A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is $K$. The child now stretches his arms so that moment of inertia of the system doubles. The kinetic energy of the system now is
A. $2K$
B. $\dfrac{K}{2}$
C. $\dfrac{K}{4}$
D. $4\,K$

Hint: The kinetic energy in rotational motion depends on the moment of inertia and square of angular velocity.
The final angular velocity when the moment of inertia is doubled can be found using the conservation of angular momentum. Angular momentum is conserved when the net torque acting on the system is zero. By equating the final angular momentum with the initial angular momentum, we can find the value of final angular velocity. Using that we can find the final kinetic energy.

Complete step by step answer:
It is given that a child is standing with folded hands at the centre of a platform rotating about its central axis.
The kinetic energy of the system is given as $K$ .
When the child stretches his arms the moment of inertia of the system doubles.
We need to find the kinetic energy of the system when the child stretches his arms.
We know that the kinetic energy in rotational motion is given as
$KE = \dfrac{1}{2}I{\omega ^2}$
Where, I is the moment of inertia and $\omega $ is the angular velocity.
Let us calculate the initial kinetic energy of a system.
 Let the initial moment of inertia be ${I_i}$ and the Initial angular velocity be ${\omega _i}$ .
So, the initial kinetic energy can be written as
$ \Rightarrow K = \dfrac{1}{2}{I_i}{\omega _i}^2$ ……………..(1)
Now when the child stretches his arms the moment of inertia becomes double the initial value.
So final moment of inertia is
$ \Rightarrow {I_f} = 2{I_i}$
In order to find the final angular velocity, ${\omega _f}$ we can use the conservation of angular momentum.
Angular momentum is the product of moment of inertia and angular velocity.
$ \Rightarrow L = I\omega $
According to conservation of angular momentum the angular momentum is conserved when the net torque acting on the system is zero.
Hence the initial angular momentum will be equal to the final angular momentum.
$ \Rightarrow {I_i}{\omega _i} = {I_f}{\omega _f}$
On substituting the $I_f$ value, we get
$ \Rightarrow {I_i}{\omega _i} = 2{I_i}{\omega _f}$
$ \Rightarrow {\omega _f} = \dfrac{{{\omega _i}}}{2}$
This is the final angular velocity.
Now we can calculate the final kinetic energy of the system.
$ \Rightarrow K' = \dfrac{1}{2}{I_f}{\omega _f}^2$
On substituting the values, we get
$ \Rightarrow K' = \dfrac{1}{2}2{I_i}{\left( {\dfrac{{{\omega _i}}}{2}} \right)^2}$ ………..(2)
Dividing equation 1 by 2 we get
$ \Rightarrow \dfrac{K}{{K'}} = \dfrac{{\dfrac{1}{2}{I_i}{\omega _i}^2}}{{\dfrac{1}{2}2{I_i}{{\left( {\dfrac{{{\omega _i}}}{2}} \right)}^2}}}$
On simplification, we get
$ \Rightarrow \dfrac{K}{{K'}} = 2$
$ \Rightarrow K' = \dfrac{K}{2}$
This is the final kinetic energy.

Hence, the correct option is option B.

Note:
Remember that the conservation law of angular momentum which states the total angular momentum of the system will be constant is applicable only if the net torque acting on the system is zero.
The individual angular momentum will not be conserved but the total angular momentum of the system will remain constant.