Verified Solution • 2025–26 Session moderate

QJEEM19Ph7e38: JEE Main 2019 Physics - Class 11 JEEM19Ph7e38 Solution: A massless spring (k = 800 N/m), attached with a mass (500 g)…

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)

Multiple Choice Options

  • A 10–3 K
  • B 10–1 K
  • C 10–5K
  • D 10–4 K

Step-by-Step Verified Solution

By law of conservation of energy

12kx2=(m1s1+m2s2)ΔT{1 \over 2}k{x^2} = \left( {{m_1}{s_1} + {m_2}{s_2}} \right)\Delta T

ΔT=16×10−24384=3.65×10−5\Delta T = {{16 \times {{10}^{ - 2}}} \over {4384}} = 3.65 \times {10^{ - 5}} K

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring