MHT CET 2024 3rd May Morning Shift | Heat and Thermodynamics Question 81 | Physics

Two spherical black bodies of radii ' $R_1$ ' and ' $\mathrm{R}_2$ ' and with surface temperature ' $\mathrm{T}_1$ ' and ' $\mathrm{T}_2$ ' respectively radiate the same power. The ratio of ' $R_1$ ' to ' $R_2$ ' will be

$\left(\frac{T_2}{T_1}\right)^4$

$\left(\frac{T_2}{T_1}\right)^2$

$\left(\frac{T_1}{T_2}\right)^4$

$\left(\frac{T_1}{T_2}\right)^2$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

Rate of flow of heat through a cylindrical rod is ' $\mathrm{H}_1$ '. The temperature at the ends of the rod are ' $T_1$ ' and ' $T_2$ '. If all the dimensions of the rod become double and the temperature difference remains the same, the rate of flow of heat becomes ' $\mathrm{H}_2$ '. Then

$\mathrm{H}_2=4 \mathrm{H}_1$

$\mathrm{H}_2=2 \mathrm{H}_1$

$\mathrm{H}_2=\frac{\mathrm{H}_1}{2}$

$\mathrm{H}_2=\frac{\mathrm{H}_1}{4}$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

A fixed mass of gas at constant pressure occupies a volume ' V '. The gas undergoes a rise in temperature so that the r.m.s. velocity of the molecules is doubled. The new volume will be

$\frac{V}{2}$

$\frac{\mathrm{V}}{\sqrt{2}}$

$2 \mathrm{~V}$

4 V

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

In an isobaric process

pressure is constant.

volume is constant.

temperature is constant.

internal energy is constant.

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