A cube of ice floats partly in water and partly in k.oil find the ratio of the volume of ice immersed in water

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Solution

Let specific gravity of water $S_{w a t e r} = \frac{1000}{1000} = 1$
specific gravity of ice $S_{i c e} = \frac{900}{1000} = 0.9$
and specific gravity of kerosene $S_{K} = \frac{800}{1000} = 0.8$
The given values are standard values.
Specific gravity $= \frac{density of object}{density of water}$
and we know density of water =1000 $k g / m^{3}$
Now volume of ice $V_{i c e} = V_{K} + V_{w a t e r}$ -------(A)
As per buoyancy flotation,
$V_{i c e} \times S_{i c e} = V_{K} \times S_{K} + V_{w a t e r} \times S_{w a t e r}$
$(V_{K} + V_{w a t e r}) S_{i c e} = V_{K} \times S_{K} + V_{w a t e r} \times S_{w a t e r}$
$0.9 V_{K} + 0.9 V_{w a t e r} = 0.8 V_{K} + V_{w a t e r}$
$0.1 V_{K} = 0.1 V_{w a t e r}$
$\frac{V_{K}}{V_{w a t e r}} = \frac{1}{1}$

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A cube of ice floats partly in water and partly in kerosene oil. Find the ratio of the volume of ice immersed in water to the volume in k.oil.

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