Kinetic Theory

JEE Main level, single correct, medium

103 of 168

Question 103 of 168

A massless spring (k=800N/m),\left(k=800 N/m\right), attached with a mass (500g)\left(500 g\right) is completely immersed in 1kg1kg of water. The spring is stretched by 2cm2cm 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 the water container and spring receive negligible heat and specific heat of mass =400J/kgK,=400J/kgK, specific heat of water =4184J/kgK=4184 J/kg K )

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Answer: (A) 10−5K{10}^{-5}K

According to the law of conservation of energy,
12kx2=(m1s1+m2s2)×ΔT\dfrac{1}{2}k{x}^{2}=\left({m}_{1}{s}_{1}+{m}_{2}{s}_{2}\right)\times \Delta T
⇒ΔT=12×800×(2×10−2)20.5×400+1×4184=16×10−24384=3.65×10−5\Rightarrow \Delta T=\dfrac{\dfrac{1}{2}\times 800\times {\left(2\times {10}^{-2}\right)}^{2}}{0.5\times 400+1\times 4184}=\dfrac{16\times {10}^{-2}}{4384}=3.65\times {10}^{-5}

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