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An object weighs 306 N in the air. When tied to a string, and immersed in water, it...

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celestabrown | Student, Grade 11 | (Level 1) eNoter

Posted February 3, 2012 at 1:38 AM via web

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An object weighs 306 N in the air. When tied to a string, and immersed in water, it weighs 227 N. When it is immersed in oil, it weighs 260 N. Find the density of the object and the density of the oil.

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najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted February 3, 2012 at 12:59 PM (Answer #1)

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Weight of object in the air = 306 N

Mass of the object = 306/9.8 = 31.224 Kg

Loss of Weight in Water = 306-227 = 79 Newton

Weght of Water Displaced = 79 Newton

Mass of 79 Kg of Water = 79/9.8 = 8.061 Kg

Density of Water = 1000 Kg/M3

Volume of Water Displaced = 8.061/1000 = 8.061*10^(-3) M3

Volume of the object = 8.061*10^(-3) M3

Therefore Density of the object = 31.224/{8.061*10^(-3)}

= 3873 Kg/M3

Loss of Weght in Oil = 46 N

Mass of Oil displaced = 46/9.8 = 4.694 Kg

Volume of Oil displaced = Volume of the object = 8.061*10^(-3) M3

Therfore Density of Oil = 4.694/{8.061*10^(-3)} = 582 Kg/M3

=

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 3, 2012 at 4:18 AM (Answer #2)

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The weight of the object in air is 306 N. Its mass is 31.22 kg. Its weight when it is immersed in water is 227 N. When it is immersed in an oil the weight decreases to 260 N.

To determine the density of the object we have to know the density of water. That is equal to 1000 kg/m^3. When an object is immersed in a fluid its weight is decreased by an amount equal to the weight of the fluid it displaces. Let the volume of the object be V. When immersed in water the mass of water displaced is 1000*V. This is equal to the decrease in the mass of the object: 1000*V = 8.06.

=> V = 8.06*10^-3 m^3

The density of the object is 31.22/8.06*10^-3 = 3.87*10^3 kg/m^3.

The object immersed in the oil loses a weight equal to 46 N or 46/9.8 = 4.69 kg.

The density of the oil is 4.69/8.06*10^-3 = 0.582*10^3 kg/m^3.

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