Muzzle speed of the bullet,
v = 150 ms^-1
= 150\( \times \) \(\frac{18}{5}\)
= 540 kmh^-1
Speed of police van, vp = 30 km/h
speed of the thief's car, vr = 192 km/h
Since the bullet is sharing the velocity of the police van, its effectively velocity is
vB = v + vP
= 540 +30
= 570  km /h
Since the speed of the bullet w.r.t the thief's car ,oving in the same direction .
v_BT = v_B - v_T
= 570 - 192
= 378  km / h
= \( \frac{378 \times 1000}{60 \times 60 } \)
= 105 ms^-1

let v km/h be the constant speed with which the buses plying between the towns A and B .The relative velociy of the bus ( for the motion A to B  ) with respect to the cyclist ( i.e. in the direction in which the cyclist is going  = (  v - 20 ) kmh^-1.The relative velocity of the bus from B to A with respect  to then cyclist ( v + 20 ) km/h.
the distance travellled by the bus  in time T(minutes)  = vT
As per question ,
\(\frac{vT}{v-20} \) = 18 or vT = 18v - 18\(\times\)20  .....(i)
and \( \frac{vT}{v-20}\)  = 6 or 6v + 20 \(\times\)6.........(ii)
By Eqs.(i) and (ii), we get 
18 v -18\(\times\)20 = 6v +20\(\times\)6
or  12 v = 20\(\times\)6 +18\(\times\)20
= 480
or   v = 40 km/h
Putting  this value of v in Eq. (i) , we get
 40T = 18\(\times\) 40 -18\(\times\)20
= 18\(\times\)20
T = 18\(\times\)20 / 40
= 9 min

The centripetal acceleration in a wheel arise due to internal elastic forces which in pairs cancel each other. So the system remains in equilibrium during rolling.
In a half wheel, the system is not symmetrical so, direction of angular momentum does not coincide with the direction of angular velocity and hence an external torque is required to maintain the rotation.

Initial KE = \(\frac { 1 }{ 2 } { I }_{ i }{ \omega }_{ i }^{ 2 }=\frac { 1 }{ 2 } \times{ I }_{ i }\times(40{ ) }^{ 2 }\)
 \(=\frac { 1 }{ 2 } \times{ I }_{ i }\times1600=800{ I }_{ i }\)
Final KE = \(\frac { 1 }{ 2 } { I }_{ f }{ \omega }_{ f }^{ 2 }=\frac { 1 }{ 2 } \times\frac { 2 }{ 5 } { I }_{ i }\times(100{ ) }^{ 2 }\)
 \(=\frac { 1 }{ 2 } \times\frac { 2 }{ 5 } \times{ I }_{ i }\times100\times100=2000{ I }_{ i }\)
Clearly, final KE > initial KE. This energy is obtained from conversion of muscular work in KE. Muscular work has to be done in folding of arms.

Moment of inertia of hoop about its centre
\(I={ MR }^{ 2 }\)  
and energy of loop = translational kinetic energy of CM and rotational kinetic energy about axis through CM.
\(=\frac { 1 }{ 2 } { mv }_{ CM }^{ 2 }+\frac { 1 }{ 2 } { I\omega }^{ 2 }\quad \quad \quad \left[ I={ mR }^{ 2 }and\quad \omega =\frac { v }{ R } \right] \)
\(=\frac { 1 }{ 2 } { mv }_{ CM }^{ 2 }+\frac { 1 }{ 2 } { MR }^{ 2 }\times \frac { { v }_{ CM }^{ 2 } }{ { R }^{ 2 } } ={ mv }_{ CM }^{2}\)                
Work done is stopping the hoop
= total KE of loop
= \({ mv }_{ CM }^{ 2 }\) = \(100\times { \left( 0.2 \right) }^{ 2 }=4J\).

Total mass of the car = 1800kg
Let m and (900-m)kg be the masses of each front wheel and each back wheel, respectively.
Distance of centre of gravity from the back axle = 1.80-1.05 = 0.75m
Taking torque about centre of gravity,
\(m\times 1.05=(900-m)\times 0.75\)
\(\\ 1.05m+0.75m=900\times 0.75\)
\(1.80m=900\times 0.75\)
\(\\ m=\frac { 900\times 0.75 }{ 1.80 } =375kg\)
\( \therefore \ (900-m)=900-375=525kg\)
\(\\ \therefore \ Weight\ of\ each\ front\ wheel(w_{ 1 })={ m }_{ 1 }g\)
\({ (w }_{ 1 })=375\times 9.8=3675N\)
Forced exerted by the level ground on each front wheel.
= force exerted by each front wheel on the level ground \({ (w }_{ 1 })=3675N\)
Weight of each back wheel (\({ w }_{ 2 }\))=525 x 9.8=  5145N
Forced exerted by the level ground on each back wheel = 5145N.

