Class 9 Physics - Gravitation

You have a bag of cotton and an iron bar. each indicating a mass of $$100\ kg$$ when measured on a weighing machine in reality, one is heavier than the other. Can you say which one is heavier and why?

Weight machine gives weight not mass.

As far as weight (and not mass) is concerned, cotton is heavier than iron as its decrease in weight due to air displaced is more than the weight of the air displaced by iron. Since the decreased weight of cotton and iron in the air are equal, the real weight of cotton in a vacuum is more than the real weight of iron in a vacuum.

Define thrust. What is its unit?

The force acting on a body perpendicular to its surface is called thrust.
The SI unit of thrust is newton (N).

Find the thrust required to exert a pressure of $$ 50000 $$ pascals on an area of $$ 0.05 m^2 $$?

$$ Pressure = Force / Area $$

$$ Force = Pressure \times Area $$

$$ = 50,000 \times 0.05 = 2,500 $$ Newton

How much thrust will be required to exert a pressure of $$ 20,000 $$ Pa on an area of $$ 1cm^2 $$

Pressure = $$ 20, 000 Pa $$

Area = $$ 1 cm^2 $$

$$ = 10^{-4} m^2 $$

Force = Pressure $$ \times $$ area

$$ = 20,000 \times 10^{-4} = 2 \quad newton $$

Define relative density. What is its SI unit?

The relative density of a substance is defined as the ratio of the density of substance to that of a given reference material.

$$Relative \, \,density=\dfrac{Density \, \,of \, \,substance}{Density \, \,of \, \, reference}$$

It is a ratio and thus has no units.

A cubical block of wood of side $$5cm$$ is placed on a table. Find the thrust by the block of wood on the table, if density of wood is $$5g$$ $${cm}^{-3}$$ and $$g$$ is $$10m$$ $${s}^{-2}$$.

Given,

Density of wood, $$\rho=5gcm^{-3}$$

$$g=10ms^{-2}$$

So, volume of cubical wood is $$V=5^3 cm^3=125cm^{3}$$

$$Density (\rho)=\dfrac{mass(m)}{Volume(V)}$$

Mass of the wood is $$m=\rho \times V=5\times 125=625g=625\times 10^{-3}kg$$

Thrust force, $$T=Weight=m\times g=625\times10^{-3}\times10=6.25N.$$

A cubical block of wood of density $$5\ g{cm}^{-3}$$ with a sides of $$10cm$$ stands on table. Find the thrust by the block of wood on the table if $$g =10\ m{s}^{-2}$$.

Side a = 10cm

So Volume = $$V=1000{ cm }^{ 3 }.$$

As we know, $$Density=\dfrac{mass}{volume}$$

$$\rho=\dfrac{m}{V}$$

$$\Rightarrow m= 5\ g/cm^3\times 1000\ cm^3$$

So mass = $$5000gm=5kg$$

Hence thrust = $$5kg\times g=50N.$$

Calculate the mass of a body in tonnes whose volume is 2 m$$^3$$ and R.D is 5.2.

R.D = Relative Density = Density of a substance / Density of water

Density of water = $$1000kg/{ m }^{ 3 }=1\quad tonne/{ m }^{ 3 }$$

Density of material = R.D. $$\times $$ 1 tonne/$${ m }^{ 3 }$$

= 5.2 tonne/$${ m }^{ 3 }$$

Volume = $$2{ m }^{ 3 }$$

So mass = density $$\times $$ volume = 5.2$$\times $$2 tonne

$$\boxed { mass\quad =\quad 10.4\quad tonne } $$

A book of mass $$20\ g$$ is placed on a table. Find the thrust exerted by the book on the table. (Take acceleration due to gravity as $$10\ m{s}^{-2}$$)

Given,

Mass of the book, $$m=20\ g = 0.02\ kg$$

Acceleration due to gravity, $$a=g=10\ m/s^2$$

Thrust is the force acting in a particular direction.

Thrust, $$F=m\times a$$

$$F=0.02\times 10$$

$$F=0.2\ N$$

Calculate the force of gravitation due to earth on a ball of mass $$60\ kg$$

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Imagine a body that is completely submerged in water, but whose depth of submergence can be varied. In which case does it experience a larger upthrust, when it is submergged deep or shallow?

According to Archimedes' principle the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces.

So clearly when it is submerged deep it will displace larger volume of water as compared to when submerged shallow so in case of deeply submerged it will experience greater buoyant force.

What force does water exert on the base of a house tank of base area $$1.5\ m^{2}$$ when it is filled with water upto a height of $$1\ m$$?
[Density of water is $$10^{3}kg\ m^{-3}$$ and $$g = 10\ m\ s^{-2}]$$

Force exerted at base of tank is equal to weight of water.

Volume of tank $$V=A\times H=1.5\times 1=1.5m^3$$

Mass of water in tank $$m=volume \times density$$

$$m=1.5\times 1000=1500kg$$

Weight, $$W=F=mg=1500\times 10=15000N$$

The force acting on a body normal to its surface is _______

Thrust

Thrust is defined as the force acting on a body normal (perpendicular) to its surface. The SI unit of thrust is Newton.

What is the mass and weight of an object on the surface of the moon, if on the earth its mass is $$12$$ kg?

The mass of the object on earth is $$12$$ kg. It will be $$12$$ kg on the moon also. The value of ‘g’ on the moon is one-sixth of its value on the earth.

Weight of the object on earth$$= mg = 12 \times 9.8 = 117.6$$

N Weight of the object on moon 1

$$=12 \times \dfrac{1}{6} \times 9.8=19.6$$

Explain buoyancy and buoyant force.

The property of a liquid to push the immersed object in the upward direction is known as buoyancy.
The net upward force acting on a body partially or completely immersed in a liquid is known as buoyant force.

Calculate the gravitational force between two metal spheres of masses $$50\,kg$$ and $$100\,kg$$ respectively and the separation between their centers is $$50\,cm$$.

Given ; $$m_1 = 50\,kg ; m_2 = 100\,kg $$

$$ r = 50\,cm = 50 \times 10^{-2} \,m $$ ; F = ?

$$ G = 6.67 \times 10^{-11} \, Nm^2kg^{-2} $$

The magnitude of gravitational force,

$$ F = G \dfrac{m_1 m_2}{r^2} $$

$$ = 6.67 \times 10^{-11} \times \dfrac{50 \times 100}{50 \times 50 \times 10^{-4}} $$

$$ = 13.34 \times 10^{-7}\,N$$

or $$ F = 1.334 \times 10^{-6} \,N $$

Two bodies are at a specific distance so as to attract each other. How many times will the mutual force of attraction be if the mass of one of them is doubled?

If the mass of one of the objects is doubled, then the force of attraction between them is doubled.

$$F=\dfrac{Gm_1m_2}{r^2}$$

If $$m_1=2m_1$$

So,

$$F_1=\dfrac{G2m_1m_2}{r^2}$$

$$F_1=2F$$

Two bodies are at a specific distance so as to attract each other.
How many times the force of attraction be if the masses of both the bodies are doubled?

By Universal law of gravitation,

$$F=\dfrac{Gm_1m_2}{r^2}$$

If $$m_1=2m_1 $$ $$m_2=2m_2$$

So,

$$F_1=\dfrac{G(2m_1)(2m_2)}{r^2}$$

$$F_1=4\dfrac{Gm_1m_2}{r^2}=4F$$

Force of attraction becomes $$4$$ times.

Explain the variation of $$'g'$$ due to the shape of the earthy?

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of $$g$$ becomes greater at the poles than at the equator.

Fill in the blanks with suitable words.
The value of ‘g’ in the poles is _________ (greater/lesser) than at the equator.

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator.

What is the effect of Non-sphericity of Earth on the value on the 'g'?

The value of $$g$$ depends on the radius of the earth: $$g=\dfrac{GM}{R^2}$$, where $$M$$ is the mass of the earth and $$R$$ is the radius of the earth.

Since the earth is not an absolute sphere, its equatorial radius is more than the polar radius. $$\therefore R_{equator}>R_{pole}$$

Thus the value of $$g$$ is less in the equatorial region as compared to that in the polar region. $$\therefore g_{equator}<g_{pole}$$

Derive the relation between acceleration due to gravity (g) and the gravitational constant (G)

Consider a body of mass 'm' on the surface of the earth of mass M. Let R be the radius of the Earth.

$$\therefore mg=\dfrac{GmM}{R^{2}}$$

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

Sate and explain Newton's law of gravitation .

Newton's law of gravitation, states that any particle of matter in the universe attracts any other object with a force varying directly as the product of the masses and inversely as the square of the distance between them.

This force is given by

$$F=\dfrac{Gm_{1}m_{2}}{r^{2}}$$

Where $$'G'$$ is called gravitational constant.

and
$$m_1$$ and $$m_2$$ are masses of both objects.

$$r$$ is the distance between them.

Write Newton's universal law of gravitation.

Newton’s universal law of gravitation states that every object of the universe attracts every other object. The force of attraction between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The direction of this force is along the line joining the centers of the objects.

When a hollow steel ball and a solid steel ball of the same size are submerged in water, do the two experience the same upthrust?

Upthrust depends on the weight of the liquid displaced.

$$F_B=V_{displaced} \ \rho_{liquid} \ g$$

$$V_{displaced}$$; displaced volume by object

Since both have the same volume when submerged inside the water, so they both will experience the same thrust.

Why some objects float on the water? And some sink?

Some object floats & some sink due to an amount of buoyant force acting on it equal to a weight of liquid displaced by the body. Hence if weight of body is more than buoyant force than body will sink & vise versa.

What do we call the gravitational force between the earth and an object?

Hint: The gravitational force due to the earth on a body pulls the body downward.

Step 1: Defining the gravitational pull due to earth.

$$\bullet$$As the force pulls the body downwards, The body exerted force proportional to the mass of the body which is known as the Weight.

Mass of an object is $$60\ kg$$ on the surface of the earth. Find its mass and weight on the surface on the moon.

Acceleration due to gravity on the moon is $$1/6$$ times as that on the earth and we know that mass is property of the material it always remains same and weight is measure of gravitational force, hence

mass of object on moon is $$60\quad kg$$ and weight $$=60g/6=10\times10=100N$$

Pressure varies with area (A) as provided thrust is same.
True =1; False=0

Pressure is defined as force per unit area; hence pressure varies with area for the same upthrust.

$$P=\dfrac{F}{A}$$

The relative density of substance is numerically equal to its density in the _____ system of units.

The relative density of a substance is numerically equal to its density in CGS system of units as density of water in CGS system of units is 1gm/cc.

The ratio of the density of the substance to the density of water is known as __________.

Relative density

The relative density of a substance is the ratio of its density to that of water:

$$Relative\ density =\dfrac{Density\ of \ substance }{Density\ of\ water}$$

The gravitational force of attraction between two masses depend on the distance between them is $$G= $$gravitaional constant $$=k \times 10^{-11} Nm^{2} kg^{-2}$$. what is the value of k?

Newton's Law of Gravitation says,

$$F_{g} = G \cfrac{m_{1} \times m_{2}}{r^{2}}$$
where F$$= Gravitational \ Force$$
$$ m_{1} , m_{2} =$$ masses of two bodies
$$ r=$$ distance
$$G= $$gravitaional constant $$=6.673 \times 10^{-11} Nm^{2} kg^{-2}$$

Hence, $$K=6.673$$

State whether weight is a scalar or vector quantity?

The weight of a body is defined as the force by which the body is attracted by the planet (say earth) towards its center.

Weight can thus be considered the same as the Gravitational force (vector) exerted by the earth on that body. Hence, weight is a vector quantity and has both direction and magnitude.

What is the SI unit of thrust?
(Newton, Joule)

Thrust is the force acting perpendicular to any surface. The SI unit of thrust will be same as that of force. Therefore, the SI unit of thrust is Newton.

Buoyant Force is the upward force that acts in the direction opposite to that of weight.Type 1 for true and 0 for false

Buoyant Force is the upward force that acts in the direction opposite to that of weight.
So, it will always act in the upward direction.

What do you understand by the term Force of gravity?

The gravitational force or the force of gravity is a force that attracts any objects with mass. This is also called as Newton's law of gravitation.

Mathematically, F = $$\dfrac{GMm}{r^{2}}$$

where F = force of gravity, G = gravitational constant, M = mass of one object, m = mass of other object, r = distance between two objects.

Explain the statement "relative density of brass is 8.4"

The above statement means the ratio of the density of brass to the density of water (at $$\displaystyle 4^{\circ}C$$) is 8.4

If the distance between the two point masses is doubled then force of attraction is decreased __________times

The force of attraction between two masses is inversely proportional to the square of the distance between them. If the distance between the two point masses is doubled then force of attraction is decreased 4 times i.e. the square of 2.

It can be seen by the universal law of gravitation as well, that

$$F_1=G\dfrac{m_1m_2}{r^2}$$

Let the distance be doubled, then

$$F_2=G\dfrac{m_1m_2}{(2r)^2}=\dfrac{1}{4}G\dfrac{m_1m_2}{r^2}=\dfrac{1}{4}F_1$$

So force of attraction is decreased by 4 times.

Is like density, the relative density of a body different in different systems of measurement ?

No, relative density is same in all systems of measurement and it has no units.

$$\text{Relative density} = \dfrac{\text{Density of unknown }}{\text{Density of known}}$$

Calculate the gravitational force between two bodies $$'a'$$ and $$'b'$$ of mass $$50\ kg$$ each at a distance of $$1\ m$$ from each other.

