$\begin{array}{1 1}(A)\;5.2\times 10^6N/m\\(B)\;4.28\times 10^6N/m\\(C)\;5.8\times 10^6N/m\\(D)\;7.3\times 10^5N/m\end{array} $

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$h=30\sin 60^{\large\circ}=25.98m$

$a=30\cos 60^{\large\circ}=15m$

$F_v=$Total weight of fluid above the curved surface(per m length)

$F_v=\rho g$(area of sector-area of triangle)

$\Rightarrow 1000\times 9.81\times [(\pi(30)^2\times \large\frac{60}{360})-(\frac{25.98\times 15}{2})]$

$\Rightarrow 2711.375KN/m$

$F_h=\large\frac{1}{2}$$\rho gh^2$

$\Rightarrow 0.5\times 1000\times 9.81\times (25.98)^2$

$\Rightarrow 3310.618KN/m$

$F=\sqrt{F_v^2+F_h^2}=\sqrt{(2711)^2+(3310)^2}KN/m$

$\;\;\;\;=4230KN/m$

Hence (B) is the correct answer.

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