The straight line $l_{1}$ with equation $x-2y+10 = 0$ meets the circle with equation $x^2 + y^2 = 100$ at B in the  first quadrant. A line through B, perpendicular to $l_{1}$ cuts the y-axis at P (0, t). Determine the value of $t$.
1. **Find the intersection point \( B \) of the line \( l_1 \) and the circle:**
   - The equation of the line \( l_1 \) is \( x - 2y + 10 = 0 \).
   - Rearrange the equation to express \( y \) in terms of \( x \):
     \[
     y = \frac{1}{2}x + 5
     \]
   - Substitute \( y = \frac{1}{2}x + 5 \) into the circle's equation \( x^2 + y^2 = 100 \):
     \[
     x^2 + \left(\frac{1}{2}x + 5\right)^2 = 100
     \]
   - Expand and simplify:
     \[
     x^2 + \left(\frac{1}{4}x^2 + 5x + 25\right) = 100
     \]
     \[
     x^2 + \frac{1}{4}x^2 + 5x + 25 = 100
     \]
     \[
     \frac{5}{4}x^2 + 5x + 25 = 100
     \]
     \[
     \frac{5}{4}x^2 + 5x - 75 = 0
     \]
     \[
     5x^2 + 20x - 300 = 0
     \]
     \[
     x^2 + 4x - 60 = 0
     \]
   - Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     x = \frac{-4 \pm \sqrt{16 + 240}}{2}
     \]
     \[
     x = \frac{-4 \pm \sqrt{256}}{2}
     \]
     \[
     x = \frac{-4 \pm 16}{2}
     \]
     \[
     x = 6 \quad \text{or} \quad x = -10
     \]
   - Since \( B \) is in the first quadrant, \( x \) must be positive:
     \[
     x = 6
     \]
   - Substitute \( x = 6 \) back into \( y = \frac{1}{2}x + 5 \):
     \[
     y = \frac{1}{2}(6) + 5 = 3 + 5 = 8
     \]
   - Therefore, the coordinates of \( B \) are \( (6, 8) \).

2. **Find the equation of the line through \( B \) perpendicular to \( l_1 \):**
   - The slope of \( l_1 \) is \( \frac{1}{2} \), so the slope of the perpendicular line is the negative reciprocal:
     \[
     \text{slope} = -2
     \]
   - The equation of the line through \( B(6, 8) \) with slope \( -2 \) is:
     \[
     y - 8 = -2(x - 6)
     \]
     \[
     y - 8 = -2x + 12
     \]
     \[
     y = -2x + 20
     \]

3. **Find the y-intercept \( t \) of the perpendicular line:**
   - The y-intercept is the value of \( y \) when \( x = 0 \):
     \[
     y = -2(0) + 20 = 20
     \]

Conclusion:
\[
\boxed{t = 20}
\]