Given:

Mass of the body $$a$$ is $$m_a=50\ kg$$

Mass of the body $$b$$ is $$m_b=50\ kg$$

Distance between $$a$$ and $$b$$ is $$d=1\ m$$

Gravitational force $$(F_G) = \dfrac{Gm_am_b}{d^2}$$

$$ = (6.67 \times 10^{-11}) \dfrac{50\times 50}{1^2}$$

$$ = 1.6675 \times 10^{-7}N$$

What is the relative density of the object immersed in water?

Relative density of object immersed in water is greater than one.

It is because when the relative density of an object is equal to 1 then the densities of both the materials are equal. If the relative density is less than 1 then the object will float ,if the relative density is greater than 1 then the object will sink in water.

How can the knowledge of relative density help in verifying the purity of substances?

Under a particular temperature and pressure conditions, the value of relative density has a fixed value. So, if relative density is calculated for a substance, it can be compared with its theoretical value to determine its purity.

State the Archimedes principle.

Hint: Archimedes’ principle states that: “The upward buoyant force that is exerted on a body immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid”.
Explanation: Archimedes determined the amount of buoyant force exerted by the liquid on the body. According to him, when a body is wholly or partially immersed in a liquid at rest, it experiences an upthrust, equal to the weight of the liquid displaced by the body.
(i) A body will sink in a liquid if its density is more than that of the liquid in which it is immersed.
(ii) The body floats being completely immersed in water if the density of the body is equal to the density of the liquid.

A block of iron of mass $$7.5kg$$ and of dimensions $$ 12cm \times 8cm \times 10cm $$ is kept on a table top on its base of side $$ 12cm \times 8cm$$. Calculate
a) Thrust
b) Pressure exerted on the table top.
Take $$1kgf = 10N$$

Thrust = $$ mg = 7.5 \times 10 = 75 N$$

Pressure = $$ \cfrac{F}{A}$$

$$ = \cfrac{75}{ 10 \times 10^{-2} \times 8 \times 10^{-2}}$$

$$ = 0.78125 \times 10^4$$

$$ = 7812.5 Pa$$

What is drag?

A drag force is the resistance force caused by the motion of a body through a fluid, such as water or air. A drag force acts opposite to the direction of the motion of the body.

When the mass of one of the two bodies is doubled the gravitational force between them increases by ________ times.

The gravitation force between two bodies given as-

$$F = \dfrac{ Gm_1m_2 }{ r^2 }$$

$$F$$; force between them

$$G$$; gravitational constant$$(6.67 \times 10^{-11}\ Nm^2\ kg^{-2})$$

$$m_1\ \text{and}\ m_2$$; masses of the two object

$$r$$; distance between them

Now suppose one mass becomes doubled-

$$m_1' = 2m_1$$

then force-

$$F' = \dfrac{ G(2m_1)m_2 }{ r^2 }$$

$$\Rightarrow F' = 2 \dfrac{ Gm_1m_2 }{ r^2 }$$

$$\Rightarrow F' = 2F$$

So force becomes double when one mass is doubled.

What is the magnitude of gravitational force between earth & $$1$$ kg object on the surface of earth?

We know gravitation force $$F=\dfrac { GMm }{ { R }^{ 2 } } =\dfrac { 6.67\times { 10 }^{ -11 }\times 6\times { 10 }^{ 24 }\times 1 }{ { (6.4\times { 10 }^{ 6 }) }^{ 2 } } =9.8\quad N$$

Where $$M$$ = mass of the earth, $$m$$ = mass of the object and $$R$$ = radius of earth

Find Pressure if 20 N of force is applied on a surface having area $$ 10 cm^2 $$.

Given,

Force $$F=20\,N$$

Area, $$A=10\ cm^2={{10}^{-3}}\,{{m}^{2}}$$

Pressure on surface, $$P=\dfrac{F}{A}=\dfrac{20}{{{10}^{-3}}}=20\ kPa$$

Define density and relative density.

Density :

Density of a substance is defined as the mass per unit volume of the substance.

Relative density :

Relative density of a substance is the ratio of its density to that of water.

Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

The gravitational force between the earth and an object on the surface of the earth is

$$F=G\dfrac{Mm}{{{d}^{2}}}$$

Where G is the universal gravitational constant

M is the mass of earth

m is the mass of object on the surface of earth

d is the distance between the two bodies.

Differentiate between $$'g'$$ and $$'G'$$.

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Distinguish between density and relative density.

Density:

The density of a substance is defined as mass of a unit volume. The density of a given substance, under specified conditions, remains the same. Therefore the density of a substance is one of its characteristic properties.
The unit of density is kilogram per metre cube ($$kg m^{-3}$$).

Relative density:

The relative density of a substance is the ratio of its density to that of water.
Since the relative density is a ratio of similar quantities, it has no unit.

Name the scientist who gave the magnitude of buoyant force acting on a solid object immersed in a liquid.

$$Archimedes$$ gave the magnitude of buoyant force acting on a solid object immersed in a liquid.

Two bodies of masses $$1\ kg$$ each is situated $$1\ m$$ apart. Find the gravitational force between them.

Given,

$$m_1 = m_2 = 1\,kg$$; masses of bodies

$$d = 1\,m$$; distance between them

Gravitational force between them-

$$F=\dfrac{Gm_1m_2}{d^2}$$

$$G$$; gravitational constant

$$\Rightarrow F =G.\dfrac{1\times 1}{1^2}$$

$$\Rightarrow F=G$$

$$\Rightarrow F=6.67 \times 10^{-11}\,N$$

On what factors does the value of gravitational force depend?

Gravitational force depends on:

The masses of the two bodies involved
The distance between the two bodies

Two spherical bodies, each of $$100$$ kg, are separated by $$1m$$. Calculate the gravitational force exerted by each on the other.

We know that,
$$F=\dfrac{Gm_1m_2}{r^2}$$
Here, $$m_1=m_2=100\,kg$$
$$r=1\,m$$
$$\therefore F=(6.67 \times 10^{-11})\Bigg(\dfrac{100 \times 100}{1^2}\Bigg)$$
$$\therefore F=(6.67 \times 10^{-7})N$$
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What will be the mass of a man of $$60$$ kg on the moon?

The man’s mass will be $$60$$ kg as the mass of an object remains constant.

It denotes amount of matter in an object which is fixed.

Two spheres each of mass $$10^5$$ kg and radius $$ 10$$ m are kept in contact. Find the force of gravitation acting between them?

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At the surface of a certain planet, the acceleration due to gravity is one-quarter of that on earth. A brass ball is transported to this planet, then find the percentage change in its weight?

Given:

Acceleration due to gravity on planet $$P$$, say $$g_p$$ is $$1/4$$ of that on earth, $$g_e$$

$$g_p=\dfrac {1}{4}g_e$$

Weight on earth,

$$W_e=mg_e$$

Weight on planet P,

$$W_p=mg_p=m \left( \dfrac {1}{4}g_e \right)$$
$$W_p=\dfrac {1}{4}W_e$$

% change in weight $$=\dfrac {W_p-W_e}{W_e}\times 100$$
$$=\dfrac {\dfrac {1}{4}W_e-W_e}{W_e}\times 100=\dfrac {-3}{4}\times 100=-75 \ \% $$

If F is the force between two bodies of masses $$m_1$$ and $$m_2$$ at certain separation. Find the force between $$\sqrt 2 m_1$$ and $$\sqrt 3 m_2$$ at same separation.

Force due to gravity is given by $$\frac{Gm_1m_2}{r^2}$$

now, if $$\dfrac{Gm_1m_2}{r^2}=F$$ is given, then

$$\dfrac{G(\sqrt2m_1)(\sqrt3m_2)}{r^2}=\sqrt6.F$$. ($$\because $$ r and G is constant for both the cases).

Find the gravitational attraction between the two atoms in a hydrogen molecule. Given that $$G=6.67\times 10^{-11} Nm^2 kg^2$$, mass of a hydrogen atom $$=1.67\times 10^{-27}$$ kg and distance between the two atoms $$=1^oA.$$

Here, $$G=6.67\times 10^{-11} N m^2 kg^2$$
$$M_1=M_2=1.67\times 10^{-27}kg;$$
$$r=1A^o=10^{-10}m$$

Now, $$F=G\dfrac {M_1M_2}{r^2}$$

$$=\dfrac {6.67\times 10^{-11}\times 1.67\times 10^{-27}\times 1.67\times 10^{-27}}{(10^{-10})^2}$$
$$=1.86\times 10^{-44}N$$

The tides formed by the rising and falling of water levels in the oceans are due to the gravitational pull of the sun and moon on the water is the importance of the law of gravitation.
Type $$1$$ for true and $$0$$ for false.

Due to the gravitational pull of mainly the moon on the spinning earth, the level of water in the sea rises and falls, and tides are formed. This phenomenon is called high and low tides and the difference in sea levels gives us tidal energy.

Define 'relative density'.

Relative density is the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.

State five differences between mass and weight.

Mass Weight

(a) Mass is the amount of matter contained in a body (a) Weight is the force with which the earth attracts a body towards itself

(b) Mass of a body is a constant quantity and does not change with the position or surrounding of the body (b) Weight of a body is a variable quantity and changes with the change in acceleration due to gravity of the earth

(c) Mass of a body is found with the help of a physical balance or beam balance (c) Weight of a body is found with the help of a spring balance

(d) The SI unit of Mass is kilogram (kg) (d) The SI unit of weight is newton (N)

What do you understand by weight?

Wight is the measure of force by which any planet attracts another body. The weight of the body can change from planet to planet due to changes in shape, size, and density.

The weight of a body on earth is $$6^{th}$$ time of weight on the moon.

Mass of any object remains the same everywhere.

What happens to the gravitational force between two objects, if
$$(i)$$ The mass of one object is doubled?
$$(ii)$$ The distance between the objects is doubled and tripled?
$$(iii)$$ The masses of both objects are doubled?
Give reasons in each case

Gravitational force $$F = \dfrac{Gm_1m_2}{r^2}$$

Case $$(i)$$ : Mass of one object is doubled i.e $$m'_1 = 2m_1$$

$$\therefore $$ $$F' = \dfrac{G (2m_1) m_2}{r^2}$$ $$\implies F' = 2F$$

Thus gravitational force gets also doubled

Case $$(ii)$$ : Distance between the objects is doubled i.e $$r' = 2r$$

$$\therefore $$ $$F' = \dfrac{G m_1 m_2}{(2r)^2}$$ $$\implies F' = \dfrac{F}{4}$$

Thus gravitational force gets reduced by $$4$$ times.

Distance between the objects is tripled i.e $$r' = 3r$$

$$\therefore $$ $$F' = \dfrac{Gm_1 m_2}{(3r)^2}$$ $$\implies F' = \dfrac{F}{9}$$

Thus gravitational force gets reduced by $$9$$ times.

Case $$(iii)$$ : Mass of both the objects are doubled i.e $$m'_1 = 2m_1$$ and $$m_2' = 2m_2$$

$$\therefore $$ $$F' = \dfrac{G (2m_1) (2m_2)}{r^2}$$ $$\implies F' = 4F$$

Thus gravitational force becomes $$4$$ times.

How does the force of gravitation between two objects change when the distance between them is reduced to half?

We know that According to the law of gravitation, the force between any two objects of mass $$M$$ and $$m $$ is directly proportional to their masses and inversely proportional to the square of the distance between them.

$$ F = G \dfrac{Mm}{r^{2}}$$

So, if Distance is halved, i.e new $$r = \dfrac{r}{2}$$

So, New $$F = G\dfrac{Mm}{\dfrac{r^{2}}{2^{2}}}$$

$$\Rightarrow F = 4G \dfrac{Mm}{r^{2}}$$

So, Force of gravitation increases by 4 times when distance reduced by half

What is the importance of universal law of gravitation?

Hint:

The gravitational force is one of the four fundamental forces of the universe which is very much responsible for the structure of the universe.

Step 1: Showing the points of importance.

$$\bullet$$ This law is responsible for the revolution of the earth around the sun and the moon around the earth.

$$\bullet$$Responsible for the weight of a body and keeps us on the ground
$$\bullet$$Phenomenon of Tides will happen due to gravitational attraction between Earth and the moon.

The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Let $$m$$ and $$M$$ be the mass of moon and earth respectively which are separated by a distance $$r$$

Force of attraction that earth exert on moon, $$F_e = \dfrac{GMm}{r^2}$$

Force of attraction that moon exert on earth, $$F_m = \dfrac{GmM}{r^2}$$

Thus earth attracts the moon with same force by which moon attracts the earth.

Also the gravitational force is an internal force of attraction between the two bodies, thus they have to be equal.

A ball is thrown vertically upwards with a velocity of $$49\ m/s$$. Calculate
$$(i)$$ The maximum height to which it rises.
$$(ii)$$ The total time it takes to return to the surface of the earth.

Given Initial velocity of ball, $$u = 49$$ $$m/s$$

Let the maximum height reached and time taken to reach that height be $$H $$ and $$t$$ respectively.

Assumption: $$g=9.8\ m/s^2$$ holds true (maximum height reached is small compared to the radius of earth)

Velocity of the ball at maximum height is zero, $$v = 0$$

$$v^2 - u^2 = 2a H$$

$$0 - (49)^2 = 2 \times (-9.8) \times H$$

$$\implies H= 122.5\ m$$

$$v = u+ at$$

$$0 = 49 - 9.8 t$$

$$\implies t = 5$$ s

$$\therefore$$ Total time taken by ball to return to the surface, $$T = 2t = 10\ s$$

In what direction does the buoyant force on an object immersed in a liquid act?

The buyont force acts in the upward direction on the object which is immersed in a liquid.

Gravitational force on the surface of the moon is only $$\displaystyle\frac{1}{6}$$ as strong as gravitational force on the earth. What is the weight in newtons of a $$10\ kg$$ object on the moon and on the earth?

