Question

a resevoir contains mercury up to a hieght of h=10cm.the atmospheric pressure at the surface of the mercury is equivalent to a pressure made by a column of mercury of hieght H=75cm.calculate in Pa the total pressure exerted at a point in the bottom of the resevoir.take g=10N/Kg and density of mercury P=13600kg/mcubed

Answer 1

A reservoir contains mercury up to a height of $h = 10cm$ . The atmospheric pressure at the surface of the mercury is equivalent to a pressure made by a column of mercury of height $H = 75cm$ . Calculate in $Pa$ the total pressure exerted at a point in the bottom of the reservoir. Take $g = 10N / Kg$ and density of mercury $P = 13600kg / m$ cubed.

Solution.

$$h = 1 0 c m = 0. 1 0 m, H = 7 5 c m = 0. 7 5 m, g = 1 0 \frac {N}{k g}, \rho = 1 3 6 0 0 \frac {k g}{m ^ {3}};$$

The total pressure exerted at a point in the bottom of the reservoir is:

$$p = p _ {o} + p _ {m};$$

$p_{o}$ - the atmospheric pressure;

$p_{m}$ - the pressure made by a column of mercury in the reservoir.

$$p _ {m} = \rho g h.$$

$p_{o} = \rho gH$ , because the atmospheric pressure at the surface of the mercury is equivalent to a pressure made by a column of mercury of height $H = 75cm$ .

$$p = \rho g h + \rho g H;$$

The total pressure exerted at a point in the bottom of the reservoir is:

$$p = \rho g (h + H).$$ $$p = 1 3 6 0 0 \frac {k g}{m ^ {3}} \cdot 1 0 \frac {N}{k g} (0. 1 0 m + 0. 7 5 m) = 1 1 5 6 0 0 P a.$$

Answer: The total pressure exerted at a point in the bottom of the reservoir is $p = 115600Pa$ .