6. (JEE Main 2021 (Online) 25th February Morning Shift)

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 :

A. 13 cm

B. 18.4 cm

C. 16.6 cm

D. 14.8 cm

Correct Answer is Option (D)

JEE Main 2021 (Online) 25th February Morning Shift Physics - Waves Question 54 English Explanation
$∴$ $l + 1.8 = \frac{λ}{4}$

Also $λ = \frac{v}{f} = \frac{336}{504}$

$\Rightarrow l + 1.8 = \frac{336}{4 \times 504}$

$\Rightarrow l = 14.86$ cm

7. (JEE Main 2021 (Online) 27th August Evening Shift)

A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be __________ cm. (Take speed of sound in air as 340 ms^{$-$ 1})

Correct Answer is 34

JEE Main 2021 (Online) 27th August Evening Shift Physics - Waves Question 39 English Explanation
$\frac{λ}{4}$ = l $\Rightarrow$ $λ$ = 4l

f = $\frac{V}{λ} = \frac{V}{4 l}$

$\Rightarrow$ 250 = $\frac{340}{4 l}$

$\Rightarrow$ l = $\frac{34}{4 \times 25}$ = 0.34 m

l = 34 cm

8. (JEE Main 2021 (Online) 16th March Evening Shift)

A closed organ pipe of length L and an open organ pipe contain gases of densities $ρ$ ₁ and $ρ$ ₂ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is $\frac{x}{3} L \sqrt{\frac{ρ_{1}}{ρ_{2}}}$ where x is ___________. (Round off to the Nearest Integer)

Correct Answer is 4

First overtone of open pipe $= \frac{v_{2}}{L_{2}}$

First overtone of closed pipe at one end $= \frac{3 v}{4 L}$

As per question,

$\frac{3 V}{4 L} = \frac{V_{2}}{L}$

$\Rightarrow \sqrt{\frac{B}{ρ_{1}}} . \frac{3}{4 L} = \sqrt{\frac{B}{ρ_{2}}} . \frac{1}{L_{2}}$ ( $∵$ $V = \sqrt{\frac{B}{ρ}}$ )

$\Rightarrow L_{2} = \frac{4 L}{3} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$ ..... (i)

According to question, the length of the open pipe is

$\frac{x}{3} L \sqrt{\frac{ρ_{1}}{ρ_{2}}}$ ..... (ii)

Comparing Eqs. (i) and (ii), we get

$x = 4$

9. (JEE Main 2020 (Online) 8th January Morning Slot)

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________ Hz.

Correct Answer is 105.75 to 106.07

Velocity of sound v = $\sqrt{\frac{B}{ρ}}$

$∴$ $\frac{v_{p i p e}}{v_{a i r}} = \sqrt{\frac{\frac{B}{2 ρ}}{\frac{B}{ρ}}}$

$\Rightarrow$ v_pipe = $\frac{v_{a i r}}{\sqrt{2}}$ = $\frac{300}{\sqrt{2}}$ = 150 $\sqrt{2}$

We know, for open organ pipe

f_n = $(n + 1) \frac{v_{p i p e}}{2 l}$

Difference between 2^nd and fundamental frequency

f₁ - f₀ = $\frac{v_{p i p e}}{2 l}$ = $\frac{150 \sqrt{2}}{2 (1)}$ = 106.06 Hz

10. (JEE Main 2020 (Online) 5th September Morning Slot)

In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is :

A. 2200 Hz

B. 3300 Hz

C. 1100 Hz

D. 550 Hz

Correct Answer is Option (A)

$l_{1} = l - 17$

$l_{2} = l - 24.5$

We know, $v = 2 f (l_{1} - l_{2})$

$\Rightarrow$ $330 = 2 \times f \times [(f \times [(l - 17) - (l - 24.5)] \times 10^{- 2}$

$\Rightarrow$ $165 = f \times 7.5 \times 10^{- 2}$

$\Rightarrow$ $f = \frac{165 \times 1000}{7.5}$

$\Rightarrow$ $f = 2200 H z$