Problem 103P

A thief uses a bag of sand to replace a gold statue that sits on a weight-sensitive, alarmed pedestal. The bag of sand and the statue have exactly the same volume, 1.75 L. (Assume that the mass of the bag is negligible.)

(a) Calculate the mass of each object, (density of gold = 19.3 g/cm3; density of sand = 3.00 g/cm3)

(b) Did the thief set off the alarm? Explain.

Problem 103P

A thief uses a bag of sand to replace a gold statue that sits on a weight-sensitive, alarmed pedestal. The bag of sand and the statue have exactly the same volume, 1.75 L. (Assume that the mass of the bag is negligible.)

(a) Calculate the mass of each object, (density of gold = 19.3 g/cm3; density of sand = 3.00 g/cm3)

(b) Did the thief set off the alarm? Explain.

Solution: Here, we are going to calculate the mass of sand and gold given in the problem.

Step1:

Density of any substance may be defined as mass per unit volume. If ‘d’ be the density, ‘m’ be the mass and ‘v’ be the volume of any substance, then, numerically,

d = -------(1)

Step2:

- Given, density of gold = 19.3 g/cm3

Volume of gold statue = 1.75 L = 1.75 x 1000 cm3 [1 L = 1000 cm3]

= 1750 cm3

Substituting the values in equation (1), we get,

19.3 g/cm3 = m / 1750 cm3

m = 19.3 x 1750 [(g/cm3) x cm3]

m = 33775 g

m = 3.38 x 104 g [rounding off up to 3 significant figures]

Thus, mass of the gold statue is 3.38 x 104 g.

Step3:

Given, density of sand = 3.00 g/cm3

Volume of the bag of sand = 1.75 L = 1.75 x 1000 cm3 [1 L = 1000 cm3]

= 1750 cm3

Substituting the values in equation (1), we get,

3.00 g/cm3 = m / 1750 cm3

m = 3.00 x 1750 [(g/cm3) x cm3]

m = 5250 g

m = 5.25 x 103 g [rounding off up to 3 significant figures]

Thus, mass of the bag of sand is 5.25 x 103 g.

Step4:

b) No, the thief didn't set off the alarm. As the mass of the gold statue and the bag of sand is different, the alarm clock will start ringing once the statue is replaced with the bag of sand. Thus, the thief screwed up the operation.

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