Q37 (2026)
A cubical block of density $\rho_{\mathrm{b}} = 600 \, \mathrm{kg/m^{3}}$ floats in a liquid of density $\rho_{\mathrm{e}} = 900 \, \mathrm{kg/m^{3}}$ . If the height of block is $H = 8.0 \, \mathrm{cm}$ then height of the submerged part is ____ cm.
- $7.3$
- $4.3$
- $6.3$
- $5.3$
### Related Formula
$$F_{\text{buoyancy}} = Mg$$
$$\rho_{\text{liquid}} V_{\text{submerged}} g = \rho_{\text{block}} V_{\text{total}} g$$
### Core Logic
For a floating block, the weight of the block equals the buoyant force exerted by the liquid.
$M g = F_b$
$$\rho_b \times A \times H \times g = \rho_e \times A \times h \times g$$
### Step 1: Calculation
$$600 \times 8 \text{ cm} = 900 \times h$$
$$h = \frac{600 \times 8}{900}$$
$$h = \frac{16}{3} \text{ cm}$$
$$h \approx 5.33 \text{ cm}$$
### Pattern Recognition
Submerged fraction is exactly the ratio of densities: $\frac{h}{H} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
