At a speed 343 m/s and frequency of 536 Hz;
the wavelength is computed by velocity / frequency = 343 /
536 = 0.640 m or 64 cm
If the tube is open at the top there is a concentrated at the top and a node at
the bottom at resonance.
i.e: at L = lambda / 4
The next resonance happens at 3 lambda/4 then 5 lambda/4 etc.
these resemble to 64 / 4 = 16 cm.
therefore,
the lengths from the open end to the water level corresponding to the first 2:
L1 = v / 4f = 343 / 4*536 = 16cm
L2 = 3v / 4f = 3*343 / 4*536 = 48 cm
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It's too much I don't understand.sorry
d = distance between the two point charges = 60 cm = 0.60 m
r = distance of the location of point "a" where the electric field is zero from charge 
between the two charges.
= magnitude of charge on one charge
= magnitude of charge on other charge
= 3
= Electric field by charge at point "a"
= Electric field by charge at point "a"
Electric field by charge at point "a" is given as
= k /r²
Electric field by charge at point "a" is given as
= k /(d-r)²
For the electric field to be zero at point "a"
=
k /(d-r)² = k /r²
/(d-r)² = 3 /r²
1/(0.60 - r)² = 3 /r²
r = 0.38 m
r = 38 cm
