CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A solid of density 5000Kg/m3 weight 0.5Kg×g in air. It is completely immersed in water of density 1000Kg/m3. Calculate the apparent weight of the solid in water.


Open in App
Solution

Step1: Given data

  1. The density of the solid is ρs=5000Kg/m3.
  2. The weight of the solid is Ws=0.5Kg×g
  3. The density of water is ρw=1000Kg/m3

Step2: The apparent weight of a body in a solid

  1. Volume is the ratio of mass and density, i.e, V=mρ, where, m is the mass of the body and ρ is the density of the body.
  2. We know the apparent weight of a body in a liquid is, Wa=Actualweightofthebody-buoyancyforceonthebody.
  3. The buoyancy force exerted on a body in a liquid is equal to the weight of the displaced water due to the body. Buoyancy is defined by the form, Fb=V×ρwg, where, V is the volume of the body and ρw is the density of the liquid, and g is the acceleration due to gravity.

Step3: Diagram

Selina Solutions For Class 9 Physics Chapter 5 Upthrust in Fluids,  Archimedes' Principle and Floatation

Step4: Finding the buoyancy force

The volume of the solid is

V=mρ=0.55000=10000orV=10-4m3................(1)

The buoyancy force exerted on the body when it is in the water is

Fb=V×ρwg=10-4×1000KggorFb=0.1Kgg.............(2)

Step5: Finding the apparent weight

As we know, Wa=Actualweightofthebody-buoyancyforceonthebody

From equations 1 and 2 we get,

The apparent weight of the body is

Wa=0.5Kgg-0.1KggorWa=0.4Kgg

Therefore, the apparent weight of the solid in water 0.4Kgg.


flag
Suggest Corrections
thumbs-up
3
BNAT
mid-banner-image