Question

A tall, open container is full of glycerine. At what depth h h below the surface of the glycerine is the pressure 2050 Pa 2050 Pa greater than atmospheric pressure? The density of glycerine is 1.26 × 10 3 kg/m 3 1.26×103 kg/m3 .

Answer 1

Answer:

The depth h is 0.166 m.

Explanation:

Given that,

Change in pressure P =2050 Pa

Density of glycerin $\rho_{g}=1.26\times10^{3}$

We need to calculate the height

Using formula for gauge pressure

$P_{2}-P_{1}=\rho gh$

Where, h = height

g = acceleration due to gravity

$\rho_{g}$ =density of glycerin

Put the value into the formula

$2050=1.26\times10^{3}\times 9.8\times h$

$h=\dfrac{2050}{1.26\times10^{3}\times 9.8}$

$h=0.166\ m$

Hence, The depth h is 0.166 m.

Answer 2

The depth below the surface of the glycerine is 0.17 m.

The given parameters;

• pressure, P = 2050 Pa = 2050 N/m²
• density of the glycerine, ρ = 1.26 x 10³ kg
• depth of glycerine, h = ?

The depth below the surface of the glycerine is calculated as follows;

$P = \rho gh\\\\h = \frac{P}{\rho g} \\\\h = \frac{2050}{1.26 \times 10^3 \times 9.8} \\\\h = 0.17 \ m$

Thus, the depth below the surface of the glycerine is 0.17 m.

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