MHT CET PYQs for Heat Transfer with Solutions: Practice MHT CET Previous Year Questions

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Shivam Yadav

Educational Content Expert | Updated on - Nov 26, 2025

Heat Transfer is an important topic in the Physics section in MHT CET exam. Practising this topic will increase your score overall and make your conceptual grip on MHT CET exam stronger.

This article gives you a full set of MHT CET PYQs for Heat Transfer with explanations for effective preparation. Practice of MHT CET Physics PYQs including Heat Transfer questions regularly will improve accuracy, speed, and confidence in the MHT CET 2026 exam.

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MHT CET PYQs for Heat Transfer with Solutions

  • 1.
    The dimensions of Stefan?? constant are

      • $[M^0 \,L^1\, T^{-3} \,K^{-4}]$
      • $[M\, L^1\, T^{-3}\, K^{-3}]$
      • $[M\, L^2\, T^{-3}\, K^{-4}]$
      • $[M \,L^0\, T^{-3} \,K^{-4}]$

    • 2.
      Sphere of color black, red, white, and yellow are heated to the same temperature. The decreasing order of cooling is:

        • \(Black > Red > Yellow > White\)
        • \(Red > Black > White > Yellow\)
        • \(White > Yellow > Red > Black\)
        • \(Yellow > White > Red > Black\)

      • 3.

        What is change in internal energy if a system gains xJ of heat and yJ work is done on it?

          • x - y
          • - x + y
          • - x - y
          • x + y

        • 4.
          The $SI$ unit and dimensions of Stefan?s constant $\sigma$ in case of Stefan?s law of radiation is

            • $\frac{J}{m^{3}\,s^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]$
            • $\frac{J}{m^{2}\,s^{4}\,K}, \left[M^{1}L^{0}T^{-3}K^{3}\right]$
            • $\frac{J}{m^{3}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{4}\right]$
            • $\frac{J}{m^{2}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]$

          • 5.
            Two spherical black bodies have radii $?r_1?$ and $'r_2'$ . Their surface temperatures are $'T_1'$ and $'T_2'$. If they radiate same power then $\frac{r_2}{r_1}$ is

              • $\frac{T_1}{T_2}$
              • $\frac{T_2}{T_1}$
              • $(\frac{T_1}{T_2})^2$
              • $(\frac{T_2}{T_1})^2$

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