11. (JEE Main 2019 (Online) 12th April Evening Slot)

A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l₁ = 30 cm and l₂ = 70 cm. Then, v is equal to -

A. 338 ms^–1

B. 384 ms^–1

C. 379 ms^–1

D. 332 ms^–1

Correct Answer is Option (B)

We know,

$λ$ = 2( $l_{2} - l_{1}$ )

given $l_{2}$ = 70 cm and $l_{1}$ = 30 cm

$∴$ $λ$ = 2(70 - 30) = 80 cm

Also we know, v = $λ$ $f$ = 0.8 $\times$ 480 = 384 m/s

12. (JEE Main 2019 (Online) 12th January Evening Slot)

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 :

A. 335 ms^–1

B. 328 ms^–1

C. 341 ms^–1

D. 322 ms^–1

Correct Answer is Option (B)

In first resonance, length of air column $= \frac{λ}{4}$ .

JEE Main 2019 (Online) 12th January Evening Slot Physics - Waves Question 82 English Explanation

So, $l_{1} + e = \frac{λ}{4}$ or $11 \times 4 + 4 e = λ$

So, speed of sound is

$\Rightarrow v = f_{1} λ = 512 (44 + 4 e)$ ...... (i)

And in second case,

$l_{1}^{'} + e = \frac{λ^{'}}{4}$ or $27 \times 4 + 4 e = λ^{'}$

$\Rightarrow v = f_{2} λ^{'} = 256 (108 + 4 e)$ ..... (ii)

Dividing both Eqs. (i) and (ii), we get

$1 = \frac{512 (44 + 4 e)}{256 (108 + 4 e)} \Rightarrow e = 5$ cm

Substituting value of e in Eq. (i), we get

Speed of sound $v = 512 (44 + 4 e)$

$= 512 (44 + 4 \times 5)$

$= 512 \times 64$ cm s^{$-$ 1} = 327.68 ms^{$-$ 1} $\approx$ 328 ms^{$-$ 1}

13. (JEE Main 2018 (Online) 16th April Morning Slot)

The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :

A. 28 cm

B. 32 cm

C. 36 cm

D. 40 cm

Correct Answer is Option (B)

Given, End correction (e) = 1 cm

For first resonance,

$\frac{λ}{4} = l_{1} + e$ = 10 + 1 = 11 cm

For second resonance,

$\frac{3 λ}{4} = l_{2} + e$

$\Rightarrow$ $l_{2}$ = 3 $\times$ 11 - 1 = 32 cm

14. (JEE Main 2016 (Offline))

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :

A. 2 f

B. f

C. $\frac{f}{2}$

D. $\frac{3 f}{4}$

Correct Answer is Option (D)

JEE Main 2016 (Offline) Physics - Waves Question 100 English Explanation
The fundamental frequency in case $(a)$ is $f = \frac{v}{2 ℓ}$

The fundamental frequency in case $(b)$ is

$f^{'} \frac{v}{4 (ℓ / 2)} = \frac{u}{2 ℓ} = f$

15. (JEE Main 2014 (Offline))

A pipe of length $85$ $c m$ is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below $1250$ $H z$ . The velocity of sound in air is $340$ $m / s$ .

A. 12

B. 8

C. 6

D. 4

Correct Answer is Option (C)

Length of pipe $= 85$ $c m$ $= 0.85 m$

Pipe is closed from one end so it behaves as a closed organ pipe

Frequency of oscillations of air column in closed organ pipe is given by,

$f = \frac{(2 n - 1) υ}{4 L}$

$f = \frac{(2 n - 1) υ}{4 L} \leq 1250$

$\Rightarrow \frac{(2 n - 1) \times 340}{0.85 \times 4} \leq 1250$

$\Rightarrow 2 n - 1 \leq 12.5 \approx 6$

Possible value of n = 1, 2, 3, 4, 5, 6

So, number of possible natural frequencies lie below 1250 Hz is 6.