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  • What percentage of an iceberg lies beneath the surface of sea. Take Density of water() = <i

What percentage of an iceberg lies beneath the surface of sea. Take Density of water(\rho) = 1.028 \times 10^3 kg/m^3  and Density of ice (d ) = 0.917 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}  

Option: 1

10.8


Option: 2

78.2


Option: 3

89.2


Option: 4

21.8


Answers (1)

best_answer

 

\begin{array}{l}{\text { Let } V \text { be the total volume of the iceberg and } v \text { be the vol- }} \\ {\text { ume inside the sea water. Then }} \\ {\qquad \begin{array}{l}{\qquad \begin{array}{l}{V d g=v \rho g} \\ {\frac{v}{V}=\frac{d}{\rho} \Rightarrow \frac{v}{V}=\frac{0.917 \times 10^{3}}{1.028 \times 10^{3}} \Rightarrow \frac{v}{V}=0.892}\end{array}} \\ {\text { About } 89.2 \% \text { of the iceberg remains inside. }}\end{array}}\end{array}

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Devendra Khairwa

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