Gravitation Test

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Gravitation Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following has highest relative density?

A
water

B
oil

C
mercury

D
glass

Solution

The relative density of a substance with density $$\rho$$ with respect to another substance with density $$\rho_0$$ is given by
$$r=\dfrac{\rho}{\rho_0}$$
Hence $$r\propto \rho$$
Since density of mercury is highest among the given options, the relative of it with respect to some other substance is also highest.
Question 2
1 / -0

A body is submerged in water. It is taken out and then submerged in mercury. What happens to its relative density with respect to the liquid in which it is submerged?

A
increases

B
decreases

C
remains constant

D
can't say

Solution

Relative density of a substance 1 with respect to some other substance 2 is
$$\rho_r=\dfrac{\rho_1}{\rho_2}$$
Since the density of mercury is greater than that of water, the relative density of body with respect to mercury is lesser than that in water.
Question 3
1 / -0

The relative density of a steel ball will be highest with respect to:

A
water

B
oil

C
mercury

D
glass

Solution

Relative density of a substance 1 with respect to some other substance 2 is defined as
$$\rho_r=\dfrac{\rho_1}{\rho_2}$$

The relative density of steel ball $$\rho_r$$ is highest when the density of the other substance $$\rho_2$$ is lowest.

The density of oil is the lowest in the given options. Hence the answer is option B.
Question 4
1 / -0

The force of gravity with which a body is attracted towards the centre of the earth is __________.

A
weight

B
mass

C
position

D
ripple

Solution

The force of gravity with which a body is attracted towards the centre of the earth is called the $$weight$$ of the body and it is equal to the product of mass and acceleration due to gravity offered by the earth

Hence correct answer is option $$A $$
Question 5
1 / -0

A body weighs $$12 N$$ on the surface of the moon. What is its weight on the surface of the earth?

A
$$72 N$$

B
$$2 N$$

C
$$24 N$$

D
Zero

Solution

The force of gravity on moon is $$\dfrac{1}{6}$$ times that of earth.

Hence a body weighing 12N on the surface of the moon will weigh six times more on the surface of the earth i.e 12N $$\ast$$ 6 = 72N.
Question 6
1 / -0

A rectangular tank of $$6 m$$ long, $$2 m$$ broad and $$2 m$$ deep is full of water, what is the thrust acting on the bottom of the tank?

A
$$23.52\times { 10 }^{ 4 }N$$

B
$$23.52 N$$

C
$$11.76\times { 10 }^{ 4 }N$$

D
$$3.92\times { 10 }^{ 4 }N$$

Solution

Volume of tank = $$6\times 2\times 2{ m }^{ 3 }=24{ m }^{ 3 }$$
of mass of water = $$1000\dfrac { kg }{ { m }^{ 3 } } \times { 24m }^{ 3 }$$
= 24000 kg
So Thrust = $$24000\times 9.8N$$
= 23.52$$\times { 10 }^{ 4 }N$$
Question 7
1 / -0

Density of a solid is $$18300 kg/m^3$$. What is its relative density?

A
1830

B
183

C
18.3

D
1.83

Solution

Density of water $$\rho_w = 1000$$ $$kg/m^3$$
Density of solid $$\rho = 18300$$ $$kg/m^3$$
$$\therefore$$ Relative density of solid $$r_s= \dfrac{\rho}{\rho_w} = \dfrac{18300}{1000} = 18.3$$
Question 8
1 / -0

Universal gravitational constant $$G$$ is proportional to distance $$r$$ between two masses as-

A
$$r^2$$

B
$$r$$

C
$$1/r$$

D
does not depend upon $$r$$

Solution

According to Newton's Law of gravitation, the gravitational force between two bodies of mass $$M$$ and $$m$$ separated by a distance $$r$$, is directly proportional to the product of the masses of the objects ( $$F \propto Mm$$) and inversely proportional to the square of distance between them $$F \propto \dfrac{1}{r^2}$$.

When we combine this information, we get
$$F \propto \dfrac{Mm}{r^2}$$
$$F= G \dfrac{Mm}{r^2}$$

where $$G$$ is the proportionality constant, called the Universal Gravitational constant.

Since $$G$$ is a constant it does not depend on $$r$$.
Question 9
1 / -0

Gravitational force

A
allows us to be in controlled motion

B
is a dissipative force

C
is a strong force having order of 5

D
none of the above

Solution
Question 10
1 / -0

Value of universal gravitational constant $$G$$ in CGS unit is-

A
$$6.67\times 10^{8}cm^3g^{1}s{}$$

B
$$6.67\times 10^{8}cm^3g^{-1}s^{-2}$$

C
$$6.67\times 10^{9}cm^3g^{1}s^{2}$$

D
$$6.67\times 10^{7}cm^3g^{-1}s^{2}$$

Solution

Newton's law of gravitation:
$$F=G \dfrac{Mm}{r^2}$$
$$G = \dfrac{F r^2}{Mm}$$
In CGS unit system:
r - distance - cm,
M,m - mass - grams,
F - force - dyne - $$g cm/s^2$$
Therefore units of G - $$\dfrac{(g \space cm/s^2)(cm^2)}{g^2} = cm^3g^{-1}s^{-2}$$

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Gravitation Test - 34

Result

Gravitation Test - 34

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SHARING IS CARING

Question 1

water

oil

mercury

glass

Solution

Question 2

increases

decreases

remains constant

can't say

Solution

Question 3

water

oil

mercury

glass

Solution

Question 4

weight

mass

position

ripple

Solution

Question 5

$$72 N$$

$$2 N$$

$$24 N$$

Zero

Solution

Question 6

$$23.52\times { 10 }^{ 4 }N$$

$$23.52 N$$

$$11.76\times { 10 }^{ 4 }N$$

$$3.92\times { 10 }^{ 4 }N$$

Solution

Question 7

1830

183

18.3

1.83

Solution

Question 8

$$r^2$$

$$r$$

$$1/r$$

does not depend upon $$r$$

Solution

Question 9

allows us to be in controlled motion

is a dissipative force

is a strong force having order of 5

none of the above

Solution

Question 10

$$6.67\times 10^{8}cm^3g^{1}s{}$$

$$6.67\times 10^{8}cm^3g^{-1}s^{-2}$$

$$6.67\times 10^{9}cm^3g^{1}s^{2}$$

$$6.67\times 10^{7}cm^3g^{-1}s^{2}$$

Solution

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