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Chapter 9: Problem 84

A stone of weight \(W\) has specific gravity \(2.50 .\) (a) When the stone is suspended from a scale and submerged in water, what is the scale reading in terms of its weight in air? (b) What is the scale reading for the stone when it is submerged in oil (specific gravity \(=0.90\) )?

Short Answer

Expert verified
(a) The scale reading for the stone submerged in water is \(W_\text{apparent, water} = \frac{1.5W}{2.5}\). (b) The scale reading for the stone submerged in oil is \(W_\text{apparent, oil} = \frac{11W}{25}\).

Step by step solution

01

Find the stone's volume

Using the given specific gravity of the stone (2.50), we can find its density: \(d_\text{stone} = 2.50 \times d_\text{water} = 2.50 \times 1000 \ \text{kg}/\text{m}^3 = 2500 \ \text{kg}/\text{m}^3\). Now, we can find the stone's volume using its weight and density: \(V = \frac{W}{d_\text{stone} \times g} = \frac{W}{2500 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2}\).
02

Calculate the buoyant force in water

In water, the buoyant force can be found using the stone's volume and the density of water: \(F_\text{buoyant, water} = V \times d_\text{water} \times g = \frac{W}{2500 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2} \times 1000 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2 = \frac{W}{2.5}\).
03

Calculate the apparent weight in water

To find the apparent weight of the stone in water, we can subtract the buoyant force from its actual weight: \(W_\text{apparent, water} = W - F_\text{buoyant, water} = W - \frac{W}{2.5} = \frac{2.5W - W}{2.5} = \frac{1.5W}{2.5}\).
04

Calculate the buoyant force in oil

For the stone submerged in oil, we need to use the given specific gravity of the oil (0.9) to find its density: \(d_\text{oil} = 0.9 \times d_\text{water} = 0.9 \times 1000 \ \text{kg}/\text{m}^3 = 900 \ \text{kg}/\text{m}^3\). Now, calculate the buoyant force in oil: \(F_\text{buoyant, oil} = V \times d_\text{oil} \times g = \frac{W}{2500 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2} \times 900 \ \text{kg}/\text{m}^3 \times 9.81 \ \text{m}/\text{s}^2 = \frac{W}{2.5} \times \frac{9}{10}\).
05

Calculate the apparent weight in oil

To find the apparent weight of the stone in oil, subtract the buoyant force from its actual weight: \(W_\text{apparent, oil} = W - F_\text{buoyant, oil} = W - \frac{W}{2.5} \times \frac{9}{10} = \frac{W}{2.5} + \frac{W}{25} = \frac{W(10 + 1)}{25}\). (a) The scale reading for the stone submerged in water is \(W_\text{apparent, water} = \frac{1.5W}{2.5}\). (b) The scale reading for the stone submerged in oil is \(W_\text{apparent, oil} = \frac{11W}{25}\).

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