JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Waves & Sound

A closed-organ pipe of length

L

and an open pipe contain gases of densities

ρ_{1}

and

ρ_{2}

respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is

\frac{x}{3} L \frac{ρ _{1}}{ρ _{2}}

where

x

is ---. (Round off to the Nearest Integer)

Answer: 4

Solution

The speed of sound in a gas is

v = B / ρ

, and the bulk modulus

B

is the reciprocal of the compressibility, which is the same for both gases, so

\frac{v _{1}}{v _{2}} = \frac{ρ _{2}}{ρ _{1}}

The first overtone of a closed pipe of length

L

is its third harmonic,

f = \frac{3 v _{1}}{4 L}

, while the first overtone of an open pipe of length

L^{'}

is its second harmonic,

f = \frac{2 v _{2}}{2 L ^{'}} = \frac{v _{2}}{L ^{'}}

. Equating the two frequencies,

\frac{3 v _{1}}{4 L} = \frac{v _{2}}{L ^{'}} \Rightarrow L^{'} = \frac{4 L}{3} \cdot \frac{v _{2}}{v _{1}} = \frac{4}{3} L \frac{ρ _{1}}{ρ _{2}}

Matching this against

\frac{x}{3} L \frac{ρ _{1}}{ρ _{2}}

gives

x = 4

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Waves & Sound

Solution

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