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Physics NEET Questions — Free MCQ Practice

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Quality of a musical note depends on :

A.

Harmonics present

B.

The amplitude of the wave

C.

Fundamental frequency

D.

The velocity of sound in the medium

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Explanation

The quality of sound depends upon the number of harmonics present. Due to different number of harmonics present in two sounds, the shape of the resultant wave is also different.

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

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

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

Physics MCQs for NEET — Practice Questions with Answers

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