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

Options

A. `0.0012^(@)C`
B. `0.0049^(@)C`
C. `0.0028^(@)C`
D. `0.0^(@)C`

Answer

B

Explanation

When the block is in equilibrium un the water the spring stretches by x
`Kx+` weight of liquid displaced `=` weight of block
`200x+(0)/(5000)xx1000xx10=2xx10 :. x=0.08m`
In equilibrium energy stored in spring `U=(1)/(2)kx^(2)`
`U=(1)/(2)xx200xx(0.08)^(2)= 0.64J`
when support in booken & block reaches bottom of the vessel
`U+Mgh=M_(W)S_(W) DeltaT+MxxS_(B) DeltaT`
Where `M` is mass of block
`:. 0.64+2xx10xx0.4=[0.3xx4200+2xx250] DeltaT`
`:. DeltaT=0.0049^(@)C`