Mass of the object, $$m = 10\ kg$$

Acceleration due to gravity at earth, $$g_e = 9.8\ ms^{-2}$$

$$\therefore$$ Weight of the object on earth, $$W = mg_e = 10 \times 9.8 = 98\ N$$

Acceleration due to gravity at moon, $$g_m =\dfrac{1}{6} \times 9.8 = 1.63$$ $$ms^{-2}$$

$$\therefore$$ Weight of the object on moon, $$W = mg_m = 10 \times 1.63 = 16.3\ N$$

Why does a block of plastic released under water comes up to the surface of water?

We know that when a body is fully or partially immersed in a liquid a buoyant force will act on it in the upward direction. When it is more than gravity, it will bring it to the surface of water. Buoyancy force is more than gravity for materials with density smaller than water which holds for plastic and hence it comes up to the surface of water.

State whether True or False :

Relative density is more easily and perhaps more accurately measured without measuring volume.

Relative density is the ratio of density of a medium w.r.t reference medium hence it is not dependent on volume.

Relative density = $$\dfrac { { S }_{ 1 } }{ { S }_{ 2 } } $$. Hence the given statement is true.

State and explain Newtons law of gravitational force.

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

$$F\propto { M }_{ 1 }{ M }_{ 2 }$$ [$${ M }_{ 1 }$$ and $${ M }_{ 2 }$$ are the respective masses of the two bodies]

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

$$\therefore \quad F\propto \dfrac { { M }_{ 1 }{ M }_{ 2 } }{ { r }^{ 2 } } $$

or $$F=\dfrac { { GM }_{ 1 }{ M }_{ 2 } }{ { r }^{ 2 } } $$ [G= Constant of proportion=Gravitation constant]

Does a sunk body at the bottom of a vessel experience an upthrust?

According to Archimedes Principle. When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

Now since the sunk body has displaced some volume of fluid so it will experience an upthrust equal to weight of fluid displaced.

Two spherical balls of mass $$10\ kg$$ each are placed with their centres $$10\ cm$$ apart. Find the gravitational force of attraction between them.

Given:

$${ M }_{ 1 }={ M }_{ 2 }=10kg$$

$$r=10cm=0.1m={ 10 }^{ -1 }\ m$$

$$\therefore $$ Gravitational force of attraction : -

$$F=\dfrac { G{ M }_{ 1 }{ M }_{ 2 } }{ { r }^{ 2 } } =\dfrac { 6.67\times { 10 }^{ -11 }\times 10\times 10 }{ { 10 }^{ -1 }\times { 10 }^{ -1 } } =6.67\times { 10 }^{ -7 }N$$

Solve:
A sphere of mass 10 kg is attracted by another sphere of mass 150 kg, with a force equal to $$1.28 \times 10^{-6} N$$, when their centers are separated by a distance of 0.28 m. Calculate the gravitational constant.

Mass of sphere 1, $${ M }_{ 1 }=10kg$$

Mass of sphere 2, $${ M }_{ 2 }=150kg$$

Force between the spheres, $$F=1.28\times { 10 }^{ -6 }N$$

DIstance between the spheres, $$R=0.28m$$

$$\therefore \quad F=\dfrac { { GM }_{ 1 }{ M }_{ 2 } }{ { R }^{ 2 } }$$

or $$G=\dfrac { { FR }^{ 2 } }{ { M }_{ 1 }{ M }_{ 2 } } =\dfrac { { 1.28\times 10 }^{ -6 }\times 0.28\times 0.28 }{ 10\times 156 } $$

$$\approx 6.7\times { 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }$$

Can you make iron to float? How?

Objects having a density less than that of the liquid in which they are immersed, float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed then it sinks into the liquid. We can make the iron float by choosing a liquid with a density higher than that of the iron. For example, mercury.

$$\displaystyle F = \cfrac{G m_1 m_2}{d^2}$$ is the mathematical form of Newton's law of gravitation. $$G$$ is gravitational constant, $$m_1$$, $$m_2$$ are the masses of two bodies separated by a distance $$d$$, then give the statement of Newton's law of gravitation.

Newton's law of gravitation :
Every object in the universe, attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$$\displaystyle F = \cfrac{G m_1 m_2}{d^2}$$.

where $$F$$ = gravitational force

$$G$$ = universal gravitational constant

$$m_1, m_2$$ = masses of the objects

$$d$$ = distance between the objects

(a) Explain Newton's first law of motion with an example.
(b) $$F=\cfrac{G{m}_{1}{m}_{2}}{{d}^{2}}$$ is the mathematical form of Newton's law of gravitation.
Give the statement of Newton's law of gravitation.

(a) Newton's first law of motion states that:

An object remains in the state of rest or of uniform motion in a straight line unless compelled to change that state by an applied unbalanced force.

All objects resist a change in their state of motion. The tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.

Eg.: A ball at rest on the ground continues to be at rest unless someone kicks it or any external force acts on it.

(b) Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$$F = G\ m_{1}\ m_{2}/ d^{2}$$ is the mathematical form of Newton's law of gravitation, $$G -$$ gravitational constant, $$m_{1} m_{2}$$, are the masses of two bodies separated by a distance $$d$$, then given the statement of Newton's law of gravitation.

According to Newton's law of gravitation, every object in the universe attracts every another object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects.

Why is it easier to swim in sea water than in river water?

It is easier to swim in sea water than in river water because we observe more buoyant force in sea water due to its hardness than in river water.

If mass of a planet is $$10$$% less than that of the Earth and radius $$20$$% greater than that of the Earth, then the acceleration due to gravity on the planet will be:

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The distance between two bodies is doubled. How is the magnitude of gravitational force between them affected?

Let $$m_1$$ and $$m_2$$ be the masses of the respective bodies.

Let the distance between the two bodies be $$d$$.

Gravitational force between them initially $$F = \dfrac{G m_1m_2}{d^2}$$

Now the distance between them is doubled i.e. $$d' = 2d$$

New gravitational force acting between them $$F' =\dfrac{Gm_1m_2}{(d')^2}$$

$$\therefore$$ $$F' = \dfrac{Gm_1m_2}{4d^2} = \dfrac{F}{4}$$

Thus gravitational force gets reduced by a factor of $$4$$.

The mass of earth is 80 times that of moon and their diameters are in the ratio 4:1 respectively.
What is the value of g on moon if g on earth $$9.8 \, m/s^2$$.

Suppose $$R_e$$ be radius of the earth and $$R_m$$ be that of the moon. so according question $$R_e=4R_m$$

And if $$M_e$$ be the mass of the earth and $$M_m$$ is the mass of the moon then $$M_e=80M_m$$

Value of g on the moon is $$g_m=GM_m/R_m^2$$...............(1)

Value of g on the earth is $$g_e=GM_e/R_e^2=\dfrac{G\times 80M_m}{16 \times R_m^2}=\dfrac{5GM_m}{R_m^2}=5g_m$$

So value of g on moon is $$g_e/5=9.8/5=1.96m/s^2$$

State and explain Universal law of Gravitation. Give its vector form.

Law of gravitation-

According to this law, any two body at some distance having mass will feel an attraction force, this law is universal because it is applicable to any body in universe having mass.

If two bodies having mass $$m_1 $$ and $$m_2$$ are at a distance $$r$$ any time then the force of gravitation $$F$$ is given as $$F=\dfrac{Gm_1m_2}{r^2} $$

its vector form is like this $$\vec F= \dfrac{-Gm_1m_2}{r^3} \vec r$$

The minus sign denote that its force of attraction that is opposite to distance vector $$\vec r$$ from one mass to second mass .

$$G$$ is universal gravitational constant having value $$6.7\times10^{-11}$$ in $$M.K.S.$$ unit.

Show that weight of the body on moon is 1/6 times its weight on earth.

$${W_e} = G\frac{{{M_e}m}}{{{R_m}^2}}$$

Weight of the object on the moon is

$${W_m} = G\frac{{{M_m} m}}{{{R_m}^2}}$$

So, the ratio of object of weight on moon to weight on earth is.

$$\begin{array}{l} \frac { { { W_{ m } } } }{ { { W_{ e } } } } =\frac { { G{ M_{ m } }m } }{ { { R_{ m } }^{ 2 } } } \times \frac { { { R_{ e } }^{ 2 } } }{ { G{ M_{ e } }m } } \\ \Rightarrow \frac { { { W_{ m } } } }{ { { W_{ e } } } } =\frac { { { M_{ m } } } }{ { { M_{ e } } } } \times { \left( { \frac { { { R_{ e } } } }{ { { R_{ m } } } } } \right) ^{ 2 } } \\ =\frac { { { M_{ m } } } }{ { 100{ M_{ m } } } } \times { \left( { \frac { { 4{ R_{ m } } } }{ { { R_{ m } } } } } \right) ^{ 2 } } \\ =\frac { { 16 } }{ { 100 } } =\frac { 1 }{ 6 } \end{array}$$

Therefore, the weight of an object on the moon is about $$\frac{1}{6}$$ of its weight on the earth.

Calculate the force of gravitation between the earth and the sun, given that the mass of the earth $$=6\times {10}^{24}kg$$ and of the Sun $$=2\times {10}^{30}kg$$. The average distance between the two is $$1.5\times {10}^{11}m$$

We know that,

Force of gravitation $$F=\dfrac{GMm}{r^2}$$

$$=\dfrac{6.67\times 10^{-11}\times6\times 10^{24}\times 2\times 10^{30}}{({1.5\times 10^{11}})^2}$$

$$=35.57\times 10^{21} N$$

$$=3.557\times 10^{22} N$$

How is the weight of an object related to its mass?

The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it.

$$ W = mg$$

In case of earth $$ g=9.8m/s^2$$

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Explain newtons law of gravitation

Newton’s law of gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

$$F=G\dfrac{mM}{{{r}^{2}}}$$

Where, F is the gravitational force and m and M is the masses of the objects and r is the distance between the objects and G is the gravitational constant

The mass of the earth is $$ 6 \times 10^ {24} $$ Kg. The distance between the earth and the sun is $$ 1.5\times 10^{11} $$ m. If the gravitational force between the two is $$3.5 \times 10^2\ N$$, What is the mass of the sun ?

Given that,

Earth of mass $$M=6\times {{10}^{24}}kg$$

Gravitational force $$F=3.5\times {{10}^{2}}\,N$$

Distance between earth and sun $$r=1.5\times {{10}^{11}}m$$

Mass of the sun = $$m$$

Now,

The gravitational force is

$$ F=\dfrac{GMm}{{{r}^{2}}} $$

$$ 3.5\times {{10}^{2}}=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}\times m}{{{\left( 1.5\times {{10}^{11}} \right)}^{2}}} $$

$$ m=\dfrac{3.5\times {{10}^{2}}\times 1.5\times {{10}^{11}}\times 1.5\times {{10}^{11}}}{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}} $$

$$ m=0.1967\times {{10}^{11}}\,kg $$

$$ m=1.97\times {{10}^{10}}\,kg $$

Hence, the mass of the sun is $$1.97\times {{10}^{10}}\,Kg$$

What is the effect of the shape of Earth on value of '$$g$$'?

As we know $$ g= \cfrac{GM} {R^2}$$

hence $$ g \propto \cfrac{1}{R^2} $$

Thus, acceleration due to gravity at a place is inversely proportional to the square of the distance of the point from the center of earth. Now, the earth is not perfectly spherical. It is flattened at the poles and elongated on the equatorial region is approximately $$21$$ km more than that at the poles. Hence $$g$$ is maximum at the poles and minimum at the equator.

Calculate the force of gravitation between two objects of masses 80 kg and 1200 kg kept at a distance of 10 m from each other.Given ($$G = 6.67 \times 10^{-11} \ Nm^{2}/kg^{2}$$)

Given : $$m_1 = 80\ kg$$ $$m_2 = 1200 \ kg$$ $$d = 10 \ m$$

Force of gravitation $$F = \dfrac{Gm_1m_2}{d^2}$$

Or $$F = \dfrac{6.67\times 10^{-11}\times 80\times 1200}{10^2} = 6.4\times 10^{-8} \ N$$

What is weight?

The weight of an object is the force with which the object is being pulled by the earth under the action of gravity.

If $$m$$ is the mass of an object and $$g$$ is the earth's gravitational then $$\text{Weight}(W)=\text{Mass}\times \text{Gravitational Acceleration}=mg$$.

Since weight is a force, its unit in SI system is $$newton$$.

The mass and weight of an object an earth are $$5\ kg$$ and $$49\ N$$ respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is $$1/6th$$ of that on the earth.

Step-1 Given data
Maas of object m= 5kg, Weight on the earth W= 49 N
Acceleration due to gravity on the moon $$g_{m}=\dfrac{1}{6}$$

Step-2 calculation of weight on the moon

What are the differences between the mass of an object and its weight?

$${\textbf{Explanation:}}$$

$$\bullet$$ The quantity of matter in a body is called mass.

$$\bullet$$ Mass = volume × density

$$\bullet$$ The force with which earth attracts objects towards center is called weight.

$$\bullet$$ Weight = mass × acceleration due to gravity

$$\bullet$$ Weight and mass are entirely different entities. The mass of an object on the earth will be same universally as the mass is the amount of matter present in the body. While, the weight of an object will change on earth and other planets due to change acceleration due to gravity.

Calculate the acceleration due to gravity on the surface of a planet whose mass and radius are double than the mass and radius of the earth.

Given that,

Mass of the earth =$$M$$

Radius of the earth = $$R$$

We know that,

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

Now, mass and radius is double

$$ g'=\dfrac{G2M}{4{{R}^{2}}} $$

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

$$ g'=\dfrac{1}{2}g $$

Hence, the acceleration due to gravity on the surface of a planet is $$\dfrac{1}{2}g$$

State two applications of universal law of gravitation.

