Question

if the specific gravity of ice is 0.92and that of sea water is 1.025 .what fraction of an ice berg float above the surface

Answer 1

The density of ice is 0.92g cm-3 & that of sea water is 1.025g cm-3.Find the total volume of an iceberg which floats with its volume 800cm3 above water?

ANSWER: 7809.5cm3

Answer 2

$\mathfrak{Question:-}$

If the specific gravity of ice is 0.92and that of sea water is 1.025 .what fraction of an ice berg float above the surface.

$\mathfrak{Answer:-}$

$\small{\textbf{\underline{\underline{Given:-}}}$

$\bold{Relative\; density\; of\; ice = 0.92}$

$\bold{Relative\; density\; of \; sea \; water = 1.025}$

$\bold{Density\; of\; water = 1000 kg/m^{3}}$

$\bold{Density \; of \; ice = Relative \; density \; of \; ice \times density\; of \; water}$

$\bold{Density \; of\; sea\; water = 1.025 \times1000}$

$\bold{Let,}$

$\bold{Mass \; of \; iceberg = m \; kg}$

$\bold{Total \; volume \; of \; iceberg = V = Mass/ Density =m/\rho_{1}}$

$\bold{V = \frac{m}{0.92\times1000} \; \; \; \; \; \; ...............(1)}$

$\bold{The \; icebergs \; floats \; on \; sea \; water.\; So,}$

$\bold{Mass \; of \; sea \; water \; displaced = Mass \; of \; iceberg = m}$

$\bold{Volume \; of \; sea \; water = V_{1} = Mass/Density = m/\rho_{2}}$

$\bold{V_{1} = \frac{m}{1.025\times1000} \; \; \; \; \; \;...............(2)}$

$\bold{This\; volume \; of \; sea \; water \; displaced \; is \; equal \; to \; volume \; of \; iceberg \; under \; water. \; So,}$

$\bold{Volume \; of \; iceberg \; under \; water = V_{1} = \frac{m}{1.025\times1000}}$

$\bold{Volume \; of \; iceberg \; which \; floats = V_{2} = V - V_{1} \; \; \; \; \; \; ................(3)}$

$\bold{Using \; values \; of \; V \; and \; V_{1} \; from \; equations \; (1) \; and (2) \; in \; (3) \; we \; have,}$

$\bold{V_{2} = \frac{m}{0.92\times1000} - \frac{m}{1.025\times1000}}$

$\bold{= \frac{0.105}{0.92\times1.025} \times \frac{m}{1000}}$

$\bold{Fraction \; of \; iceberg \; which \; floats = \frac{Volume \; of \; iceberg \; which \; floats}{Total \; volume \; of \; iceberg}}$

$\bold{\frac{V_{2} }{V} = \frac{\frac{\frac{0.105}{0.92\times1.025}\times\frac{m}{1000}  }{m}}{0.92\times1000}}$

$\bold{= 0.105/1.025}$

$\bold{\boxed{\boxed{= 0.1024}}}$