NEET Questions Solved

NEET - 2012

A slab of stone of area of 0.36 m2 and 

thickness 0.1 m is exposed on the lower 

surface to steam at 100°C. A block of ice 

at 0°C rests on the upper surface of the slab.

In one hour 4.8 kg of ice is melted. The thermal

conductivity of slab is

(Given latent heat of fusion of ice 

=3.36x105 J kg-1)

(a) 1.24 J/m/s/°C        (b) 1.29 J/m/s/°C

(c) 2.05 J/m/s/°C        (d) 1.02 J/m/s/°C

δQδt=KAL(T1-T2)   Q=KAL(T1-T2)t    Q=mLfKAL(T1-T2)t=mLf                  K=mLfLA(T1-T2)t                  K=4.8×3.36×105×0.10.36×100×3600                    =4.8×3.360.36×36                    =1.24 J/m/s/°C

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