AIPMT MAINS2012Physics-Thermal Properties of Matter

AIPMT MAINS 2012 Physics Thermal Conductivity MCQ Question

Type: MCQ-numerical-Medium-Class 11

A slab of stone of area 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 = 3.36 × 10⁵ 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

Correct Answer

Option A

Detailed Explanation

The thermal conductivity can be calculated using the formula Q = kA(T₁-T₂)t/d, where Q is the heat transferred, A is the area, T₁ and T₂ are the temperatures, t is the time, and d is the thickness. Solving for k gives the correct answer.

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