JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 15)
A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is :
13 cm
18.4 cm
16.6 cm
14.8 cm
Penjelasan
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$$ \therefore $$ $$l + 1.8 = {\lambda \over 4}$$
Also $$\lambda = {v \over f} = {{336} \over {504}}$$
$$ \Rightarrow l + 1.8 = {{336} \over {4 \times 504}}$$
$$ \Rightarrow l = 14.86$$ cm