Question

An object hangs from a spring balance. The balance registersin air,when this object is immersed in water and 24 N when the object is immersed in another liquid of unknown density. What is the density of that other liquid?

Short answer

The density of the unknown liquid is$6\times {10}^2\mathrm{kg}/{\mathrm{m}}^3$.

Step-by-step solution

Listing the given quantities

• Weight in air $=30\mathrm{N}$.
• Weight in water $=20\mathrm{N}$.
• Weight in the unknown liquid $=24\mathrm{N}$.

Understanding the concept of buoyant force

By using the relation between apparent weight, actual weight, and force of buoyancy we find the density of the unknown liquid.

Formula:

$F_b=M_{f}g$

$={\rho }_{f}Vg$

Where${\rho }_f$is the density of fluid, V is the volume, and g is the acceleration due to gravity.

$W_{\mathrm{app}}=W_{Act}-F_b$

(a) Calculation of density of the unknown liquid

The weight in air of an object is its actual weight,$W_{\text{act}}=30\mathrm{N}$

For the weight of an object when it is immersed in water, we can write

$W_{app}=W_{Act}-F_b$

$20=30-{\rho }_{w}Vg$

$Vg=\frac{10}{{\rho }_w}$ (1)

Applying the formula 2 to the object when it is immersed in an unknown liquid

$W_{app}=W_{Act}-F_b$

$24\mathrm{N}=30\mathrm{N}-{\rho }_{u}Vg$

${\rho }_u=\frac{6\mathrm{N}}{Vg}$

Using equation (1) and rearranging

${\rho }_u=\frac{6\mathrm{N}\times \rho \mathrm{w}}{10}$

$=\frac{6\mathrm{N}\times 1\times {10}^3}{10}$

$=6.0\times {10}^2\mathrm{kg}/{\mathrm{m}}^3$

The density of the unknown liquid is $6\times {10}^2\mathrm{kg}/{\mathrm{m}}^3$.