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Chapter 15: Problem 32

A steel drum has volume \(0.23 \mathrm{m}^{3}\) and mass \(16 \mathrm{kg} .\) Will it float in water when filled with (a) water or (b) gasoline (density \(\left.860 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

Short Answer

Expert verified
Yes, the steel drum will float in both water and gasoline.

Step by step solution

01

Derive the formula for floating

According to the principle of buoyancy, an object will float if the weight of the fluid it displaces is equal to or greater than the weight of the object. Mathematically, this can be expressed as \( \rho_{f} \times V \times g \geq m \times g \), where \( \rho_{f} \) is the density of the fluid, \( V \) is the volume of the object, \( g \) is the gravitational acceleration, and \( m \) is the mass of the object.
02

Establish the given quantities

Given that the steel drum weighs 16 kg and its volume is 0.23 cubic meters. The density of water is given as 1000 kg/m³, and that of gasoline is 860 kg/m³. The gravitational acceleration g is approximately 9.8 m/s².
03

Calculate for water

Substitute the given values into the equation derived in step 1, using the density of water: \( 1000 \times 0.23 \times 9.8 \geq 16 \times 9.8 \). Simplifying, we find that the steel drum will float in water as 2254 ≥ 156.8.
04

Calculate for gasoline

Substitute the given values into the equation derived in step 1, this time using the density of gasoline: \( 860 \times 0.23 \times 9.8 \geq 16 \times 9.8 \). Simplifying, we find that the steel drum will also float in gasoline as 1945.4 ≥ 156.8.

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