#### Answer

$|r|$ is also constant.

#### Work Step by Step

As we are given that $r \dfrac{dr}{dt}=0 \\ \implies (2r) \dfrac{dr}{dt}=0 \\ \implies r \dfrac{dr}{dt}+r \dfrac{dr}{dt}=0$
Thus, we have $\dfrac{d(r\cdot r)}{dt}=0\\ \implies
\dfrac{d(|r^2|)}{dt}=0$
This implies that $|r^2|$ is constant .
Hence, $|r|$ is also constant.