Question

Astronauts visiting a new planet find a lake filled with an unknown liquid. They have with them a plastic cube, $6.0\mathrm{cm}$on each side, with a density of $840\mathrm{Kg}/{\mathrm{m}}^3$. First they weigh the cube with a spring scale, measuring a weight of $21\mathrm{N}$. Then they float the cube in the lake and find that two-thirds of the cube is submerged. At what depth in the lake will the pressure be twice the atmospheric pressure of$85\mathrm{kPa}$ ?

Short answer

The depth value for the lake will the pressure be twice is 0.5827 m

Step-by-step solution

Given information

Side of the cube is$6\mathrm{cm}$.

Density value is $840\mathrm{Kg}/{\mathrm{m}}^3$.

Weight is $21\mathrm{N}$.

Atmospheric pressure is $85\mathrm{kPa}$.$6\mathrm{cm}$

Calculation for depth

Mass of the cube is,

$=(0.06\times (840)$

$=0.18144\mathrm{Kg}$

It weighs 21 N so

we can find the value of g

mg = 21 N

=> 0.18144 kg x g = 21 N

=> g = 115.740 m/s²

The cube is immersed two-thirds of the way.

So,

$(0.06m{)}^3\left(\frac{2}{3}\right)\times (115.740m/s^2)\rho =21N$

$\mathrm{\rho }=1260.18\mathrm{kg}/{\mathrm{m}}^3$

Depth is,

$\mathrm{p}=\mathrm{\rho gd}$

localid="1651160111576" $\Rightarrow 1260.18\times (115.740)\times \mathrm{d}=85\times {10}^3$

From solving that,

localid="1651160159914" $\mathrm{d}=0.5827\mathrm{m}$