Question

The temperature at which speed of sound in air becomes double its value at 0${}^o$C is

• A. 546${}^o$C
• B. 819${}^o$C
• C. 273${}^o$C
• D. 1092${}^o$C

Answer 1

Answer: (B)

819${}^o$C

We know that speed of velocity in air $v\propto \sqrt{T}$ or $v^2\propto T$

Thus, $\frac{v_1^2}{v_2^2}=\frac{T_1}{T_2}$

or $\frac{v_1^2}{(2v_1{)}^2}=\frac{0+273}{T_2}$

or $T_2=1092K=(1092-273{)}^{o}C={819}^{o}C$