InterviewSolution
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InterviewSolution
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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 isA. `0.0012^(@)C`B. `0.0049^(@)C`C. `0.0028^(@)C`D. `0.0^(@)C` |
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Answer» Correct Answer - B 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` |
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