According to the law of conservation of energy,
21kx2=(m1s1+m2s2)×ΔT
⇒ΔT=0.5×400+1×418421×800×(2×10−2)2=438416×10−2=3.65×10−5
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…
JEE Main 2019 — Physics Thermodynamics
2019mcqhard
A massless spring (k=800N/m), attached with a mass (500g) is completely immersed in 1kg of water. The spring is stretched by 2cm 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, specific heat of water =4184J/kgK )
Official previous-year question
Held on 9 Apr 2019 · Verified 6 Jul 2026.
Options
- A
10−5K
- B
10−1K
- C
10−3K
- D
10−4K
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