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NEET NEET Questions — Free MCQ Practice

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If T is the reverberation time of an auditorium of volume V then :

A.

$T \propto \frac{1}{V}$

B.

$T \propto \frac{1}{V^{2}}$

C.

$T \propto V^{2}$

D.

$T \propto V$

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Explanation

Reverberation time $T = \frac{k V}{α S}$ ⇒ T ∝ V.

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The intensity level due to two waves of the same frequency in a given medium are 1 bel and 5 bel. Then the ratio of amplitudes is :

A.

1 : 4

B.

1 : 2

C.

1 : 10⁴

D.

1 : 10²

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Explanation

By using $L = \log_{10} \frac{I}{I_{0}}$

$L_{2} - L_{1} = \log_{10} \frac{I_{2}}{I_{0}} - \log_{10} \frac{I_{1}}{I_{0}}$

$5 - 1 = \log_{10} \frac{I_{2}}{I_{1}}$ ⇒ $4 = \log_{10} \frac{I_{2}}{I_{1}}$ ⇒ $\frac{I_{2}}{I_{1}} = 10^{4}$

⇒ $\frac{a_{2}^{2}}{a_{1}^{2}} = 10^{4}$ ⇒ $\frac{a_{2}}{a_{1}} = \frac{10^{2}}{1}$ ⇒ $\frac{a_{1}}{a_{2}} = \frac{1}{10^{2}}$

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Of the following the one which emits the sound of a higher pitch is :

A.

Mosquito

B.

Lion

C.

Man

D.

Woman

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Explanation

Pitch of mosquito is higher among all given options.

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A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distances of 2m and 3m respectively from the source. The ratio of the intensities of the waves at P and Q is :

A.

9 : 4

B.

2 : 3

C.

3 : 2

D.

4 : 9

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Explanation

Intensity $\propto \frac{1}{{(Distance)}^{2}}$

⇒ $\frac{I_{1}}{I_{2}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{3}{2})}^{2} = \frac{9}{4}$ .

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Two waves having sinusoidal waveforms have different wavelengths and different amplitudes. They will be having :

A.

Same pitch and different intensity

B.

Same quality and different intensity

C.

Different quality and different intensity

D.

Same quality and different pitch

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Explanation

The pitch depends upon the frequency of the source. As the two waves have different amplitude therefore they having different intensity. While quality depends on number of harmonics/overtone produced and their relative intensity

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The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is $y_{1} = A \sin (π x / L) \sin ω t$ and energy is E₁, and in another experiment its displacement is $y_{2} = A \sin (2 π x / L) \sin 2 ω t$ and energy is E₂. Then :

A.

E₂ = E₁

B.

E₂ = 2E₁

C.

E₂ = 4E₁

D.

E₂ = 16E₁

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Explanation

Energy (E) ∝ (Amplitude)² (Frequency)²

Amplitude is same in both the cases, but frequency 2ω in the second case is two times the frequency (ω) in the first case. Hence E₂ = 4E₁.

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In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. when this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction :

A.

0.012m

B.

0.025m

C.

0.05m

D.

0.024m

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Explanation

Let x be the end correction then according to question.

$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (l_{2} + x)} \Rightarrow x = 2 .5 c m = 0.023 m$ .

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Two identical stringed instruments have a frequency 100 Hz. If the tension in one of them is increased by 4% and they are sounded together then the number of beats in one second is :

A.

1

B.

8

C.

4

D.

2

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Explanation

Frequency of vibration in tight string

$n = \frac{p}{2 l} \sqrt{\frac{T}{m}} \Rightarrow n \propto \sqrt{T}$ ⇒ $\frac{Δ n}{n} = \frac{Δ T}{2 T} = \frac{1}{2} \times (4 %) = 2 %$

⇒ Number of beats = $Δ n = \frac{2}{100} \times n = \frac{2}{100} \times 100 = 2$

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The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. If the velocity of sound in air is 300 m/sec, the velocity of the source is : (It is given that velocity of source << velocity of sound)

A.

6 m/sec

B.

3 m/sec

C.

1.5 m/sec

D.

12 m/sec

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Explanation

When the source approaches the observer

Apparent frequency $n' = \frac{v}{v - v_{s}} . n = n [\frac{1}{1 - \frac{v_{s}}{v}}]$

= $n {[1 - \frac{v_{s}}{v}]}^{- 1} = n [1 + \frac{v_{s}}{v}]$

(Neglecting higher powers because of v_S << v)

When the source recedes the observed apparent frequency $n^{''} = n [1 - \frac{v_{s}}{v}]$

Given $n' - n'' = \frac{2}{100} n, v = 300 m / s e c$

∴ $\frac{2}{100} n = n [1 + \frac{v_{s}}{v}] - n [1 - \frac{v_{s}}{v}] = n [2 \frac{v_{s}}{v}]$

$\Rightarrow \frac{2}{100} = 2 \frac{v_{s}}{v} \Rightarrow v_{s} = \frac{v}{100} = \frac{300}{100} = 3 m / s e c$

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Two whistles A and B produce notes of frequencies 660 Hz and 596 Hz respectively. There is a listener at the mid-point of the line joining them. Now the whistle B and the listener start moving with speed 30 m/s away from the whistle A. If the speed of sound be 330 m/s, how many beats will be heard by the listener :

A.

2

B.

4

C.

6

D.

8

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Explanation

For observer note of B will not change due to zero relative motion.

Observed frequency of sound produced by A

= $660 \frac{(330 - 30)}{330} = 600 H z$

∴ No. of beats = 600 – 596 = 4

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