A closed organ pipe of length 20 cm is sounded with tuning fork in...

If we adjust the length of air-column in closed organ pipe as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance. At first resonance $l=\frac{\lambda}{4}$ So, frequency of tuning fork $f=\frac{v}{\lambda}=\frac{v}{4 l}$ Given, $l=20 \mathrm{~cm}=0.2 \mathrm{~m}, v=332 \mathrm{~m} / \mathrm{s}$ Hence, $\quad f=\frac{0.332}{4 \times 0.2}=415 \mathrm{~Hz}$