Question

when an object of mass 850 g and density 8000 kg/m³ is submerged in methanol, which has a density of 0.79 g/cm³, what would be its apparent weight? (use g = 10 m/s²)

Explanation:

Step1: Convert mass to SI - unit

The mass of the object $m = 850g=0.85kg$. The density of the object $
ho_{object}=8000kg/m^{3}$. First, find the volume of the object using the density formula $
ho=\frac{m}{V}$. So, $V=\frac{m}{
ho_{object}}$.
$V = \frac{0.85}{8000}m^{3}=1.0625\times 10^{-4}m^{3}$

Step2: Find the buoyant - force

The density of methanol $
ho_{methanol}=0.79g/cm^{3}=790kg/m^{3}$. The buoyant - force $F_b$ acting on the object when submerged in methanol is given by Archimedes' principle $F_b=
ho_{methanol}gV$.
$F_b = 790\times9.8\times1.0625\times 10^{-4}N\approx0.82N$

Step3: Find the actual weight

The actual weight of the object $W = mg$, where $g = 9.8m/s^{2}$.
$W=0.85\times9.8N = 8.33N$

Step4: Calculate the apparent weight

The apparent weight $W_{apparent}=W - F_b$.
$W_{apparent}=8.33 - 0.82N=7.51N$
To convert the apparent weight back to grams (since the original mass was given in grams), we use $W = mg$. So, $m_{apparent}=\frac{W_{apparent}}{g}=\frac{7.51}{9.8}kg\approx0.766kg = 766g$

Answer:

766g