Two applications of universal law of gravitation are

(1) It is used to calculate the force or pull of gravity of the planets in universe.

(2) It is also used in calculating the trajectory of astronomical bodies and to predict their motion and path.

State the universal law of gravitation.

$$F\,\alpha \,\dfrac{{{M}_{1}}{{M}_{2}}}{{{R}^{2}}}$$

Where,

$${{M}_{1}}\,and\,{{M}_{2}}$$ is masses of object.

$$R=$$ separation between them.

The mass of the earth is $$6\times {10}^{24}kg$$. The distance between the earth and the Sun is $$1.5\times {10}^{11}m$$. If the gravitational force between the two is $$3.5\times {10}^{22}N$$, what is the mass of the Sun?
Use $$G=6.7\times {10}^{-11}N{m}^{2}{kg}^{-2}$$

The gravitational force between the Sun and the Earth can be found using the formula,

$$F=G\dfrac{M_e M_s}{R^2}$$

where $$M_e$$ and $$M_s$$ are the masses of the Earth and the Sun, respectively.

Using all the given values, we have

$$3.5 \times 10^{22}=(6.7 \times 10^{-11})\dfrac{(6 \times 10^{24}) \times M_s}{(1.5 \times 10^{11})^2}$$

$$\implies M_s = \dfrac{(3.5 \times 10^{22})\times (1.5 \times 10^{11})^2}{(6.7 \times 10^{-11}) \times (6 \times 10^{24})} = 1.96 \times 10^{30}kg$$

(1) Mass of a book is 500g on the surface of the earth . what will be its mass at a height equal to radius of earth
(2) find the weight of the book at the surface of the earth

(1) Mass is the quantity of matter present in it. So, mass at a height equal to surface of earth will remains the same i.e. $$500\,g$$

(2) Mass, $$m=500\,Kg=0.5\,g$$

Weight,

$$ w=mg $$

$$ w=0.5\times 10 $$

$$ w=5\,N $$

Calculate the gravitational force on a object of $$40\ kg$$ at the surface of earth, if radius of earth is $$R = 6400\ km$$, mass of earth = $$6\times { 10 }^{ 24 } \ kg.$$

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The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force the moon attracts the earth? Why?

When two objects attract each other then gravitational force of attraction applied by the first object on the second object is same as the force applied by the second object on the first object. The force is given by the Newton's law of gravitation.

$$F=\dfrac{GMm}{R^2}$$

Newton's Third Law of Motion explains that forces always come in action-reaction pairs. The Third Law states that for every action force, there is an equal and opposite reaction force. ... The two forces create an interaction pair on different objects and are equal in strength and opposite in direction. So, the force of attraction of Earth on Moon is equal and opposite to the force of attraction of Moon on Earth

Why does a block of plastic released under water come up to the surface of water ?

According to Archimedes' principle

“The upward buoyant force that is exerted on a body immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction”.

Since the density of plastic is less than that of water, so the force of buoyancy(upward) on plastic block will be greater than the weight of plastic block(downward) . Hence, the plastic block will experience net force in upward direction, and comes up to the surface of water.

Why is the weight of an object on the moon is $$\frac{1}{6}$$ th its weight on the earth?

Mass of moon is lesser than that of earth. Due to this, the moon exerts a lesser force of attraction on objects. Hence the weight of an object on the moon is $$\dfrac 16 th$$ its weight on the earth.

The radius of planet A is half the radius of planet B. If the mass of A is $$M_{A}$$, what must be the mass of B so that value of g on B is half that of its value on A?

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

$$\implies M=\dfrac{gR^2}{G}$$

Here, $$\dfrac{g_B}{g_A}=\dfrac{1}{2}$$

and $$\dfrac{R_B}{R_A}=2$$

$$\implies \dfrac{M_B}{M_A}=\dfrac{g_B}{g_A}\left(\dfrac{R_B}{R_A}\right)^2$$

$$\implies \dfrac{M_B}{M_A}=\dfrac{1}{2}\left(2\right)^2$$

$$\implies M_B=2M_A$$

State universal law of gravitation and derive the expression for force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance 'd'.

Universal Law of Gravitation : It states that the force of attraction between two

bodies is directly proportional to the

product of their masses and inversely proportional to the square of the

distance

between them.

Let the two bodies 'A' and 'B' be of masses 'M' and 'm' respectively,

which are separated by a distance According

to Universal Law of Gravitation,

$$F=\dfrac{GM\ m}{r^2}$$

Where, 'G' is called universal gravitation constant.

The numerical value of$$G=6.67\times { 10 }^{ -11 }Nm²kg².$$

Two electrons each of mass $$9.1\times 10^{-31}$$ kg are at a distance of $$10A^0$$. Calculate the gravitational force of attraction between them. Given $$1A^0=10^{-10}m.$$

$$F=\dfrac { GMm }{ d² } $$

$$=\dfrac { Gm² }{ d² } $$

$$=G{ \left( \dfrac { m }{ d } \right) }^{ 2 }$$

$$=\left( 6.67\times { 10 }^{ -11 }Nm^{ 2 }/kg^{ 2 } \right) \times { \left[ \dfrac { \left( 9.1\times { 10 }^{ -31 }kg \right) }{ \left( 10\times { 10 }^{ -10 }m \right) } \right] }^{ 2 }$$

$$=5.52\times { 10 }^{ -53 }Newtons$$

Force of gravity on a mass of $$15$$ kg on the surface of a planet is $$25$$N. Find acceleration due to gravity due to this planet.

force of gravity $$F=25N$$

mass of body $$m=15kg$$

We know that

$$F=mg$$

$$25=15g$$

$$g=1.67m/{ s }^{ 2 }$$

An object weight $$10N$$when measured on the surface of the earth. what would be its weight when measured on the surface of the moon?

$$\textbf{Step 1 - Calculate mass of object on earth}$$

Given, $$Weight = 10\ N$$

$$W_{e} = mg$$

$$m = \dfrac {W_{e}}{g} = \dfrac {10N}{10\dfrac {m}{s^{2}}} = 1\ kg$$

$$\textbf{Step 2 - Calculate weight of object on moon}$$

$$W_{m} = mg_{m} = m\times \dfrac {g_{e}}{6}$$

$$= 1\ kg \times \dfrac {10}{6} \dfrac {m}{s^{2}}$$ $$(\because \text {mass on object on moon is same as on earth)}$$

$$\Rightarrow W_{m} = 1.67\ N$$

Gravitational force between two bodies is 1 newton.If the distance between them twice, what will be the force?

We know, Newton's law of gravitation in mathematical form is

$$F=\dfrac{GMm}{r^2}$$.........................(1)

Here, F=1 N (Given)

If distance between them is made twice , then,

$$F'=\dfrac{GMm}{(2r)^2}$$

=$$\dfrac{GMm}{4r^2}$$.......................(2)

Comparing (1) and (2),

$$F'=\dfrac{F}{4}$$=$$0.25 *1$$ =$$0.25N$$

Write the another name of the buoyant force.

Buoyant force is also known by the name 'upthrust'.

The weight of a person on the moon is about $$\dfrac { 1 }{ 6 } $$ times that on the earth . He can lift a mass of 15 kg on the earth . What will be maximum mass which can be lifted by the same force applied by the person on the moon ?

Since the weight of any person on the moon is about $$1/6$$ times that on the earth, hence acceleration due to gravity at the moon is $$1/6$$ of that on earth. This means that by applying the same force a person can lift six times heavier object on the moon than what he could lift on the earth. So, the maximum mass which can be lifted by the same force applied by the person on the moon is $$6\times 15\ kg=90\ kg$$.

Mahendra and virat are sitting at a distance $$1$$ metre from each other. Their masses are $$75\ kg$$ and $$80\ kg$$ respectively. What is the gravitational force between them?

Given : $$r = 1$$ $$m$$, $${m}_{1} = 75$$ kg, $${m}_{2}=80$$ kg and $$G = 6.67\times{10}^{-11} N{m}^{2}/{kg}^{2}$$

According to Newton’s law of gravitation:

$$F=\dfrac{G{m}_{1}{m}_{2}}{{r}^{2}}$$

$$=\dfrac{6.67\times{10}^{-11}\times 75\times 80}{1^2}$$$$N$$

$$=4.002\times{10}^{-7}$$$$N$$

The gravitational force between Mahendra and Virat is $$4.002\times{10}^{-7}$$ $$N$$

Differentiate between acceleration due to gravity and universal gravitational constant. Derive a relation between 'g' and 'G'.

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If the ratio of the final velocities of two bodies falling freely is $$4:3$$, then the ratio of the heights from which they fall is ______.

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The relative density of lead is 11.Its density is _ $$gcm^{-3}$$ or ___ $$kgm^{-3}$$.

Given,

Relative density of lead, $$R =11.3$$

We know,

Relative density, $$R=\dfrac{Density\ of\ lead\ (\rho_{lead})}{Density\ of\ water\ (\rho_{water})}$$

Density of water, $$\rho_{water}=1000\ kg/m^3=1\ g/cm^3$$

$$\implies \rho_{lead}=11.3\times \rho_{water}$$

$$\rho_{lead}(in\ g/cm^3)=11.3\times 1$$

$$\rho_{lead}(in\ g/cm^3)=11.3\ g/cm^3$$

And,

$$\rho_{lead}(in\ kg/m^3)=11.3\times 1000$$

$$\rho_{lead}(in\ kg/m^3)=11300\ kg/m^3$$

Find the gravitational force between the sun and the earth. Mass of the sun is $$2.0 \times 10^{30} \ kg$$ , mass of the earth is $$6 \times 10^{24} \ kg$$, and distance between the sun and the earth is $$1.5 \times 10^{11} \ m$$. $$(G = 6.67 \times 10^{-11} \ Nm^2/kg^2$$).

$$\large \textbf{Explanation :}$$

$$\text{Given:}$$

$$\text{Mass of the Sun }\ m_1 = 2 \times 10^{30} Kg$$

$$\text{Mass of the Earth}\ m_2 = 6 \times 10^{24} Kg$$

$$\text{Distance between Sun and Earth }\ r = 1.5 \times 10^{11} m$$

$$\text{Gravitational Force F = } \dfrac{Gm_1m_2}{r^2}$$

$$\Rightarrow F = \dfrac{6.67 \times 10^{-11} (Nm^2/Kg^2) \times 2 \times 10^{30} (Kg)\times 6 \times 10^{24}(Kg)}{(1.5 \times 10^{11}m)^2}$$

$$\Rightarrow F = 3.557 \times 10^{22}N$$

How is the gravitational forces between two points masses affected when they are dipped in water keeping the separation between them the same ?

The Newton's Universal law of gravitational forces of attraction $$ (F) $$ between two bodies of masses $$ m_1, m_2 $$ separated by distance $$ r $$ is $$ F = \frac {Gm_1m_2}{r^2} $$
$$ G $$ does not depend upon the medium. So the force of attraction does not change if the masses are kept in water or any other medium.

Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them.

Gravitational force of attraction.
$$ F = \dfrac{Gm_1m_2}{r^2} $$

where $$G \rightarrow$$ gravitational constant
G=6.67 × 10-11 m3 kg-1 s-2

$$m_1, m_2 \rightarrow$$ mass of objects

$$r \rightarrow$$ separation between the object r=10 cm = 10-2 m

$$ = \dfrac{6.67 \times 10^{-11} \times 10 \times 10}{(0.1)^2} = 6.67 \times 10^{-7} N $$

State and explain Newton's law of gravitation?

Newton's law of gravitation states that, anybody of certain mass $$M_!$$ attracts another body of mass $$M_2$$ such that force varying directly proportional to the product of their masses and inversely proportional to their distance 'R' between them.

Mathematically, it can be expressed as, $$ F=G \dfrac{ (M_1 \ M_2)}{R^2}$$

where $$G$$ is the gravitational constant.

State Newton's law of gravitation. Obtain the relation between universal gravitational constant and gravitational acceleration on the surface of the earth.

According to Newton's Law of gravitational, every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. [Fig. (1)]

$$F=\left(\dfrac{Gm_1m_2}{r^2}\right)$$ (where $$G$$ is the universal gravitational constant)

Now, If $$m_1$$ is earth's mass $$M$$ and $$m_2$$ is the mass of block $$m$$ kept on the earth's surface. The distance between the center of the earth to the block is equal to the radius of the earth $$R$$.

$$\therefore$$ Force on the block, $$F=\dfrac{GMm}{R^2}$$
Where $$R$$ is the radius of the earth. [Fig. (2)]

As we know that, acceleration due to gravity is the force by the earth on any block per unit mass of the block.
Acceleration due to gravity, $$=g=\dfrac{F}{m}=\dfrac{GMm}{R^2\times m}$$
$$\Rightarrow \boxed{g=\dfrac{Gm}{R^2}}$$

This is the relation between universal gravitational constant and gravitational acceleration on the surface of the earth.

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Give scientific reasons:
The weight of an object changes from place to place though its mass is constant.

The mass of an object is the amount of matter contained in it which always remains constant but the weight of a body depends on gravity because $$W = mg$$

So, the value of "g" changes from place to place hence the weight of body also changes.

If the same body is taken to places having different gravitational field strength, then what will vary: its weight or mass?

Weight of an object $$W=mg$$.

When the gravitational field strength varies, the gravitational acceleration($$g$$) also varies. So, the weight of the body will vary, not its mass.

Can we apply Newton's third law to the gravitational force? Explain your answer.

Yes, Newton's third law of motion holds good for the force of gravitation. This means that when earth exerts a force of attraction on an object, then the object also exerts an equal force on the earth, in the opposite direction.

Give reason for the following:
The force of gravitation between two cricket balls is extremely small but that between a cricket ball and the earth is extremely large.

