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New York University Archimedes Principle Density Lab Questions

New York University

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“Archimedes’ Principle

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Archimedes’ Principle. Background knowledge
Questions’ Solutions

1.
The sphere is fully submerged in Mercury, so the volume of displaced Mercury equals the volume of
a sphere of radius 𝑅 = 2 𝑐𝑚 = 0.02 𝑚, so;
4
4
𝑉𝐷𝑖𝑠𝑝 = 𝜋𝑅 3 = 𝜋(0.02 𝑚)3 = 3.351 × 10−5 𝑚3
3
3
The mass of displaced Mercury is:
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝑀𝑒𝑟𝑐𝑢𝑟𝑦
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝐷𝑖𝑠𝑝
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔/𝑐𝑚3 ) × (3.351 × 10−5 𝑚3 )
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔⁄𝑐𝑚3 ) × (33.51 𝑐𝑚3 ) = 459.09 𝑔 = 0.45909 𝑘𝑔
Then;
𝑊𝑒𝑖𝑔ℎ𝑡𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑔 = (0.45909 𝑘𝑔) × (9.80 𝑚⁄𝑠 2 ) = 4.4991 𝑁

2.
Half of the Gold sphere is submerged in Mercury, so the volume of displaced Mercury is;
1 4
4
𝑉𝐷𝑖𝑠𝑝 = ( 𝜋𝑅 3 ) = 𝜋(0.02 𝑚)3 = 1.6755 × 10−5 𝑚3
2 3
6
The mass of displaced Mercury is:
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝑀𝑒𝑟𝑐𝑢𝑟𝑦
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝜌𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑉𝐷𝑖𝑠𝑝
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔/𝑐𝑚3 ) × (1.6755 × 10−5 𝑚3 )
𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = (13.7 𝑔⁄𝑐𝑚3 ) × (16.755 𝑐𝑚3 ) = 229.54 𝑔 = 0.22954 𝑘𝑔
Then;
𝑊𝑒𝑖𝑔ℎ𝑡𝑀𝑒𝑟𝑐𝑢𝑟𝑦 = 𝑀𝑀𝑒𝑟𝑐𝑢𝑟𝑦 × 𝑔 = (0.22954 𝑘𝑔) × (9.80 𝑚⁄𝑠 2 ) = 2.2495 𝑁
𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 = 2.2495 𝑁

3.
The maximum possible water volume to be displaced by the box is;
𝑉𝐷𝑖𝑠𝑝 = 𝑉𝐵𝑜𝑥 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡
𝑉𝐷𝑖𝑠𝑝 = (2 𝑚) × (2 𝑚) × (0.5 𝑚) = 2.0 𝑚3
The mass of displaced Water is:
𝑀𝑊𝑎𝑡𝑒𝑟 = 𝜌𝑊𝑎𝑡𝑒𝑟 × 𝑉𝑊𝑎𝑡𝑒𝑟
𝑀𝑊𝑎𝑡𝑒𝑟 = 𝜌𝑊𝑎𝑡𝑒𝑟 × 𝑉𝐷𝑖𝑠𝑝
𝑀𝑊𝑎𝑡𝑒𝑟 = (1000 𝑘𝑔⁄𝑚3 ) × (2.0 𝑚3 ) = 2000 𝑘𝑔
Then;
𝑊𝑒𝑖𝑔ℎ�... University of Maryland    Review Anonymous
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