Question

A closed tank with length 10 m, breadth 8 m and depth 6 m is filled with water to the top. If g = 10 $m{s}^{-2}$ and density of water is 1000 kg $m^{-3}$, then the thrust on the bottom is

• A. 6 $\times$ 1000 $\times$ 10 $\times$ 80 N
• B. 3 $\times$ 1000 $\times$ 10 $\times$ 48 N
• C. 3 $\times$ 1000 $\times$ 10 $\times$ 60 N
• D. 3 $\times$ 1000 $\times$ 10 $\times$ 80 N

Answer 1

Answer: (A)

6 $\times$ 1000 $\times$ 10 $\times$ 80 N

The pressure of a liquid at a height is given by $P=P_0+\rho gh$, where
$P_0$ = pressure at the surface of the liquid
$\rho$ = density of the liquid
$h$ = depth if the point from the surface of water.
Since the tank is closed, $P_0=0$
$\rho =1000kg{m}^{-3}$
$h=6m$
$P=1000\times 6\times 10N{m}^{-2}$
$Area=8\times 10=80m^2$
$F=P\times Area=1000\times 6\times 10\times 80N$