Gravitation Test

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Gravitation Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The mass of the moon is $$\left(\dfrac{1}{8}\right)$$ of the earth but the gravitational pull is $$\left(\dfrac{1}{6}\right)$$ of the earth. It is due to the fact that:

A
moon is the satellite of the earth

B
the radius of the earth is $$\left(\dfrac{8}{6}\right)$$ of the moon

C
the radius of the earth is $$\left(\sqrt{\dfrac{8}{6}}\right)$$ of the moon

D
the radius of the moon in $$\left(\dfrac{6}{8}\right) $$ of the earth

Solution

We know that, $$g = \dfrac{GM}{R^2}$$

$$\dfrac{g_{M}}{g_{E}}=\dfrac{\dfrac{G\left({M_{E}}/{8}\right)}{R_{M}^{2}}}{\dfrac{GM_{E}}{R_{E}^{2}}}=\dfrac{R_{E}^{2}}{8R_{M}^{2}}$$ ....(i)
where the sunscripts M and E are for the moon and the earth respectively.
Given, $$\dfrac{g_{M}}{g_{E}}=\dfrac{1}{6}$$ .....(ii)
From (i) and (ii), we get
$$\dfrac{R_{E}^{2}}{8R_{M}^{2}}=\dfrac{1}{6}$$ or
$${R}_{E}=\sqrt{\dfrac{8}{6}}R_{M}$$
Question 2
1 / -0

If density of a planet is double that of the earth and the radius 1.5 times that of the earth, the acceleration due to gravity on the surface of the planet is:

A
$$\dfrac{3}{4}$$ times that on the surface of the earth

B
3 times that on the surface of the earth

C
$$\dfrac{4}{3}$$ times that on the surface of the earth

D
6 times that on the surface of the earth

Solution

The correct option is B

We have,

The density of the planet is double of the earth

So, $$D_p=2D_e$$

The radius of the planet is equal to 1.5 times the radius of the earth

So, $$R_p=1.5R_e$$

Since$$ D_p=2D_e$$

$$\dfrac{M_p}{\dfrac{4}{3}\pi R_p^3}=2\times\dfrac{M_e}{\dfrac{4}{3}4R_e^3}$$

$$M_p=2\times(1.5)^3M_e$$

So,

$$\dfrac{g_p}{g_e}=\dfrac{\dfrac{GM_p}{R_P^2}}{\dfrac{GM_e}{R_e}^2}$$

$$g_P=3g_e$$
Question 3
1 / -0

If the radius of the earth is made three times, keeping the mass constant, then the weight of a body on the earth's surface will be as compared to its previous value is

A
one third

B
one ninth

C
three times

D
nine times

Solution

Weight is nothing but mass multiplied by the acceleration due to gravity
ie $$W=mg$$ ............(1)
If Mass of earth is $$M$$ and radius is $$r$$ then gravitational acceleration, $$g=GM/r^2$$

According to question, radius is increased by three times
ie $$r'=3r$$
$$g'=GM/(3r)^2=GM/9r^2=\dfrac{g}{9}$$
$$g'=\dfrac{g}{9}$$
As gravity became one-ninth so from equation (1) the weight will also become one-ninth because mass of the body remain same.
Question 4
1 / -0

Two spheres of radius $$r=0.5 \ m$$ and mass $$ m = 1 \ kg$$ are placed in contact with each other, what will be the gravitational force between them:

A
$$0 N$$, since the spheres are touching each other and r=0

B
$$6.67 \times 10^{-11}N$$

C
$$6.67 \times 10^{-4}N$$

D
$$6.67 \times 10^{-6}N$$

Solution

The centre of mass of each sphere lies at the centre of the sphere. Thus, the spheres can be considered to be equal to point masses and they are separated by a distance equal to $$R=2 \times r=1 \ m$$

The gravitational force of attraction is given by $$F = \dfrac{Gm_1m_2}{R^2}$$

Hence,
The force of attraction between them is $$F=\dfrac{Gm^2}{R^2}= \dfrac{G1^2}{1^2} = 6.67 \times 10^{-11}N$$

The correct option is (B)
Question 5
1 / -0

A body of weight $$W_1$$ is suspended from the ceiling of a room through a chain of weight $$W_2$$. The ceiling pulls the chain by a force equal to:

A
$$W_2$$

B
$$(W_1 + W_2)/2$$

C
$$(W_1 + W_2)$$

D
$$W_1 \times W_2$$

Solution

The ceiling pulls the chain with a force equal to the sum of weights of chain and the body. So the pulling force = (w1 + w2)
Question 6
1 / -0

The ratio between masses of two planets is 2 : 3 and the ratio between their radii is 3 : 2. The ratio between acceleration due to gravity on these two planets is:

A
4 : 9

B
8 : 27

C
9 : 4

D
27 : 8

Solution

Hint:Use the formula of acceleration due to gravity.

