In a resonance tube with tuning fork of frequency 512 Hz, first resonance occurs at water . . .

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

a) 51.2 cm/s
b) 102.4 cm/s
c) 204.8 cm/s
d) 153.6 cm/s

Doubt by Ananya.

Solution : 

We know, 

In resonance experiment the speed is given by 

V=2f(l₂-l₁) 

Here V = Speed 

f = frequency of the tuning fork
l₂=water level in 2nd resonance. 
l₁=water level in 1st resonance.

V=2f(l₂-l₁) — (1)

ΔV/V = Δ(l2-l₁)/(l2-l₁)
ΔV = [Δ(l₂-l₁)/(l₂-l₁)]×V
ΔV = [Δ(l₂-l₁)/(l₂-l₁)]×[2f(l₂-l₁)] 
[Using eq (1)]

ΔV = 2f Δ(l₂-l₁)
ΔV = 2f (Δl₂+Δl₁)
[Remember that when two quantities are subtracted even then their errors are always get added]

ΔV = 2f (Δl₂+Δl₁)
ΔV = 2×512(0.1+0.1)

[Remember when error is not given then we use least count of the instrument as the the error]

ΔV = 2×512×0.2
ΔV = 512×0.4
ΔV = 204.8 cm/s

Hence, c) 204.8 cm/s is the correct option.