Answer:
It will feel 4.85% lighter.
Explanation:
Normal weight of the gold bar = 5.0 × 9.81 = 49.05 N
The weight of the gold bar would be reduced because of the buoyant force acting upwards.
weight of the gold bar under water = normal weight - weight of the water displaced by the gold bar.
[tex]W_r=\rho_{G} V g - \rho_w V g = (\rho_G -\rho_w) V g[/tex]
first, we need to find the volume (V) of the gold bar.
[tex]V = \frac{M}{\rho_G} = \frac{5.0 kg}{19300 kg/m^3} = 2.6\times 10^{-4}m^3[/tex]
[tex]W_r = (\rho_G -\rho_w) V g = (19300-1000)(2.6\times 10^{-4})(9.81) = 46.67N [/tex]
Percentage of decrease in weight [tex]= \frac{W-W_r}{W} = \frac{49.05-46.67}{49} \times 100 \% =4.85\% [/tex]
