Question

a coin is dropped in a beaker containing mercury settles down at the bottom after 3 seconds. if the depth of mercury in the beaker is 1.5 feet. calculate the pressure on the country at that point​

Answer 1

Given:

A coin is dropped in a beaker containing mercury settles down at the bottom after 3 seconds. if the depth of mercury in the beaker is 1.5 feet.

To find:

Pressure on the coin at the bottom.

Calculation:

Depth = 1.5 feet = 1.5 × 0.305 = 0.457 m,

Density of mercury = 13600 kg/m³.

General expression of pressure exerted by liquid column of height "h" is given as:

$\boxed{ \sf{ P = \rho \times g \times h}}$

Putting available values in SI unit:

$= > \sf{ P = 13600 \times 10 \times 0.457}$

$= > \sf{ P = 136\times 457}$

$= > \sf{ P = 62152 \: pascal}$

$= > \sf{ P = 62.2 \: kilopascal}$

So, final answer is:

$\boxed{\bf{ Pressure \: on \: coin = 62.2 \: kilopascal}}$