Question

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What is the density of water at a depth where pressure is 80 atm,given that its density at the surface is 1.03×10³ kg/m³???

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Answer 1

Answer:

FLUID DYNAMICS

Step-by-step explanation:

★ DENSITY IS NOT A CHANGING quantity.

» SO ; DENSITY REMAINS SAME THROUGH OUT THE FLUID .

SO ; DENSITY OF WATER AT DEPTH = 1.03 × 10³ KG / M³

[ same as Surface ]

★ DENSITY IS INDEPENDENT OF PRESSURE AND DEPTH ★

Answer 2

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Given:

• Pressure of water at a certain depth = $80 \: atm$

• At the surface of water, density of water (d) = $1.013 \times {10}^{13} \: \: \frac{kg}{ {m}^{3} }$

To Find:

• Density of water at a certain depth

$\huge{\red{\star}} {\blue{\underline{\huge{\textsf{Solution}}}}} \huge{\red{\star}}$

Let Volume at surface of water = $\frac{m}{ d_{1} }$

Let volume at a certain depth = $\frac{m}{ d_{2} }$

Change in volume =

$v_{2} - v_{1} \\ \\ = \frac{m}{ d_{2} } - \frac{m}{ d_{1} } \\ \\ = m( \frac{1}{ d_{2} } - \frac{1}{ d_{1} } )$

We know Bulk modulus (B) =

$\frac{change \: \: in \ \: pressure \: \times \: volume}{change \: \: in \: \: volume}$

Compressibility of water = $\frac{1} {b} \\ \\ = \frac{change \: \: in \: \: volume \: \: }{change \: \: in \: \: pressure \: \times \: volume} \\ \\ = 45.8 \times {10}^{ - 11 \:} pascals$

Let change in pressure = dp

Now,

$45.8 \times {10}^{ - 11} = \frac{m( \frac{1}{ d_{2} } - \frac{1}{ d_{1}} ) }{dp \times \frac{m}{ d_{1} } } \\ \\ 45.8 \times {10}^{ - 11} = \frac{ d_{1} }{dp}( \frac{1}{ d_{2} } - \frac{1}{ d_{1} } ) \\ \\ 45.8 \times {10}^{ - 11} = \frac{1}{dp} (1 - \frac{ d_{1} }{d _{2} } )$

Substitute all the given values,

$45.8 \times {10}^{ - 11} = 1 - (\frac{1.03 \: \times \: {10}^{3} }{ d_{2} } ) \: \times \: \frac{1}{80 \times 1.013 \: \times \: {10}^{5} } \\ \\ by \: \: solving \\ \\ d_{2} = 1.034 \times {10}^{3} \: \: \frac {kg}{ {m}^{3} }$

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