A bat can hear sound of frequencies upto 120 kHz. The minimum wavelength of sound which it can hear is : ( Take speed of sound in air as 344 $m s^{- 1}$ )

2.87 mm

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3.87 mm

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4.87 mm

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5.87 mm

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Solution

The correct option is A 2.87 mm
Given, frequency, $F = 120 k H z = 120 \times 10^{3} H z$ and speed of sound in air, $V = 344 m s^{- 1}$ .
Let wavelength be $λ$ .
As we know, $V = F λ$
$\Rightarrow λ = \frac{V}{F} = \frac{344}{120 \times 10^{3}} = 2.87 \times 10^{- 3} m = 2.87 m m$ .

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A bat can hear sound of frequencies upto 120 kHz. The minimum wavelength of sound which it can hear is : ( Take speed of sound in air as 344 ms−1)

The correct option is A 2.87 mmGiven, frequency, F=120 kHz=120×103 Hz and speed of sound in air, V=344 ms−1. Let wavelength be λ. As we know, V=Fλ ⇒λ=VF=344120×103=2.87×10−3 m=2.87 mm.

A bat can hear sound of frequencies upto 120 kHz. The minimum wavelength of sound which it can hear is : ( Take speed of sound in air as 344 $m s^{- 1}$ )

The correct option is A 2.87 mm
Given, frequency, $F = 120 k H z = 120 \times 10^{3} H z$ and speed of sound in air, $V = 344 m s^{- 1}$ .
Let wavelength be $λ$ .
As we know, $V = F λ$
$\Rightarrow λ = \frac{V}{F} = \frac{344}{120 \times 10^{3}} = 2.87 \times 10^{- 3} m = 2.87 m m$ .