Question

what fraction of a ship lies below the surface of water given that the density of water is 1.03×10^3 kg/m^3 ,density of ship is 0.92×10^3kg/m^3

Answer 1

Answer: The fraction of a ship lies below the surface of water is $f = \dfrac{11}{1000}$.

Explanation:

Given that,

Density of water $\rho_{w}= 1.03\times10^{3}\ kg/m^3$

Density of ship $\rho_{s}= 0.92\times10^{3}\ kg/m^3$

Using buoyant force

Weight of displaced water= weight of ship

$m_{w}g=m_{s}g$

$\rho_{w}V_{w}g=\rho_{s}V_{s}g$

$\dfrac{V_{w}}{V_{s}}=\dfrac{\rho_{s}}{\rho_{w}}$

The fraction of a ship lies below the surface of water

$f = 1-\dfrac{V_{w}}{V_{s}}$

$f = 1-\dfrac{\rho_{s}}{\rho_{w}}$

$f = 1-\dfrac{0.92\times10^{3}}{1.03\times10^{3}}$

$f = \dfrac{11}{1000}$

Hence, The fraction of a ship lies below the surface of water is $f = \dfrac{11}{1000}$.

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) = $\begin{lgathered}1.013 \times {10}^{13} \: \: \frac{kg}{ {m}^{3} } \\\end{lgathered}$

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 = $\begin{lgathered}\frac{m}{ d_{1} } \\\end{lgathered}$

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

Change in volume =

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

We know Bulk modulus (B) =

$\begin{lgathered}\frac{change \: \: in \ \: pressure \: \times \: volume}{change \: \: in \: \: volume} \\\end{lgathered}$

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

Let change in pressure = dp

Now,

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

Substitute all the given values,

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

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