Step 1:Writing the acceleration due to gravity on both the planet,
Acceleration due to gravity on the first planet,
$${g_1} = \dfrac{GM_1}{R_1^2}$$ ..................(i)
Acceleration due to gravity on the second planet
$${g_1} = \dfrac{GM_2}{R_2^2}$$ ....................(ii)

Step 2:Dividing equation (i) by equation (ii),
$$\dfrac{g_1}{g_2} = \dfrac{M_1}{M_2} \times \dfrac{R_2^2}{R_1^2}$$
$$ = \dfrac{2}{3} \times {\left( {\dfrac{2}{3}} \right)^2}$$
$$ = \dfrac{8}{27}$$
$$\textbf{Option B is correct.}$$
Question 7
1 / -0

A cricket ball is thrown up with a speed of $$19.6\ ms^{-1}$$. The maximum height it can reach is

A
$$9.8\ m$$

B
$$19.6\ m$$

C
$$29.4\ m$$

D
$$39.2\ m$$

Solution

Given,
$$u=19.6m/s$$
$$v=0m/s$$
$$g=9.8m/s^2$$
From 3rd equation of motion,
$$2gh=v^2-u^2$$
$$-2\times 9.8\times h=0^2-19.6\times 19.6$$
$$h=\dfrac{19.6\times 19.6}{2\times 9.8}=19.6m$$
The correct option is B.
Question 8
1 / -0

The SI unit of the universal gravitational constant G:

A
$$Nm{{Kg}^{ - 2}}$$

B
$$N{m^2}{{Kg}^{ - 2}}$$

C
$$N{m^2}{{Kg}^{ - 1}}$$

D
$$Nm{{Kg}^{ - 1}}$$

Solution

Force $$=\dfrac{G\ M_{1}\ M_{2}}{R^{2}}$$

$$[G]=\dfrac{[Force][R]^{2}}{[M_{1}][M_{2}]}$$

$$=\dfrac{N\ m^{2}}{(kg)^{2}}$$

$$=N\ m^{2}\ kg^{-2}$$

Option $$B$$.
Question 9
1 / -0

$$F = G\dfrac{m_1m_2}{R^2}$$ is the formula to prove ______.

A
Newton's First law of motion

B
Newton's Second law of motion

C
Newton's Third law of motion

D
Newton's law of Gravitation

Solution

According to Newton's Law of Gravitation, every body attracts every other body with a constant force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$$F=G\dfrac{m_1m_2}{R^2}$$
Question 10
1 / -0

The gravitational force $$F_{g}$$ between two objects does not depend on

A
sum of the masses

B
product of mass

C
graviational constant

D
distance between the masses

Solution

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Gravitation Test - 56

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Gravitation Test - 56

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SHARING IS CARING

Question 1

moon is the satellite of the earth

the radius of the earth is $$\left(\dfrac{8}{6}\right)$$ of the moon

the radius of the earth is $$\left(\sqrt{\dfrac{8}{6}}\right)$$ of the moon

the radius of the moon in $$\left(\dfrac{6}{8}\right) $$ of the earth

Solution

Question 2

$$\dfrac{3}{4}$$ times that on the surface of the earth

3 times that on the surface of the earth

$$\dfrac{4}{3}$$ times that on the surface of the earth

6 times that on the surface of the earth

Solution

Question 3

one third

one ninth

three times

nine times

Solution

Question 4

$$0 N$$, since the spheres are touching each other and r=0

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

$$6.67 \times 10^{-4}N$$

$$6.67 \times 10^{-6}N$$

Solution

Question 5

$$W_2$$

$$(W_1 + W_2)/2$$

$$(W_1 + W_2)$$

$$W_1 \times W_2$$

Solution

Question 6

4 : 9

8 : 27

9 : 4

27 : 8

Solution

Question 7

$$9.8\ m$$

$$19.6\ m$$

$$29.4\ m$$

$$39.2\ m$$

Solution

Question 8

$$Nm{{Kg}^{ - 2}}$$

$$N{m^2}{{Kg}^{ - 2}}$$

$$N{m^2}{{Kg}^{ - 1}}$$

$$Nm{{Kg}^{ - 1}}$$

Solution

Question 9

Newton's First law of motion

Newton's Second law of motion

Newton's Third law of motion

Newton's law of Gravitation

Solution

Question 10

sum of the masses

product of mass

graviational constant

distance between the masses

Solution

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