(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.

(i) Before the disc is brought in contact with table, it is only rotating about its horizontal axis. So, its centre of mass is at rest, i.e. \({ v }_{ CM }=0\).
(ii) When rim is placed in contact with the table, then its linear velocity at any point on the rim of disc will reduce due to kinetic friction.
(iii) If rotating disc is placed in contact with the table, its centre of mass acquires some velocity (which was zero before contact) due to kinetic friction. So, linear velocity of CM will increase.
(iv) kinetic friction.
(v) Rolling will begin when \({ v }_{ CM }=R\omega \).
(vi) Acceleration produced in the centre of mass due to friction \({ a }_{ CM }=\frac { F }{ m } =\frac { { \mu }_{ k }mg }{ m } ={ \mu }_{ k }g\)
Angular acceleration produced by the torque due to friction. \(\alpha =\frac { \tau }{ I } =\frac { { \mu }_{ k }mgR }{ I } \)     
\(\therefore \ { v }_{ CM }={ u }_{ CM }+{ a }_{ CM }t\) 
\(\Rightarrow { v }_{ CM }={ \mu }_{ k }gt\) 
and       \(\omega ={ \omega }_{ 0 }+\alpha t\Rightarrow \omega ={ \omega }_{ 0 }-\frac { { \mu }_{ k }{ mg }R }{ I } t\)           
For rolling without slipping,
\(\frac { { v }_{ CM } }{ R } ={ \omega }_{ 0 }-\frac { { \mu }_{ k }{ mg }R }{ I } t\)  
\(\Rightarrow \ \frac { { \mu }_{ k }{ gt } }{ R } ={ \omega }_{ 0 }-\frac { { \mu }_{ k }{ mg }R }{ I } \) 
\(t=\frac { { R\omega }_{ 0 } }{ { { \mu }_{ k }{ g } }\left( 1+\frac { { { mR }^{ 2 } } }{ I } \right) } \).

(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