(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) \(g=\cfrac { GM }{ { R }^{ 2 } } \)

(i) The SI unit of thrust is newton (N). The SI unit of pressure is N m^-2or pascal (Pa).
(ii) We exert more pressure on ground when we stand on one foot than the both feet, as the area of one foot is half than that of two feet and \(p\propto \cfrac { 1 }{ A } \)

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

Let AB be the window and suppose the roof is at a height y above A.Also, suppose it takes a time t₁, for the ball to reach A. The velocity of the ball at A is
v₁= 0 + gt₁ = 9.8t₁ 
Now, consider the motion of the ball from A to B.
Here, the initial velocity is v₁ the distance covered is 2 m and the time taken is 0.1 s .
From secon d eeqcuuaattiion 0f moontion, s = ut + \(\cfrac { 1 }{ 2 } \)gt
\(\Rightarrow\) 2.0 =v₁ (0.1) +\(\cfrac { 1 }{ 2 } \) - X 9.8 X (0.1)² = 9.8t₁ (0.1)+0.049
\(\Rightarrow\) t₁=199\(\approx \)2s
The height y is \(y=\cfrac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }\)
= \(\cfrac { 1 }{ 2 } \) x 9.8 X (2)²  =19.6m
The roof is at a height 19.6 m above the top of the window.

When the ball is thrown upwards, then it will reach certain height h and starts falling. At maximum height h, the final velocity will be v = 0.
 Maximum height reached by the ball,
v²- u² = 2gh [using equation]
0- u² = - 2gh [\(\therefore\) acceleration = - g]
\(\Rightarrow\) \(h=\cfrac { { u }^{ 2 } }{ 2g } \)
In second case, when the ball starts to fall, then the initial velocity u = 0. It will accelerate due to gravity, i.e. a = g and reach ground with speed (sayv₂).
Using equation,
\({ v }_{ 2 }^{ 2 }-{ u }^{ 2 }=2gh\)
\(\Rightarrow\) \({ v }_{ 2 }^{ 2 }-0=2gh\)
\({ v }_{ 2 }^{ 2 }=2g\left( \cfrac { { u }^{ 2 } }{ 2g } \right) ={ u }^{ 2 }\)          [from Eq. (i)]
\(\Rightarrow\) v₂=u
Thus, the ball reach on the ground with same speed

For the upward motion
v =u - gt₁ > 0 = u - gt_1, \({ t }_{ 1 }=\cfrac { u }{ g } \) and the downward motion,
v = u + gt₂, v = 0 + gt₂
The body falls back to the earth with the same speed as it was thrown vertically upwards.
\(\therefore\) v=u, u=0+gt2\(\Rightarrow \) \({ t }_{ 2 }=\cfrac { u }{ g } \)
From Eqs. (i) and (ii), we get
tl = t2 => Time of ascent = Time of descent

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

Both will reach the ground at the same time as we know that an object experiences acceleration during free fall. This acceleration experienced by an object is independent of its mass because \(g=\cfrac { GM }{ R^{ 2 } } \)
As they are dropped at the same time, they will reach the ground at the same time.

As we know,\(F=\cfrac { G{ m }_{ 1 }{ m }_{ 2 } }{ { 4r }^{ 2 } } \)
According to question, r' = 6r
So, \(F=\cfrac { G{ m }_{ 1 }{ m }_{ 2 } }{ \left( 6r \right) ^{ 2 } } =\cfrac { G{ m }_{ 1 }{ m }_{ 2 } }{ 36r^{ 2 } } =\cfrac { F }{ 6 } \)
To maintain same force one of the mass is to be increased by 36 times.

No, it is not necessary that their weights in air should also be the same. This is because the two bodies have undergone the same loss in weight on completely immersing in water due to their equal volumes, not due to their equal weights. So, they may have different weights in air.

\(\therefore\) Volume of cubical tub = Length x Breadth X Height
Here, length = breadth = height = 2 m
\(\therefore\) Volume=2 x 2 x 2 =8 m³
Mass of water = Volume x Density of water
= 8 x 1000 = 8000 kg
\(\therefore\) Weight of water = Mass x Acceleration due to
gravity = 8000 x 10 = 8 x 104 N
\(\therefore\) Total thrust = Weight of water = 8 x 104 N
Area of bottom = Area of square shape of tub
= 2 x 2 = 4m²
\(\therefore\) Pressure of water at the bottom = \(\frac{Force}{Area}=\frac{Weight}{Area}\)
= \(\frac{8\times 10^4}{4}\) = 2 x 10⁴ Pa

(i) The value of g depends on
(a) shape of the earth,
(b) rotation of the earth,
(c) altitude, and
(d) depth.
(ii) Acceleration due to gravity decreases.
(iii) The value of g decreases due to rotation of the earth.

The average density of a fresh egg is more than of pure water but less than that of water in which salt is, dissolved. So a fresh egg sinks in pure water while it floats in salty water.

Archimedes' principle. This principle states that 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.
Applications of Archimedes' principle
(i) Archimedes' principle is used in designing ships and submarines.
(ii) Lactometers based on Archimedes' principle are used to measure purity of a sample of milk.
(iii) Hydrometers used to measure density off liquids are based on Archimedes' principle.