The temperature at which speed of sound in air becomes double of its value at 0° C is:

Question

The temperature at which speed of sound in air becomes double of its value at 0° C is: (A)  1092 text K  (B)  1092° C  (C)  819 text K  (D)  819° C

Tardigrade Admin · Accepted Answer

Using the relation ( upsilon 1/ upsilon 0) =√(T/T0) [Here: upsilon t=2 upsilon , upsilon 0= upsilon , T0 =273 K ] ∴ ( ( upsilon /2 upsilon ) ) =(√T/273) or T=273× 4=1092 K

Q. The temperature at which speed of sound in air becomes double of its value at $0^{\circ} C$ is:

$1092 K$

$1092^{\circ} C$

$819 K$

$819^{\circ} C$

Using the relation $\frac{υ _{1}}{υ _{0}} = \frac{T}{T _{0}}$ [Here: $υ_{t} = 2 υ, υ_{0} = υ, T_{0} = 273 K$ ] $∴$ $(\frac{υ}{2 υ}) = \frac{T}{273}$ or $T = 273 \times 4 = 1092 K$

Q. The temperature at which speed of sound in air becomes double of its value at 0∘C is:

1092K

1092∘C

819K

819∘C