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9-1.Fluid Mechanics
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A beaker containing water is placed on the platform of a spring balance. The balance reads $1.5$ $kg$. A stone of mass $0.5$ $kg$ and density $500$ $kg/m^3$ is immersed in water without touching the walls of beaker. What will be the balance reading now ? ..... $kg$
A
$2$
B
$2.5$
C
$1 $
D
$3 $
Solution
Volume of stone $=\frac{m}{\rho}=\frac{0.5}{500}=10^{-3} \,m ^3$
Force of Buoyancy $=\rho V g=1000 \times 10^{-3} \times 10=10\, N$
There will be a reaction force to the buoyant force, by the stone on the water.
Therefore, net force on spring balance is sum of force of buoyancy and weight
of water and beaker $= F _{ B }+ mg =10+1.5 g =10+15=25 \,N$
Net reading of spring balance $=2.5 \,kg$
Standard 11
Physics
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