$\begin{array}{1 1}(A)\;1561KN\\(B)\;1563.29KN\\(C)\;1326.3KN\\(D)\;1389KN\end{array} $

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$h_2=17m$,so $h_1=17-7.5=9.5m$

$x=\large\frac{9.5}{\tan 60^{\\large\circ}}=$$5.485$

Vertical force =Wt.of water above the surface

$F_V=\rho g(h-2\times x+0.5 h_1 x)$

$\Rightarrow 9810\times (7.5\times 5.485+0.5\times 9.5\times 5.485)$

$\Rightarrow 659.123KN/m$

$F_H=$Force on the projection of the surface on to a vertical plane

$\;\;\;\;\;\;=\large\frac{1}{2}$$\rho g h^2$

$\;\;\;\;\;\;=1417.545KN/m$

$F_R=1563.29KN/m$

Hence (B) is the correct answer.

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