Let the acceleration of the centre of mass of disc be a, then , Ma = F - f 
The angular acceleration of the disc is a =a/ R (if there is no sliding).
Then, \(\left( \frac { 1 }{ 2 } M{ R }^{ 2 } \right) \alpha \) = Rf \(\Rightarrow \) Ma =2f
Thus, f = F /3.Since, there is no sliding.
\(\Rightarrow \) f <\(\mu \) F\(\le \) 3\(\mu \) Mg

Gain in PE,
\(\triangle U={ U }_{ f }-{ U }_{ i }\)
\(=-\frac { GMm }{ 2R } -\left( -\frac { GMm }{ R } \right)\)
\( =\frac { GMm }{ R } \left( -\frac { 1 }{ 2 } +1 \right) =\frac { GMm }{ 2R } \)
\(\\ =\frac { (g{ R }^{ 2 })m }{ 2R } =\frac { 1 }{ 2 } mgR\ \left[ as\ g=\frac { GM }{ R^{ 2 } } \right] \)

\(Work\ done\ in\ strtching\ a\ wire\ is\ given\ by\)
\(\\ W=\frac { 1 }{ 2 } F\times \triangle l\)
\(\\ As\ springs\ of\ steel\ and\ copper\ are\ equally\ streched.Therefore,for\ same\ force(F).\)
\(\\ W\propto \triangle l\)
\(\\ Young's\ modulus\ (Y)=\frac { F }{ A } \times \frac { l }{ \triangle l } \)
\(\\ or\ \triangle l=\frac { F }{ A } \times \frac { l }{ Y } \)
\(\\ As\ both\ springs\ are\ identical\)
\( \triangle l\propto \frac { 1 }{ Y } \)
\(\\ From \ Eqs.(i) \ and \ (ii),\ we \ get \ W\propto \frac { 1 }{ Y }\)
\( \\ \therefore \frac { { W }_{ steel } }{ { W }_{ copper } } =\frac { { Y }_{ copper } }{ { Y }_{ steel } } <1\ [as\ { Y }_{ steel }>{ Y }_{ copper }]\)
\(\\ or\ { W }_{ steel }\ <\ { W }_{ copper }\)
\(\\ Therefore,\ more\ work\ will\ be\ done\ for\ stretching\ copper\ spring.\)

\(Given,{ V }_{ 1 }=2.0\ L,{ V }_{ 1 }=3.0\ L,{ \mu }_{ 1 }=4.0\ moles,{ \mu }_{ 2 }\ =5.0\ moles\)
\(\\ { p }_{ 1 }=1.00atm,{ p }_{ 2 }=2.00atm\)
\(\\ { p }_{ 1 }{ V }_{ 1 }={ \mu }_{ 1 }{ RT }_{ 1 },{ p }_{ 2 }{ V }_{ 2 }={ \mu }_{ 2 }{ RT }_{ 2 }\)
\(mu ={ \mu }_{ 1 }+{ \mu }_{ 2 },V={ V }_{ 1 }+{ V }_{ 2 }\)
\(\\ For\quad 1\ \ mole,\ pV=\frac { 2 }{ 3 } E\ \)
\( For\ { \mu }_{ 1 }\ mole,\ { p }_{ 1 }{ V }_{ 1 }=\frac { 2 }{ 3 } { \mu }_{ 1 }{ E }_{ 1 }\)
\(\\ For\ { \mu }_{ 2 }\ mole,\ { p }_{ 2 }{ V }_{ 2 }=\frac { 2 }{ 3 } { \mu }_{ 2 }{ E }_{ 2 }\)
\(\\ Total\ energy\ is\ ({ \mu }_{ 1 }{ E }_{ 1 }+{ \mu }_{ 2 }{ E }_{ 2 })=\frac { 3 }{ 2 } ({ p }_{ 1 }{ V }_{ 1 }+{ p }_{ 2 }{ V }_{ 2 })\)
\(\\ p\left( { V }_{ 1 }+{ V }_{ 2 } \right) =\frac { 2 }{ 3 } \times \frac { 3 }{ 2 } ({ p }_{ 1 }{ V }_{ 1 }+p_{ 2 }{ V }_{ 2 })\)
\(\\ p=\frac { { p }_{ 1 }{ V }_{ 1 }+{ p }_{ 2 }{ V }_{ 2 } }{ { V }_{ 1 }+{ V }_{ 2 } } \ ---\ (i)\)
\(=\left( \frac { 1.00\times 2.0+2.00\times 3.0 }{ 2.0+3.0 } \right) =\frac { 8.0 }{ 5.0 } =1.60\ atm\)

