Distance of moon from the earth (r_ME ) = 3.84 \(\times \) 10⁸ m
Distance of sun from the earth (r_SE) = 1.496\(\times \) 10¹¹m
Diameter of Sun (D) = 1.39 \(\times \) 10⁹ m
During the total solar eclipse, the sun is completely covered by the disk of the moon.
\(\therefore\)    Angular diameter of the moon = Angular diameter of the sun 
  \(\frac { d }{ { r }_{ ME } } =\frac { D }{ r_{ SE } } \)

\(\therefore\)   d = D\(\times \) \(\frac { { r }_{ ME } }{ r_{ SE } } \) = 1.39\(\times \)10⁹\(\times \)\(\frac { 3.84\times 10^{ 8 } }{ 1.496\times { 10 }^{ 11 } } \)
 = 3.5679\(\times \)10⁶ m = 3567.9\(\times \)10³ = 3567.9 km

(a) When an object is projected upwards, its velocity at the top-most point is zero even though the acceleration on it is 9.8m/s²(g).
(b) When a stone tied to a string is whirled in a circular path, the acceleration acting on it is always at right angles i.e. perpendicular to the direction of motion of stone (we will study about it in chapter 'motion in a plane').

Using relations 
Consider upwards as negative and downwards as positive.
\(h=-u{ t }_{ 1 }+\frac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }.............(i)\)
\(\\ h=u{ t }_{ 2 }+\frac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }...............(ii)\)
\(\\ On \ substracting \ Eqs. \ (i) \ from \ (ii), \ we \ get\)
\(\\ or \ 0=({ t }_{ 2 }+{ t }_{ 1 })+\frac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }-\frac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }\)
\(\\ or\ u({ t }_{ 2 }+{ t }_{ 1 })+\frac { 1 }{ 2 } g({ t }_{ 2 }+{ t }_{ 1 })({ t }_{ 2 }-{ t }_{ 1 })=0\)
\(\\ or\ u+\frac { 1 }{ 2 } g({ t }_{ 2 }+{ t }_{ 1 })=0\)
\(\\ or \ u=-\frac { g }{ 2 } ({ t }_{ 2 }-{ t }_{ 1 }).........(iii)\)
\(\\ From \ Eqs \ (i) \ and \ (iii), \ we \ get\)
\(\\ Now,\quad h=\frac { g{ t }_{ 1 } }{ 2 } ({ t }_{ 2 }-{ t }_{ 1 })+\frac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }=\frac { 1 }{ 2 } g{ t }_{ 1 }{ t }_{ 2 }.....(iv)\)
\(\\ Again, \ when \ the \ body \ falls \ freely.\)
\(\\ h=\frac { 1 }{ 2 } g{ t }^{ 2 };\frac { 1 }{ 2 } g{ t }_{ 1 }{ t }_{ 2 }=\frac { 1 }{ 2 } g{ t }^{ 2 }\quad [fron\quad eq.iv]\)
\(\\ or\quad t=\sqrt { { t }_{ 1 }{ t }_{ 2 } } \quad Hence\quad proved\) 

As, maximum height, H = \(\frac { { u }^{ 2 } }{ 2g } { sin }^{ 2 }\theta \)
Consider \(\triangle \)H b ethe increase in H when u changes by \(\triangle \)u, it can be obtained by differentiating the above equation, we get
\(\triangle H=\frac { 2u\triangle u{ sin }^{ 2 }\theta }{ 2g } =\frac { 2\triangle u }{ u } H\)
\(\\ \Rightarrow \frac { \triangle H }{ H } =\frac { 2\triangle u }{ u } \)
Given % increasing in H is 10% so 
\(\frac { \triangle H }{ H } =\frac { 10 }{ 100 } =0.1\Rightarrow \frac { 2\triangle u }{ u } =0.1\)
\(\\ As,\quad R=\frac { { u }^{ 2 }sin2\theta }{ g } \)
\(\\ \because \quad \triangle R=\frac { 2u\triangle u }{ u } sin2\theta \Rightarrow \frac { \triangle R }{ R } =\frac { 2\triangle u }{ u } =0.1\)
\(\\ \therefore \ increase \ in \ horizontal \ range=\frac { \triangle R }{ R } \times 100\)
= 0.1 x 100
= 10%

