Question

An object weighing 300 N in air is immersed in water after being tied to a string connected to a balance. The scale now reads 262 N. Immersed in oil, the object appears to weigh 273 N. Find the density of the oil.

Answer 1

Approximately
`711\; {\rm kg \cdot m^(-3)}`, (approximately
`0.711\; {\rm g \cdot cm^(-3)}`,) assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.

Step-by-step explanation:

The density of water is
`\rho(\text{water}) =10^(3)\; {\rm kg \cdot m^(-3)}`.

The buoyancy force on an object immersed in a liquid is equal to the weight of the liquid that object displaces.

While in water, the buoyancy force on the object in this question is
`(300 - 262)\; {\rm N} = 38\; {\rm N}`. In other words, this object displaces
`38\; {\rm N}` of water. Divide weight by
`g` to find mass:

`\begin{aligned}m(\text{water displaced}) &= \frac{38\; {\rm N}}{9.81\; {\rm N\cdot kg^(-1)}} \approx 3.8736\; {\rm kg} \end{aligned}`.

Since the object is fully immersed, the volume of water displaced would be equal to the volume of the object. Divide the mass of the water displaced by the density of water to find this volume:

`\begin{aligned}V(\text{water displaced}) &= \frac{m(\text{water displaced})}{\rho(\text{water})} \\ &\approx \frac{3.8736\; {\rm kg}}{10^(3)\; {\rm kg \cdot m^(-3)}} \\ &= 3.8736 * 10^(-3)\; {\rm m^(3)}\end{aligned}`.

`V(\text{object}) = V(\text{water displaced}) \approx 3.8736 * 10^(-3)}\; {\rm m^(3)}`.

When the object is immersed in oil, the volume of oil displaced would also be equal to the volume of this object:

`V(\text{oil displaced}) = V(\text{object}) \approx 3.8736 * 10^(-3)}\; {\rm m^(3)}`.

It is given that the buoyancy force on this object is
`(300 -273)\; {\rm N} =27\; {\rm N}` when immersed in oil. Similar to the case when the object is in water, the object would displace
`27\; {\rm N}` of oil. The mass of that much oil would be:

`\begin{aligned}m(\text{oil displaced}) &= \frac{27\; {\rm N}}{9.81\; {\rm N\cdot kg^(-1)}} \approx 2.7523\; {\rm kg} \end{aligned}`.

Divide mass of the oil displaced by volume to find density:

`\begin{aligned}\rho(\text{oil}) &= \frac{m(\text{oil displaced})}{V(\text{oil displaced})} \\ &\approx \frac{2.7523\; {\rm kg}}{3.8736 * 10^(-3)\; {\rm m^(3)}} \\ &\approx 711\; {\rm kg\cdot m^(-3)}\end{aligned}`.