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Archimedes’ Principle
The density and mass of a metal block are 3.2 ×103 kg m-3 and 5.0 kg respectively. Find the upthrust that act on the metal block when it is fully immersed in water. [ Density of water = 1000kg m-3 ]
Answer
The upthrust can be found from the formula F = ρVg. From the question, we know that
Density of the water, ρ
= 1000kg m-3
Gravitational field strength, g = 10 ms-2
Since the metal block is fully immersed in water, the volume of the displaced water = volume of the block.
The volume of the block, V can be found be using the equation of density, ρ = m/V.
$\begin{gathered}
{\text{Density of Block, }}\rho_{metal} {\text{ = 3200kgm}}^{{\text{ - 3}}} \hfill \\
{\text{Mass of Block, }}m = 5kg \hfill \\
{\text{Hence, Volume of Block,}} \hfill \\
V = \frac{m}
{\rho } = \frac{{(5)}}
{{(3200)}} = 0.0015625m^3 \hfill \\
\end{gathered} $
Upthrust exerted on the block,
$\begin{gathered}
F = \rho Vg \hfill \\
F = (1000)(0.0015625)(10) \hfill \\
F = 14.625N \hfill \\
\end{gathered} $
Principle of Floatation:
- Displaced volume of fluid = volume of the object that immerse in the fluid.
- If weight of the object > upthrust, the object will sink into the fluid.
- If weight of the object = upthrust, the object is in balance and therefore float on the surface of the fluid.
A block that has volume of 0.2 m3 is hanging in a water tank as shown in the figure above. Find the tension of the string? [ Density of the metal = 8 × 103 kg m-3, density of water = 1 × 103 kg m-3]
Answer:
Volume of metal block, Vblock = 0.2 m3
Density of metal block, ρblock = 8 × 103 kg m-3
Density of metal block, ρwater = 1 × 103 kg m-3
The system is in equilibrium, hence
Upthrust + Tension = Weight
ρwaterVwaterg + T = ρblockVblockg
T = ρblockVblockg - ρwaterVwaterg
T = (8000)(0.2)(10) - (1000)(0.2)(g)
T = 14000 N
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