Given, R = 1000m
Horizontal range of the bullet fired at an angle \(\theta \) is
\(R=\frac { { u }^{ 2 }sin2\theta }{ g } \Rightarrow 1000=\frac { { u }^{ 2 }sin\theta cos\theta }{ g } ....(i)\)
Bullet is fired from the car moving with 36km/h
i.e.10m/s, then horizontal component of the velocity of = usin\(\theta \) + 10
Vertical component of the velocity of the bullet=usin​\(\theta \) 
Then, new range of the bullet is
\({ R }_{ 1 }=\frac { 2 }{ g } (usin\theta )(ucos\theta +10)\)
\(\\ =\frac { 2 }{ g } { u }^{ 2 }sin\theta cos\theta +\frac { 20 }{ g } usin\theta \Rightarrow { R }_{ 1 }=R+\frac { 20 }{ g } usin\theta\)
\( \\ \Rightarrow { R }_{ 1 }-R=\frac { 20 }{ g } usin\theta ....(ii)\)
\(\\ From\ Eq.(i),we \ have \ u=\sqrt { \frac { 1000\times g }{ 2sin\theta cos\theta } } .....(iii)|\)
\(\\ Now, \ substituting \ the \ value \ of \ u \ in \ Eq.(ii), \ we \ get\ \)
\(\\ { R }_{ 1 }-R=\frac { 20 }{ g } \sqrt { \frac { 1000\times g }{ 2sin\theta cos\theta } } sin\theta =20\sqrt { \frac { 500\times sin\theta }{ gcos\theta } }\)
\( \\ =20\sqrt { \frac { 500 }{ 9.8 } tan\theta } =142.9\sqrt { tan\theta } \) ​​​​

When the motorcyclist is at the uppermost point of the death well, then weight of the cyclist, as well as the normal reaction R of the ceiling of the chamber, is in downward direction. These forces are balanced by the outward centrifugal force acting on the motorcyclist
\(\therefore \\ \)  \(Rg+mg=\frac { { mv }^{ 2 } }{ r }\)
Where, v = Speed of the motorcyclist
m = mass of (motorcycle + driver)
r = radius of the death well.
As the forces acting on the motorcyclist are balanced, therefore, motorcyclist does not fall down.
The minimum speed required to perform a vertical loop is given by
 \(mg=\frac { mv^{ 2 }min }{ r }\)
[\(\because \\ \) In the case weight of the object = centripetal force]
or    vmin =\(\sqrt { rg } =\sqrt { 25\times 9.8 } =15.65{ m }/{ s }\) 

Given  
 \({ M }_{ e }=6\times { 10 }^{ 24 }kg,{ r }_{ e }=6400km\)  
\( { r }_{ 1 }=6400+200=6600km=6.6\times { 10 }^{ 6 }m\) 
\({ r }_{ 2 }=6400+199=6599km=6.599\times { 10 }^{ 6 }m\)
Change in energy =   \(GMm\left( \frac { 1 }{ { r }_{ 1 } } -\frac { 1 }{ { r }_{ 2 } } \right) \)
\(\\ =6.67\times `{ 10 }^{ -11 }\times 6\times { 10 }^{ 24 }\times 500\)
\( \left( \frac { 1 }{ 6.6\times { 10 }^{ 6 } } =\frac { 1 }{ 6.599\times { 10 }^{ 6 } } \right) \)
\(\\ =2\times { 10 }^{ 17 }(1.5152\times { 10 }^{ -7 }-1.5154\times { 10 }^{ 7 })J\)
\(=-4\times { 10 }^{ 6 }J\)
If this occurs during one orbit,then the energy lost = \(force\times distance\) .If we take the sistances as being the circumference of one orbit Then
 Retarding force
\(=\frac { 4\times { 10 }^{ 6 } }{ 2\pi \times 6.6\times { 10 }^{ 6 } } =\frac { 4\times { 10 }^{ 6 } }{ 2\times 6.6\times 3.14\times { 10 }^{ 6 } } =0.1N\) 

As the brass scale of a barometer gives correct reading at T₁ = 0⁰ C , hence at temperature T₂ = 27⁰ C., the scale will expand and will not give correct reading
In such, true value
= Observed scale reading x\((1+\alpha \triangle T)\)
True pressure = 75.00 cm x [1 + 2.0 x 10^-5 x (27 - 0)]
= 75 x (1 + 2.0 x 10^-5 x 27]
= 75.00(1 + 54 x 10^-5) cm = 75.04 cm

At 273 K temperature and 1 atm pressure mean STP condition.
\(\therefore \) Number of H₂ molecules on 22400cm³ at STP = 6.02 x 10²³
Number of H₂ molecules on 1cm³ at STP = \(\frac { 6.02\times { 10 }^{ 23 } }{ 22400 } =2.6875\times { 10 }^{ 19 }\)
\(\therefore \) Number of degrees of freedom associated with each H₂ (diatomic) molecule = 5
\(\therefore \) Total number of degrees of freedom associated with 1cm³ of gas = 2.6875 x 10¹⁹ x 5 = 1.34375 x 10²⁰