Physical Quantities and Measurement

Step 1: Convert mass to grams: $3.9\text{ kg} = 3900\text{ grams}$.
• (a) Density in $\text{g/cm}^3$:

$$d = \frac{M}{V} = \frac{3900\text{ g}}{500\text{ cm}^3} = \mathbf{7.8\text{ g/cm}^3}$$

• (b) Density in $\text{kg/m}^3$:

$$d = 7.8 \times 1000 = \mathbf{7,800\text{ kg/m}^3}$$

• Relative Density ($RD$):

$$RD = \frac{7.8\text{ g/cm}^3}{1.0\text{ g/cm}^3} = \mathbf{7.8}$$

(No units!).

• Relative Density is defined as the ratio of the density of a substance to the density of pure water at $4^{\circ}\text{C}$:

$$RD = \frac{\text{Density of Substance}}{\text{Density of Water}}$$

• Because both numerator and denominator have the exact same units (either $\text{g/cm}^3$ or $\text{kg/m}^3$), the units cancel out completely.
• Therefore, Relative Density is a pure numerical ratio (dimensionless) with no physical units.

Step 1: Apply the Law of Floatation fraction formula:

$$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{water}}}$$

Step 2: Substitute $\rho_{\text{body}} = 0.8\text{ g/cm}^3$ and $\rho_{\text{water}} = 1.0\text{ g/cm}^3$:

$$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{0.8}{1.0} = \frac{8}{10} = \mathbf{\frac{4}{5} \quad (\text{or } 80\%)}$$

• Exactly $\frac{4}{5}$ (or $80\%$) of the wooden block is submerged, while $\frac{1}{5}$ ($20\%$) projects above the water surface.

1. Measure and record the mass of the clean, dry empty density bottle with its stopper: $M_1$.
   2. Fill the bottle completely with the given liquid, insert the capillary stopper (excess liquid overflows through the capillary tube), wipe the outside dry, and measure mass: $M_2$.
   

$$\text{Mass of liquid} = M_2 - M_1$$

3. Empty and rinse the bottle, fill it completely with pure water, insert stopper, wipe dry, and measure mass: $M_3$.

$$\text{Mass of equal volume of water} = M_3 - M_1$$

4. Calculate Relative Density and Density:

$$RD = \frac{M_2 - M_1}{M_3 - M_1} \implies \text{Density} = RD \times 1000\text{ kg/m}^3$$

.

Apply the flotation ratio:

$$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{ice}}}{\rho_{\text{seawater}}} = \frac{0.92}{1.025} \approx \mathbf{0.8975 \approx 90\%}$$

• Because the density of ice is almost $90\%$ that of seawater, approximately $90\%$ of the iceberg remains submerged underwater, while only $10\%$ is visible above the sea surface.
• This massive hidden underwater mass poses extreme hazards to shipping navigation.

• The Law of Floatation states that a floating body displaces a volume of liquid whose weight is exactly equal to the total weight of the floating body.
• Mathematically:

$$\mathbf{\text{Weight of Body } (W) = \text{Weight of Liquid Displaced } (F_B)}$$

• Narrow Stem: A narrow stem increases the sensitivity of the hydrometer: a small difference in liquid density produces a large, easily readable vertical displacement of the liquid meniscus on the scale.
• Weighted Bulb: Lead shots or mercury at the bottom lower the center of gravity well below the center of buoyancy, ensuring the hydrometer floats stably in an upright vertical position without tilting or toppling.

• This is known as the Anomalous Expansion of Water.
• From $0^{\circ}\text{C}$ to $4^{\circ}\text{C}$, instead of expanding, water contracts: its volume decreases, reaching a minimum at $4^{\circ}\text{C}$. Hence, its density increases and reaches a maximum of $1000\text{ kg/m}^3$ at $4^{\circ}\text{C}$.
• Above $4^{\circ}\text{C}$ up to $10^{\circ}\text{C}$, water behaves normally: it expands, and its density decreases continuously.
• This biological anomaly ensures ponds freeze only from the top, allowing aquatic life to survive in $4^{\circ}\text{C}$ liquid water at the bottom.