Question
Derive the condition of floatation of a body.
✓
Answer
When a body floats in a liquid with a part submerged in the liquid, the weight of the liquid displaced by the submerged part is always equal to the weight of the body. Let V = volume of the body$\sigma=$ density of its material
$\rho=$ density of the liquid in which the body floats such that its volume V' is outside the liquid.
Then volume of the body inside the liquid = V - V’ Weight of the displaced liquid $=(\text{V}-\text{V}')\rho\text{g}$ Also weight of the body $=\text{V}\sigma\text{ g}$ For the body to float, Weight of the liquid displaced by the sumberged part = weight of the body, i.e., $(\text{V}-\text{V}')\rho\text{g}=\text{V}\sigma\text{ g}$ or $\text{V}'=\frac{(\rho-\sigma)\text{V}}{\rho}$
$\rho=$ density of the liquid in which the body floats such that its volume V' is outside the liquid.
Then volume of the body inside the liquid = V - V’ Weight of the displaced liquid $=(\text{V}-\text{V}')\rho\text{g}$ Also weight of the body $=\text{V}\sigma\text{ g}$ For the body to float, Weight of the liquid displaced by the sumberged part = weight of the body, i.e., $(\text{V}-\text{V}')\rho\text{g}=\text{V}\sigma\text{ g}$ or $\text{V}'=\frac{(\rho-\sigma)\text{V}}{\rho}$
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