According to ideal gas equation, we get
\(\frac { { p }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { p }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \)
\(\\ \frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { { p }_{ 2 }{ T }_{ 1 } }{ { p }_{ 1 }{ T }_{ 2 } } =\frac { 2\times 300 }{ 1\times 400 } =\frac { 3 }{ 2 } \)
\(\\ { p }_{ 1 }=\frac { 1 }{ 3 } \frac { M }{ { V }_{ 1 } } \left( { v }_{ rms } \right) _{ 1 }^{ 2 },\quad { p }_{ 2 }=\frac { 1 }{ 3 } \frac { M }{ { V }_{ 2 } } \left( { v }_{ rms } \right) _{ 2 }^{ 2 }\)
\(\\ \left( { v }_{ rms } \right) _{ 2 }^{ 2 }=\left( { v }_{ rms } \right) _{ 1 }^{ 2 }\times \frac { { V }_{ 2 } }{ { V }_{ 1 } } \times \frac { { p }_{ 2 } }{ { p }_{ 1 } }\)
\( \\=\left( 100 \right) ^{ 2 }\times \frac { 2 }{ 3 } \times 2\)
\(\\ or\ \left( { v }_{ rms } \right) _{ 2 }=\frac { 200 }{ \sqrt { 3 } } \ { ms }^{ -1 }\)

Loss in kinetic energy of gas = \(\triangle E=\frac { 1 }{ 2 } (mn){ v }_{ 0 }^{ 2 }\)
where, n is number of moles mass.
If its temperature changes by \(\triangle T\) , then
\(n\frac { 3 }{ 2 } R\triangle T=\frac { 1 }{ 2 } mn{ v }_{ 0 }^{ 2 }\)
\(\therefore \) Change in temperature,\(\triangle T=\frac { m{ v }_{ 0 }^{ 2 } }{ 3R } \)

(i) The moon  has small gravitational; force and hence the escape velocity is small .As the moon is in tyhe proximity of the earth as seen from the sun, the moon has the same amount of heat per unit area  as that of the earth , The air molecules have l;arge range of speeds.
Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number of  molecules have speed greater than escape velocity and they escape.
Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape sppeed. Hence, over a long time the moon has lost most of its atmosphere.
(ii) As the molecules move higher , their potential energy increases and hence kinetic energy decreases and hence temperature reduces.
At greater height, more volume is available and gas expands and hencde some cooling takes place.

From  Kinetic theory of gases, 
\(p=\frac { 1 }{ 3 } pc^{ 2 }\), where c is rms speed of molecules of gas.
\(\Rightarrow c=\sqrt { \frac { 3p }{ p } } \)
v = speed of sound in the gas = \(\sqrt { \frac { p }{ \Upsilon \rho } } \)
\(\Rightarrow from\ eqs.(i)\ and\ (ii)\)
\( \frac { c }{ v } =\sqrt { \frac { 3p }{ \rho } \times \frac { \rho }{ \Upsilon \rho } } =\sqrt { \frac { 3 }{ \Upsilon } } \)
Fer diatomic gases, 
\(\Upsilon =1.4=\ constant\)
\(\\ \Rightarrow \frac { c }{ v } =\sqrt { \frac { 3 }{ 1.4 } } =1.46 =\ constant\) 

Here, R = 2m M = 100 kg
v = 20 cm/s = 0.2 m/s
Total energy of the hoop = \(\frac { 1 }{ 2 } { Mv }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }\)
= \(\frac { 1 }{ 2 } {Mv }^{ 2 }+\frac { 1 }{ 2 } (MR^{ 2 }){ \omega }^{ 2 }\)
=\(\frac { 1 }{ 2 } { Mv }^{ 2 }+\frac { 1 }{ 2 } { Mv }^{ 2 }={ Mv }^{ 2 }\)
Work required to stop the hoop = total energy of the hoop
W = Mv² = 100 (0.2)² = 4 Joule.