Question

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The height of mercury column in a barometer in a laboratory was recorded to be 75cm. Calculate this pressure in SI and CGS units using the following data:

Specific gravity of mercury = 13.6
Density of water (p) = 10³kg/m³
g = 9.8m/s²
Pressure = hpg in usual symbols.


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Answer 1

Pressure = hpg = (75 cm)(13.6 * 1000 kg/m3)(9.8 m/s2)

SI units

We only need to convert cm to m, since all other units are already SI.

So, pressure in SI units = (0.75 m)(13.6 * 1000 kg/m³)(9.8 m/s²) = 99960 bar

CGS units

Here, we need to convert kg/m³ to g/cm³ and m/s² to cm/s²

1 kg/m³ = 1000 g/m³ * (1 m/100 cm)³ = 1000 g / 1000000 cm³ = 0.001g/cm³ .

So 1000 kg/m³ = 0.001 * 1000 g/cm³

= 1 g/cm³

1 m/s² = 100 cm/s²

So, 9.8 m/s² = 980 cm/s² .

Therefore, Pressure in CGS units = (75 cm)(13.6 * 1 g/cm³)(980 cm/s²) = 999600 g/cm-s²

For easy conversion between units, remember that multiplying by 1 does not make any difference. So to convert 1 m to 100 cm, you can multiply 1 m by (100cm/1 m), since (100cm/1m) is 1. This is useful when you have big units. eg. to convert kg-m/torr to g-mm/Pa, just multiply with (1000g/1kg)*(1000mm/1m)*(760 torr/10125 Pa) etc.

Answer 2

Given:

Height of the column = 75 cm

That is: 0.75 m

Density of mercury = 13600 $\sf{kg/m^{3}}$

g = 9.8 $\sf{m/s^{2}}$

Now:

In SI units:

Pressure (P) = hpg

$\implies$ 0.75 × 13600 × 9.8

$\implies$ 0.75 × 133280

$\implies$ 99960 $\sf{N/m^{2}}$

Or: $\sf{10 \times 10^{4} N/m^{2}}$ approx.

In CGS unit:

$\implies$ $\sf{P = \frac{10 \times {10}^{4} \times {10}^{5} \: dyne }{ {10}^{4} {cm}^{2}}}$

$\implies$ $\sf{P = 10 \times {10}^{5} \: dyne/ {cm}^{2}}$

Therefore:

Final answer: $\sf{10 \times {10}^{5} \: dyne/ {cm}^{2}}$