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gaurav pandvia Grade:
        

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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

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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7 years ago
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