The time now is 7:00.

If the minute hand moves a distance of $x$ , the hour hand moves a distance of $\dfrac{x}{12}$ . From the figure,

$x=35+\dfrac{x}{12}$ $\color{#D61F06}\implies$ $x=38 \frac{2}{11}$

Thus, the minute hand will overtake the hour hand at $\boxed{7:38 \frac{2}{11}}$ .

Let $\omega_h$ and $\omega_m$ be the angular frequency of the hour hand and the minute hand respectively.
The hour hand completes one full revolution ( $2\pi$ rad) every 12h=12*60m. So: $\omega_h=\frac{2\pi}{12*60}=\frac{\pi}{360}\frac{rad}{m}$
The minute hand complete one full revolution ( $2\pi$ rad) every 60m. So: $\omega_m=\frac{2\pi}{60}=\frac{\pi}{30}\frac{rad}{m}$
Now, we let $\theta_h$ be the angle (counting clockwise) between the vertical position and the hour hand and $\theta_m$ the angle between the vertical position and the minute hand. If we consider $t=0$ the moment at which the clock shows 7:00 o'clock then the starting angles are: $\theta_h(0)=\frac{7}{12}2\pi=\frac{7\pi}{6}rad$ and $\theta_m(0)=0rad$ . Therefore we have:
$\theta_h(t)=\theta_h(0)+\omega_h*t=\frac{7\pi}{6}+\frac{\pi}{360}t$
$\theta_m(t)=\theta_m(0)+\omega_m*t=\frac{\pi}{30}t$
We want to find the moment at which the two hands are on top of one another. That is the moment at which $\theta_m=\theta_h$ . So:
$\frac{\pi}{30}t=\frac{7\pi}{6}+\frac{\pi}{360}t$
Working out some tedious algebra we conclude that:
$t=\frac{7*60}{11}=38+\frac{2}{11}m$
So the answer is: $\boxed{7:38\frac{2}{11}}$