Physics - Fluid Question with Solution | TestHub

(A) Ice alone:
$$V_D=\frac{m_i}{{\rho }_w}$$
$$V_F=\frac{m_i}{{\rho }_w}$$
The level remains the same (R).

(B) Ice with steel:
$$V_D=\frac{m_i}{{\rho }_w}+\frac{m_s}{{\rho }_w}$$
$$V_F=\frac{m_i}{{\rho }_w}+\frac{m_s}{{\rho }_s}$$
$${\rho }_s>{\rho }_w\Rightarrow V_F<V_D$$
The level decreases (Q).

(C) Ice with wood:
$$V_D=\frac{m_i}{{\rho }_w}+\frac{m_{\mathrm{wood}}}{{\rho }_w}$$
After melting, the wood floats:
$$m_{\mathrm{wood}}g={\rho }_{w}Vg$$
$$V_F=\frac{m_i}{{\rho }_w}+\frac{m_{\mathrm{wood}}}{{\rho }_w}$$
$$V_F=V_D$$
The level remains the same (R).