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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?
γwater= 1.5 * 10-4            αsteel=1 * 10-5```
7 years ago

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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