Question

1. A hollow steel sphere, weighing 200 kg is floating
on water. A weight of 10 kg is to be placed on it
in order to submerge when temperature is 20°C.
How much less weight (in kg) is to be placed when
temperature increases to 25°C?
(/water = 1.5 * 10-41°C, Osteel = 1 * 10-5°C)​

Answer 1

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.