JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2019
- Hard
- MCQ
A massless spring \((k = 800\, N/m)\), attached with a mass \((500\, g)\) is completely immersed in \(1\, kg\) of water. The spring is 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 the water container and spring receive negligible heat and specific heat of mass \(= 400\, J/kg\, K\), specific heat of water \(= 4184\, J/kg\, K\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(10^{-5}\,K\)
Step-by-step Solution
Detailed explanation
By law of conservation of energy
\(\frac{1}{2} \mathrm{kx}^{2}=\left(\mathrm{m}_{1} \mathrm{s}_{1}+\mathrm{m}_{2} \mathrm{s}_{2}\right) \Delta \mathrm{T}\)
\(\Delta T=\frac{16 \times 10^{-2}}{4384}=3.65 \times 10^{-5}\)
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