An 81-kg person puts on a life jacket, jumps into water, and floats. The jacket's volume is $3.1*10^-2m^3$ and is completely under water. The volume of the person’s body that is under water is $6.2*10^-2m^3$. What is the density of the life jacket?

Md Ariful Islam

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The idea here is that you need to find the mass of the life jacket by using the total mass of water displaced by the person and his life jacket.

The volume of water displaced by the jacket is equal to the volume of the jacket under water. Likewise, the volume of water displaced by the person is equal to the volume of the person under water.

The total volume of water displaced by the person and his life jacket will be

$V_"water" = V_"jacket" + V_"person"$

$V_"water" = 3.1 * 10^(-2)"m"""^3 + 6.2 * 10^(-2)"m"""^3 = 9.3 * 10^(-2)"m"""^3$

If you take water's density to be equal to $"1000 kg/m"""^3$, you can say that the mass of water displaced by the person and his jacket will be

$rho = m_"water"/V_"water" implies m = rho_"water" * V_"water"$

$m_"water" = 1000"kg"/color(red)(cancel(color(black)("m"^3))) * 9.3 * 10^(-2)color(red)(cancel(color(black)("m"^3))) = "93 kg"$

Since the mass of the person and his jacket will be equal to

$m_"water" = m_"person" + m_"jacket"$

you can find the mass of the jacket

$m_"jacket" = m_"water" - m_"person" = "93 kg" - "81 kg" = "12 kg"$

The density of the jacket will thus be

$rho = m_"jacket"/V_"jacket" = "12 kg"/(3.1 * 10^(-2)"m"""^3) = "387.1 kg/m"""^3$

Rounded to two , the answer will be

$rho_"jacket" = color(green)("390 kg/m"""^3)$

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