Solution:

2.50×103 kg/m3 and 880 kg/m3.

Given data:

The mass of metal piece in air is 5.00 g.

The mass of metal piece in water is 3.00 g.

The mass of metal piece in benzene is 3.24 g.

Formula used:

The density of a body is calculated using the formula:

ρ=mV

Here, V is the volume of the body, ρ is the density, and m is the mass.

Archimedes’ principle states that the apparent upward force experienced by the solid immersed in a fluid is equal to the weight of the fluid displaced by it. This upward buoyant force on the solid is expressed as

FB=ρlVSg

Here, FB is the buoyant force, ρl is the density of the liquid in which the solid is submerged, VS is the volume of the solid submerged in the liquid, and g is the acceleration due to gravity.

The weight of a body is expressed as

W=mg

Here, W is the weight of the body.

The apparent weight of a body submerged in a liquid is calculated as

WA=W−FB

Here, WA is the apparent weight of the body in the liquid.

Explanation:

Consider the situation when the metal piece is submerged in water and draw its free body diagram:

Here, W is the force exerted due to the weight of the metal piece and FB1 is the buoyant force on the metal piece exerted by water.

Recall the expression for the weight of the metal piece:

W=mg

Consider, the standard value of acceleration due to gravity as 9.81 m/s2.

Substitute 5.00 g for m and 9.81 m/s2 for g

W=(5.00 g)(9.81 m/s2)=(5.00 g)(0.001 kg1 g)(9.81 m/s2)=0.04905 N

Recall the expression for buoyant force exerted by water on the metal piece:

FB1=ρwVSg

Here, ρw is the density of water.

Consider the standard value of density of water as 1000 kg/m3.

Substitute 1000 kg/m3 for ρw and 9.81 m/s2 for g

FB1=(1000 kg/m3)VS(9.81 m/s2)=9810VS

Write the expression for the apparent weight of the metal piece in water in terms of apparent mass:

WA1=mA1g

Here, WA1 is the apparent weight of themetal piece in water and mA1 is the apparent mass of the metal piece in water.

Substitute 3.00 g for mA1 and 9.81 m/s2 for g

WA1=(3.00 g)(9.81 m/s2)=(3.00 g)(0.001 kg1 g)(9.81 m/s2)=0.02943 N

Recall the expression for the apparent weight of the metal piece in water, in terms of buoyant force:

WA1=W−FB1

Substitute 0.02943 N for WA1, 9810VS for FB1, and 0.04905 N for W

0.02943 N=0.04905 N−9810VS9810VS=0.04905 N−0.02943 N9810VS=0.01962 NVS=0.01962 N9810

Further solve as

VS=2×10−6 m3

It is understood that the metal piece is completely submerged in water, therefore, the expression for calculation of the volume of the metal piece submerged in water becomes:

V=VS

Substitute 2×10−6 m3 for VS

V=2×10−6 m3

Recall the expression for the density of the metal piece:

ρ=mV

Substitute 2×10−6 m3 for V and 5.00 g for m

ρ=5.00 g2×10−6 m3=5.00 g(0.001 kg1 g)2×10−6 m3=2.50×103 kg/m3

Thus, the density of the metal piece is 2.50×103 kg/m3.

Consider the situation when themetal piece is submerged in benzene and draw its free body diagram:

Here, FB2 is the buoyant force on the metal piece, exerted by benzene.

Recall the expression for buoyant force exerted by benzene, on the metal piece:

FB2=ρlVSg

Here, ρl is the density of benzene.

Substitute 2×10−6 m3 for VS and 9.81 m/s2 for g

FB2=ρl(2×10−6 m3)(9.81 m/s2)=(19.62×10−6)ρl

Write the expression for apparent weight of themetal piece in benzene in terms of apparent mass:

WA2=mA2g

Here, WA2 is the apparent weight of themetal piece in benzene and mA2 is the apparent mass of themetal piece in benzene.

Substitute 9.81 m/s2 for g and 3.24 g for mA2

WA2=(3.24 g)(9.81 m/s2)=(3.24 g(0.001 kg1 g))(9.81 m/s2)=0.03178 N

Recall the expression for the calculation of the apparent weight of the metal piece in benzenein terms of buoyant force:

WA2=W−FB2

Substitute 0.03178 N for WA1, (19.62×10−6)ρl for FB2, and 0.04905 N for W

0.03178 N=0.04905 N−(19.62×10−6)ρl(19.62×10−6)ρl=0.04905−0.03178(19.62×10−6)ρl=0.01727ρl=0.01727(19.62×10−6)

Further solve as

ρl≃880 kg/m3

The density of benzene is 880 kg/m3.

Conclusion:

Thus, thedensity of the metal pieceis 2.50×103 kg/m3 and the density of benzene is 880 kg/m3.