(a) Here, height(h)=1 km = 1000m, g = 10m/s²
Velocity attained by the rain drop in freely falling through a height h.
\(v=\sqrt { 2gh } =\sqrt { 2\times 10\times 1000 }\)
\( \\ =100\sqrt { 2 } m/s\)
\(\\ =100\sqrt { 2 } \times \frac { 60\times 60 }{ 1000 } Km/h\)
\(\\ =360\sqrt { 2 } km/h\approx 510km/h\) 
(b) Diameter of the drop(d) = 2r = 4mm
Radius of the drop(r) = 2mm = 2 x 10^-3m
Mass of a rain drop(m)=Vx\(\rho \)
\(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\rho =\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (2\times { 10 }^{ -3 })^{ 3 }\times { 10 }^{ 3 }\)
\( \approx 3.4\times { 10 }^{ -5 }kg\)
\(\\ Momentum \ of \ the \ rain \ drop(\rho )=mv \)
\(\\ =3.4\times { 10 }^{ -5 }\times 100\sqrt { 2 }\)
\(\\ \approx 4.7\times { 10 }^{ -3 }{ kg }^{ -m/s }\)
(c) Time required to flatten the drop
= time taken by the drop to travel the distance equal to the diameter of the drop near the ground
\(t=\frac { d }{ v } =\frac { 4\times { 10 }^{ -3 } }{ 100\sqrt { 2 } } =0.028\times { 10 }^{ -3 }s\)
\(\\ =2.8\times { 10 }^{ -5 }s\)
(d) Force exerted by a rain drop
\(F=\frac { Change \ in \ momentum }{ Time } \)
\(=\frac { \rho -0 }{ t } =\frac { 4.7\times { 10 }^{ -3 } }{ 2.8\times { 10 }^{ -5 } } \approx 168N\)
(e) Radius of the umbrella(R) = \(\frac { 1 }{ 2 } m\)
\(\therefore \ Area \ of \ the \ umbrella(A)=\pi { R }^{ 2 }\)
\(\\ =\frac { 22 }{ 7 } \times { (\frac { 1 }{ 2 } ) }^{ 2 }=\frac { 22 }{ 28 } =\frac { 11 }{ 14 } \approx 0.8{ m }^{ 2 }\)
Number of drops striking the umbrella simultaneously with average of 5cm or 5 x 10^-2m
\(=\frac { 0.8 }{ { (5\times { 10 }^{ -2 }) }^{ 2 } } =320\)
Net force exerted on umbrella 
= 320 x 168 = 53760 N

Total mass of girl, bicycle and stone, m₁ = ( 50 + 0.5 ) kg = 50.5 kg
Velocity of bicycle, u₁ = 5 m/ s
Mass of stone, m₂ = 0.5 kg
Velocity of stone, u₂ = 15 m /s
Mass of girl and bicycle, m = 50 kg
Yes, the speed of the bicycle changes after the stone is thrown.
Let, after throwing the stone the speed of bicycle be \(\nu \) m /s. 
According to the law of conservation of linear momentum,
m₁u₁ = m₂u₂ + m\(\nu \) 
50.5 x 5 = 0.5 x 15 + 50 x \(\nu \)
252.5 - 705 = 50\(\nu \)   or  \(\nu \) = \(\frac { 245.0 }{ 50 } \) 
\(\nu \) = 4.9 m/ s
Change in speed = 5 - 4.9 = 0.1 m / s

Given, mass of the particle (m) = 500 g = 0.5 kg
x-t graph of the particle is a straight line.
Hence, particle is moving with a uniform velocity along x-axis, i. e. its acceleration along x-axis is zero and hence, force acting along x-axis is zero.
y- t graph of particle is a parabola. Therefore, particle is in accelerated motion along y-axis.
At     t = 0, u_y= 0
Along y-axis, at t = 2s, y =4m
Using equation of motion,
 \(y={ u }_{ y }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\) 
\(\\ 4=0\times 2+\frac { 1 }{ 2 } \times { a }_{ y }\times { (2) }^{ 2 }\)
\(\\ { a }_{ y }=2m/{ s }^{ 2 }\)
Force acting y-axis (f_y) = ma_y
 \(0.5\times 2=1.0N\)   (along y-axis)

(a) As no external torque is involved with child + turntable system, so angular momentum of system remains constant.
or   \({ I }_{ i }{ \omega }_{ i }={ I }_{ f }{ \omega }_{ f }\)
\({ I }_{ i }{ \omega }_{ i }\) = initial angular momentum, \({ I }_{ f }{ \omega }_{ f }\) = final angular momentum
Here,      \({ \omega }_{ i }=40\quad rpm\Rightarrow { I }_{ f }=\frac { 2 }{ 5 } { I }_{ i }\)
Substituting, we get
\({ I }_{ i }\times40=\frac { 2 }{ 5 } { I }_{ i }\times{ \omega }_{ f }\)
or  \({ \omega }_{ f }=\frac { 5\times40 }{ 2 } =100\quad rpm\)
(b) \(\frac{\text { Final Kinetic Energy of rotation }}{\text { Initial Kinetic Energy of rotation }}=\frac{\frac{1}{2} I_{2} \omega_{2}^{2}}{\frac{1}{2} I_{1} \omega_{1}^{2}}=\frac{\frac{1}{2} I_{2}\left(2 \pi v_{2}\right)^{2}}{\frac{1}{2} I_{1}\left(2 \pi v_{1}\right)^{2}}\)
\(=\frac{I_{2} v_{2}^{2}}{I_{1} v_{1}^{2}}=\frac{\frac{2}{5} I_{1} \times(100)^{2}}{\frac{2}{5} I_{1} \times(40)^{2}}\)
= 2.5
Clearly, final (K.E) becomes more because the child used his internal energy when he folds his hands to increase the kinetic energy.

