For an object floating in a fluid, Archimedes' principle states that the buoyant force acting on the object equals its weight. The buoyant force is also equal to the weight of the fluid displaced.
Key information:
First, calculate the total volume (\(V_{total}\)) of the cubical block:
\(V_{total} = s^3\) \(V_{total} = (10\text{ cm})^3 = 1000\text{ cm}^3\)The weight of the block (\(W_{block}\)) is \(m \times g\). The buoyant force (\(F_B\)) equals the weight of the displaced water (\(m_w \times g\)).
Since the block is floating:
\(W_{block} = F_B\) \(m \times g = m_w \times g\)The mass of the displaced water (\(m_w\)) can be expressed as \(\rho_w \times V_{sub}\), where \(V_{sub}\) is the submerged volume.
\(m = \rho_w \times V_{sub}\)Substitute the given mass and water density:
\(600\text{ g} = (1\text{ g/cm}^3) \times V_{sub}\)Solve for the submerged volume (\(V_{sub}\)):
\(V_{sub} = \frac{600\text{ g}}{1\text{ g/cm}^3} = 600\text{ cm}^3\)Finally, calculate the percentage of the block's volume that is submerged:
\(\text{Percentage Submerged} = \frac{V_{sub}}{V_{total}} \times 100\%\) \(\text{Percentage Submerged} = \frac{600\text{ cm}^3}{1000\text{ cm}^3} \times 100\%\) \(\text{Percentage Submerged} = 0.6 \times 100\% = 60\%\)Thus, 60% of the block's volume is submerged.
The apparent mass of a piece of metal when fully immersed in water is 60 gm. If the relative density of this metal piece is 2.5, find its actual mass (in gm)?
Which of the following instruments is based on Archimedes principle?
The apparent mass of a piece of metal when fully immersed in water is 60 gm. If the relative density of this metal piece is 2.5, find its actual mass (in gm)?
Which of the following statement(s) is/are true?
1. Archimedes Principle is not an independent principle.
2. Archimedes Principle is an independent principle.
3. Archimedes Principle can be deduced from Newton's law of Motion.
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A piece of copper of density 8.8 g/cm 3 having an internal cavity weight 264 g in air and 221 g in water. the volume of cavity is: