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**Hint:**The speed of a sound is directly proportional to the square root of temperature of its medium i.e., with the increase in temperature, speed of sound increases and vice-versa.

**Step by step answer:**The speed of sound v in a gas (air) increases with the increase in temperature of the gas. The reason is that with the increase of temperature, there is a decrease in density as the volume of the gas increases. From the relation,

$v = \sqrt {\dfrac{{\gamma P}}{\rho }} $

where γ is the ratio of specific heat at constant pressure to the specific heat at constant volume, P is the pressure of the gas and ρ is the density of the gas. So, v is inversely proportional to the square root of density of the gas i.e. v is proportional to$\dfrac{1}{{\sqrt \rho }}$ .

Initially let the speed of sound be v₁ , temperature be T₁ .

As v α√T. [v is speed of sound and T is temperature]

${v^2} = kT$ [ where k is proportionality constant]

Now, we put the value of T₁ and v in the above equation,

${v_1}^2 = k \times 27$

${v_1}^2 = k\left( {27 + 273} \right)$ [ changing the temperature from ⁰C to K]

$

{v_1}^2 = 300k \\

k = \dfrac{{{v_1}^2}}{{300}} \\

$

Now, when the speed of sound is 2 v₁ and temperature be T₂. On substituting the value in the same equation,

${\left( {2{v_1}} \right)^2} = k{T_2}$

$4{v_1}^2 = k{T_2}$

Substituting the value of k in the equation,

$4{v_1}^2 = \left( {\dfrac{{{v_1}^2}}{{300}}} \right){T_2}$

$

{T_2} = 4{v_1}^2 \times \dfrac{{300}}{{{v_1}^2}} \\

\therefore {T_2} = 1200K \\

$

Temperature in ⁰C = $1200 - 273 = 927$⁰C

Hence Option (D) is correct.

**Note:**The temperature in degree Celsius should be converted to Kelvin. To convert temperature in degree Celsius to Kelvin, we have to add 273 ⁰C to the given temperature and to convert Kelvin to degree Celsius, subtract 273 ⁰C from the temperature in Kelvin.

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