Chapter 4, Problem 41Q

To determine

The buoyant forces on the block from largest to smallest.

Answer to Problem 41Q

Thus, the buoyant forces will be arranged as

$0.05>0.0375>0.03>0.02$

$(a)>(b)>(c)=(e)>(d)$.

Explanation of Solution

Given data:

1. M_b = 0.04kg; M_l = 0.20kg.
2. M_b = 0.03kg; M_l = 0.15kg.
3. M_b = 0.04kg; M_l = 0.12kg.
4. M_b = 0.02kg; M_l = 0.08kg.
5. M_b = 0.03kg; M_l = 0.12kg.

Formula used:

The buoyant force is calculated by the following formula:

$F_b={\rho }_l{V}_{d}g$.

Calculation:

Having the known values, buoyant force is calculated as:

$\begin{array}{l}{F}_b={\rho }_l{V}_{d}g\\ \left(a\right) M_b=0.04kg;M_l=0.20kg\\ {\rho }_l=\frac{massofliquid}{volumeofliquid}=\frac{0.20}{V_l}\\ V_d=voulumeofblock\\ V_d=V_b=\frac{V_l}{4}\\ F_b=\left(\frac{0.20}{V_l}\right)\left(\frac{V_l}{4}\right)g=\frac{0.20}{4}g\\ F_b=0.05kg\\ \left(b\right) M_b=0.03kg;M_l=0.15kg\\ {\rho }_l=\frac{0.15}{V_l}\\ F_b=\left(\frac{0.15}{V_l}\right)\left(\frac{V_l}{4}\right)g=\frac{0.15}{4}g\\ F_b=0.0375kg\end{array}$

$\begin{array}{l}\left(c\right) M_b=0.04kg;M_l=0.12kg\\ {\rho }_l=\frac{0.12}{V_l}\\ F_b=\left(\frac{0.12}{V_l}\right)\left(\frac{V_l}{4}\right)g=\frac{0.12}{4}g\\ F_b=0.03kg\end{array}$

$\begin{array}{l}\left(d\right) M_b=0.02kg;M_l=0.08kg\\ {\rho }_l=\frac{0.08}{V_l}\\ F_b=\left(\frac{0.08}{V_l}\right)\left(\frac{V_l}{4}\right)g=\frac{0.08}{4}g\\ F_b=0.02kg\end{array}$

$\begin{array}{l}\left(e\right) M_b=0.03kg;M_l=0.12kg\\ {\rho }_l=\frac{0.12}{V_l}\\ F_b=\left(\frac{0.12}{V_l}\right)\left(\frac{V_l}{4}\right)g=\frac{0.12}{4}g\\ F_b=0.03kg\end{array}$.

Conclusion:

Thus, the buoyant forces will be arranged as:

$0.05>0.0375>0.03>0.02$

$(a)>(b)>(c)=(e)>(d)$.