Two bodies are in equilibrium when suspended in water from the arms of a balan

Discussion Forum : Mechanical Properties Of Fluids

Question -

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / $c m^{3}$ . If the mass of the other is 48 g, its density in g / $c m^{3}$ is

Options:

A . 4/3

B . 3/2

C . 3

D . 5

Answer: Option C
:
C

Apparent weight = V $(ρ - σ) g$ = $\frac{m}{ρ} (ρ - σ) g$

where m = mass of the body,

$ρ$ = density of the body

$σ$ = density of water

If two bodies are in equilibrium then their apparent weight must be equal.

$∴ \frac{m_{1}}{ρ_{1}} (ρ_{1} - σ)$ = $\frac{m_{2}}{ρ_{2}} (ρ_{2} - σ)$

$\Rightarrow \frac{36}{9} (9 - 1)$ = $\frac{48}{ρ_{2}} (ρ_{2} - 1)$

By solving we get $ρ_{2}$ = 3

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