Relative density of salty water
= \(\cfrac { Loss\quad of\quad weight\quad salary\quad water }{ Loss\quad of\quad weight\quad in\quad pure\quad water } \)
= \(\cfrac { \left( 280-190 \right) g-wt }{ \left( 280-240 \right) g-wt } \)
= \(\cfrac { 90g-wt }{ 40g-wt } =\cfrac { 9g-wt }{ 4g-wt } =2.25\)

The object will float with its maximum volume submerged under the liquid R because it has minimum density, out of the given liquids.

When we immerse an empty plastic bottle in a bucket of water, the upward force (upthrust or buoyant force) exerted by water on the bottle is greater than its own weight, therefore it comes above the surface of water.
To keep the bottle completely immersed, an external force which is equal to the difference between the upward force and the weight of the bottle, must be applied on the botde in downward direction. This is because the upthrust on the bottle due to water must be balanced.

Since, density \(\left( \rho \right) =\cfrac { mass\left( m \right) }{ volume\left( V \right) } \)
Therefore, if the volume of a body is compressed to half of its previous volume, then the density will be doubled.

Given, mass on the earth = mass on the moon = 80 kg
Man's weight on the earth,
w_e=9.8 x 80= 784 N
Man's weight on the moon,
w_m = 1.63 X 8O= 130.4 N
Thus, the weight on the moon is 130.4 N.

Given, initial velocity, u = 0
Time taken, t = 2s
Acceleration due to gravity, a = g
From second equation of motion,
\(s=ut+\cfrac { 1 }{ 2 } { gt }^{ 2 }=0+\cfrac { 1 }{ 2 } \times 10\times \left( 2 \right) ^{ 2 }\)                                         
The height of body above the ground after 2 s of its fall, h = 100 - 20 = 80m·

When a body is immersed partially or wholly in a fluid (liquid or gas), it displaces fluid. The displaced fluid tends to regain its original condition. In doing so, it exerts an upward force on the body.
TRhe upward force acting on a body immersed in a fluid is called upthrust or force of buoyancy and phenomenon is called buoyancy.
For example, a cork taken inside water experiences an upwards thrust and comes to the surface. Similarly, while drawing water from a well, a bucket is found too much lighter when it is inside water than when it comes out of it.
The buoyancy acts through the centre of gravity of the displaced fluid which is called centre of buoyancy.

(a) The upward force acting on a body immersed in a fluid is called force of buoyancy and the phenomenon is called buoyancy.
(b) An object floats on water when its density is less than that of water.
(c) When a body is immersed fully or partially in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it. This is Archimedes' principle.
(d) The upthrust or the buoyant force exerted by water bring as the plastic block to the surface of water.

Suppose at height h, the value of g reduces by 36 % i.e., it becomes 64 % of that at the surface. Then

\({ g }_{ h }=64\cdot /\cdot \quad of\quad g=\frac { 64 }{ 100 } g\) 
\(But\ { g }_{ h }=g{ \left( \frac { R }{ R+h } \right) }^{ 2 }\) 
\(\therefore \frac { 64 }{ 100 } g\quad =g{ \left( \frac { R }{ R+h } \right) }^{ 2 }\quad or\quad \frac { 8 }{ 10 } =\frac { R }{ R+h } \) 
\(or\ h=\frac { R }{ 4 } =\frac { 6400 }{ 4 } =1600\quad km.\) 

mass of the object on the moon = 6 Kg.
mass of the object on the earth = 6 kg.
Weight of the object on the moon = 1/6 x 98 = 16.3 N
Weight of the object on the moon = 1/6 X 16.3 N

Let M_E be the mass of the Earth and m be an object on the surface of the Earth. Let R_E be the radius of the Earth. According to the universal law of gravitation, weight W_E of the object on the surface of the Earth is given by,
\(W_{E}=\frac{G M_{E} m}{R_{E}^{2}}\) 
Let M_M and R_M be the mass and radius of the moon. Then, according to the universal law of gravitation, weight W_M of the object on the surface of the moon is given by:-
\(\begin{aligned} &W_{E}=\frac{G M_{M} m}{R_{M}^{2}} \end{aligned}\) 
\(\frac{W_{M}}{W_{E}}=\frac{M_{M} R_{E}^{2}}{M_{E} R_{M}^{2}}\) 
Where, M_E = 5.98 x 10²⁴ Kg, M_M = 7.36 x 10²² Kg
R_E = 6.4 x 10⁶ m, Rm = 1.74 x 10⁶ m
\(\therefore \frac{W_{M}}{W_{E}}=\frac{7.36 \times 10^{22} \times\left(6.37 \times 10^{6}\right)^{2}}{5.98 \times 10^{24} \times\left(1.74 \times 10^{6}\right)^{2}}=0.165 \approx \frac{1}{6}\) 
Therefore, weight of an object on the moon is \(\frac{1}{6}\)^th of its weight on the Earth.

Weight of an object on the moon. The weight of an object on the moon is the force with which it is attracted towards the centre of the moon.

The mass and radius of the moon is less than that of the earth. Due to this, the moon exerts lesser force of attraction on the object. Hence, the weight of an object on the moon is less than that on the earth. The gravitational force of the moon is about one-sixth of that on the earth.

As the weight of body is a force acting vertically downwards towards the centre of the earth, it has both magnitude and direction. Therefore, weight is a vector quantity.

As 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 same as that of force, i.e., Newton (N)

A heavy coin and a piece of paper were placed inside it. The ends of the tube were closed. Air of the tube was removed by a vacuum pump. When the tube was quickly inverted, it was observed that both the coin and the paper hit the bottom at the same time. When the experiment was repeated with air inside the tube, it was observed that the piece of paper falls slowly while the coin hits the bottom immediately. This proves the Galileo assertion that in vacuum all bodies irrespective of their mass fall towards the earth with the same acceleration.

Mass of the earth, M = 6x10²⁴
Mass of the sun, M_s = 2x10³⁰
Mass of the Moon, M _m  = 7.3x10²²
Distance from the sun, r_s = 1.5x10¹¹m
Distance from the moon, r_m  = 3.84x10⁸m
Gravitational force exerted by the sun on the earth. \({ F }_{ s }=G\frac { { M }_{ s }M }{ { r }_{ s }^{ 2 } } \)
Gravitational force exerted by the moon on the earth, \({ F }_{ m }=G\frac { { M }_{ m }M }{ { r }_{ m }^{ 2 } } \)

\(\therefore \ \frac { { F }_{ s } }{ { F }_{ m } } =\frac { { GM }_{ s }M }{ { r }_{ s }^{ 2 } } \times \frac { { r }_{ m }^{ 2 } }{ { M }_{ m }M } =\frac { { M }_{ s } }{ { M }_{ m } } \times { \left( \frac { { r }_{ m } }{ { r }_{ s } } \right) }^{ 2 }\) 
\(=\frac { 2\times { 10 }^{ 30 } }{ 7.3\times { 10 }^{ 22 } } \times { \left( \frac { 3.84\times { 10 }^{ 8 } }{ 1.5\times { 10 }^{ 11 } } \right) }^{ 2 }=179.55\)
Thus the sun exerts force, about 180 times the exerted by the moon on the earth.