Answer
$P=7.3atm$
Work Step by Step
The required pressure can be determined as follows:
First, we convert the given density into $Kg/m^3$
$d=\frac{0.867g}{cm^3}\times\frac{1Kg}{1000g}\times \frac{(100)^3cm^3}{(1)^3m^3}$
This simplifies to:
$d=867Kg/m^3$
Now $P=hdg$
We plug in the known values to obtain:
$P=87m\times 867Kg/m^3\times 9.807m/s^2$
This simplifies to:
$P=7.3973\times 10^5Pa$
This pressure in $atm$ can be converted as
$P=7.3973\times 10^5Pa\times\frac{1KPa}{1000Pa}\times \frac{1atm}{101.325KPa}$
This simplifies to:
$P=7.3atm$
