An ice ballerina rotates at a constant angular velocity at one particular point. That is, she does not translationally move. Her arms are fully extended as she rotates. Her moment of inertia is $I$. Now, she pulls her arms in and her moment of inertia is now $\frac{7}{10}I$. What is the ratio of the new kinetic energy (arms in) to the initial kinetic energy (arms out)? 
	
	$ \textbf {(A) } \dfrac {7}{10} \qquad \textbf {(B) } \dfrac {49}{100} \qquad \textbf {(C) } 1 \qquad \textbf {(C) } \dfrac {100}{49} \qquad \textbf {(E) } \dfrac {10}{7} $ 

[i]Problem proposed by Ahaan Rungta[/i]
1. **Conservation of Angular Momentum:**
   Since there is no external torque acting on the ballerina, her angular momentum is conserved. This can be expressed as:
   \[
   L_i = L_f
   \]
   where \(L_i\) is the initial angular momentum and \(L_f\) is the final angular momentum. Angular momentum \(L\) is given by:
   \[
   L = I \omega
   \]
   where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

2. **Initial and Final Angular Momentum:**
   Let \(I_i = I\) and \(\omega_i = \omega\) be the initial moment of inertia and angular velocity, respectively. When the ballerina pulls her arms in, her moment of inertia changes to \(I_f = \frac{7}{10} I\). Let \(\omega_f\) be the final angular velocity. Using the conservation of angular momentum:
   \[
   I \omega = \left(\frac{7}{10} I\right) \omega_f
   \]
   Solving for \(\omega_f\):
   \[
   \omega_f = \frac{10}{7} \omega
   \]

3. **Initial and Final Kinetic Energy:**
   The rotational kinetic energy \(K\) is given by:
   \[
   K = \frac{1}{2} I \omega^2
   \]
   - Initial kinetic energy \(K_i\):
     \[
     K_i = \frac{1}{2} I \omega^2
     \]
   - Final kinetic energy \(K_f\):
     \[
     K_f = \frac{1}{2} \left(\frac{7}{10} I\right) \left(\frac{10}{7} \omega\right)^2
     \]
     Simplifying \(K_f\):
     \[
     K_f = \frac{1}{2} \left(\frac{7}{10} I\right) \left(\frac{100}{49} \omega^2\right)
     \]
     \[
     K_f = \frac{1}{2} \left(\frac{700}{490} I \omega^2\right)
     \]
     \[
     K_f = \frac{1}{2} \left(\frac{10}{7} I \omega^2\right)
     \]
     \[
     K_f = \frac{5}{7} I \omega^2
     \]

4. **Ratio of Final to Initial Kinetic Energy:**
   The ratio of the new kinetic energy to the initial kinetic energy is:
   \[
   \frac{K_f}{K_i} = \frac{\frac{5}{7} I \omega^2}{\frac{1}{2} I \omega^2}
   \]
   Simplifying the ratio:
   \[
   \frac{K_f}{K_i} = \frac{\frac{5}{7}}{\frac{1}{2}} = \frac{5}{7} \times \frac{2}{1} = \frac{10}{7}
   \]

Conclusion:
\[
\boxed{\frac{10}{7}}
\]