6. On the face of a circular clock, $O A$ represents the second hand, $O B$ represents the minute hand ($O$ is the center of rotation for both hands). If the current time is exactly 12 o'clock, then after $\qquad$ seconds, the area of $\triangle O A B$ will reach its maximum for the first time.
Ni.6. $15 \frac{15}{59}$.
Let the height on side $O A$ be $h$, then $h \leqslant O B$, so,
$S_{\triangle O A B}=\frac{1}{2} O A \cdot h \leqslant \frac{1}{2} O A \cdot O B$.
The equality holds when $O A \perp O B$.
At this moment, the area of $\triangle O A B$ is maximized.
Suppose after $t \mathrm{~s}$, $O A$ and $O B$ are perpendicular for the first time. Since the second hand rotates $6^{\circ}$ per $1 \mathrm{~s}$, and the minute hand rotates $0.1^{\circ}$ per $1 \mathrm{~s}$, we have $(6-0.1) t=90$. Solving this, we get $t=15 \frac{15}{59}$.