Air is blown at the mouth of an open tube of length 25 cm and diameter 2 cm. If the velocity of sound in air is 330 ms-1, then the emitted frequencies are (in Hz)

Question

Air is blown at the mouth of an open tube of length 25 cm and diameter 2 cm. If the velocity of sound in air is 330 ms-1, then the emitted frequencies are (in Hz) (A) 660, 1320, 2640 (B) 660, 1000, 3300 (C) 302, 664,1320 (D) 330, 990 1690

Tardigrade Admin · Accepted Answer

v = n λ n = (v/2l) = (330/2 × 25) = 660 Hz (∵ l = (λ/2) ) n1 = 2 × 660=1320 Hz n2 = 2 × 1320=2640 Hz

Q. Air is blown at the mouth of an open tube of length $25 c m$ and diameter $2 c m$ . If the velocity of sound in air is $330 m s^{- 1},$ then the emitted frequencies are (in $Hz$ )

660, 1320, 2640

660, 1000, 3300

302, 664,1320

330, 990 1690

330,660,990

$v = nλ$
$n = \frac{v}{2 l} = \frac{330}{2 \times 25} = 660 Hz$
$(∵ l = \frac{λ}{2})$
$n_{1} = 2 \times 660 = 1320 Hz$
$n_{2} = 2 \times 1320 = 2640 Hz$

Q. Air is blown at the mouth of an open tube of length 25cm and diameter 2cm. If the velocity of sound in air is 330ms−1, then the emitted frequencies are (in Hz)