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

Dear gaurav

γ_water= 1.5 * 10^-4           

_^αsteel=1 * 10^-5

γ_steel = 3 * 10^-5

let initial volume of sphere is V_o

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+γ_{steel *5)}

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

let now w weight is placed

Vdg = (200 +w)g

Vd =200+w

Vo(1+γ_{steel *5)  X} do(1-γ_water *5)  =200 + w

210 (1+γ_{steel *5)}  (1-γ_water *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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