A vessel with a symmetrical hole in its bottom is fastened on a cart. The mass of the vessel and the cart is $1.5 \mathrm{kg}$. With what force F $\left(\text{ in }\times {10}^2\mathrm{N}\right)$ should the cart be pulled so that the maximum amount of water remains in the vessel. The dimensions of the vessel are as shown in the figure. Given that $\mathrm{b}=50 \mathrm{cm}, \mathrm{c}=10 \mathrm{cm}$, area of base $\mathrm{A} = 40 {\mathrm{cm}}^2, \mathrm{L}=20 \mathrm{cm}$, $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$

As the cart is drawn by a force F, the water in the vessel takes up a slant position rising upward at the back wall of the vessel. To prevent water flowing out of the hole H, the acceleration of the vessel should have such a value that it occupies a face area DBH and a width of vessel given by $\frac{A}{L}$

$\text{ Area of }\mathrm{\Delta DBH}=\frac{1}{2}\mathrm{bc}$

Volume of liquid retained $=\frac{1}{2}\mathrm{bc}\times \frac{\mathrm{A}}{\mathrm{L}}$

Mass of cart and water $=\mathrm{M}+\frac{\mathrm{bcA}\rho }{2\mathrm{L}}$

$\tan \theta =\frac{\mathrm{ma}}{\mathrm{mg}};\mathrm{a}=\mathrm{gtan}\theta =\mathrm{g}\times \frac{\mathrm{b}}{\mathrm{c}}$

Required force

$=\left(\mathrm{M}+\frac{\mathrm{bcA}\rho }{2\mathrm{L}}\right)\frac{\mathrm{gb}}{\mathrm{c}}=[1.5+0.5]\times 50$

$=2.0\times 50=100\mathrm{N}$