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
\({ W }^{ ' }=\frac { GMm }{ { (2R) }^{ 2 } } =\frac { 1 }{ 4 } \frac { GMm }{ { R }^{ 2 } } =\frac { 1 }{ 4 } \quad W=\frac { 1 }{ 4 } \times 600\quad N=150\quad N.\) 

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

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

In both cases, the force exerted on the sand is equal to the weight of our body. When we stand on loose sand, our weight acts on a smaller area of the sand. Hence, a large pressure (=force/area) is exerted on sand and our feet go deep into the sand.
When we lie down, our weight acts on a larger area of the sand. Hence, a smaller pressure is exerted on sand and our feet do not go that deep in the sand.

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.

Here \({ m }_{ 1 }=40kg,\quad { m }_{ 2 }=60kg,\quad F=4\times { 10 }^{ -5 }N,\quad G=6\times { 10 }^{ -11 }\quad { Nm }^{ 2 }/{ kg }^{ 2 }.\)

\(F=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } }\) 
\(r=\sqrt { \frac { G{ m }_{ 1 }{ m }_{ 2 } }{ F } } =\sqrt { \frac { 6\times { 10 }^{ -11 }\times 40\times 60 }{ 4\times { 10 }^{ -5 } } } =\sqrt { 36\times { 10 }^{ -4 } } =6\times { 10 }^{ -2 }m\) 
\(=6\times { 10 }^{ -2 }\times 100cm=6m.\) 

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

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

\(\therefore\) 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

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.

(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) (a) g on the moon is given by
\({ g }^{ ' }=\cfrac { g }{ 6 } =\cfrac { 9.8 }{ 6 } \)
= 1.63 m/s²
(b) Mass of the person on the moon
=\(\cfrac { 110.84 }{ 1.63 } =68kg\)
(c) Weight of person on the earth = mg
= 68 x9.8
= 666.4 N- m²/kg²
(ii) Weight of a person on the earth will be
\(w=\cfrac { GMm }{ { R }^{ 2 } } \)
where, M = mass of the earth
R = radius of the earth
m = mass of person
and G =6.67 X 10 ^-11 N-m²/kg²

(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.