Young's modulus \((Y)=\frac { Stress }{ Longitudinal\ strain } \)
For same longitudinal strain \(Y\alpha stress\)
\(\frac { { Y }_{ stress } }{ { Y }_{ rubber } } =\frac { { (stress) }_{ steel } }{ { (stress) }_{ rubber } } \)
But          \({ Y }_{ steel }>{ Y }_{ rubber }\)
\(\frac { { Y }_{ steel } }{ { Y }_{ rubber } } >1\)
Therefore from Eq(i) we get
\(\frac { { (stress) }_{ steel } }{ { (stress) }_{ rubber } } >1\)
\(or { (stress) }_{ steel }>(stress)_{ rubber }\)

Here, T₁ = 500K, T₂ = 300K, W = 1KJ = 1000J
As, efficiency, \(\eta =\frac { W }{ Q } =\frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \)
So, heat transferred to engine by the reservoir in cycle
\(Q_{ 1 }=\frac { W{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } =\frac { 1000\times 500 }{ 500-300 } =2500J\quad or\quad 2.5kJ\)
and heat transferred by the engine to the cold reservoir in one cycle
Q₂ = Q₁ - W  = 2.5kJ-  1kJ = 1.5kJ

O₂ has five degrees of freedom.
Therefore, energy pre mole = \(\frac { 5 }{ 2 } RT\)
For 2 moles of O_2, energy = 5RT
Neon has three degrees of freedom
Energy per mole = \(\frac { 3 }{ 2 } RT\)
For 4 moles of neon, energy
\(=4\times \frac { 3 }{ 2 } RT=6RT\)
Total eneryg = 5RT + 6RT = 11RT.

Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)  
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right)\) 
\( y=a\cos { \left( \omega t-kx+\phi \right) } \)
\(a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\quad T=0.1s\)
\(\\ k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
When x = \(\frac { 3\lambda }{ 4 } \); \({ \phi }_{ 4 }=\frac { 2\pi }{ \lambda } \times \frac { 3\lambda }{ 4 } =\frac { 3\pi }{ 2 } rad\).

(a) Using law of conservation of energy

\(\frac { 1 }{ 2 } mv^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }\)=mgh
or  \(\frac { 1 }{ 2 } mv^{ 2 }+\frac { 1 }{ 2 } \left( \frac { 2 }{ 5 } m{ R }^{ 2 } \right) \frac { { v }^{ 2 } }{ { R }^{ 2 } } =mgh\)
or \(\frac { 7 }{ 10 } v^{ 2 }=gh\) or v =\(\sqrt { \frac { 10gh }{ 7 } } \)
Since h is same for both the inclined planes therefore v is the same
(b) 

\(l=\frac { 1 }{ 2 } \left( \frac { g\quad sin\quad \theta }{ 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } } \right) { t }^{ 2 }=\frac { g\quad sin\theta }{ 2\left( 1+\frac { 2 }{ 5 } \right) } { t }^{ 2 }=\frac { 5gsin\theta }{ 14 } { t }^{ 2 }\)
or \(t=\sqrt { \frac { 14l }{ 5g\quad sin\quad \theta } } \)
Now, sin \(\theta\)=\(\frac { h }{ l } \) or l =\(\frac { h }{ sin\theta } \)
∴ \(t=\frac { 1 }{ sin\quad \theta } \sqrt { \frac { 14h }{ 5g } } \)
Lesser the value of,\(\theta\) more will be t.
(c) Clearly, the solid sphere will take longer to roll down the plane with smaller inclination.

Here, m = 5.30\(\times\)10^-26 kg
I = 1.94\(\times\)10^-46kg m²
v =500 m/s
If \(\frac { m }{ 2 } \)is mass of each atom of oxygen and 2r is distance between the two atoms as shown in Fig.then
\(I=\frac { m }{ 2 } { r }^{ 2 }+\frac { m }{ 2 } { r }^{ 2 }={ mr }^{ 2 }\)
\(r=\sqrt { \frac { 1 }{ m } } =\sqrt { \frac { 1.94\times { 10 }^{ -46 } }{ 5.30\times { 10 }^{ -26 } } } \)
= 0.61\(\times\)10^-10 m
As K.E. of rotation= \(\frac { 2 }{ 3 } \)K.E of translation
∴ \(\frac { 1 }{ 2 } I{ \omega }^{ 2 }=\frac { 2 }{ 3 } \times \frac { 1 }{ 2 } { m\omega }^{ 2 }\)
\(\frac { 1 }{ 2 } ({ mr }^{ 2 }){ \omega }^{ 2 }=\frac { 1 }{ 2 } { mv }^{ 2 }\)
\(\omega =\sqrt { \frac { 2 }{ 3 } } \frac { \upsilon }{ r } =\sqrt { \frac { 2 }{ 3 } } \times \frac { 500 }{ 0.61\times { 10 }^{ -10 } } \) = 6.7 10¹² rad/s