JEE PYQ: Waves And Sound Question 44
Question 44 - 2019 (12 Jan Shift 1)
A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to:
(1) 322 ms$^{-1}$
(2) 341 ms$^{-1}$
(3) 335 ms$^{-1}$
(4) 328 ms$^{-1}$
Show Answer
Answer: (4)
Solution
For the first fork: $\frac{\lambda_1}{4} = l_1 + e = 11 + e$ …(i)
For the second fork: $\frac{\lambda_2}{4} = l_2 + e = 27 + e$ …(ii)
Since $f_2 = \frac{f_1}{2}$, we have $\lambda_2 = 2\lambda_1$.
From (i) and (ii): $27 + e = 2(11 + e)$
$27 + e = 22 + 2e \Rightarrow e = 5$ cm
$v = f_1\lambda_1 = 512 \times 4 \times (11 + 5) \times 10^{-2} = 512 \times 0.64 = 327.7 \approx 328$ m/s
BETA
