The Amazing World of Solutes, Solvents, and Solutions — NCERT Solutions

Given:

• Mass of the object $= 400\ \text{g}$
• Volume of the object $= 40\ \text{cm}^3$

Formula:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

Calculation:
$$\text{Density} = \frac{400\ \text{g}}{40\ \text{cm}^3} = 10\ \text{g/cm}^3$$

Answer: The density of the object is $\mathbf{10\ g/cm^3}$.

(Note: The figures show an unpeeled orange floating in water and a peeled orange sinking in water.)

Explanation:

An object floats in water if its density is less than the density of water ($1\ \text{g/cm}^3$), and sinks if its density is greater than $1\ \text{g/cm}^3$.

• Unpeeled orange (floats): The peel (skin) of an orange is thick and spongy. It contains a large number of tiny air pockets (pores). These air pockets increase the overall volume of the orange significantly without adding much mass. As a result, the average density of the unpeeled orange (orange flesh + peel with air pockets) becomes less than the density of water, so it floats.

• Peeled orange (sinks): When the peel is removed, the air pockets are gone. The remaining orange flesh is compact and denser. The average density of the peeled orange becomes greater than the density of water, so it sinks.

Conclusion: The peel with its air pockets lowers the overall density of the orange below that of water, causing it to float. Removing the peel removes the air pockets, increasing the average density above that of water, causing it to sink.

Given:

• Object A: Mass $= 200\ \text{g}$, Volume $= 40\ \text{cm}^3$
• Object B: Mass $= 240\ \text{g}$, Volume $= 60\ \text{cm}^3$

Formula:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

Density of Object A:
$$\text{Density}_A = \frac{200\ \text{g}}{40\ \text{cm}^3} = 5\ \text{g/cm}^3$$

Density of Object B:
$$\text{Density}_B = \frac{240\ \text{g}}{60\ \text{cm}^3} = 4\ \text{g/cm}^3$$

Comparison:
$$\text{Density}_A = 5\ \text{g/cm}^3 > \text{Density}_B = 4\ \text{g/cm}^3$$

Answer: Object A is denser than Object B.

Given:

• Mass of modeling clay $= 120\ \text{g}$ (remains constant throughout)
• Volume when moulded into a cube $= 60\ \text{cm}^3$

Density of the compact cube:
$$\text{Density} = \frac{120\ \text{g}}{60\ \text{cm}^3} = 2\ \text{g/cm}^3$$

When flattened into a thin sheet:

• The mass of the clay does not change — it is still $120\ \text{g}$.
• The volume of the clay also does not change — clay is a solid and reshaping it does not compress or expand the material itself, so the volume remains $60\ \text{cm}^3$.

Therefore:
$$\text{Density (thin sheet)} = \frac{120\ \text{g}}{60\ \text{cm}^3} = 2\ \text{g/cm}^3$$

Prediction: The density of the modeling clay will remain the same ($2\ \text{g/cm}^3$) even after flattening, because both mass and volume remain unchanged when the shape is changed. Density is an intrinsic property of a substance and does not depend on its shape.

Given:

• Mass of iron block $= 600\ \text{g}$
• Density of iron $= 7.9\ \text{g/cm}^3$

Formula:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

Rearranging for Volume:
$$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$

Calculation:
$$\text{Volume} = \frac{600\ \text{g}}{7.9\ \text{g/cm}^3} \approx 75.95\ \text{cm}^3$$

Answer: The volume of the iron block is approximately $\mathbf{75.95\ cm^3}$ (or $\approx 76\ \text{cm}^3$).

(Note: The figures show a test tube with a glass tube inserted; when placed in hot water, the water level in the glass tube rises, indicating that the water inside the test tube expands.)

Observation: When the test tube is placed in hot water (~70 °C), the water inside the test tube gets heated and its level rises in the glass tube. This indicates that the volume of water increases (water expands) on heating.

Effect on Density:

Using the formula:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

• On heating, the mass of the water remains the same (no water is added or removed).
• However, the volume increases because water expands on heating.
• Since density $= \dfrac{\text{Mass}}{\text{Volume}}$, and the volume increases while mass stays constant, the density decreases.

Conclusion: When water is heated, its volume increases (as seen by the rising water level in the glass tube), while its mass remains unchanged. Therefore, the density of water decreases with an increase in temperature. This is consistent with the general principle that density decreases with an increase in temperature.