Atuning fork with a frequency of f = 536 hz is placed near the top of the tube shown below. the water level is lowered so that the length l slowly increases from an initial value of 20.0 cm. determine the next two values of l that correspond to resonant modes. (assume that the speed of sound in air is 343 m/s.)

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

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

it was given as a lesson and guide for prayer.