The force of gravitation between two bodies is directly proportional to the product of their masses.
$$F \ \propto \ m \times M$$
Since the mass of cricket balls is very small as compared to that of the earth, so the force of gravitation between two cricket balls is extremely small while that between a ball and the earth is extremely large.

On what factor/ factors does the weight of a body depend?

The weight of a body is directly proportional to its mass. It also depends on the acceleration due to gravity which varies from place to place.

Describe how the gravitational force between two objects depends on the distance between them.

The gravitational force F between two bodies of masses M and m kept at a distance d from each other is:
$$F = G \times \dfrac{m \times M}{d^2}$$
The force between two bodies is inversely proportional to the square of the distance between them. This is,
$$F \propto \dfrac{1}{d^2}$$
Therefore, if we double the distance between two bodies, the gravitational force becomes one-fourth and if we halve the distance between two bodies, then the gravitational force becomes four times.

If the weight of a body on the earth is 6 N, what will it be on the moon?

The gravitational acceleration of the earth is $$g$$.

The gravitational acceleration of the moon is $$\dfrac{g}{6}$$

We know that $$Weight=mass\times g$$.

Since, the gravitation acceleration of the moon is one-sixth of that of the earth, the weight of the body on the moon will be $$\dfrac{1}{6}\times 6= 1\ N$$

Give the formula for the gravitational force F between two bodies of masses M and m kept at a distance d from each other.

The gravitational force F between two bodies of masses M and m kept at a distance d from each other is:
$$F = G \times \frac{m \times M}{d^2}$$
Here, Gravitational constant, $$G = 6.7 \times 10^{-11} Nm^2 kg^{-2}$$

What name has been given to the force with which two objects lying apart attract each other?

Gravitational force causes two objects lying apart attract each other.

A loaded cargo ship from sea water to river water. State and explain your observation.

The cargo ship dips into water as it enters into river water from sea water.
Explanation: Density of sea water is more than that of river water. Consequently, as per the law of floatation, in order to balance the weight of the ship, more volume of water is required for the river water having a lower density to displace it.

Universal law of gravitational states that every object exerts a gravitational force of attraction on every other object. If this is true, why don't we notice such forces? Why don't the two objects in a room move towards each other due to this force?

In order to be able to notice the gravitational force of attraction between any two objects, at least one of the objects on the earth should have an extremely large mass. Since no object on the earth have an extremely large mass, we cannot notice such forces.
The two objects in a room do not move towards each other because due to their small masses, the gravitational force of attraction between them is very, very weak.

How much is the weight of an object on the moon as compared to its weight on the earth?
Give reason for your answer.

The weight of an object on the moon is about one-sixth of its weight on the earth. This is because the value of acceleration due to gravity on the moon is about one-sixth of that on the earth.

State two applications of the Universal Law of Gravitation.

(i) Universal law of gravitational is used to determine the masses of the sun, the earth and the moon accurately.
(ii) Universal law of gravitational helps in discovering new stars and planets.

The mass of sun is $$2 \times 10^{30}$$ kg and the mass of earth is $$6 \times 10^{24}$$ if the average distance between the sun and the earth be $$1.5 \times 10^8 km$$, calculate the force of gravitation between them.

Distance $$d = 1.5 \times 10^8 \ km = 1.5 \times 10^{11} \ m$$
Mass of the sun, $$ m = 2 \times 10^{30} \ kg$$
Mass of the earth, $$M = 6 \times 10^{24} \ kg$$
Force of gravitation, $$F = G \times \frac{ m \times M}{d^2}$$
$$F = 6.7 \times 10^{-11} \times \frac{2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{11})^2} = 3.57 \times 10^{22} \ N$$

An object has mass of $$20 \ kg$$ on earth. What will be its (i) mass, and (ii) weight, on the moon? ($$g_{moon} = 1.6 \ m/s^2$$)

(i) Its mass will be $$20 \ kg$$ as mass is a constant quantity.
(ii) Weight, $$W = m \times g = 20 \times 1.6 = 32 \ N$$

(a) State the universal law of gravitation. Name the scientist who gave this law.
(b) Define gravitational constant. What are the units of gravitational constant?

(a) According to universal law of gravitation: Everybody in the universe attracts every other body with a force (F) which is directly proportional to the product of their masses (m and M) and inversely proportional to the square of the distance (d) between them.
$$F = G \times \dfrac{m \times M}{d^2}$$
Sir Issac Newton gave this law.
(b) The gravitational constant G is numerically equal to the force of gravitation which exists between two bodies of unit masses kept at a unit distance from each other.
$$G = F \times \dfrac{d^2}{m \times M}$$
Units of gravitational constant = $$Nm^2/kg^2$$

What happens to the gravitational force between two objects when the distance between them is:
(i) doubled?
(ii) halved?

(a) If we double the distance between two bodies, the gravitational force becomes one-fourth.
(b) If we halve the distance between two bodies, then the gravitational force becomes four times.

If the distance between two masses is increased by a factor of 5, by what factor would the mass of one of them have to be altered to maintain the same gravitational force? Would this be an increase or decrease in the mass?

Gravitational force is given by:
$$F = G \times \frac{m \times M}{d^2}$$

Thus, F is inversionly proportional to square of "d"

Distance between two masses is increased 5 times. So, new distance is D= 5 d

So, to maintain the same force between given masses any one of the two masses should be increased by 25 times.

Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively kept at a distance of 10 m from one another. (Gravitational constant, $$G = 6.7 \times 10^{-11} Nm^2 kg^{-2}$$)

$$F = G \times \frac{m \times M}{d^2}$$
m = 50 kg
M = 120 kg
Distance, d = 10 m
$$G = 6.7 \times 10^{-11}Nm^2/kg^2$$
$$F = 6.67 \times 10^{-11} \times \frac{50 \times 120}{10^2}$$
$$F = 6.67 \times 60 \times 10^{-11}$$
$$F = 4.02 \times 10^9 N$$

Which is more fundamental, the mass of a body or its weight? Why?

Mass is more fundamental because mass is defined as the constituents or components of any body which does not changes with time or in space.

Whereas, weight consist of mass and acceleration due to gravity.

This cannot be a fundamental quantity as it depends upon other physical quantity.

As in moon g decreases so weight decreases but mass remains constant.

Name the principle which gives the magnitude of buoyant force acting on an object immersed in a liquid.

The principle which gives the magnitude of buoyant force acting on an object immersed in a liquid is Archimedes' principle.

Archimedes' principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

The upward force exerted by the water on the bottle is known as upthrust or buoyant

force.

Why does a block of wood held under water rise to the surface when released?

Because the weight of the block of wood is less than the weight of an equal volume of water. So when it is completely submerged in water, the upward buoyant force (equal to weight of volume of water displaced) on it is greater than the downward gravitational force on it. Hence, the block rises to the surface.

How does the density $$\rho_s$$ of a substance determines whether a solid piece of density $$ρ_s$$ of that substance will float or sink in a given liquid of density $$\rho_L$$?

Density is a measure of how heavy something is compared to its size. If an object is more dense than liquid it will sink when placed in liquid, and if it is less dense than liquid it will float.

The body will float if $$\rho_s \le \rho_L$$ and it will sink if $$\rho_s > \rho_L$$.

Give reason for the following:
If the force of gravity acts on all bodies proportional to their masses, why does not a heavy body fall faster than a light body ?

Let us consider two bodies of mass $$m_A$$ and $$m_B$$ such that $$m_A>m_B$$.

We know the gravitational force $$F \propto m$$

The corresponding gravitational force on them are $$F_A$$ and $$F_B.$$

$$F_A=m_A \times g$$

$$F_B=m_B \times g$$

Since $$m_A>m_B \implies F_A>F_B$$,

But the greater force on the heavy body A doesn't mean the object A will fall faster. It is because the rate of falling is determined by the acceleration caused by gravitational force, i.e. $$g.$$ The gravitational acceleration for both the light and the heavy body is same, i.e.

$$g=9.8 m/s^2$$ downwards.

So both the bodies fall at the same rate irrespective of their different masses.

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Give reason for the following :
As we go higher above the surface of the earth the weight of the body decreases

We know that weight of a body is the force with which the body is attracted towards the centre of earth.

$$ W = m \times g $$

Where, m = Mass of the body

And, $$g$$ = Acceleration due to gravity

The mass of a body or object is constant and does not change from place to place.

But acceleration due to gravity is inversely proportional to the square of the distance from the center of earth.

So, as we go up from surface of the earth, the distance from the center of earth increases and hence value of g decreases.

Since, $$ W = m \times g $$

So, as the $$g$$ decreases, weight of the body also decreases with increase in height above the earth.

Thus, as we go higher above the surface of the earth the weight of the body decreases.

Give reason for the following :
Why is the weight of an object on the moon one-sixth its weight on the earth ?

Suppose we have an object of mass = m also let the weight on moon be $$ W_m $$ and weight on earth be $$ W_e $$ .

Now , weight of object on moon will be :

$$ W_m = G \dfrac{M \times m}{r^2} $$ .........1

Now the weight is actually the force with which moon attract the object.

M = Mass of Moon

r = radius of Moon

Now, we know

Mass of earth = $$100$$ times mass of moon

Radius of earth = $$4$$ times radius of moon

So , weight of object on earth will be given by :

$$ W_e = \dfrac{G \times 100 \times M \times m}{(4 \times r)^2} $$ ....... 2

Now , dividing equation $$1$$ by $$2$$ , we get ,

$$ \dfrac{W_m}{W_e} = \dfrac{{\text{Weight on Moon}}}{{\text{Weight on Earth}}} = \dfrac{16}{100} \approx \dfrac{1}{6} $$

Thus we can say that weight of an object on the moon one-sixth its weight on the earth.

(a) What is the difference between the density and relative density of a substance?
(b) If the relative density of a substance is 7.1, what will be its density in SI units?

(a) The Density of a substance is defined as mass of the substance per unit volume.
Density of substance = $$\frac{mass \ of \ substance}{volume \ of \ substance}$$
SI unit of density is $$kg/m^3$$.
The Relative density of a substance is the ratio of its density to that of water.
Relative density of substance = $$\frac{Density \ of \ substance}{Density \ of \ water}$$
It has no units.
(b)
Relative density of substance = $$\frac{Density \ of \ substance}{Density \ of \ water}$$
$$7.1 = \frac{Density \ of \ substance}{1000 kg/m^3}$$
density of substance $$ = 7.1 \times 10^3 kg/m^3$$

State whether the following statements are true or false:
(a) The buoyant force depends on the nature of object immersed in the liquid
(b) 'Archimedes' principle can also be applied to gases.

(a) The buoyant force depends on the nature of the object immersed in the liquid - False. Since buoyancy depends only on the volume of the object and the density of the fluid, thus it is not dependent on the nature of the object.

(b) YES. Archimedes' Principle is applicable to gases also. Gases and liquids are classified as fluids. ... Gases also exert buoyant force on objects but it is very small.

(a) What are fluids? Name two common fluids.
(b) State Archimedes ' principle.
(c) When does an object float or sink when placed on the surface of a liquid?

(a) Those substances which can flow easily are called fluids. All the liquid and gases are fluids, like water, air, etc.

(b) Archimedes' Principle :
When an object is wholly (or partially) immersed in a liquid, it experiences a buoyant force (or upthrust) which is equal to the weight of the liquid displaced by the object.
The buoyant force on an object = weight of the liquid displaced by that object

(c) If the buoyant force exerted by the liquid is less than the weight of the object, the object will sink in the liquid. If the buoyant force exerted by the liquid is equal to or greater than the weight of the object, the object will float in the liquid.

Give reason for the following :
The weight of a body is zero at the centre of the earth

The weight of the body is the force with which it is attracted towards the centre of the earth.
$$ W = m \times g $$

Weight of the body changes from place to place, as the value of g changes from place to place.

Acceleration due to gravity g is zero at the centre of earth, as at the centre of earth we are surrounded by equal masses in all the direction, hence equivalent gravitational force acting on us due to earth is zero and thus making acceleration due to gravity to zero.

So , $$ W = m\times g = 0 $$

Hence , weight at the centre of earth is zero .

State whether thrust is a scalar or vector?

Thrust is the force acting on a body perpendicular to its surface. Since thrust is defined in the terms of both direction(as it is only applicable to the body perpendicular to its surface) and magnitude, so it is considered as a Vector quantity.

State Archimedes' principle.

Archimedes' principle states that when a body is immersed partially or completely in a liquid, it experience an upthrust which is equal to the weight of the liquid displaced by it.

Give a reason for the following :
Why does the moon not actually fall on the surface of the earth?

The moon never actually falls on the surface of the earth.

Now, to understand this, we first need to understand the nature of the force between the moon and earth and also the movement of the moon under this force.

So, the gravitational force is the force that acts between the earth and the moon, and the moon moves in the circular orbital due to the centripetal force generated by the gravitational force.

Similarly, the earth and sun also experience this gravitational force between themselves.

So, yes the moon does fall towards the earth, but the earth also moves in its orbit and covers more distance away from the moon relative to the distance covered by the moon.

Hence, the moon never falls on the surface of the earth.

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Define Thrust and Pressure.

Thrust: The force acting on an object perpendicular to the surface is called thrust.

Its SI unit is Newton(N).

Pressure: It is the thrust exerted per unit area of a body. Its SI unit is Pascal (Pa).

On what principle does the submarines based?

Submarines are based on Archimedes' Principle. A submarine can dive into the water or rise to the surface as needed.
The compartments of a submarine that help in its flotation are the ballast tank and the compressed air tank. To sink, the ballast tanks are filled with water so that the average density of the submarine becomes greater than the density of sea water, and the submarine sinks. To rise, the water from the ballast tanks is forced into the sea by allowing air from the compressed air tank to enter the ballast tank. As a result, the average density of the submarine decreases, and the submarine rises.
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A person weighs $$600 \ N$$ on Earth and $$100 \ N$$ on the moon, what is the acceleration due to gravity on the moon?

