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 A cubical block of side 0.5 m floats on water with 30 % of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take density of water = 10³ kg/m³]
A cubical block of side 0.5 m floats on water with 30 % of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take density of water = 10³ kg/m³]
Discover the maximum weight a floating cubical block can bear without complete submersion. Learn the physics behind buoyancy as 30% of its volume dips beneath the water's surface.
by Maivizhi A
Updated Feb 24, 2024
A cubical block of side 0.5 m floats on water with 30 % of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take density of water = 10³ kg/m³]
A mass of 87.5 kg is the maximum weight that can be put on the block without fully submerging it under water.

Given Information:
 The block is floating on water with 30% of its volume submerged.
 The density of water (ρw) is 1000 kg/m³.
 The volume of the block (V) is calculated as (0.5 m)³ = 0.125 m³.
 According to Archimedes' principle, the weight of the block is equal to the weight of the water displaced.

Determination of Density of the Block (ρb):
 Archimedes' principle states that the density of the block (ρb) is 30% of the density of water (ρw).
 Therefore, ρb = 0.3 × ρw.

Equilibrium Condition:
 When a mass (m) is placed on the block such that it is just fully submerged, the system reaches equilibrium.
 At equilibrium, the weight of the water displaced is equal to the combined weight of the block and the additional mass (m).

Application of Archimedes' Principle:
 The weight of the water displaced (V × ρw × g) equals the weight of the block (V × ρb × g) plus the weight of the additional mass (m × g), where g is the acceleration due to gravity.

Calculation of Additional Mass (m):
 Substituting the given values and derived relationships into the equilibrium condition equation, we can solve for m.
 We already have V = 0.125 m³ and ρw = 1000 kg/m³.
 Therefore, m = V × 0.7 × ρw.

Final Result:
 After substituting the values, we find that m = 0.125 × 0.7 × 1000 = 87.5 kg.
This calculation confirms that a mass of 87.5 kg is the maximum weight that can be put on the block without fully submerging it under water.
Define Archimedes' Principle
Archimedes' Principle is a fundamental principle in physics, particularly in fluid mechanics, named after the ancient Greek mathematician and inventor Archimedes. It states that when a body is partially or wholly immersed in a fluid (liquid or gas), it experiences an upward buoyant force equal to the weight of the fluid it displaces. In simpler terms, this principle explains why objects float or sink in fluids.
Mathematically, Archimedes' Principle can be expressed as:
F_buoyant = ρ_fluid × V_displaced × g
Where:
 F_buoyant is the buoyant force acting on the object,
 ρ_fluid is the density of the fluid,
 V_displaced is the volume of fluid displaced by the immersed object, and
 g is the acceleration due to gravity.
This principle is commonly applied in various fields, including naval architecture, engineering, and the design of flotation devices, boats, and submarines.
A cubical block of side 0.5 m floats on water with 30 % of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water  FAQs
1. What is Archimedes' Principle?
Archimedes' Principle states that when an object is partially or wholly immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces.
2. How is Archimedes' Principle mathematically expressed?
Archimedes' Principle is expressed as: F_buoyant = ρ_fluid × V_displaced × g, where F_buoyant is the buoyant force, ρ_fluid is the density of the fluid, V_displaced is the volume of fluid displaced, and g is the acceleration due to gravity.
3. What is the density of water?
The density of water is 1000 kg/m³.
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