JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 7)
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 :
335 ms–1
328 ms–1
341 ms–1
322 ms–1
Explanation
In first resonance, length of air column $$ = {\lambda \over 4}$$.
_12th_January_Evening_Slot_en_7_1.png)
So, $${l_1} + e = {\lambda \over 4}$$ or $$11 \times 4 + 4e = \lambda $$
So, speed of sound is
$$ \Rightarrow v = {f_1}\lambda = 512(44 + 4e)$$ ...... (i)
And in second case,
$$l{'_1} + e = {{\lambda '} \over 4}$$ or $$27 \times 4 + 4e = \lambda '$$
$$ \Rightarrow v = {f_2}\lambda ' = 256(108 + 4e)$$ ..... (ii)
Dividing both Eqs. (i) and (ii), we get
$$1 = {{512(44 + 4e)} \over {256(108 + 4e)}} \Rightarrow e = 5$$ cm
Substituting value of e in Eq. (i), we get
Speed of sound $$v = 512(44 + 4e)$$
$$ = 512(44 + 4 \times 5)$$
$$ = 512 \times 64$$ cm s$$-$$1 = 327.68 ms$$-$$1 $$\approx$$ 328 ms$$-$$1
Comments (0)
