal to the radius of the 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 the radius of earth be $$R$$ and density $$\rho$$
The value of $$g$$
for $$0<r<R$$ =$$\frac{G\rho r}{3}$$ (directly proportional to $$r$$)
$$r>R$$ =$$\frac{G\rho R^{3}}{3r^{2}}$$ (inversely proportional to r)

maximum value of $$g$$ is at $$r=R$$
$$g=\frac{G \rho R}{3}$$
Question 5
1 / -0

In the motion of the planets,

A
The angular velocity is constant

B
The linear momentum is constant

C
The angular momentum is constant

D
None of the above

Solution

In the motion of the planets the angular momentum is constant which leads to Kepler's second Law of motion.
Linear momentum and angular velocity cannot be conserved as motion of planets is elliptical.
Question 6
1 / -0

Maximum weight of the body is

A
at the centre of the eath

B
inside the earth

C
on the surface of the earth

D
above the the surface of the earth

Solution

$$\textbf{Hint: Formula for gravitational acceleration is g}$$ =$$\dfrac{GM}{{{R}^{2}}}$$

$$\textbf{Step 1: Weight of the body}$$

Weight of a body is (W)=mg.

Since, mass (m) is constant, ‘g’ is the variable force, also called gravitational force.

$$\textbf{Step 2: Utilising the formula for gravitational acceleration}$$

The effect on gravitational force can be found by considering the formula

g = $$\dfrac{GM}{{{R}^{2}}}$$

where;

G = Gravitational constant ($$6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$$)

M = Mass of Earth

R = distance between centre of the earth and the body.

$$\textbf{Step 3: Finding the effect of ‘g’ above the surface}$$

Now, when the body goes away from the centre of the Earth, ‘R’ increases.

Since $$g\propto \dfrac{1}{R}$$, the gravitational force decreases.

Hence, gravitational force above the surface of the Earth is lesser than at the surface.

$$\textbf{Step 4: Finding the effect of ‘g’ below the surface}$$

When the body goes towards the centre of the Earth, although ‘R’ decreases; ‘M’ decreases as the mass of earth below the body is lesser/decreasing.

Since $$g\propto M$$, the gravitational force decreases.

Hence, gravitational force below the surface of the Earth is lesser than at the surface.

$$\textbf{Step 5: Finding the effect of ‘g’ at the centre}$$

When the body is at the centre of the Earth, M=0.

⸫ g = 0

Hence, gravitational force at centre of Earth is 0.

$$\textbf{Step 6: Conclusion}$$

Since ‘g’ is maximum for body at the surface of the earth, then maximum weight of the body (W) will also be at the surface of the Earth.

$$\textbf{Answer:}$$

$$\textbf{Hence, the correct option is (C) on the surface of the Earth}$$
Question 7
1 / -0

If a body is sent with a velocity of ............... km $$\displaystyle { sec }^{ -1 }$$, it would leave the earth forever.

A
$$11.9$$

B
$$11.6$$

C
$$11.4$$

D
$$11.2$$

Solution

$$Answer:-$$ D
Escape velocity is the minimum speed needed for an object to escape from the gravitational attraction of a massive body. The escape velocity from Earth is about 11.186 km/s (40,270 km/h; 25,020 mph) at the surface. More generally, escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero.
And it can be calculated by :-
$$v_e=\sqrt{\dfrac{2Gm}{R}}$$
Question 8
1 / -0

From the centre of the earth to the surface of the earth, the relation between the value of g and distance (r) represented as a proportionality, is given by

A
$$g \propto \dfrac {1}{r^2}$$

B
$$g \propto r$$

C
$$g \propto r^2$$

D
$$g \propto r^o$$

Solution

The variation of g with the depth(r) $$=g_r=\dfrac{GM}{R^2}(1- \dfrac{r}{R}) $$

$$\Rightarrow g_r$$ $$\alpha$$ $$r$$

Hence correct answer is option $$B $$
Question 9
1 / -0

The maximum weight of a body on earth is

A
at the centre of the earth

B
inside the earth

C
on the surface of the earth

D
above the surface of the earth

Solution

the maximum weight of body on earth is on the surface of the earth because either you go down or up the surface of earth the gravity decreases.
so the answer is C.
Question 10
1 / -0

The value of $$g $$

A
Increases with increase in depth

B
Decreases with increase in depth

C
Remains same regardless of the depth

D
Depends upon the mass of the object

Solution

$$Answer:-$$ B
Variation of g with depth is given by:
$${ g }_{ d }=g(1-\dfrac { d }{ R } )$$
where,
$$g$$ is acceleration due to gravity on the surface,
$$d$$ is depth,
$$R$$ is radius of earth
so we can see that as value of d increases $$g_d$$ decreases.

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Gravitation Test - 38

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Gravitation Test - 38

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Question 1

Solution

Question 2

$$g \propto \dfrac {1}{r^2}$$

$$g \propto r$$

$$g \propto \ r^2$$

$$g \propto r^o$$

Solution

Question 3

$$N^2m^2/kg$$

$$Nm^2/kg$$

$$N \ ml \ kg$$

$$Nm^2/kg^2$$

Solution

Question 4

at the surface of earth

at the centre of earth

at the highest point

none of the above

Solution

Question 5

The angular velocity is constant

The linear momentum is constant

The angular momentum is constant

None of the above

Solution

Question 6

at the centre of the eath

inside the earth

on the surface of the earth

above the the surface of the earth

Solution

Question 7

$$11.9$$

$$11.6$$

$$11.4$$

$$11.2$$

Solution

Question 8

$$g \propto \dfrac {1}{r^2}$$

$$g \propto r$$

$$g \propto r^2$$

$$g \propto r^o$$

Solution

Question 9

at the centre of the earth

inside the earth

on the surface of the earth

above the surface of the earth

Solution

Question 10

Increases with increase in depth

Decreases with increase in depth

Remains same regardless of the depth

Depends upon the mass of the object

Solution

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