Question

An object weighs 79.1 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 21.8 N. Determine the density of the object. (b)When the object is immersed in oil, the force scale reads 48.4 N. Calculate the density of the oil.

Answer 1

Answer:

(a). The density of the object is 1382 kg/m³.

(b). The density of the oil is 536.4 kg/m³.

Explanation:

Given that,

Weight in air = 79.1 N

Weight in water = 21.8 N

Weight in oil = 48.4 N

We need to calculate the volume of object

Using formula of buoyant force

$F_{b}=W_{air}=W_{water}$

$F_{b}=79.1-21.8$

$F_{b}=57.3\ N$

$F_{b}=\rho g h$

Put the value into the formula

$57.3=1000\times V\times 9.8$

$V=\dfrac{57.3}{1000\times9.8}$

$V=5.84\times10^{-3}\ m^3$

We need to calculate the density

Using formula of buoyant force

$F_{b}=\rho Vg$

$79.1=\rho\times5.84\times10^{-3}\times9.8$

$\rho=\dfrac{79.1}{5.84\times10^{-3}\times9.8}$

$\rho=1382\ kg/m^3$

The density of the object is 1382 kg/m³.

(b). We need to calculate the volume of object

Using formula of buoyant force

$F_{b}=W_{air}=W_{oil}$

$F_{b}=79.1-48.4$

$F_{b}=30.7\ N$

We need to calculate the density

Using formula of buoyant force

$F_{b}=\rho_{oil} Vg$

$30.7=\rho_{oil}\times5.84\times10^{-3}\times9.8$

$\rho_{oil}=\dfrac{30.7}{5.84\times10^{-3}\times9.8}$

$\rho=536.4\ kg/m^3$

The density of the oil is 536.4 kg/m³.

Hence, This is the required solution.