Question
There are two identical small holes $$P$$ and $$Q$$ of area of cross-section $$a$$ on the opposite sides of a tank containing a liquid of density $$\rho .$$ The difference in height between the holes is $$h.$$ Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is
There are two identical small holes $$P$$ and $$Q$$ of area of cross-section $$a$$ on the opposite sides of a tank containing a liquid of density $$\rho .$$ The difference in height between the holes is $$h.$$ Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is
A.
$$gh\rho a$$
B.
$$\frac{{2gh}}{{\rho a}}$$
C.
$$2\,\rho agh$$
D.
$$\frac{{\rho gh}}{a}$$
Answer :
$$2\,\rho agh$$
Solution :
Force $$F = \rho Qv = \rho a{v^2} = \rho a \times 2gh$$
Thus net force
$$\eqalign{ & = {F_2} - {F_1} \cr & = \rho av_2^2 - \rho av_1^2 \cr & = \rho a\left( {v_2^2 - v_1^2} \right) \cr & = \rho a\left[ {2g\left( {y + h} \right) - 2gy} \right] \cr & = \rho a \times 2gh \cr} $$
Force $$F = \rho Qv = \rho a{v^2} = \rho a \times 2gh$$
Thus net force
$$\eqalign{ & = {F_2} - {F_1} \cr & = \rho av_2^2 - \rho av_1^2 \cr & = \rho a\left( {v_2^2 - v_1^2} \right) \cr & = \rho a\left[ {2g\left( {y + h} \right) - 2gy} \right] \cr & = \rho a \times 2gh \cr} $$
