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Physics · PYQ NEET UG 2019(2)

हिन्दी

A deep rectangular pond of surface area \( A \), containing water (density = \( \rho \), specific heat capacity = \( s \)), is located in a region where the outside air temperature is at a steady value of \( -26^\circ\text{C} \). The thickness of the frozen ice layer in this pond, at a certain instant is \( x \). Taking the thermal conductivity of ice as \( K \), and its specific latent heat of fusion as \( L \), the rate of increase of the thickness of ice layer, at this instant, would be given by

  1. A
    \( \frac{26K}{\rho x(L + 4s)} \)
  2. B
    \( \frac{26K}{\rho x(L - 4s)} \)
  3. C
    \( \frac{26K}{\rho x^2 L} \)
  4. D
    \( \frac{26K}{\rho x L} \)
View Answer & Solution

Correct answer · Option D

\( \frac{26K}{\rho x L} \)

Explanation

Heat conducted per second \( \frac{dQ}{dt} = \frac{KA(0 - (-26))}{x} = \frac{26KA}{x} \). Rate of mass formation \( \frac{dm}{dt} = \rho A \frac{dx}{dt} \). Since \( \frac{dQ}{dt} = L \frac{dm}{dt} \), we get \( \frac{26KA}{x} = L \rho A \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{26K}{\rho x L} \).
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