1
JEE Main 2023 (Online) 11th April Evening Shift
+4
-1

The Thermodynamic process, in which internal energy of the system remains constant is

A
Isobaric
B
Isochoric
C
D
Isothermal
2
JEE Main 2023 (Online) 11th April Evening Shift
+4
-1

The root mean square speed of molecules of nitrogen gas at $$27^{\circ} \mathrm{C}$$ is approximately : (Given mass of a nitrogen molecule $$=4.6 \times 10^{-26} \mathrm{~kg}$$ and take Boltzmann constant $$\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{JK}^{-1}$$ )

A
91 m/s
B
1260 m/s
C
27.4 m/s
D
523 m/s
3
JEE Main 2023 (Online) 11th April Morning Shift
+4
-1

$$1 \mathrm{~kg}$$ of water at $$100^{\circ} \mathrm{C}$$ is converted into steam at $$100^{\circ} \mathrm{C}$$ by boiling at atmospheric pressure. The volume of water changes from $$1.00 \times 10^{-3} \mathrm{~m}^{3}$$ as a liquid to $$1.671 \mathrm{~m}^{3}$$ as steam. The change in internal energy of the system during the process will be

(Given latent heat of vaporisaiton $$=2257 \mathrm{~kJ} / \mathrm{kg}$$, Atmospheric pressure = $$\left.1 \times 10^{5} \mathrm{~Pa}\right)$$

A
+ 2090 kJ
B
$$-$$ 2426 kJ
C
+ 2476 kJ
D
$$-$$ 2090 kJ
4
JEE Main 2023 (Online) 11th April Morning Shift
+4
-1

On a temperature scale '$$\mathrm{X}$$', the boiling point of water is $$65^{\circ} \mathrm{X}$$ and the freezing point is $$-15^{\circ} \mathrm{X}$$. Assume that the $$\mathrm{X}$$ scale is linear. The equivalent temperature corresponding to $$-95^{\circ} \mathrm{X}$$ on the Farenheit scale would be:

A
$$-148^{\circ} \mathrm{F}$$
B
$$-48^{\circ} \mathrm{F}$$
C
$$-63^{\circ} \mathrm{F}$$
D
$$-112^{\circ} \mathrm{F}$$
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