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States the Archimedes' principle. List its two applications.

Archimedes' principle states that when an object is fully or partially immersed in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.
The two applications of Archimedes principle are:
i) It is used to design ships and submarines.
ii) It is used to design hydrometer which is used to measure the density of different liquids.

(a) State Newton's universal law of gravitation.
(b) Why is it called a universal law?

(a) According to Newton's law of gravitation, the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The law applies to objects anywhere in the universe.

$$F = \dfrac{Gm_{1}m_{2}}{d^{2}}$$

(b) Newton's law of gravitation is called a universal law because it applies to objects anywhere in the universe.

The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:

(i) the force that binds us to the earth

(ii) the motion of the moon around the earth

(iii) the motion of planets around the Sun

(iv) the tides due to the moon and the Sun

A balloon filled with hydrogen rises in air. explain the reason.

When a ballon is filled with hydrogen gas the weight of the ballon is less than the weight of air displaced by the ballon. Due to this extra upthrust, the balloon rises up. But as it rises higher, the density of air decreases and upthrust becomes less and ultimately upthrust becomes equal to the weight of the balloon and it stops riding further.

Is it possible for a body to have inertia but no weight ?

Each body has a mass or inertia, the weight of any body is measured by a spring balance or a weighing machine, and is equal to mg. g in the space can be zero or weight of a freely falling body is zero. The $$ g$$ at the center of the earth is zero . So, the weight $$ (mg) $$ of a body can be zero, but mass $$ (m) $$ can never be zero, and thereby inertia is also never zero.

The separation between the two masses is reduced to half. How is the magnitude of the gravitational force between them affected?

Consider two objects of masses $$m_1$$ and $$m_2$$, separated by a distance $$r$$.

According to Newton's law of gravitation,

$$F=G\dfrac{m_1m_2}{r^2}$$ (where, $$G=$$ Gravitational constant)

If distance between the two masses is reduced to half.

$$r'=\dfrac{r}{2}$$

Then,

$$F'=G\dfrac{m_1m_2}{r'^2}$$

$$F'=G\dfrac{m_1m_2}{(r/2)^2}$$

$$F'=4G\dfrac{m_1m_2}{r^2}$$

$$F'=4F$$

If distance between the two masses is reduced to half, gravitational force between them will become four times.

Explain the meaning of the statement 'relative density of aluminium is 2.7 ?

The statement 'Relative density of aluminium is 2.7 means .
A piece of aluminium of any volume has mass 2.7 times that off equal volume of water.
i.e Aluminium is 2.7 times heavier than water .

How does the weight of an object vary with respect to mass and radius of the earth?
In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass become four times of its present value, then how would the weight of any object on the surface of the earth be affected?

The weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth. i.e.
Weight of a body $$\propto \dfrac{M}{R^2}$$
Original weight $$W_o =mg=mG\dfrac{M}{R^2}$$
when hypothetically $$M$$ becomes $$4\ M$$ and $$R$$ becomes $$\dfrac{R}{2}$$
Then weight becomes
$$W_n=mG\dfrac{4M}{\left( \dfrac R2\right)^2}=(16mG)\dfrac{M}{R^2}=16\times W_o$$
The weight will become $$16$$ times the present weight.

In what direction and at what point does the buoyant force on a body due to liquid act?

The buoyant force on a body due to a liquid acts in the direction at the center of buoyancy and in opposite direction of weight of the body
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State Newton's law of gravitation.

Newton’s law of gravitation states that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

If $$F$$ is the gravitational force between two masses $$m_1$$ and $$m_2$$, separated at a distance d then,

$$F\propto m_1m_2$$

$$F\propto d^2$$

$$\therefore F\propto \dfrac{m_1m_2}{d^2}$$

$$\Rightarrow F=G\dfrac{m_1m_2}{d^2}$$, where $$G$$ is the universal gravitational constant.

A body of mass $$70\ kg$$ when completely immersed in water, displaces $$20,000\ cm^{3}$$ of water. Find: the relative density of material of the body.

Given:
mass of the body $$=70\ kg$$
Volume of water displaced $$=20,000\ cm^3$$ or $$0.02\ m^3$$
Density in C.G.S = mass/ volume $$=70\times 1000/20000=3.5\ g\ cm^3$$
Relative density of the solid = density in C.G.S. without unit
$$=3.5$$

Define the term thrust .State its S.I unit.

Thrust is the force acting normally on a surface
The S.I unit of thrust is newton (N).

Differentiate between thrust and pressure.

The total force acting perpendicular on a given surface is called thrust. Thrust acting upon a unit area is called pressure.

State whether the gravitational force between two masses is attractive or repulsive?

According to Newton's law of gravitation, any two bodies in this universe attract each other with the force of gravity. So, the gravitational force between two masses is always attractive.

How are g and G related?

$$g$$ is the gravitational acceleration and $$G$$ is the universal gravitational constant.

$$\text{Relation}$$:

The gravitational force acting on a body of $$m$$ due to another body of mass $$M$$ is $$F_g = \dfrac{GMm}{R^2}$$, where $$R$$ is the distance between the bodies.

Let $$g$$ the acceleration of the mass $$m$$ due to this gravitational force.

Then, $$F_g = mg$$

$$\therefore$$ $$mg = \dfrac{GMm}{R^2}$$ $$\Rightarrow g = \dfrac{GM}{R^2}$$, this is the relation between $$g$$ and $$G$$.

Distinguish between mass and weight.

$$Mass$$	$$Weight$$
It is the quantity of matter it contains	It is the force with which earth attracts a body
Mass is a scalar quantity	Weight is a vector quantity
Mass of a body is always constant	Weight of a body varies from place to place
S.I. unit of mass is kg	S.I. unit of weight is newton

Write an expression for the gravitational force of attraction two bodies of masses $$m_{1}$$ and $$m_{2}$$ separated by a distance r.

$$F\propto \dfrac{Gm_1m_2}{d^2}$$

Here $$G$$ is a constatnt of proportionally called the unniversal gravitational constant.

What some uses of the law of gravitation?

The law of gravitation can be used to explain the followings:

$$\bullet$$ The motion of planets around the sun

$$\bullet$$ The motion of the moon(satellite) around the earth

$$\bullet$$ The motion of a freely falling body.

How is the gravitational force between two masses affected if the separation between them is doubled?

$$F\propto \dfrac{Gm_1m_2}{d^2}$$

If the distance between the masses becomes half, the force reduces to one -fourth.

Define the terms mass and weight.

$$Mass:$$ mass of a body is the quantity of matter it contains. It is a scalar quantity.

$$Weight:$$ Weight of a body is the force with which the earth attracts it. It is the force of gravity on it. Weight is a vector quantity.

How does the gravitational force of attraction between two masses depend on the distance between them?

The gravitational force of attraction between two masses is inversely proportional to the square of the distance between them.

The gravitational force $$F=G\dfrac{m_1\times m_2}{r^2}$$, where $$r$$ is the distance between two masses $$m_1$$ and $$m_2$$.

A body falls freely under gravity from rest and reaches the ground in time t. Write an expression for the height fallen by the body.

Given: acceleration due to gravity$$=’g’,$$

initial velocity, $$u=0, time=t.$$

Let $$‘h’$$ be the height fallen by the body

$$h=ut+\dfrac{1}{2} gt^{2}$$

$$u = 0 m/s, $$ therefore the equation becomes;

$$h = \dfrac{1}{2} gt^2$$

A pebble is dropped freely in a well from its top. It takes the $$20s$$ for the pebble to reach the water surface in the well. Taking g = $$10 m / s^{2}$$ and speed of sound $$= 330m/s.$$ find the time when the sound is heard after the pebble is dropped.

Given:$$ t=20s, g=10m/s^2$$

Assume ‘h’ to be the depth of the well,

We know from the equation of motion,

$$H=ut + \dfrac{1}{2} gt^2 $$

$$= 0 + \dfrac{1}{2} (10)(20)^2 $$

$$ = 2000m$$

$$\underline{\text{Finding the time when the sound is heard after the pebble is dropped}}:$$

Given: speed of sound $$= 330 m/s$$

$$Time\, taken = \dfrac{depth}{speed}$$

$$= \dfrac{2000}{330 }=6.1s$$

For the pebble to reach the water surface, it takes $$20s,$$

Hence the total time taken when the sound is heard after the pebble is dropped

$$= 20 + 6.1 = 21.6s$$

State the S.I. units of
weight.

The S.I. unit of weight is newton and symbol is $$N$$

Find the gravitational force between two masses $$15$$ kg each separated by a distance of $$2m$$.
$$(G = 6.67 \times 10^{-11} Nm^{2} Kg^{-2})$$

Given, $$m_{1} = m_{2} = 15 kg$$

$$d = 2m $$

$$G = 6.67 \times 10^{-11} Nm^{2} Kg^{-2}$$

$$\therefore F = \dfrac{m_{1}m_{2}}{d^{2}}$$

$$F= \dfrac{6.67 \times 10^{-11} \times 15 \times 15}{2^{2}}$$

$$F= 3.75 \times 10^{-9} N$$.

ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m/s. the ball reaches the ground after 5s. Calculate:
the velocity of ball on reaching the ground. Take g = $$9.8m / s^{2}$$

Let $$t_1 $$ be the time taken by the ball to reach the greatest point from the top of the tower

To calculate the time for which the ball remains in air;

We know the from the equation of motion,

$$v=u-gt$$

$$0 = 19.6 – (9.8)(t_1)$$

$$t_1 = 2s$$

Assume motion for $$(h+d)$$ part;

Time taken for the ball to reach from the highest point of the tower to the ground is

$$t-t_1 = 5-2 = 3s$$

From the equation of motion,

$$s = ut+ \dfrac{1}{2} gt^2$$

$$‘s’$$ here is the distance from the top of the tower to the highest point, i.e., $$h+d$$

$$h+d = 0 + \dfrac{1}{2} (9.8)(3)^2$$

$$d+19.6 = 44.1m$$

$$d=24.5m$$

The height of the tower is $$24.5m$$

Assume $$‘v’$$ be the velocity of the ball when it strikes the ground

We know from the relation;

$$v = u +gt$$

$$v = 0 + (9.8)(3)$$

$$ = 29.4m/s$$

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The weight of a body on earth is $$98\ N$$ where the acceleration due to gravity is $$9.8 ms^{-2}$$. What will be its weight on the moon where the acceleration due to gravity is $$1.6 m s^{-2}$$?

Given: weight$$=98N, g=9.8m/s^2$$

If $$‘m’$$ is the mass of the body,

$$m = \dfrac{W}{g}$$

$$ = \dfrac{98}{9.8 }= 10kg$$

Mass remains constant

Mass on the moon is the same as the mass on earth $$= 10kg$$

If weight on the moon is $$W_1,$$

$$W_1 = mass \times g'$$,

on moon $$g' = 1.6\ ms^{-2}$$

$$= 10 \times 1.6 = 16N$$

A block of iron of mass 7.5 kg and of dimensions 12cm $$12cm \times 8cm \times 10cm$$ is kept on a table top on its base of side $$12cm \times 8cm$$. Calculate :
thrust,

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A ball is thrown vertically upwards with an initial velocity of 49 m/s. Calculate:
the time taken by it before it reaches the ground again. (Take g = $$9.8 m / s^{2}$$)

Consider upward direction positive

then Initial velocity $$u=49m/s$$

$$t$$ be the time taken to return

Since it come back to same point ,displacement=$$s=0$$

Given:

Acceleration $$a=-g=-9.8m/s^2$$ because gravity is downwards

$$s=ut+\dfrac{1}{2}at^2$$

$$0=ut-\dfrac{1}{2}gt^2$$

$$ut=\dfrac{1}{2}gt^2$$

$$2u=gt$$

$$t=\dfrac{2u}{g}$$

$$t=\dfrac{2(49)}{9.8}$$

$$= 10s$$

A man can jump higher on the moon. What is the cause of it?

When you're on the Moon, you are 1 / 6th as heavy. So if you weigh 72 kg on Earth, you would only weigh about 12 kg on the Moon. so we can say the gravitational force of moon is less, that's why a man can jump higher on the moon than on the earth.

At which place on the surface of the earth, the acceleration due to gravity is more and why?

The value of g varies from point to point in the universe, according to the values of M and r. Since the earth is flattened at poles and V reduces. Now since.

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

Therefore, the value of g is more at the poles of the earth than the equator.

What is Archimedes principle ?

When an object is immersed completely or partly in a fluid (liquid or gas), it experiences an upward force, called buoyant force which is equal to the weight of the fluid displaced by it.

What changes will arise in force of gravitation between the masses, if the distance between them is doubled?

Gravitational force between two masses is given by

$$F=\dfrac{Gm_1m_2}{r^2}$$

$$G$$ is gravitational constant

$$m_1$$ and $$m_2$$ are both masses respectively and $$r$$ is distance between them.

Now $$r$$ is doubled So it becomes $$2r$$

$$F=\dfrac{Gm_1m_2}{(2r)^2}$$

$$F=\dfrac{Gm_1m_2}{4r^2}$$=$$\dfrac{F}{4}$$

Hence force becomes one-fourth of initial value.

A man moves from earth poles to equatorial line, then what changes will be there in mans weight and why?

The gravitational acceleration by the earth is $$g=\dfrac{GM}{R^2}$$, where $$M$$ is the mass of the earth and $$R$$ is the radius of the earth.

As the earth is not an absolute sphere, its radius at the equator is more than its radius at the pole. $$\therefore R_{equator} > R_{pole}$$

$$\therefore g_{pole} > g_{equator}$$

Now we know that the weight of an object of mass m is $$W=mg$$.

