Difference between mass and weight:

(i) In Cartesian sign convention, upwards velocity is taken positive and acceleration due to gravity is taken negatively.
\(\therefore\)   u = + 49 ms^-1, g = -9.8 ms^-2
At the height point, v = o
\(\therefore\)  As  v² - u² = 2gs
O² - 49² = 2(-9.8) x s
Maximum height, s = \(\frac { 49\times 49 }{ 2\times 2.9 } =122.5\)
(ii) let t be the time taken by the stone to reach the height point.
As,v = u + gt
0 = 49 - 9.8 x t
t = \(49\over9.8\) = 5 s
\(\therefore\)  time of ascent = Tme of descent
Time taken by the stone to return earth's surface
= 2t = 2 x 5 = 10 s

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) 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.

Height of object A,\({ h }_{ 1 }=\cfrac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }\)
Height of object B,\({ h }_{ 2 }=\cfrac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }\)
\(\therefore \) \({ h }_{ 1 }{ h }_{ 2 }={ t }_{ 1 }^{ 2 }:{ t }_{ 2 }^{ 2 }\)
or  \({ t }_{ 1 }:{ { t }_{ 2 }=\sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } } }\)
(i) Acceleration due to gravity is independent of mass of falling body. So, ratio remains the same
(ii) If bodies are hollow, then also ratio remains the  same, i.e.\({ t }_{ 1 }:{ { t }_{ 2 }=\sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } } }\)
 

((t) Formula to find the magnitude of gravitational
Force, \(F=\cfrac { GMm }{ { R }^{ 2 } } \)
where, M = mass of the earth
m = mass of the object
R = radius of the earth
and universal gravitational constant,
G = 6.67 X 10^-11 N-m²/kg²
(ii) (a) Let gravitational force be F when the distance between them is R
\(F=\cfrac { GMm }{ { R }^{ 2 } } \)
Now, when the distance reduces to half
\({ F }^{ ' }=\cfrac { GMm }{ \left( \cfrac { R }{ 2 } \right) ^{ 2 } } =\cfrac { 4GMm }{ { R }^{ 2 } } \)
On dividing Eq. (i) by Eq. (ii), we get
\(\cfrac { F }{ { F }^{ ' } } =\cfrac { GMm }{ { R }^{ 2 } } \times \cfrac { { R }^{ 2 } }{ 4GMm } \)
F' = 4F
(b) When the mass becomes 4 times
\(\cfrac { F }{ { F }^{ ' } } =\cfrac { GMm }{ { R }^{ 2 } } \times \cfrac { { R }^{ 2 } }{ 4GMm } \)
\(\Rightarrow\) F' = 4F