The speed of sound in air under ordinary condition is around $330 m s^{- 1}$ .The speed of sound in hydrogen under similar conditions will be $(m s^{- 1})$ nearest to :

330

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1200

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600

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900

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Solution

The correct option is A 1200
Speed of sound is
$v \propto \frac{1}{\sqrt{ρ}}$
So
$\frac{v_{H_{2}}}{v_{a i r}} = \sqrt{\frac{ρ_{a i r}}{ρ_{H_{2}}}}$
Now $ρ_{a i r} = 1.2$ gm/lt and $ρ_{H_{2}} = 0.089$ gm/lt.
$v_{H_{2}} = 330 \times \sqrt{\frac{1.2}{0.089}} = 1205$ m/s

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Similar questions

Q. If

v_{m}

is the speed of sound in moist air and

v_{d}

is the speed of sound in dry air under identical conditions of pressure and temperature, then:

How long will sound take to travel in (a) an iron rail and (b) air, both 3-3 km in length ? Take speed of sound in air to be 330 m $s^{- 1}$ and in iron to be 5280 m $s^{- 1}$ .

The speed of sound in air at $0^{o} C$ is nearly :

(a) 1450 m $s^{- 1}$

(b) 450 m $s^{- 1}$

(d) 330 m $s^{- 1}$

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The speed of sound in air under ordinary condition is around 330ms−1.The speed of sound in hydrogen under similar conditions will be (ms−1) nearest to :

The correct option is A 1200Speed of sound isv∝1√ρSovH2vair=√ρairρH2Now ρair=1.2 gm/lt and ρH2=0.089 gm/lt.vH2=330×√1.20.089=1205 m/s

The speed of sound in air under ordinary condition is around $330 m s^{- 1}$ .The speed of sound in hydrogen under similar conditions will be $(m s^{- 1})$ nearest to :

The correct option is A 1200
Speed of sound is
$v \propto \frac{1}{\sqrt{ρ}}$
So
$\frac{v_{H_{2}}}{v_{a i r}} = \sqrt{\frac{ρ_{a i r}}{ρ_{H_{2}}}}$
Now $ρ_{a i r} = 1.2$ gm/lt and $ρ_{H_{2}} = 0.089$ gm/lt.
$v_{H_{2}} = 330 \times \sqrt{\frac{1.2}{0.089}} = 1205$ m/s