The density of ice is $918 k g / m^{3}$ and that of seawater $1030 k g / m^{3}$ . An ice block floats in water with a portion of volume $9.18 \times 10^{- 4} m^{3}$ below the surface of the water. Find the volume of the ice block.

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Solution

Step 1: Given data
The density of ice is = $918 k g / m^{3}$
The density of seawater is = $1030 k g / m^{3}$
The volume of ice immersed in seawater = $9.18 \times 10^{- 4} m^{3}$
Step 2: Finding the volume of ice block,
$R e l a t i v e d e n s i t y = \frac{d e n s i t y o f i c e}{d e n s i t y o f s e a w a t e r} = \frac{918 k g / m^{3}}{1000 k g / m^{3}} = 0.918 A l s o, r e l a t i v e d e n s i t y = \frac{v o l u m e o f i c e i m m e r s e d i n w a t e r}{T o t a l v o l u m e o f i c e} S o, T o t a l v o l u m e o f i c e = \frac{v o l u m e o f i c e i m m e r s e d i n w a t e r}{r e l a t i v e d e n s i t y} = \frac{9.18 \times 10^{- 4} m^{3}}{0.918} = 0.001 m^{3}$
Hence, the volume of the ice block is $0.001 m^{3}$ .

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