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You are provided with a hollow iron ball A of volume 15 cm3 and mass 12 g and a solid iron ball B of mass 12 g. Both are placed on the surface of water contained in a large tub. - Physics

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Question

You are provided with a hollow iron ball A of volume 15 cm3 and mass 12 g and a solid iron ball B of mass 12 g. Both are placed on the surface of water contained in a large tub.

  1. Find upthrust on each ball.
  2. Which ball will sink? Give a reason for your answer (Density of iron = 8.0 g cm-3) 
Sum

Solution

(a) Volume of hollow iron ball A = 15 cm3

Mass of hollow iron ball A = 12 g

Mass of solid iron ball B = 12 g

Density of iron = `8.0  "g cm"^-3`

Upthrust due to liquid = volume of the solid × density of fluid × acceleration due to gravity           ...[Equation 1]

Volume = `"Mass"/"density"`           ...[Equation 2]

Substituting the values in equation 2, to find volume of solid iron ball B,

Volume of solid iron ball B = `12/8` = 1.5 cm3

Substituting the values in equation 1 to get upthrust on hollow ball A

Upthrust on hollow ball A = 15 × 1 × g = 15 gf

Hence, Upthrust on hollow ball A = 15 gf

Substituting the values in equation 1 to get upthrust on solid iron ball B

Upthrust on solid iron ball B = 1.5 × 1 × g = 1.5 gf

Hence, Upthrust on solid iron ball B = 1.5 gf

(b) Solid iron ball B will sink.

Ball B experiences an upthrust of 1.5 gf in water that is less than it's weight of 12 gf. Hence ball B will sink. Ball A experiences an upthrust of 15 gf that is greater than it's weight of 12 gf. Hence it will float with its that much part submerged for which upthrust becomes equal to its weight of 12 gf.

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Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation - Exercise 5 (A) [Page 111]

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Selina Concise Physics [English] Class 9 ICSE
Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (A) | Q 3 | Page 111
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