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Fluid Mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsFluid Mechanics2 Marks
Question
A metal piece of mass $160g$ lies in equilibrium inside a glass of water The piece touches the bottom of the glass at a small number of points. If the density of the metal is $8000kg/m^3$, find the normal force exerted by the bottom of the glass on the metal piece.
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Answer

Given mass of the metal piece, $m = 160g = 0.160kg$
Density of metal = $8000kg/m^3$
Here we can say that the weight of the metal piece = buoyant force(FB) + Normal force
i.e. weight of the metal piece = mg
$\text { Buoyant force, } F_B=\text { Volume of water displaced by metal } \times P_{\text {water }} \times g$
$m g=F B+\text { Normal force }$
$\text { Normal force }=m g-F_B$
$=m \times g-V \times P_{\text {water }} \times g$
Volume of water displaced by metal. = $=\frac{\text{m}}{\text{pwater}}=\frac{0.160}{8000}=0.00002$
Normal force = (0.160 × 9.8) - (0.00002 × 1000 × 9.8)
= 1.568 - 0.196
= 1.372N
= 1.4N

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