Q

# 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  $2\:cm$and 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)

Views

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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