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NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    A slab of stone of area of \[0.36{{m}^{2}}\] and thickness 0.1 m is exposed on the lower surface to steam at\[100{}^\circ C\]. A block of ice at \[0{}^\circ C\] 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 \[=3.36\times {{10}^{5}}J\,k{{g}^{-1}}\])

    A)  \[1.24J/m/s/{}^\circ C\]              

    B)  \[1.29\text{ }J/m/s/{}^\circ C\]

    C)  \[2.05\text{ }J/m/s/{}^\circ C~\]

    D)  \[1.02\text{ }J/m/s/{}^\circ C\]

    Correct Answer: A

    Solution :

    \[\frac{\delta Q}{\delta t}=\frac{KA}{L}({{T}_{1}}-{{T}_{2}})\] \[Q=\frac{KA}{L}({{T}_{1}}-{{T}_{2}})\] \[Q=m{{L}_{f}}\]                 \[\frac{KA}{L}({{T}_{1}}-{{T}_{2}})t=m{{L}_{f}}\]                                 \[K=\frac{m{{L}_{f}}L}{A({{T}_{1}}-{{T}_{2}})t}\]                                 \[K=\frac{4.8\times 3.36\times {{10}^{5}}\times 0.1}{0.36\times 100\times 3600}\]                                 \[=\frac{4.8\times 3.36}{0.36\times 36}\]                                 \[\text{=1}\text{.24J/m/}{{\text{s}}^{\text{o}}}\text{C}\]


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