Physics Q95 – AP-EAMCET 2026

Home Physics Mechanical Properties of FluidsQuestion #252

An iceberg floats on sea water of density

ρ = 1.03 \times 10^{3} {kg m}^{- 3}

. The percentage of total volume of ice above the surface of water is about (Density of ice

σ = 0.92 \times 10^{3} {kg m}^{- 3}

)

Clue 1 of 3: Archimedes' Principle

For a floating object, the buoyant force acting on it is equal to the weight of the object itself.

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Step 1: Apply Archimedes' Principle for Floating Objects✦ Active

When an object floats, its total weight is balanced by the buoyant force exerted by the fluid. The weight of the iceberg ( $W_{i c e}$ ) is given by $W_{i c e} = m_{i c e} g = ρ_{i c e} V_{t o t a l} g$ , where $ρ_{i c e}$ is the density of ice and $V_{t o t a l}$ is the total volume of the iceberg. The buoyant force ( $F_{B}$ ) is given by $F_{B} = ρ_{w} V_{s u b m e r g e d} g$ , where $ρ_{w}$ is the density of seawater and $V_{s u b m e r g e d}$ is the volume of the iceberg submerged in water. Since the iceberg is floating, the weight of the iceberg must be equal to the buoyant force ( $W_{i c e} = F_{B}$ ). Substituting the expressions for weight and buoyant force:

ρ_{i c e} V_{t o t a l} g = ρ_{w} V_{s u b m e r g e d} g

The acceleration due to gravity ( $g$ ) cancels out from both sides:

ρ_{i c e} V_{t o t a l} = ρ_{w} V_{s u b m e r g e d}

💡 Teacher's Secret Hint

Remember that 'g' (acceleration due to gravity) cancels out, which simplifies the calculation significantly. This means the result is independent of the gravitational field.

Step 2: Calculate the Fraction of Volume Submerged○ Expand

From the equality $ρ_{i c e} V_{t o t a l} = ρ_{w} V_{s u b m e r g e d}$ , we can find the ratio of the submerged volume to the total volume:

\frac{V_{s u b m e r g e d}}{V_{t o t a l}} = \frac{ρ_{i c e}}{ρ_{w}}

Given the density of ice $ρ_{i c e} = 0.92 \times 10^{3} {kg m}^{- 3}$ and the density of sea water $ρ_{w} = 1.03 \times 10^{3} {kg m}^{- 3}$ . Substitute these values:

\frac{V_{s u b m e r g e d}}{V_{t o t a l}} = \frac{0.92 \times 10^{3}}{1.03 \times 10^{3}} = \frac{0.92}{1.03} \approx 0.8932

This means approximately $89.32 %$ of the iceberg's volume is submerged below the water surface.

Step 3: Calculate the Percentage of Volume Above the Surface○ Expand

The fraction of the iceberg's volume that is above the water surface ( $V_{a b o v e}$ ) is the total volume minus the submerged volume, divided by the total volume:

\frac{V_{a b o v e}}{V_{t o t a l}} = 1 - \frac{V_{s u b m e r g e d}}{V_{t o t a l}}

Using the calculated ratio from Step 2:

\frac{V_{a b o v e}}{V_{t o t a l}} = 1 - \frac{0.92}{1.03} = \frac{1.03 - 0.92}{1.03} = \frac{0.11}{1.03}

Now, perform the division and convert to a percentage:

\frac{V_{a b o v e}}{V_{t o t a l}} \approx 0.106796

To express this as a percentage, multiply by $100\%:

Percentage above surface = 0.106796 \times 100 % \approx 10.68 %

Rounding to one decimal place, the percentage of the iceberg's volume above the surface is approximately $10.7 %$ . This matches option 1.

💡 Teacher's Secret Hint

Be careful with rounding until the final step to maintain accuracy. Always convert the fraction to a percentage by multiplying by $100 %$ . If the problem asks for 'about', a slight rounding is expected.

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Physics Question 95 – AP-EAMCET 2026

Clue 1 of 3: Archimedes' Principle

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