Question

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) $1.4$

Answer 1

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 $v=\sqrt{{\displaystyle \frac{\gamma P}{\rho }}}$, (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 ($C_p\mathrm{\&}{C}_v$) 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:
$f={\displaystyle \frac{v}{4l}}$,
On putting values and solving, we get,
$84={\displaystyle \frac{v}{4\times 1}}$
On solving it we get v as,
$v=84\times 4=336m{s}^{-1}$

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

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 $f={\displaystyle \frac{v}{4l}}$, as the wavelength becomes 4 times of length of tube.