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Chapter 15: Problem 30

A 5.4 -g jewel has apparent weight 32 mN when submerged in water. Could the jewel be a diamond (density \(3.51 \mathrm{g} / \mathrm{cm}^{3}\) )?

Short Answer

Expert verified
No, the jewel could not be a diamond.

Step by step solution

01

Calculate the volume of the jewel

First, calculate the volume of the jewel in cubic centimeters (cm^3) using the known mass (5.4 g) and the density of a diamond (3.51 g/cm^3). Use the formula for volume \(V = m / p\), where \(m\) is mass and \(p\) is density. So, the volume of the jewel is \(V = 5.4 g / 3.51 g / cm^3 = 1.54 cm^3\)
02

Determine the weight of an equivalent volume of water

Because the buoyant force is equal to the weight of the fluid displaced by the object, we need to figure out the weight of the water (density = 1 g/cm^3) displaced by the jewel. To do this, simply multiply the volume of the jewel by the density of water and the acceleration due to gravity (9.8 m/s^2). The weight of the water displaced is therefore \(1.54 cm^3 * 1 g/cm^3 * 9.8 m/s^2 = 15.1 mN\)
03

Compare the apparent weight with the weight of the displaced water

Finally, compare the calculated weight of the displaced water (15.1 mN) with the given apparent weight of the jewel (32 mN). If the apparent weight is greater than the weight of the displaced water, the jewel could be a diamond. In this case, however, the apparent weight (32 mN) is much greater than the weight of the displaced water (15.1 mN), which suggests that the jewel is not a diamond because a diamond of the given dimensions would displace less water and therefore have a lower apparent weight.

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