A two hinged parabolic arch of span $$l$$ and rise h carries a load varying from zero at the left end to $$\omega $$ per unit run at the right end. The horizontal thrust is A. $$\frac{{\omega {l^2}}}{{4{\text{h}}}}$$ B. $$\frac{{\omega {l^2}}}{{8{\text{h}}}}$$ C. $$\frac{{\omega {l^2}}}{{12{\text{h}}}}$$ D. $$\frac{{\omega {l^2}}}{{16{\text{h}}}}$$

The correct answer is $\boxed{\frac{{\omega {l^2}}}{{8{\text{h}}}}}$.

The horizontal thrust is the force that the arch exerts on the supports at its ends. It is equal to the total load on the arch divided by the rise of the arch.

The load on the arch is a uniform load of $\omega$ per unit run, which means that the total load is $\omega l$. The rise of the arch is $h$.

The horizontal thrust is therefore:

$$T = \frac{\omega l}{h} = \frac{{\omega {l^2}}}{{8{\text{h}}}}$$

The other options are incorrect because they do not divide the total load by the rise of the arch.