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 kg`. With what force `F` (in `xx 10^(2)N)` should the cart be pulled that the maximum amount of water remains in the vessel. The dimensions of the vessel are as shown in the figure. Given that `b = 50 cm, c = 10 cm`, area of base `A= 40cm^(2), L = 20 cm, g = 10 m//s^(2)`.

Answer

1

Explanation

As the cart is drawn by a force F, te water in the vessel takes up a slant position rising upward at the back of the vessel. To prevent water from flowing out of the gole G, the acceleration of the vessel should have such a value that it occupies a face area `DBH` and a width of vessel given by `A//L`.
Area of `DeltaDBH=(1)/(2)bc`
Volume of the liquid retained `=(1)/(2)bcxx(A)/(2)`
Mass of the cart and water `=M+(bcArho)/(2L)`
`tan theta=(ma)/(mg),a=g tan theta=gxx(b)/(c)`
Required force is
`(M+(bcArho)/(2L))(gb)/(c)=[1.5+0.5]xx50`
`=2.0xx50=100N`