1
MHT CET 2023 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A monoatomic ideal gas initially at temperature '$$\mathrm{T}_1$$' is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly the gas is allowed to expand to adiabatically to a temperature '$$\mathrm{T}_2$$'. If '$$\mathrm{L}_1$$' and '$$\mathrm{L}_2$$' are the lengths of the gas columns before and after expansion respectively, then $$\frac{\mathrm{T}_2}{\mathrm{~T}_1}$$ is

A
$$\frac{\mathrm{L}_1}{\mathrm{~L}_2}$$
B
$$\frac{\mathrm{L}_2}{\mathrm{~L}_1}$$
C
$$\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{2 / 3}$$
D
$$\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}$$
2
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $$\gamma_1$$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $$\gamma_2$$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio $$\frac{\gamma_2}{\gamma_1}$$ is

A
$$\frac{37}{21}$$
B
$$\frac{27}{35}$$
C
$$\frac{21}{25}$$
D
$$\frac{35}{27}$$
3
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The molar specific heat of an ideal gas at constant pressure and constant volume is $$\mathrm{C}_{\mathrm{p}}$$ and $$\mathrm{C}_{\mathrm{v}}$$ respectively. If $$\mathrm{R}$$ is universal gas constant and $$\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$$ then $$\mathrm{C}_{\mathrm{v}}=$$

A
$$\frac{1-\gamma}{1+\gamma}$$
B
$$\frac{1+\gamma}{1-\gamma}$$
C
$$\frac{\gamma-1}{\mathrm{R}}$$
D
$$\frac{\mathrm{R}}{\gamma-1}$$
4
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A composite slab consists of two materials having coefficient of thermal conductivity $$\mathrm{K}$$ and $$2 \mathrm{~K}$$, thickness $$\mathrm{x}$$ and $$4 \mathrm{x}$$ respectively. The temperature of the two outer surfaces of a composite slab are $$\mathrm{T}_2$$ and $$\mathrm{T}_1\left(\mathrm{~T}_2 > \mathrm{T}_1\right)$$. The rate of heat transfer through the slab in a steady state is $$\left[\frac{\mathrm{A}\left(\mathrm{T}_2-\mathrm{T}_1\right) \mathrm{K}}{\mathrm{x}}\right] \cdot \mathrm{f}$$ where '$$\mathrm{f}$$' is equal to

A
1
B
$$\frac{2}{3}$$
C
$$\frac{1}{2}$$
D
$$\frac{1}{3}$$
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