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Physics For Scientists & Engineers
Found in: Page 16

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Short Answer

Let ρAl represent the density of aluminium and ρFe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius ρFe on an equal-arm balance.

The radius of aluminum sphere is rFeρFeρAl1/3.

See the step by step solution

Step by Step Solution

Step 1: Identify given data

The density of aluminium is ρAl

The mass of iron is ρFe.

Step 2: Determine radius of aluminum sphere

The density is the ratio of the mass of an object tothe unit volume of the space. It describes the heavier or lighter among the different objects.

The volume of the aluminum sphere is given by

VAl=43πrAl3

The volume of the iron sphere is given by

VFe=43πrFe3

The condition for the equal-arm balanceis given by

ma=miρAlVAl=ρFeVFe

Substitute all the values in the above equation.

role="math" localid="1663665328044" ρAl43πrAl3=ρFe43πrFe3rAl=rFeρFeρAl1/3

Therefore, the radius of the aluminum sphere that satisfies the equal-armbalance criterion is rFeρFeρAl1/3.

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