$\begin{array}{1 1}(A)\;791461N\\(B)\;153.4N\\(C)\;141N\\(D)\;165544N\end{array} $

Density of oil $\rho_{oil}=0.9$

$\rho_{water}=900Kg/m^3$

Force/length,F=Force under graph

$\Rightarrow$ Sum of the areas

$\Rightarrow f_1+f_2+f_3$

$f_1=\large\frac{(900\times 9.81\times 2)\times 2}{2}$$\times 3=52974N$

$f_2=(900\times 9.81\times 2)\times 1.5\times 3=79461N$

$f_3=\large\frac{(1000\times 9.81\times 1.5)\times 1.5}{2}$$\times 3=33109N$

$F=f_1+f_2+f_3$

$F=(52974+79461+33109)N$

$F=165544N$

Hence (D) is the correct answer.

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