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(D) Consider a strip of width $dy$ at a height $y$ from the bottom. The depth of this strip below the surface is $(h-y)$.The pressure at this depth is

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) Consider a strip of width $dy$ at a height $y$ from the bottom. The depth of this strip below the surface is $(h-y)$.The pressure at this depth is $P = ho_w g(h-y)$.The force on the strip of area $dA = L \cdot dy$ is $dF = P \cdot dA = ho_w g(h-y) L \cdot dy$.The torque $d	au$ about the bottom axis (at $y=0$) is $d	au = dF \cdot y = ho_w g L (h-y) y \cdot dy$.Integrating from $y=0$ to $y=h$ for the total torque $	au_1$:$	au_1 = \int_{0}^{h} ho_w g L (hy - y^2) dy = ho_w g L \left[ \frac{hy^2}{2} - \frac{y^3}{3} ight]_{0}^{h} = ho_w g L \left( \frac{h^3}{2} - \frac{h^3}{3} ight) = \frac{ho_w g L h^3}{6}$.For the second case,the height is $h' = h/2$ and length is $L' = L/2$. The torque $	au_2$ is:$	au_2 = \int_{0}^{h/2} ho_w g L' (h' - y) y \cdot dy = \int_{0}^{h/2} ho_w g \left(\frac{L}{2}ight) \left(\frac{h}{2} - yight) y \cdot dy$$= \frac{ho_w g L}{2} \left[ \frac{h}{2} \frac{y^2}{2} - \frac{y^3}{3} ight]_{0}^{h/2} = \frac{ho_w g L}{2} \left[ \frac{h}{4} \left(\frac{h^2}{4}ight) - \frac{1}{3} \left(\frac{h^3}{8}ight) ight]$$= \frac{ho_w g L}{2} \left[ \frac{h^3}{16} - \frac{h^3}{24} ight] = \frac{ho_w g L}{2} \left[ \frac{3h^3 - 2h^3}{48} ight] = \frac{ho_w g L h^3}{2 \cdot 48} = \frac{ho_w g L h^3}{96}$.Therefore,$\frac{	au_1}{	au_2} = \frac{ho_w g L h^3 / 6}{ho_w g L h^3 / 96} = \frac{96}{6} = 16$.

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