Question

question 7 (1 point)
a spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (de
13.6 g/cm3). which answer best describes the result?

a) the lead ball will float with its top exactly even with the surface of the
mercury.

b) the lead will sink to the bottom of the mercury.

c) the lead ball will float with about 83% of its volume above the surface of
mercury.

d) the lead ball will float with about 17% of its volume above the surface of th
mercury.

Explanation:

Step1: Recall Archimedes' Principle

Archimedes' Principle states that the buoyant force on an object is equal to the weight of the fluid displaced. For a floating object, the weight of the object equals the weight of the displaced fluid. Let \(
ho_{lead} = 11.3\ \text{g/cm}^3 \), \(
ho_{mercury}=13.6\ \text{g/cm}^3 \), \( V_{total} \) be the volume of the lead ball, and \( V_{submerged} \) be the volume of the lead ball submerged in mercury. Then \(
ho_{lead} V_{total} g=
ho_{mercury} V_{submerged} g \), so \( \frac{V_{submerged}}{V_{total}}=\frac{
ho_{lead}}{
ho_{mercury}} \).

Step2: Calculate Submerged Volume Fraction

Substitute the densities: \( \frac{V_{submerged}}{V_{total}}=\frac{11.3}{13.6}\approx0.8309 \). This means the submerged volume is about 83% of the total volume. Therefore, the volume above the surface is \( 1 - 0.8309 = 0.1691\approx17\% \).

Step3: Analyze Options

• Option a: Incorrect, because the lead's density is less than mercury's, but it won't have the top even with the surface (that would require \(

ho_{lead}=
ho_{mercury} \)).

• Option b: Incorrect, because \(

ho_{lead}<
ho_{mercury} \), so it will float, not sink.

• Option c: Incorrect, the submerged volume is ~83%, so above is ~17%, not 83%.
• Option d: Correct, as calculated, the volume above the surface is about 17%.

Answer:

d) The lead ball will float with about 17% of its volume above the surface of the mercury.