Question

a) A long vertical tube contains oil to a height of 42.5 cm. The density of the oil is 0.8 g cm-3 . Calculate the maximum lateral pressure exerted by the oil on the sides of the tube.

Answer 1

Given:

A long vertical tube contains oil to a height of 42.5 cm. The density of the oil is 0.8 g cm-3 .

To find:

Maximum lateral pressure on the sides of the tube.

Calculation:

Considering Pascal's Law, the maximum lateral pressure on the side of a tube is exerted at the level of lowest point (max depth) of the tube.

$\sf{ \therefore \: P_{max} = \rho \times g \times h}$

Putting available values in SI unit:

$\sf{ = > \: P_{max} = 800 \times 10 \times \dfrac{42.5}{100} }$

$\sf{ = > \: P_{max} = 8000 \times \dfrac{42.5}{100} }$

$\sf{ = > \: P_{max} = 80 \times 42.5 }$

$\sf{ = > \: P_{max} = 3400 \: pascal}$

So, final answer is:

$\boxed{ \bf{\: P_{max} = 3400 \: pascal}}$

Answer 2

Answer:

$3.4 \times {10}^{3} pascal$

Explanation:

$height \: = 42.5 \: m \: = \frac{42.5}{100} cm \: = 0.425 \: cm$

$density = 0.8 \: g \: {cm}^{3} = 800 \: kg \: {m}^{ - 3}$

$acceleration \: due \: to \: gravity \: \\ = 10 \: m \: {s }^{ -2}$

$as \: we \: know. \: \\ pressure \: = hdg \: \\ \: \: \: \: \: \: \: \: \: \: = 0.425 \times 800 \times 10 \\ = 3400 \: pascal \: \\ = 3.4 \times {10}^{3} pascal$