Question

A piece of metal of density 7.8x10 ^3 kgm^-2 weight 20n in air calculate the apparent weight of the metal when completely immersed in a liquid of 8.3x10^-2gm^-3 and y=(10ms^-2)

Answer 1

Task:

A piece of metal of density $7.8 \cdot 10^{3} \frac{kg}{m^{3}}$ weight $20N$ in air. Calculate the apparent weight of the metal when completely immersed in a liquid of $8.3 \cdot 10^{-2} \frac{g}{m^{3}}$ and $g = 10 \frac{m}{s^{2}}$

Solution:

$F_{net} = F_P - F_A = mg - \rho_{liquid}gV = \rho_{body}Vg - \rho_{liquid}gV = Vg(\rho_{body} - \rho_{liquid})$

$F_{P} = mg = \rho_{body}Vg = 20N$

$Vg = \frac{F_P}{\rho_{body}}$

$F_{net} = Vg\left(\rho_{body} - \rho_{liquid}\right) = \frac{F_P}{\rho_{body}}\left(\rho_{body} - \rho_{liquid}\right) = F_P\frac{\rho_{body} - \rho_{liquid}}{\rho_{body}}$

$F_{net} = F_P\frac{\rho_{body} - \rho_{liquid}}{\rho_{body}} = 20N\cdot \frac{7.8\cdot 10^3\frac{kg}{m^3} - 8.3\cdot 10^{-5}\frac{kg}{m^3}}{7.8\cdot 10^3\frac{kg}{m^3}}\approx 20N\cdot 1 = 20N$

Answer:

$F_{net}\approx 20N$