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A hollow steel sphere ,weighing 200kg is floating on water.A weight 10 kg is to be placed on it in order to submerge when the temperature is 20°C. How much less weight is to be placed when temperature increases to 25°C? γ water = 1.5 * 10 -4 α steel =1 * 10 -5

A hollow steel sphere ,weighing 200kg is floating on water.A weight 10 kg is to be placed on it in order to submerge when the temperature is 20°C. How much less weight is to be placed when temperature increases to 25°C?


γwater= 1.5 * 10-4            αsteel=1 * 10-5

Grade:

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear gaurav

γwater= 1.5 * 10-4           

αsteel=1 * 10-5

γsteel = 3 * 10-5

let initial volume of sphere is Vo

and initial density of water in do

so initiall Vo volume of water replaced by sphere

so Vo do g = (200 + 10)g

    Vo do   = 210 ....................1

when temperature incresed

volume of sphere changes to V =Vo(1+γsteel *5)

density of water changes  to d = do(1-γwater *5)

let now w weight is placed

Vdg = (200 +w)g

Vd =200+w

Vo(1+γsteel *5)  X do(1-γwater *5)  =200 + w

210 (1+γsteel *5)  (1-γwater *5)  = 200 +w

210 (1+1.5 * 10-4(1-7.5 *10-4)  = 200 +w

  210( 1 -6 *10-4) = 200 +w

 10 -.1260 = w

w =9.874 kg

 difference in weight = 10 -w

                              = .1260 kg

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