If the angular momentum of a rotating body is increased by 200%, then its kinetic energy of rotation will be increased by

The correct answer is 800 %.

To understand the relationship between the angular momentum and the rotational kinetic energy of a rotating body, we can start with their standard physical definitions.

The rotational kinetic energy (represented as K) of a body with moment of inertia I and angular velocity $\omega$ is given by:
$K=\frac{1}{2}I{\mathrm{��}}^2$
The angular momentum (represented as L) of the body is defined as:
$L=I\omega$

By expressing the angular velocity as $\omega =\frac{L}{I}$ and substituting it into the equation for rotational kinetic energy, we get:
$K=\frac{1}{2}I{\left(\frac{L}{I}\right)}^2=\frac{L^2}{2I}$

Since the rotating body itself does not change, its moment of inertia I remains constant. Therefore, the rotational kinetic energy is directly proportional to the square of its angular momentum:
$K\propto L^2$

Let the initial angular momentum be $L_1=L$ and the initial rotational kinetic energy be $K_1$.
If the angular momentum is increased by 200%, the new angular momentum $L_2$ becomes:
$L_2=L+200\%\text{of}L=L+2L=3L$

Since kinetic energy is proportional to the square of angular momentum, the new rotational kinetic energy $K_2$ is:
$K_2\propto L_2^2={\left(3L\right)}^2$
This means the new kinetic energy is 9 times the initial kinetic energy:
$K_2=9K_1$

The percentage increase in the rotational kinetic energy can be calculated as follows:
$\text{Percentage Increase}=\frac{K_2-K_1}{K_1}\times 100$
$\text{Percentage Increase}=\frac{9K_1-K_1}{K_1}\times 100=8\times 100=800\%$

Thus, the kinetic energy of rotation increases by 800%.