Since, $$ g_{pole} > g_{equator}$$, we get $$ W_{pole} > g_{equator}$$

So, the weight of the man will decrease as he moves from the pole to the equator.

Write the formula of the gravitational force between two bodies.

Let the mass of two interacting bodies be $$m_1$$ and $$m_2$$ and $$d$$ be the distance between them, then the force of gravitation/gravitational force is given by:

$$F=G\dfrac{m_1m_2}{d^2}$$

where G is called gravitational constant.

State Archimedes principle. Explain the law of floatation of bodies in a liquid.

Archimedes determined the amount of buoyant force exerted by the liquid on the body. According to him, when a body is wholly or partially immersed in a liquid at rest, it experiences an upthrust, equal to the weight of the liquid displaced by the body.
(i) A body will sink in a liquid if its density is more than that of the liquid in which it is immersed.
(ii) The body floats being completely immersed in water if the density of the body is equal to the density of the liquid.
(iii) The body floats being partially immersed if the density of the body is less than the density of liquid.

Calculate weight of a body of $$6$$ kg mass on the earth and the moon.($$g_e=9.8 m/s^{2}$$ , $$g_m=1.63m/s^{2}$$)

We know that,
Weight = mass $$\times$$ acceleration due to gravity
Here, mass $$= 6\,kg$$
Acceleration due to gravity on the Earth $$= 9.8 m/s^2$$
Acceleration due to gravity on the Moon $$= 1.63 m/s^2$$

Therefore,

Weight on the Earth $$= 6 \times 9.8 = 58.8\,N$$

And,

Weight on the Earth $$= 6 \times 1.63 = 9.8\,N$$

A body of mass $$50\ kg$$ and another body of mass $$60\ kg$$ are separated by a distance of $$2\ m$$.
What is the force of attraction between them?

$$m_1=50\ kg, m_2= 60\ kg$$
$$d=2\ m, G=6.67\times 10-11\ Nm^{2}/kg^{2}$$
$$F=G\dfrac{m_1 m_2}{d^{2}}$$
$$=\dfrac{G\times 50\times 60}{2^2}=750\times G$$
$$=750\times 6.67\times 10^{-11}=5.0025\times 10^{-8}N$$
Force of attraction by a body on earth $$GMm$$
$$F=\left(\dfrac{GMm}{R^{2}}\right)$$
$$G-$$ Gravitational constant
$$M-$$ Mass of earth
$$m-$$ Mass of body
$$R-$$ Radius of earth
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State universal law of gravitation. Establish the relation $$M_{e}=\dfrac{gR_{e}^{2}}{G}$$, where $$M_{e} \ and \ R_{e}$$ are the mass and radius of Earth respectively .

Gravitational force acting on the mass $$m$$ kept at earth surface, $$F_g = \dfrac{GM_em}{R_e^2}$$.......(1)

also, $$F_g = mg$$.........(2)

From above equations

$$\therefore$$ $$mg = \dfrac{GM_em}{R_e^2}$$

$$ { M_e = \dfrac{gR_e^2}{G}}$$ Proved.

Where will the value of g be the minimum on the earth's surface?

The value of ‘g’ is minimum at the equator because earth is not a perfect sphere. Its radius is maximum at the equator. Hence, according to the equation $$g = \dfrac{GM}{R^2}$$, the equator is the place where g will be minimum. The value of g is at the equator is 9.78 m/s^2.

If a body of mass $$40\ kg$$ is kept at a distance of $$0.5\ m$$ from a body of mass $$60\ kg$$, what is the mutual force of attraction between them?

$$F=\dfrac{Gm_{1}m_{2}}{d^{2}}$$ $$m_{1}=40\ kg, m_{2}=60\ kg, d=0.5\ m$$
$$=\dfrac{G\times 40\times 60}{(0.5)^{2}}$$
$$=\dfrac{G\times 2400}{0.25}$$
$$=\dfrac{6.67\times 10^{-11}\times 2400}{0.25}$$
$$=64032\times 10^{-11}$$
$$=6.4032\times 10^{-7}N$$

The earth is not a perfect sphere, its radius is not the same everywhere. If so, will the value of $$g$$ be the same everywhere on earth?

The magnitude of the gravitational force $$F$$ will be equal to the product of mass and acceleration due to the gravitational force, that is,

$$F = m g$$

$$mg=G \ \dfrac{m \times M}{R^2}$$

$$\Rightarrow g=G\ \dfrac{M}{R^2}$$ . . . . (i)

Where $$M$$ is the mass of the earth.

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of $$g$$ according to the eq. $$(i)$$ becomes greater at the poles than at the equator.

Mass of the earth is $$6\times 10^{24}kg$$ and that of the Moon is $$7.4\times 10^{22}\ kg$$. The distance between Earth and Moon is $$3.84\times 10^{5}\ km$$. Calculate the force of attraction of Earth of Moon. $$(G=6.7\times 10^{-11}\ Nm^{2}kg^{-2})$$

The gravitational force between two objects

$$F=\left(\dfrac{GMm}{d^{2}}\right)$$
$$\Rightarrow F=\dfrac{6.7\times 10^{-11}\times 6\times 10^{24}\times 7.4\times 10^{22}}{(3.84\times 10^{5})^{2}}$$
$$\Rightarrow F=2017\times 10^{25}N$$

Find the factors affecting the value of $$g$$?

The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of $$g$$ becomes greater at the poles than at the equator. For most calculations, we can take $$g$$ to be more or less constant on or near the earth. But for objects far from the earth, the acceleration due to the gravitational force of earth $$g$$ is given by

$$g=G\dfrac {M}{d^2}$$

where $$G=$$ universal gravitational constant, $$M=$$ mass of the earth, and $$d=$$ distance between the object and the earth.

If the distance between two bodies that attract each other is trepled, how many times will their mutual force of attraction be ?

$$F=\left(\dfrac{GMm}{d^{2}}\right)$$

Now $$d=3d$$

$$F_1=\left(\dfrac{GMm}{(3d)^{2}}\right)$$

$$F_1=\dfrac{F}{9}$$

How the mass and distance between objects affect gravitational force?

According to the Newton's law of gravitation,

$$F_{gravitation} = \dfrac{Gm_1m_2}{r^2}$$

where $$m_1$$ and $$m_2$$ are masses of the two bodies and $$r$$ is the distance between them.

Among the two bodies if the mass of one body is doubled gravitational force becomes two times. When the mass of both the bodies are doubled. Gravitational force becomes four times. When the distance increases gravitational force decreases, Gravitational force varies in reciprocal of the square of the distance between the bodies.

Mass and weight of a body are determined at the pole and at the equator.
Is there any change in the weight? Justify your answer.

If a body has the mass $$m$$ then its weight is $$W=mg$$.

The mass of the body is constant everywhere but the value of $$g$$ is different at different latitudes.

We know that $$g=\dfrac{GM}{R^2}$$, where $$M$$ is the mass of the earth and $$R$$ is the radius of the earth.

Our planet earth is not an absolute sphere. It is bulged out at the equator. Thus, the radius of the earth is more at the equator as compared to that of the earth at the poles.

Since, $$R_{equator}>R_{poles}$$, $$\therefore g_{equator}<g_{poles}$$

So, $$W_{equator}<W_{poles}$$

The mass of a body is 30 kg. What is its weight on the moon? $$(g=1.62\ m/s^{2})$$

Given:

Mass of body = $$30\ kg$$

gravitational acceleration on the moon = $$1.62\ m/s^2$$

Weight of the body on the moon = $$\text{Mass of the body}\times \text{gravitational acceleration on the moon} = 30\times 1.62=48\ N$$

Calculate the weight of a body on the moon if it weights $$150\ kg$$ in earth?

$$\text{Weight of a body on the moon} = \dfrac{1}{6}\times \text{Weight of the body on the earth}$$

Given: Weight of the body on the earth = $$150\ kg$$

Weight on the moon $$=150\times 1/6=25\ kg$$

Two identical isolated particles, each of mass $$2.00 kg$$ are separated by a distance of $$30.0\ cm$$. What is the magnitude of the gravitational force exerted by one particle on the other?

Given

$$m_{1} = m_{2} = 2\ kg$$; mass of the each particle

$$r = 30\ cm $$

$$\Rightarrow r= \dfrac {30}{100}\ m $$

$$\Rightarrow r= 0.3\ m$$

Gravitational force on one particle due to other

$$F = \dfrac {Gm_{1}m_{2}}{r^{2}}$$

$$\Rightarrow F = \dfrac {6.637 \times 10^{-11} Nm^{2} \times 2\ kg \times 2\ kg}{0.09 m^{2} \times kg^{2}}$$

$$\Rightarrow F= 294.9\times 10^{-11} N$$

$$\Rightarrow F = 29.5 \times 10^{-11} N$$

Hence, the gravitational force on one particle due to the other is $$29.5\times 10^{-11}N$$ towards the other particles.

$$F=\dfrac{G.m_{1}m_{2}}{d^{2}}$$
Mention each letter indicates to what?

The above formula gives force of interaction between two masses according to newton law of gravitaion.

$$m_{1}, m_{2}-$$ mass of the bodies
$$d-$$ Distance between the objects
$$F-$$ Gravitational force
$$G-$$ Gravitational constant.

State Universal Law of gravitation.

Explanation :

The universal law of gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
For two objects of masses, $$m_1$$and $$m_2$$ and the distance between them r, the force (F) of attraction acting between them is given by

$$F=G \frac{m_{1} m_{2}}{r^{2}}$$

Where G is the universal gravitation constant.

The value of G is found to be G $$=$$ $$6.673$$ X $$10^{-11}$$$$\mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}$$

Tabulate some situations in your daily life where buoyant forces are experienced in liquids and gases.

$$(i)$$A hydrogen-filled balloon rises in the air.
$$(ii)$$Ships float on the surface of water.
$$(iii)$$Submarines work by controlling the upthrust of water.
$$(iv)$$While drawing water from well, weightlessness is felt when the bucket is underwater level.
$$(v)$$ Life jacket floats on the surface of water.

How is relative density of kerosene calculated?

Relative density of a liquid $$=\dfrac{Density\,of\,liquid}{Density\,of\,water}$$

$$Density\ of\ kerosene = 810 kg/m^3$$

$$Density\ of\ water = 1000 kg/m^3$$

$$Relative\ density\ of\ kerosene=\dfrac{810\,kg/m^{3}}{1000\,kg/m^{3}}$$

$$=0.81$$

The mass of a body can never be _________.

The mass of a body cannot be zero because mass is a quantity of matter contained in a body.

Two identical copper spheres of radius R are in contact with each other. If the gravitational attraction between them is F, find the relation between F and R.

Instead of two big spheres of radius $$R$$ assume . $$2$$ small sphere of mass $$m$$ at their respective COM.

then $$F=\cfrac { Gm\times m }{ { (2R) }^{ 2 } } \\ F=\cfrac { G{ m }^{ 2 } }{ 4{ R }^{ 2 } } $$

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Assume that life has existed on the surface of the moon, without changing its present acceleration due to gravity i.e., one sixth that on the surface of the earth. If 5 kg weight of sugar is purchased on the Earth and the Moon, how many cups of tea can be made out of it on the Earth and the Moon respectively?
Note, From 100 g of sugar 10 cups of tea can be made on the earth. (g = 10 m s$$^{-2}$$)

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What happens to the gravitational force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?

$$F=\dfrac{GMm}{R^2}$$

$$(i)$$

If one mass say, $$m$$ is doubled, then force is

$$F'=\dfrac{2GMm}{R^2}=2F$$ the force gets doubled.

$$(ii)$$

If the distance is doubled and tripled-

$$F'=\dfrac{2GMm}{(2R)^2}=\dfrac{F}{4}$$

and,

$$F'=\dfrac{2GMm}{(3R)^2}=\dfrac{F}{9}$$

$$(iii)$$

If both masses are doubled, then force is

$$F'=\dfrac{4GMm}{R^2}=4F$$ the force gets quadrupled.

What will be the weight of an object on the surface of the moon for a body of mass $$m$$.

The weight of object on the surface of moon will be = $$ mg_{moon}$$

$$ g_{moon}$$ = $$ g_{earth}/6$$ = $$1.66 \ m/s^2$$

Weight on moon = $$1.66m$$

The force with which the earth attracts an object is called weight of the object. Calculate the weight of the moon from the following data : The universal constant of gravitation ($$G=6.67 \times 10^{-11} N-m^2/ kg^2 $$ , mass of the moon $$ = 7.36 \times 10^{22} kg $$, mass of the earth $$ = 6 \times 10^{24} kg $$ and the distance between the earth and the moon $$ = 3.8 \times 10^5 km. $$)

Given:

$$G=6.67×10^{−11} N−m^2/kg^2\\M_m=7.36×10^{22}kg\\M_e=6×10^{24}kg\\r=3.8×10^5km$$

The attracting force is given as,

$$F = G\dfrac{{{m_m}{m_e}}}{{{r^2}}}$$

$$ = 6.67 \times {10^{ - 11}} \times \dfrac{{6 \times {{10}^{24}} \times 7.36 \times {{10}^{22}}}}{{{{\left( {3.8 \times {{10}^8}} \right)}^2}}}$$

$$ = 2.04 \times {10^{20}}\;{\rm{N}}$$

Thus, the weight of the moon is $$2.04 \times {10^{20}}\;{\rm{N}}$$.

A Metal has weight in Air=210 g in water = 180 g in liquid =120 g.
Find relative density of metal and liquid.

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How the acceleration due to gravity(g) related to the gravitational constant(G)

Gravitational force acting on the mass $$m$$ $$F_g = \dfrac{GMm}{R^2}$$

But $$F_g = mg$$

$$\therefore$$ $$mg = \dfrac{GMm}{R^2}$$

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

What is the change in the gravitational force when the distance between two masses is doubled?

