PHYSICS:F/SCI.+ENGRS.,V.1
PHYSICS:F/SCI.+ENGRS.,V.1
10th Edition
ISBN: 9781337553575
Author: SERWAY
Publisher: CENGAGE L
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Chapter 1, Problem 13P
To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

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Answer to Problem 13P

The radius of a solid aluminum sphere is 2.86cm that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is 2.70×103kg and the mass of one cubic meter iron is 7.86×103kg. The radius of the solid iron sphere is 2.0cm.

Write the formula to calculate the density of aluminum sphere

    ρalu=m1V1

Here, ρalu is the density of the aluminum sphere, m1 is the mass of one cubic meter aluminum sphere and V1 is the volume of the aluminum sphere.

Substitute 2.70×103kg for m1 and 1.0m3 for V1 to find ρalu.

    ρalu=2.70×103kg1.0m3=2.70×103kg/m3

Write the formula to calculate the density of iron sphere

    ρiron=m2V2

Here, ρiron is the density of the iron sphere, m2 is the mass of one cubic meter iron sphere and V2 is the volume of the iron sphere.

Substitute 7.86×103kg for m1 and 1.0m3 for V1 to find ρiron.

    ρiron=7.86×103kg1.0m3=7.86×103kg/m3

Write the formula to calculate the mass of an aluminum sphere

    malu=ρaluValu

Here, malu is the mass of the aluminum sphere, Valu is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

    Valu=43πralu3

Here, ralu is the radius of the aluminum sphere.

Substitute 43πralu3 for Valu.

    malu=ρalu43πralu3=43πralu3ρalu                                                    (I)

Write the formula to calculate the mass of a solid iron sphere of radius 2.0cm

    miron=ρironViron

Here, miron is the mass of the aluminum sphere, Viron is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

    Viron=43πriron3

Here, riron is the radius of the iron sphere.

Substitute 43πriron3 for Viron.

    miron=ρiron43πriron3=43πriron3ρiron                                                    (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

    malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3riron

Conclusion:

Substituting 2.0cm for riron, 7.86×103kg/m3 for ρiron and 2.70×103kg/m3 for ρalu in the above equation to find ralu.

    ralu=(7.86×103kg/m32.70×103kg/m33)(2.0cm)=1.427×2.0cm=2.86cm

Therefore, the radius of a solid aluminum sphere is 2.86cm that balance a solid iron sphere on an equal arm balance.

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