JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 22)
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)
10–3 K
10–1 K
10–5K
10–4 K
Explanation
By law of conservation of energy
$${1 \over 2}k{x^2} = \left( {{m_1}{s_1} + {m_2}{s_2}} \right)\Delta T$$
$$\Delta T = {{16 \times {{10}^{ - 2}}} \over {4384}} = 3.65 \times {10^{ - 5}}$$ K
$${1 \over 2}k{x^2} = \left( {{m_1}{s_1} + {m_2}{s_2}} \right)\Delta T$$
$$\Delta T = {{16 \times {{10}^{ - 2}}} \over {4384}} = 3.65 \times {10^{ - 5}}$$ K
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