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Physics MCQs for NEET — Practice Questions with Answers

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A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio :

1 : 2

2 : 3

3 : 4

4 : 5

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Explanation

It is given that the first overtone of closed pipe = First overtone of the open pipe ⇒ $3 (\frac{v}{4 l_{1}}) = 2 (\frac{v}{2 l_{2}})$ ; where l₁ and l₂ are the lengths of closed and open organ pipes hence $\frac{l_{1}}{l_{2}} = \frac{3}{4}$ .

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An empty vessel is getting filled with water, then the frequency of vibration of the air column in the vessel

Remains the same

Decreases

Increases

First increases then decrease

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Explanation

For closed pipe in general $n = \frac{v}{4 l} (2 N - 1)$ ⇒ $n \propto \frac{1}{l}$

i.e. if length of air column decreases, frequency increases.

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If the velocity of sound in air is 350 m/s. Then the fundamental frequency of an open organ pipe of length 50 cm, will be

350 Hz

175 Hz

900 Hz

750 Hz

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Explanation

Fundamental frequency of open pipe $n_{1} = \frac{v}{2 l} = \frac{350}{2 \times 0 .5} = 350 H z$ .

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The fundamental note produced by a closed organ pipe is of frequency f. The fundamental note produced by an open organ pipe of same length will be of frequency

$\frac{f}{2}$

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Explanation

$n_{closed} = \frac{v}{4 l}, n_{open} = \frac{v}{2 l}$ ⇒ $n_{open} = 2 n_{closed} = 2 f$

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If the velocity of sound in air is 336 m/s. The maximum length of a closed pipe that would produce a just audible sound will be :

3.2 cm

4.2 m

4.2 cm

3.2 m

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Explanation

Minimum audible frequency = 20 Hz.

$\Rightarrow \frac{v}{4 l} = 20 \Rightarrow l = \frac{336}{4 \times 20} = 4 .2 m$

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A cylindrical tube, open at both ends, has a fundamental frequency f₀ in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is

3 f₀/4

f₀

f₀/2

2 f₀

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Explanation

$n_{open} = \frac{v}{2 l_{open}}$

$n_{closed} = \frac{v}{4 l_{closed}} = \frac{v}{4 l_{open} / 2} = \frac{v}{2 l_{open}}$

$(A s l_{c l o s e d} = \frac{l_{o p e n}}{2})$ , i.e. frequency remains unchanged.

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If the length of a closed organ pipe is 1.5 m and the velocity of sound is 330 m/s, then the frequency for the second note is

220 Hz

165 Hz

110 Hz

55 Hz

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Explanation

For closed pipe second note = $\frac{3 v}{4 l} = \frac{3 \times 330}{4 \times 1 .5} = 165 H z$ .

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A pipe 30 cm long is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1.1 kHz source? (Take the speed of sound in air = 330 ms^–1)

First

Second

Third

Fourth

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Explanation

Fundamental frequency of open pipe

$n_{1} = \frac{v}{2 l} = \frac{330}{2 \times 0 .3} = 550 H z$

Second harmonic = $2 \times n_{1} = 1100 H z = 1 .1 k H z$

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A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $ρ_{0}$ inside the tube. If the atmospheric pressure is $ρ_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be :

$\frac{(ρ_{A} + ρ_{0})}{(ρ_{A} - ρ_{0})}$

$\frac{(ρ_{A} + 2 ρ_{0})}{(ρ_{A} - 2 ρ_{0})}$

$\frac{ρ_{A}}{ρ_{A}}$

$\frac{(ρ_{A} + \frac{1}{2} ρ_{0})}{(ρ_{A} - \frac{1}{2} ρ_{0})}$

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Explanation

Maximum pressure at closed-end will be atmospheric pressure adding with acoustic wave pressure

So $ρ_{\max} = ρ_{A} + ρ_{0}$ and $ρ_{\min} = ρ_{A} - ρ_{0}$

Thus $\frac{ρ_{\max}}{ρ_{\min}} = \frac{ρ_{A} + ρ_{0}}{ρ_{A} - ρ_{0}}$

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Two closed pipe produce 10 beats per second when emitting their fundamental nodes. If their length are in ratio of 25 : 26. Then their fundamental frequency in Hz, are :

270, 280

260, 270

260, 250

260, 280

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Explanation

$n_{1} - n_{2} = 10$ ..…(i)

Using $n_{1} = \frac{v}{4 l_{1}}$ and $n_{2} = \frac{v}{4 l_{2}}$

⇒ $\frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} = \frac{26}{25}$ …..(ii)

After solving these equation $n_{1} = 260 H z$ , $n_{2} = 250 H z$

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