Gravitation Test

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Gravitation Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The weight of a person on Earth is $$600\ N$$. Find his weight on the moon.

A
Zero

B
$$100\ N$$

C
$$600\ N$$

D
$$6300\ N$$

Solution

Given,
Weight of a person on earth, $$W_{earth}=600\ N$$
We know,
$$\text{Weight of the object on the moon}=\dfrac 16 \times \text{its weight on the earth}$$

$$\text{Weight of the object on the moon}=\dfrac 16 \times 600$$

$$\text{Weight of the object on the moon}=100\ N$$
Question 2
1 / -0

The force of gravitation between two bodies of mass 1 kg each separated by a distance of 1 m in vaccum is

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

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

C
$$6.67 \times 10^{-9} N$$

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

Solution

Given : $$m_1 =m_2 = 1 \ kg $$, $$r =1 \ m$$
Force of gravitation $$F = \dfrac{Gm_1m_2}{r^2}$$
$$F = \dfrac{6.67 \times 10^{-11}\times 1 \times 1 }{1^2} = 6.67 \times 10^{-11} N$$
Question 3
1 / -0

Which of the following units can be used to express G?

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

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

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

D
Nm$$^{-2}$$ Kg$$^{-3}$$

Solution

Gravitational force between two masses $$m$$ and $$M$$, separated by a distance $$r$$ is given by $$F = \dfrac{GMm}{r^2}$$
Thus, $$G = \dfrac{Fr^{2}}{Mm}$$
We know that the unit of force ($$F$$) is newton (N), the unit of mass ($$m\ or\ M$$) is kg and the unit of distance ($$r$$) is meter(m).
$$\text{The unit of G} = \dfrac{Nm^2}{kg^2} = Nm^2kg^{-2}$$
Question 4
1 / -0

During a planned manoeuvre in a space flight, a free-floating astronaut A pushes another free floating astronaut B, the mass of A being greater than that of B. Then, during a push

A
the acceleration of A is greater than that of B

B
the acceleration of A is less than that of B

C
neither is accelerated

D
their accelerations are equal in magnitude but opposite in direction

Solution

Forces exerted by two bodies are action-reaction pairs. So, the magnitude of the force exerted by a body A on B will be the same as the magnitude of force exerted by body B on A, irrespective of their masses.
So, $$F_{AB} = F_{BA}$$
$$m_Aa_A = m_Ba_B$$
Since, mass of A is greater than B, the acceleration of A is less than that of B.
Question 5
1 / -0

The gravitational force between two bodies is

A
repulsive at large distances

B
attractive at all places

C
attractive at short distances

D
repulsive at short distances

Solution

Gravitation is a natural phenomenon by which all things with mass are brought towards one another. It is an attractive force and behaves uniformly for all values of $$r$$ (distance between the two bodies).
Question 6
1 / -0

Specific gravity of the body is numerically equal to ________ of the body.

A
apparent weight in water

B
density

C
relative density

D
thermal capacity
Solution
- Specific gravity and relative density are numerically same.
Question 7
1 / -0

The total force exerted by the body perpendicular to the surface is called:

A
pressure

B
thrust

C
impulse

D
none of these

Solution

The total force exerted by any object perpendicular to the surface is called thrust.
Question 8
1 / -0

The density of a metal block is $$2.7 \times {10}^{3} {kg}/{{m}^{3}}$$. It is immersed in water. State the situation.

A
Sinks

B
Floats

C
Partially immersed

D
No part immersed

Solution

Any solid floats on water if its density is less than the density of water.

Or we can say the object will float on water if the relative density is less than 1.
Relative density = $$\dfrac{Density\ of\ metal}{Density\ of\ water}=\dfrac{2.7\times 10^3}{10^3}$$
In this case relative density is 2.7, so it will sink.
Question 9
1 / -0

A solid weighs $$32 g f$$ in air and $$28.8 g f$$ in water. The R.D. of the solid is

A
$$8.9$$

B
$$10$$

C
$$29.12$$

D
$$20$$

Solution

Given,
Weight of solid in air $$W=32\ gf$$
Weight of solid in water $$W_w=28.8\ gf$$
Relative density $$=\dfrac { weight\quad of\quad solid \quad in \quad air}{ decrease\quad in\quad weight\quad of\quad solid\quad in\quad water } $$

$$=\dfrac { 32 }{ 32-28.8 } $$

Relative density $$=\dfrac { 32 }{ 3.2 } =10$$
Question 10
1 / -0

Apparent loss of weight of a body when immersed in a liquid can be explained on the basis of

A
molecular theory

B
electron theory

C
Archimedes' principle.

D
Bernoulli's principle.

Solution

Archimede's principle states that when a body is immersed partly or wholly in a liquid, there is an apparent loss in the weight of the body which is equal to the weight of the liquid displaced by the body.
Hence, option C is correct.

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Gravitation Test - 27

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Gravitation Test - 27

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SHARING IS CARING

Question 1

Zero

$$100\ N$$

$$600\ N$$

$$6300\ N$$

Solution

Question 2

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

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

$$6.67 \times 10^{-9} N$$

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

Solution

Question 3

N Kg$$^{-2}$$

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

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

Nm$$^{-2}$$ Kg$$^{-3}$$

Solution

Question 4

the acceleration of A is greater than that of B

the acceleration of A is less than that of B

neither is accelerated

their accelerations are equal in magnitude but opposite in direction

Solution

Question 5

repulsive at large distances

attractive at all places

attractive at short distances

repulsive at short distances

Solution

Question 6

apparent weight in water

density

relative density

thermal capacity

Solution

Question 7

pressure

thrust

impulse

none of these

Solution

Question 8

Sinks

Floats

Partially immersed

No part immersed

Solution

Question 9

$$8.9$$

$$10$$

$$29.12$$

$$20$$

Solution

Question 10

molecular theory

electron theory

Archimedes' principle.

Bernoulli's principle.

Solution

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