JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

The r.m.s speed of oxygen molecules at 47°C is equal to that of the hydrogen molecules kept at______°C. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

A
−235
B
−100
C
−253
D
−20

Solution

The rms speed is $v_{r m s} = \frac{3 R T}{M}$ . Equating rms speeds for $O_{2}$ and $H_{2}$ gives $\frac{T _{O_{2}}}{M _{O_{2}}} = \frac{T _{H_{2}}}{M _{H_{2}}}$ .

Convert: $T_{O_{2}} = 47 + 273 = 320$ K. Using the mass ratio $\frac{M _{O_{2}}}{M _{H_{2}}} = \frac{32}{2}$ : $\frac{320}{32} = \frac{T _{H_{2}}}{2}$ , so $T_{H_{2}} = 20$ K.

Back to Celsius: $T_{H_{2}} = 20 - 273 = - 253$ °C.

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