Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1662: NEET 2012 Physics - Class 11 AIPM-2012-Ph-1662 Solution: A slab of stone of area \(0.36\ {m}^2\) and thickness 0.1 m is…

A slab of stone of area 0.36 m20.36\ \text{m}^2 and thickness 0.1 m is exposed on the lower surface to steam at 100∘C100^\circ C. A block of ice at 0∘C0^\circ 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 = 3.36×105 J kg−13.36 \times 10^5\ \text{J kg}^{-1}) 1.24 J/m/s/∘C^\circ C 1.29 J/m/s/∘C^\circ C 2.05 J/m/s/∘C^\circ C 1.02 J/m/s/∘C^\circ C

Multiple Choice Options

  • A 1.24 J/m/s/ o C
  • B 1.29 J/m/s/ o C
  • C 2.05 J/m/s/ o C
  • D 1.02 J/m/s/ o C

Step-by-Step Verified Solution

Rate of heat given by steam = Rate of heat taken by ice where K = Thermal conductivity of the slab m = Mass of the ice L = Latent heat of melting/fusion A = Area of the slab d Q d t = K A ( 100 _ 0 ) 1 = m d L d t , K _ 100 _ 0.36 0.1 = 4.8 _ 3.36 _ 10 5 60 _ 60 K =1.24 J/m/s/�C

Related Learning Resources & Exam Preparation

NEET PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring