A metal block of density `6000 kg m^(-3)` and mass `1.2 kg` is suspended through a spring of spring constant `200Nm^(-1)` .The spring - block system is dipped in water kept in a vessel .The water has a mass of `250g` and the block is at a height `40cm` above the bottom of the vessel .If the support to the spring is broken , what will be the rise in the temperature of the water specific heat capacity of the block is `250J kg^(-1)K^(-1)` and that of water is `4200J kg^(-1)K^(-1)` .Heat capacities of the vessel and the spring are negligible .

Answer

C

Explanation

volume of the block `= (12)/(600) = 2 xx 10^(-4)m^(3)`
When the mass in diped in water the block experience a huoyant force and spring experience PE whih is counteracted by its own weight
`Kx + Vpg = mg`
`rArr 200tau + 2 xx 10^(-1) xx 1000 xx 10= 12`
`rArr x = ((12 - 2))/(200)`
`= (10)/(200)= 0.05`
Now the heat is equally transfered to both block and water
`So, (1)/(2)Kx^(2) + mgh - Vpgh = m_(1)s_(1) Delta theta + m_(2)s_(2) Delta theta`
rArr (1)/(2) xx 200 xx 0.0025 + 1.2 xx 10 xx ((40)/(100))`
`= ((260)/(1000)) xx 6200 xx Delta t + 1.2 xx 250 xx Delta t`
`rArr 0.25 + 4.8 - 0.8= 1092 = 200 Delta t`
rArr 1392 Delta t = 4.25`
`rArr Delta t = (4.25)/(1392) = 0.0030531`
`= 3 xx 10^(-3)^(@)C`