JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Elasticity & SHM

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)

(A) $1 0^{- 3} K$
(B) $1 0^{- 4}$
(C) $1 0^{- 1} K$
(D) $1 0^{- 5} K$ ✓ Correct

Answer: (D)

Solution

When the vibrations finally die out, all the elastic energy that was stored in the stretched spring has turned into heat shared by the block and the water. That stored energy is

\frac{1}{2} k x^{2} = \frac{1}{2} (800) (0.02)^{2} = 0.16

J. This warms the

0.5

kg block (specific heat

400

J/kg K) together with the

1

kg of water (specific heat

4184

J/kg K), so

0.16 = (0.5 \times 400 + 1 \times 4184) Δ T = 4384 Δ T

. That gives

Δ T \approx 3.6 \times 1 0^{- 5}

K, which is of the order of

1 0^{- 5}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Elasticity & SHM →

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Elasticity & SHM

Solution

More JEE Main questions on Elasticity & SHM