Gravitational force is given as,

$$F = \dfrac{Gm_1m_2}{d^2}$$

$$G$$; gravitational constant

$$m_1,m_2$$; masses of the two object

$$d$$; distance between two objects

If distance is doubled-

$$d' = 2d$$

So gravitational force,

$$F' = \dfrac{Gm_1m_2}{(d')^2}$$

$$\Rightarrow F' = \dfrac{Gm_1m_2}{(2d)^2}$$

$$\Rightarrow F' = \dfrac{Gm_1m_2}{4(d)^2}$$

$$\Rightarrow F' = \dfrac{F}{4}$$

The weight of any person on the moon is about $$1/6$$ times that on the earth. He can lift a mass of $$15$$ $$kg$$ on the earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon?

$$Weight=mass\times g$$

Since mass of person is constant so the reduction in weight on moon will be due to reduction in value of $$g$$.

Weight is directly proportional to $$g$$.

So weight reduced to $$1/6$$ times mean that $$g$$ reduced to $$1/6$$ times.

Force required is nothing but weight of object being lifted.

Since the applied force will be same on moon it means weight is going to be same.That is product of $$m$$ and $$g$$ is going to be same.So when $$g$$ reduces to $$1/6$$ times , $$m$$ change oppositely, that is , it becomes $$6$$ times to keep the product same.

So new mass that can be lifted=$$15 \times 6=90kg$$

On the moon's surface the acceleration due to gravity is 1.67 $${ ms }^{ -2 }$$ If the radius of the moon is $$1.74\times { 16 }^{ 6 }m$$, Calculate the mass of moon

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(a) Define mass of a body. What is the SI unit of mass?
(b) Define weight of a body. What is the SI unit of weight?
(c) What is the relation between mass and weight of a body?

(a) The mass of a body is the quantity of matter contained in it. The SI unit of mass is kilogram (kg).
(b) The weight of a body is the force with which it is attracted towards the centre of the earth. The SI unit of weight is newton (N).
(c) Weight, $$W = m \times g$$, i.e. the weight of a body is directly proportional to its mass.

Why the mass of an object on Earth be same as that on Mars but the weight on both the planets are different?

This is because the weight ($$W$$) of an object at a place depends on the acceleration due to gravity of that place i.e. $$W=mg$$ or $$W\ \alpha\ g$$ and since the values of acceleration due to gravity on both the planets differ, thus the weight of the object will be different for both the planets.

Gravitation - Class 9 Physics - Extra Questions

You have a bag of cotton and an iron bar. each indicating a mass of $$100\ kg$$ when measured on a weighing machine in reality, one is heavier than the other. Can you say which one is heavier and why?

Define thrust. What is its unit?

Find the thrust required to exert a pressure of $$ 50000 $$ pascals on an area of $$ 0.05 m^2 $$?

How much thrust will be required to exert a pressure of $$ 20,000 $$ Pa on an area of $$ 1cm^2 $$

Define relative density. What is its SI unit?

A cubical block of wood of side $$5cm$$ is placed on a table. Find the thrust by the block of wood on the table, if density of wood is $$5g$$ $${cm}^{-3}$$ and $$g$$ is $$10m$$ $${s}^{-2}$$.

A cubical block of wood of density $$5\ g{cm}^{-3}$$ with a sides of $$10cm$$ stands on table. Find the thrust by the block of wood on the table if $$g =10\ m{s}^{-2}$$.

Calculate the mass of a body in tonnes whose volume is 2 m$$^3$$ and R.D is 5.2.

A book of mass $$20\ g$$ is placed on a table. Find the thrust exerted by the book on the table. (Take acceleration due to gravity as $$10\ m{s}^{-2}$$)

Calculate the force of gravitation due to earth on a ball of mass $$60\ kg$$

Imagine a body that is completely submerged in water, but whose depth of submergence can be varied. In which case does it experience a larger upthrust, when it is submergged deep or shallow?

What force does water exert on the base of a house tank of base area $$1.5\ m^{2}$$ when it is filled with water upto a height of $$1\ m$$?[Density of water is $$10^{3}kg\ m^{-3}$$ and $$g = 10\ m\ s^{-2}]$$

The force acting on a body normal to its surface is _______

What is the mass and weight of an object on the surface of the moon, if on the earth its mass is $$12$$ kg?

Explain buoyancy and buoyant force.

Calculate the gravitational force between two metal spheres of masses $$50\,kg$$ and $$100\,kg$$ respectively and the separation between their centers is $$50\,cm$$.

Two bodies are at a specific distance so as to attract each other. How many times will the mutual force of attraction be if the mass of one of them is doubled?

Two bodies are at a specific distance so as to attract each other.How many times the force of attraction be if the masses of both the bodies are doubled?

Explain the variation of $$'g'$$ due to the shape of the earthy?

Fill in the blanks with suitable words.The value of ‘g’ in the poles is _________ (greater/lesser) than at the equator.

What is the effect of Non-sphericity of Earth on the value on the 'g'?

Derive the relation between acceleration due to gravity (g) and the gravitational constant (G)

Sate and explain Newton's law of gravitation .

Write Newton's universal law of gravitation.

When a hollow steel ball and a solid steel ball of the same size are submerged in water, do the two experience the same upthrust?

Why some objects float on the water? And some sink?

What do we call the gravitational force between the earth and an object?

Mass of an object is $$60\ kg$$ on the surface of the earth. Find its mass and weight on the surface on the moon.

Pressure varies with area (A) as provided thrust is same.True =1; False=0

The relative density of substance is numerically equal to its density in the _____ system of units.

The ratio of the density of the substance to the density of water is known as __________.

The gravitational force of attraction between two masses depend on the distance between them is $$G= $$gravitaional constant $$=k \times 10^{-11} Nm^{2} kg^{-2}$$. what is the value of k?

State whether weight is a scalar or vector quantity?

What is the SI unit of thrust?(Newton, Joule)

Buoyant Force is the upward force that acts in the direction opposite to that of weight.Type 1 for true and 0 for false

What do you understand by the term Force of gravity?

Explain the statement "relative density of brass is 8.4"

If the distance between the two point masses is doubled then force of attraction is decreased __________times

Is like density, the relative density of a body different in different systems of measurement ?

Calculate the gravitational force between two bodies $$'a'$$ and $$'b'$$ of mass $$50\ kg$$ each at a distance of $$1\ m$$ from each other.

What is the relative density of the object immersed in water?

How can the knowledge of relative density help in verifying the purity of substances?

State the Archimedes principle.

A block of iron of mass $$7.5kg$$ and of dimensions $$ 12cm \times 8cm \times 10cm $$ is kept on a table top on its base of side $$ 12cm \times 8cm$$. Calculate a) Thrust b) Pressure exerted on the table top.Take $$1kgf = 10N$$

What is drag?

When the mass of one of the two bodies is doubled the gravitational force between them increases by ________ times.

What is the magnitude of gravitational force between earth & $$1$$ kg object on the surface of earth?

Find Pressure if 20 N of force is applied on a surface having area $$ 10 cm^2 $$.

Define density and relative density.

Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Differentiate between $$'g'$$ and $$'G'$$.

Distinguish between density and relative density.

Name the scientist who gave the magnitude of buoyant force acting on a solid object immersed in a liquid.

Two bodies of masses $$1\ kg$$ each is situated $$1\ m$$ apart. Find the gravitational force between them.

On what factors does the value of gravitational force depend?

Two spherical bodies, each of $$100$$ kg, are separated by $$1m$$. Calculate the gravitational force exerted by each on the other.

What will be the mass of a man of $$60$$ kg on the moon?

Two spheres each of mass $$10^5$$ kg and radius $$ 10$$ m are kept in contact. Find the force of gravitation acting between them?

At the surface of a certain planet, the acceleration due to gravity is one-quarter of that on earth. A brass ball is transported to this planet, then find the percentage change in its weight?

If F is the force between two bodies of masses $$m_1$$ and $$m_2$$ at certain separation. Find the force between $$\sqrt 2 m_1$$ and $$\sqrt 3 m_2$$ at same separation.

Find the gravitational attraction between the two atoms in a hydrogen molecule. Given that $$G=6.67\times 10^{-11} Nm^2 kg^2$$, mass of a hydrogen atom $$=1.67\times 10^{-27}$$ kg and distance between the two atoms $$=1^oA.$$

The tides formed by the rising and falling of water levels in the oceans are due to the gravitational pull of the sun and moon on the water is the importance of the law of gravitation.Type $$1$$ for true and $$0$$ for false.

Define 'relative density'.

State five differences between mass and weight.

What do you understand by weight?

What happens to the gravitational force between two objects, if $$(i)$$ The mass of one object is doubled?$$(ii)$$ The distance between the objects is doubled and tripled?$$(iii)$$ The masses of both objects are doubled?Give reasons in each case

How does the force of gravitation between two objects change when the distance between them is reduced to half?

What is the importance of universal law of gravitation?

The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

A ball is thrown vertically upwards with a velocity of $$49\ m/s$$. Calculate$$(i)$$ The maximum height to which it rises.$$(ii)$$ The total time it takes to return to the surface of the earth.

In what direction does the buoyant force on an object immersed in a liquid act?

Gravitational force on the surface of the moon is only $$\displaystyle\frac{1}{6}$$ as strong as gravitational force on the earth. What is the weight in newtons of a $$10\ kg$$ object on the moon and on the earth?

Why does a block of plastic released under water comes up to the surface of water?

State whether True or False :Relative density is more easily and perhaps more accurately measured without measuring volume.

State and explain Newtons law of gravitational force.

Does a sunk body at the bottom of a vessel experience an upthrust?

Two spherical balls of mass $$10\ kg$$ each are placed with their centres $$10\ cm$$ apart. Find the gravitational force of attraction between them.

Solve: A sphere of mass 10 kg is attracted by another sphere of mass 150 kg, with a force equal to $$1.28 \times 10^{-6} N$$, when their centers are separated by a distance of 0.28 m. Calculate the gravitational constant.

Can you make iron to float? How?

What force does water exert on the base of a house tank of base area $$1.5\ m^{2}$$ when it is filled with water upto a height of $$1\ m$$?
[Density of water is $$10^{3}kg\ m^{-3}$$ and $$g = 10\ m\ s^{-2}]$$

Two bodies are at a specific distance so as to attract each other.
How many times the force of attraction be if the masses of both the bodies are doubled?

Fill in the blanks with suitable words.
The value of ‘g’ in the poles is _________ (greater/lesser) than at the equator.

Pressure varies with area (A) as provided thrust is same.
True =1; False=0

What is the SI unit of thrust?
(Newton, Joule)

A block of iron of mass $$7.5kg$$ and of dimensions $$ 12cm \times 8cm \times 10cm $$ is kept on a table top on its base of side $$ 12cm \times 8cm$$. Calculate
a) Thrust
b) Pressure exerted on the table top.
Take $$1kgf = 10N$$

The tides formed by the rising and falling of water levels in the oceans are due to the gravitational pull of the sun and moon on the water is the importance of the law of gravitation.
Type $$1$$ for true and $$0$$ for false.

What happens to the gravitational force between two objects, if
$$(i)$$ The mass of one object is doubled?
$$(ii)$$ The distance between the objects is doubled and tripled?
$$(iii)$$ The masses of both objects are doubled?
Give reasons in each case

A ball is thrown vertically upwards with a velocity of $$49\ m/s$$. Calculate
$$(i)$$ The maximum height to which it rises.
$$(ii)$$ The total time it takes to return to the surface of the earth.

State whether True or False :

Relative density is more easily and perhaps more accurately measured without measuring volume.

Solve:
A sphere of mass 10 kg is attracted by another sphere of mass 150 kg, with a force equal to $$1.28 \times 10^{-6} N$$, when their centers are separated by a distance of 0.28 m. Calculate the gravitational constant.

(a) Explain Newton's first law of motion with an example.
(b) $$F=\cfrac{G{m}_{1}{m}_{2}}{{d}^{2}}$$ is the mathematical form of Newton's law of gravitation.
Give the statement of Newton's law of gravitation.

The mass of earth is 80 times that of moon and their diameters are in the ratio 4:1 respectively.
What is the value of g on moon if g on earth $$9.8 \, m/s^2$$.

The mass of the earth is $$6\times {10}^{24}kg$$. The distance between the earth and the Sun is $$1.5\times {10}^{11}m$$. If the gravitational force between the two is $$3.5\times {10}^{22}N$$, what is the mass of the Sun?
Use $$G=6.7\times {10}^{-11}N{m}^{2}{kg}^{-2}$$

(1) Mass of a book is 500g on the surface of the earth . what will be its mass at a height equal to radius of earth
(2) find the weight of the book at the surface of the earth

The relative density of lead is 11.Its density is _ $$gcm^{-3}$$ or ___ $$kgm^{-3}$$.

Give scientific reasons:
The weight of an object changes from place to place though its mass is constant.

Give reason for the following:
The force of gravitation between two cricket balls is extremely small but that between a cricket ball and the earth is extremely large.

How much is the weight of an object on the moon as compared to its weight on the earth?
Give reason for your answer.

(a) State the universal law of gravitation. Name the scientist who gave this law.
(b) Define gravitational constant. What are the units of gravitational constant?

What happens to the gravitational force between two objects when the distance between them is:
(i) doubled?
(ii) halved?

Give reason for the following:
If the force of gravity acts on all bodies proportional to their masses, why does not a heavy body fall faster than a light body ?

Give reason for the following :
As we go higher above the surface of the earth the weight of the body decreases

Give reason for the following :
Why is the weight of an object on the moon one-sixth its weight on the earth ?