For air at room temperature the atmospheric pressure is $1.0 \times {10^5}N{m^{ - 2}}$ and density of air is $1.2kg{m^{ - 3}}$. For a tube of length 1.0m closed at one end the lowest frequency generated is $84Hz$. The value of γ (ratio of two specific heats) for air is:
A) $2.1$
B) $1.5$
C) $1.8$
D) $1.4$

Hint: Use the Newton’s formula for speed of sound and laplace’s correction that establishes a relationship between speed of sound wrt pressure, temperature and ratio of specific heats.

Complete step by step answer:
We know that $v = \sqrt {\dfrac{{\gamma P}}{\rho }} $, (According to Newton’s formula for speed of sound)
Where v is the velocity of sound,
$\gamma $is the ratio of the two specific heats (${C_p}\& {C_v}$) of air,
P is the pressure and
$\rho $is the density of air.
Now, for calculation v, we can use the relation among frequency(f), tube length(l), and velocity of air(v), and that relation is:
$f = \dfrac{v}{{4l}}$,
On putting values and solving, we get,
$84 = \dfrac{v}{{4 \times 1}}$
On solving it we get v as,
$v = 84 \times 4 = 336m{s^{ - 1}}$

Putting all values in Newton’s formula, we get,
$336 = \sqrt {\dfrac{{\gamma \times 1.0 \times {{10}^5}}}{{1.2}}} $
Squaring on both sides, we get,
\[{\left( {336} \right)^2} = \dfrac{{\gamma \times {{10}^5}}}{{1.2}}\]
On further solving we get gamma as,
$
  \gamma = \dfrac{{{{\left( {336} \right)}^2} \times 1.2}}{{{{10}^5}}} \\
  \gamma \approx 1.35 \approx 1.4 \\
 $

So, the correct option is (D).

Note: In the tubes that are open from one end and close at other end, at close end, amplitude of the wave will be minimum, that is, 0 and at the open end, amplitude of wave will be maximum, that’s why we use the formula $f = \dfrac{v}{{4l}}$, as the wavelength becomes 4 times of length of tube.