First slide

Fluid statics

Question

ln figure, block A hangs by a cord from spring balance D and is submerged in a liquid C contained in a beaker B. The mass of the beaker is 1 kg. The mass of the liquid is 1.5 kg. Balance D reads 7.5 kg. The volume of block is 0.003 $m^{3}$ . The mass per unit volume of the liquid is

Moderate

A

$2500 kg m^{- 3}$
B

5000 $kg m^{- 3}$
C

$1 kg m^{- 3}$
D

$\frac{5000}{3} kg m^{- 3}$

Solution

7.5g = 1g + 1.5g + Downward reaction force of block
or 5g = |Downward reaction force|
= |Upthrust| = 0.003 $ρ_{1} g$
$∴ ρ_{1} = \frac{5}{0.003} kg m^{- 3} = \frac{5000}{3} kg m^{- 3}$

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