Difference between mass and weight:

Force of gravitation, F = \(F'=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \)
(i) When mass of one body (m1 or m2) is doubled, the force gets doubled.
\(F'=G\frac { { (2m }_{ 1 }){ m }_{ 2 } }{ { r }^{ 2 } } =2G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } =2F\)
(ii) when the distance between the bodies is doubled, 
\(F'=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ (2{ r }^{ 2 }) } =\frac { 1 }{ 4 } G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \frac { 1 }{ 4 } F\)i.e. the force becomes one-fourth of the original force.
(iii) When the masses of both bodies are doubled,
\(F'=G\frac { { (2m }_{ 1 }){ (2m }_{ 2 }) }{ { r }^{ 2 } } =4G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } =4F\)
i.e., the force becomes four times the original force.
(iii) When the distance between the two bodies is tripled,
\(F'=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ (3{ r }^{ 2 }) } =\frac { 1 }{ 9 } G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \frac { 1 }{ 9 } F\)
i.e., the force becomes one-ninth of the originals force.

(i) \(u=0,\quad g=-9.8\quad m/s^{ 2 },\quad s=-4.9m\) 
\(s=ut+\frac { 1 }{ 2 } { gt }^{ 2 }\) 
\(\therefore -4.9=0-\frac { 1 }{ 2 } \times 9.8\times { t }^{ 2 }\quad or\quad t=\sqrt { \frac { 2\times 4.9 }{ 9.8 } } =1\ s.\) 
(ii) \(u=0,\quad s=-4.9m,\quad g=-9.8\quad m/s^{ 2 },\) 
\({ v }^{ 2 }-{ u }^{ 2 }=2gs\) 
\(\therefore \ { v }^{ 2 }-0=2\times (-9.8)\times (-4.9)\quad or\quad v=\sqrt { 2\times 9.8\times 4.9 } =9.8\ m/s.\) 
(iii) \(u=0,\quad g=-9.8\quad m/s^{ 2 },\quad s--7.9m\)
\(\because { v }^{ 2 }-{ u }^{ 2 }=2gs\) 
\(\\ \therefore { \ v }^{ 2 }-0=2\times (-9.8)\times (-7.9)\) 
\(\\ v=\sqrt { 2\times 9.8\times 7.9 } m/s=10.46\quad m/s\)
(iv) Acceleration after 1 s and after 2 s will be 9.8 m/s² , because acceleration due to gravity remains almost same for small heights.

Let acceleration due to gravity at equator,
\({ g }_{ e }=\cfrac { { GM }_{ e } }{ { R }_{ e }^{ 2 } } \)
and acceleration due to gravity at poles
\({ g }_{ p }=\cfrac { { GM }_{ e } }{ { R }_{ p }^{ 2 } } \)
The variation of acceleration due to gravity
\(\Delta \) g=g_p -g_e =GM_e\(\left( \cfrac { 1 }{ { R }_{ p }^{ 2 } } -\cfrac { 1 }{ { R }_{ e }^{ 2 } } \right) \)
Percentage variation in g =\(\cfrac { { GM }_{ e }\left( \cfrac { 1 }{ { R }_{ p }^{ 2 } } -\cfrac { 1 }{ { R }_{ e }^{ 2 } } \right) }{ \cfrac { GM_{ e } }{ { R }_{ p }^{ 2 } } } \times 100\)
=\(\cfrac { { R }_{ e }^{ 2 }-{ { R }_{ p }^{ 2 } } }{ { R }_{ e }^{ 2 }{ R }_{ p }^{ 2 } } \times 100\times { R }_{ e }^{ 2 }=\cfrac { { R }_{ e }^{ 2 }-{ R }_{ p }^{ 2 } }{ { R }_{ p }^{ 2 } } \times 100=0.7%\)
\(\therefore\) % variation in the weight of a body = % change in g = 0.7%

(i) The relative density of a substance is the ratio of its density to that of water.
Relative density of a substance
= \(\cfrac { Density\quad of\quad the\quad substance }{ Density\quad of\quad water } \)
Relative density of a substance
=\(=\cfrac { Mass\quad of\quad the\quad substance }{ Volume\quad of\quad the\quad substance } \times \cfrac { Volume\quad of\quad water }{ Mass\quad of\quad water } \)  \(\left[ \therefore Density=\cfrac { mass }{ volume } \right] \)
Given that, mass of the cube = 50 g
Side of cube = 1.5 cm
\(\therefore\) Volume of cube = (1.5)³ cm³ = 3.375 cm³
\(\therefore\) Density =\(\cfrac { mass }{ Volume } \)
=\(\cfrac { 50 }{ 3.375 } \) = 14.81 g cm^-3
(iil) Given that, mass, m = 250 g
Volume, V= 400 cc
\(\therefore \) Density=\(\cfrac { mass }{ Volume } \)
= \(\cfrac { 250 }{ 400 } \)g(cc)^-1= 0.625 g (cc)^-1
As we know that, density of water = 1 g (cc)^-1. So, density of tin is less than that of water and hence tin will float.

(i) Buoyancy and gravitational force are the two forces which act on a body, when it is immersed in a liquid.The conditions for a body required to float or sink in a liquid are as follows:
(a) If weight of the body is more than the upthrust act on it, then body will sink. 
[\(\therefore\)W > U (sink)]
(b) If weight of the body is less than the upthrust act on it, then body will float. 
[\(\therefore\) W < U (floatl]
(ii) Volume of the water displaced by an iron nail is less than that of piece of wood. That is why, more buoyancy force act on piece of wood and it floats on the surface of water.

(i) Ship displaces more water than needle as volume of ship is more than that of needle. Since, upthrust depend on volume of object (U =Vdg), so more the volume of object, more upthrust act on it and object floats.
(ii) Since, pressure act on the body is inversely proportional to the surface area of contact, i.e. \(p\propto \cfrac { 1 }{ A } \)
It means that more the area of contact, less pressure will act on the body. As the broad and the thick handle of our suitcase has large area, due to which less pressure acts on our hand and it is very easy to take from one place to another.

() Difference between thrust and pressure are

The pressure exerted by 1 N of force, acting perpendicular on the s~rface of 1m 2 area is called 1pascal.
1 Pa= 1Nm^-2
(1)
The force required to accelerate 1 kilogram of mass at the rate of 1 metre per second square is called 1newton.
1N = 1 kg ms^-2
Since, pressure is inversely proportional ro the area 0 contact, Le. \(p\propto \cfrac { 1 }{ A } \)
Therefore, pressure will reduce to half, if area of contact is doubled.

(t) Let the two bodies have masses m) and mz and they are placed at the same distance R from the centre of the earth. According to the question, if the sameforce  acts on both of them, then
\({ F }_{ 1 }=\cfrac { GM{ m }_{ 1 } }{ { R }^{ 2 } } \)
and \({ F }_{ 2 }=\cfrac { GM{ m }_{ 2 } }{ { R }^{ 2 } } \)
As, F₁=F₂
Hence,\(\cfrac { GM{ m }_{ 1 } }{ { R }^{ 2 } } =\cfrac { GM{ m }_{ 2 } }{ { R }^{ 2 } } \)
So, m₁ = m₂, their masses will be same
(ii) Mathematically,\(g=\cfrac { GM }{ { R }^{ 2 } } \)
where, g= acceleration due to gravity
G = universal gravitational constant
M = mass of the earth
and R = radius of the earth 
(ii) G is known as the universal gravitational constant because its value remains same all the time everywhere in the universe, applicable to all bodies whether celestial or terrestrial.