1
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The ratio of the specific heats $\frac{C_p}{C_v}=\gamma$, in terms of degrees of freedom ( n ) is

A
$\left(1+\frac{1}{n}\right)$
B
$\left(1+\frac{2}{n}\right)$
C
$\left(1+\frac{\mathrm{n}}{3}\right)$
D
$\left(1+\frac{\mathrm{n}}{2}\right)$
2
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is

A
$\left(\frac{3}{4}\right)^{\text {th }}$ of kinetic energy per unit volume of a gas.
B
$\left(\frac{2}{3}\right)^{\text {rd }}$ of kinetic energy per unit volume of a gas.
C
$\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of kinetic energy per unit volume of a gas.
D
$\left(\frac{3}{2}\right)^{\text {rd }}$ of kinetic energy per unit volume of a gas.
3
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If heat energy $\Delta \mathrm{Q}$ is supplied to an ideal diatomic gas, the increase in internal energy is $\Delta U$ and the amount of work done by the gas is $\Delta \mathrm{W}$. The ratio $\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}$ is

A
$2: 3: 5$
B
$2: 5: 7$
C
$7: 5: 9$
D
$1: 2: 5$
4
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The power radiated by a black body is P and it radiates maximum energy around the wavelength $\lambda_0$. Now the temperature of the black body is changed so that it radiates maximum energy around wavelength $\left(\frac{\lambda_0}{2}\right)$. The power radiated by it will now increase by a factor of

A
2
B
8
C
16
D
32
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