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  • Confused! kindly explain, - Properties of Solids and Liquids - JEE Main-2

A massless spring (k=800\:N/m) , attached with a mass (500\:g) is completely immersed in 1\:kg of water .The spring stretched  by  2\:cmand released so that it starts vibrating . What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely ? ( Assume that water container and spring receive negligible heat and specific  heat of mass = 400\:J/kg\:K , specific heat of water 4184\:J/kg\:K )

 

  • Option 1)

    10^{-4}\:K

     

  • Option 2)

    10^{-5}\:K

  • Option 3)

  • Option 4)

Answers (1)

best_answer

Given

 K=800\:\:N/m

m=0.5\:\:kg

\frac{1}{2}KA^{2}=m_{1} \:s_{1}\Delta T+ m_{2} \:s_{2}\Delta T

\Rightarrow \frac{1}{2}\times 800\times \left ( \frac{2}{100} \right )^{2}=\frac{1}{2}\times 400\times\Delta T+4184\Delta T

\Rightarrow \Delta T=\frac{400\times\left ( \frac{2}{100} \right )^{2}}{200+4184}=3.64\times10^{-5}K

 

 


Option 1)

10^{-4}\:K

 

Option 2)

10^{-5}\:K

Option 3)

Option 4)

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solutionqc

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