A metal block of density `5000kg//m^(3)` and mass `2kg` in suspended by a spring of force constant `200 N//m`. The spring block system is submerged in water vessel. Total mass of water in vessel is `300 gm` and in equilibrium the block is at a height `40` cm above the bottom of vessel. The specific heat of material of block is `250J//kg//k` and that of water is `4200J//kg//k`. Neglect the heat capacities of vessel and the spring. If the support is broken the rise in temperature of water, when block reaches botton of vessel is

Explanation

When the block is in equilibrium in the water and spring is stretched by a distance x and the spring force balances the effective weight of block i.e. weight of blockminus the buoyant force on block. Thus for eqilibrium of block in water, we have
`kx+` weight of liquid displced `=` weight of block
`200x + 2/5000 xx1000xx10=2xx10`
or `x=(20-4)/200=16/200=0.08m=8cm`
Thus in equilibrium energy stored in spring is
`U=1/2 kx^(2)`
`=1/2xx200xx(0.08)^(2)=0.64` Joule
When the support is broken, the mass fall down to the bottom of vessel and the potential energy stored in spring and the grativational potential eneryg of block is released and used in heating the water and block. Thus we have
`0.64+mg(0.4)=m_(omega)xxs_(omega)xxDeltaT+m_(B)xxs_(B)xxDeltaT`
or `0.64+2xx10xxx0.4=[0.3xx4200+2xx250]DeltaT`
ro `DeltaT=8.64/1760=0.0049^(@)`