An ice cube of size l = 50 cm is floating in a tank (base area A = 2m x 2m) partially filled with water. Density of water is ρw=1000kgm−3 and that of ice is ρice =900kg m−3. Calculate change increase in gravitational potential energy (in J) when ice melts completely. (g = 10 ms-2)
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 28.125.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
As the density of the ice ρice is less than the density of the water ρwater , hence the ice cube float partially immersed.In case of floatation Buoyancy force = weight of cube B=mg or l2h0ρwater g=l3ρice g⇒ h0=l×ρice ρwater =50×9001000=45cmCentre of mass of ice cube is at a depth of 10 cm from its upper surface. But the upper surface is at a height of 5 cm from water surface. Therefore, depth of the centre of the mass of ice cube from free water surface is h1=25−5=20cm from free water surface. When ice melts, water is formed and that water occupies the space which was displaced by the cube. Therefore, depth of centre of mass of the water body formed after the melting of ice cube is h2=452=22.5cmit means that the centre of mass move down through distance Δh=h2−h1=22.5−20=2.5cmIt means gravitational potential energy of the system will decrease. Decrease in gravitational potential energy of the system ΔU=mgΔh=(0.5)3×900×10×2.5100=28.125JSince, level of free water surface does not rise or fall, hence size of tank is of no concern.