Given, density of iron = 7900 kg m^-3
Density of water = 1000 kg m^-3
\(\therefore\) Relative density =\(\cfrac { Density\quad of\quad iron }{ Density\quad of\quad water } =\cfrac { 7900\quad kgm^{ -3 } }{ 1000kgm^{ -3 } } \)
=7.9
As the relative density of an iron object is greater than density of water, so it will sink to bottom.

The two forces are
(i) Weight of the body acting downwards.
(ii) Buoyant force acting upwards.

Weight on the surface of the earth is
\(W=\frac{GMm}{R^2}\)
At a height equal to the radius of the earth, the distance from the centre of the earth = R+R=2R
Therefore, the weight will become
\({ W }^{ ' }=\frac { GMm }{ { (2R) }^{ 2 } } =\frac { 1 }{ 4 } \frac { GMm }{ { R }^{ 2 } } =\frac { 1 }{ 4 } \quad W=\frac { 1 }{ 4 } \times 600\quad N=150\quad N.\) 

P=10 pascal \(10N{ m }^{ -2 },\quad A-4cm\times 4cm=16\times 10N{ m }^{ -2 }\)
\(P=\frac { F }{ A } =\frac { W }{ A }\) 
\(\therefore \quad W=P\times A\) 
\(=\ 10N{ m }^{ -2 }\times 16\times { 10 }^{ -4 }{ m }^{ 2 }=1.6\times { 10 }^{ -2 }N\)  

Importance of the universal law of gravitation. The universal law of gravitation successfully explained many phenomena occurring in nature. Some of these phenomena are as follows:
1. The force that binds us to the earth.
2. The motion of the moon around the earth.
3. The motion of planets around the sun.
4. The tides due to the moon.

(I) We know that, pressure =\(\cfrac { force }{ area } \)
The pressute acting on the surface is inversely proportional to the area of the surface on which force is applied, i.e
Pressure \(\left( p \right) \propto \cfrac { 1 }{ Area\left( A \right) } \)
(ii) Shruti and Kriti observed that pressure is more, if the area of cuboid in contact with the sand is less.
On the other hand, pressure is less, if the area of contact is large, as pressure\(\left( p \right) \propto \cfrac { 1 }{ Area\left( A \right) } \)
(iii) Shruti is intelligent, scientific and enthusiastic.

(I) Both balls reach the ground simultaneously. Because both of them have been dropped from the same height. 
(ii) Mohit and Shobhit are inquisitive, logical, experimental and competitive.
(iii) \(g=\cfrac { GM }{ { R }^{ 2 } } \)

Let the weight of the particle on the surface of the earth,
\(w=120=\cfrac { GMm }{ { R }^{ 2 } } \)
where, R = 6400 km = 6.4 X 10⁶ m
Hence, 120=\(\cfrac { GMm }{ \left( 6.4\times { 10 }^{ 6 } \right) ^{ 2 } } \)
Let the height h above the earth's surface, where its weight will be 30 N.
Hence,\(30=\cfrac { GMm }{ \left( h+R \right) ^{ 2 } } \)
\(30=\cfrac { GMm }{ \left( h+6.4\times 10^{ 6 } \right) ^{ 2 } } \)                         ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
\(\cfrac { 120 }{ 30 } =\cfrac { GMm }{ \left( 6.4\times 10^{ 6 } \right) } \times \cfrac { \left( h+6.4\times { 10 }^{ 6 } \right) ^{ 2 } }{ GMm } \)
\(\cfrac { 4 }{ 1 } =\cfrac { \left( h+R \right) ^{ 2 } }{ { R }^{ 2 } } \Rightarrow 2=\cfrac { h+R }{ R } \)
\(\Rightarrow\) 2R = h + R
\(\Rightarrow\) h = 2R - R = R = 6400 km
= 6.4 x 10⁶m

As we know, acceleration due to gravity,\(g=\cfrac { GM }{ R^{ 2 } } \)
For the satelite,R = 1.74 X1 0⁶ cm = \(\cfrac { 1.74\times { 10 }^{ 6 } }{ 100 } \)
=1.74 X 10⁴m
M = 7.4 X10²² kg
\(\therefore\) g= \(\cfrac { 6.67\times 10^{ -11 }\times 7.4\times 10^{ 22 } }{ 1.74\times 10^{ 4 }\times 1.74\times { 10 }^{ 4 } } \)
= \(\cfrac { 6.67\times 7.4 }{ 1.74\times 1.74 } \times 10^{ 3 }\)  g=16.30 X10³m/s²

(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

(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.

(a) Gravitational force acting on the 50 kg,
mg= 50x 9.8 = 490N 
(b) Gravitational force acting on the 50 kg mass due to jupiter,
\({ F }_{ jupiter }=\cfrac { G\times { M }_{ jupiter }\times { { M }_{ person } } }{ \left( distance\quad of\quad jupiter\quad from\quad the\quad earth \right) ^{ 2 } } \)
\({ F }_{ jupiter }=\cfrac { 6.67\times { 10 }^{ -11 }\times 2\times { 10 }^{ 27 }\times 50 }{ 6.3\times { 10 }^{ 11 }\times 6.3\times { 10 }^{ 11 } } \)
\({ F }_{ jupiter }=\cfrac { 6.67\times 2\times 50\times { 10 }^{ -11+27-22 } }{ 6.3\times 6.3 } \)
F_Jupiter = 1.68 X 10^-5 N
\({ F }_{ saturn }=\cfrac { G\times M_{ saturn }\times { M }_{ person } }{ \left( distance\quad of\quad saturn\quad from\quad the\quad earth \right) ^{ 2 } } \)
\({ F }_{ saturn }=\cfrac { 6.67\times { 1 }0^{ -11 }\times 6\times { 10 }^{ 26 }\times 50 }{ 1.28\times 10^{ 12 }\times 1.28\times 10^{ 12 } } \)
\({ F }_{ saturn }=\cfrac { 6.67\times 6\times 50 }{ 1.28\times 1.28 } \times { 10 }^{ -11+26-24 }\)
F_saturn = 0.12x 10^-5N
\(\therefore\) Total gravitational force due to the jupiter and the saturn = (1.68x 10^-5+0.12x 10^-5)N
= 1.8x 10^-5 N
Thus, the combined force due to the planets jupiter and saturn (1.8 x10^-5) N is negligible as compared to the gravitational force due to the earth.
(ii) We know that, g at equator is less than g at poles (Antarctica). Thus, weight at equator is less than weight at pole (Antarctica). A bag of sugar weighs w at a certain place on the equator. If this bag is taken to Antarctica, then it will weigh more due to greater value of g.