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A solid of density 2700 kgm’ and of volume 0.0015 m3 ¡s completely immersed in alcohol of density 800 kgm’.Calculate: - Physics

Sum

A solid of density 2700 kgm’ and of volume 0.0015 m3 ¡s completely immersed in alcohol of density 800 kgm’.Calculate:

  1. Weight of solid in SI system.
  2. Upthrust on solid in SI system.
  3. Apparent weight of solid in alcohol.
  4. Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, [g = 10 ms-2]
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Solution

Density of solid = ρ= 2700 kgm3
Volume of solid = V = 0.0015 m3
Density of alcohol = ρ’ = 800 kgm3

  1. Mass of solid = m = V x p
    m = 0.0015 x 2700 = 4.05 kg
    Weight of solid = mg = 4.05 x 10 = 40.5 N
  2. Volume of alcohol displaced = Volume of solid
    V = 0.0015 m3
    Mass of alcohol displaced = m’ = V x p’
    m’ = 0.0015 x 800 = m’ = 1.2 kg
    Upthrust = Weight of alcohol displaced
    = m’g= 1.2 x 10= 12N
  3. Apparent weight of solid in alcohol
    = Actual weight of solid – Upthrust
    = 40.5 -12 = 28.5 N
  4. When a solid is immersed completely in brine solution, then upthrust acts on it in an upward direction, as a result, its apparent weight of solid will be less than actual weight of solid will be less than the actual weight of solid.
Concept: Archimedes' Principle
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APPEARS IN

Goyal Brothers Prakashan Class 9 A New Approach to ICSE Physics Part 1
Chapter 5 Upthrust and Archimedes’ Principle
Practice Problem 1 | Q 1
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