In a resonance tube with tuning fork of frequency `512 Hz`, first resonance occurs at water level equal to `30.3cm` and second resonance ocuurs at `63.7cm`. The maximum possible error in the speed of sound is

Options

A. `5.12 cm//s`
B. `102.4 cm//s`
C. `204.8 cm//s`
D. `153.6 cm//s`

Answer

C

Explanation

For first the resonance `l_(1) + r = (lamda)/(4)`
But `v = f lamda`
`:. v = f4 (l_(1) + e)`
`rArr l_(1) + e = (v)/(4 f)`…(i)
For the second resonance
`l_(2) + e = (3 lamda)/(4)`
`:. v = f(4)/(3) (l_(2) +e)`
`rArr l_(2) + e =(3v)/(4f)` ....(ii)
From Eqs. (i) and (ii), we get
`v = 2f(l_(2) -l_(1))`
`:. Delta v = 2f(Delta l_(2) + Delta l_(1))`
=`2 xx 512 xx (0.1 + 0.1) cm//s`
=`204.8 cm//s`.