Answer :


Given:
Density of metal block, d= 6000 kg m−3
Mass of metal block, m = 1.2 kg
Spring constant of the spring,
k = 200 N m−1


Volume of the block, V


=1.2/6000=2×10-4 m3


When the mass is dipped in water, it experiences a buoyant force and in the spring there is potential energy stored in it.


the net force on the block is zero before breaking of the support of the spring, then


balancing forces
kx + Vρg = mg
200x + (2 × 10−4)× (1000) × (10) = 12


=x=12-2200x=10200=0.05 m


The mechanical energy of the block is transferred to both block and water. Let the rise in temperature of the block and the water be ΔT.


Applying conservation of energy, we get


1/2kx2+mgh-Vρgh=m1s1∆T+m2s2∆T


=1/2×200×0.0025+1.2×10×40× 10-2- 2× 10-4×1000×10×40×10-2


=260× 1000×4200×∆T+1.2×250×∆T


=0.25+4.8-0.8=1092 ∆T+300∆T


=1392 ∆T=4.25


=∆T=4.251392=0.0030531


∆T=3×10-3°


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