Question

What are the volume and the density of an object with apparent weights of 403 N in air and 264 N when immersed in alcohol? The density of alcohol is 0.70 g/cm^3 and the density of air is 0.0012 g/cm^3.

Answer 1

Answer:

$V=2.026*10^4cm^3$

$\rho=2.03\frac{g}{cm^3}$

Explanation:

By the Archimedes principle we know that every body submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object.

Due to this, the object has a loss weight when immersed in alcohol of:

$403N-264N=139N$

So, we can find the displaced mass of alcohol:

$m=\frac{W}{g}\\m=\frac{139N}{9.8\frac{m}{s^2}}\\m=14.18kg$

Now, we calculate the displaced volume of alcohol, which is the same volume of the object:

$V=\frac{m}{\rho}\\V=\frac{14.18*10^3g}{0.7\frac{g}{cm^3}}\\V=2.026*10^4cm^3$

Object mass is:

$m=\frac{W}{g}\\m=\frac{403N}{9.8}=41.12kg$

Finally, we can find the density of the object:

$\rho=\frac{m}{V}\\\rho=\frac{41.12*10^3g}{2.025*10^4cm^3}\\\rho=2.03\frac{g}{cm^3}$