2 Marks Questions - Physics STD 11 Science Questions - Vidyadip

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Question 12 Marks

A steel tube of length 1.00m is struck at one end. A Find the time gap between the two hearings. Use the person with his ear close to the other end hears the table in the text for speeds of sound in various sound of the blow twice, one travelling through the body substances. of the tube and the other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for speeds of sound in various substances.

Answer

$\text{V}_\text{air}=230\text{m/s},\ \text{V}_\text{s}=5200\text{m/s}$

Here $\text{S}=7\text{m}$

So,

$\text{t}=\text{t}_1-\text{t}_2=\Big(\frac{1}{330}-\frac{1}{5200}\Big)$

$=2.75\times10^{-3}\text{sec}=2.75\text{ms}.$

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Question 22 Marks

Find the minimum and maximum wavelengths of sound in water that is in the audible range (20-20000Hz) for an average human ear. Speed of sound in water = 1450m/s.

Answer

For minimum wavelength $\text{n}=20\text{KHz}$

$\Rightarrow\text{v}=\text{n}\lambda$

$\Rightarrow\lambda=\Big(\frac{1450}{20\times10^3}\Big)=7.25\text{cm}.$

For maximum wavelength n should be minium

$\Rightarrow\text{v}=\text{n}\lambda$

$\Rightarrow \lambda =\frac{\text{v}}{\text{n}}$

$\Rightarrow\frac{1450}{20}=72.5\text{m}.$

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Question 32 Marks

A bullet passes past a person at a speed of 220m/s. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. Speed of sound in air = 330m/s.

Answer

Let the frequency of the bullet will be f
Given, $\text{u}=330\text{m/s},\ \text{v}_\text{s}=220\text{m/s}$
Apparent frequency before crossing = $\text{f}'\Big(\frac{330}{330-220}\Big)\text{f}=3\text{f}$
Apparent frequency after crossing = $\text{f}'\Big(\frac{330}{330+220}\Big)\text{f}=0.6\text{f}$
So, $\Big(\frac{\text{f}'}{\text{f}'}\Big)=\frac{0.6\text{f}}{3\text{f}}=0.2$
Therefore, fractional change = 1 - 0.2 = 0.8.

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Question 42 Marks

Two point sources of sound are kept at a separation of 10cm. They vibrate in phase to produce waves of wavelength 5.0cm. What would be the phase difference between the two waves arriving at a point 20cm from one source-

On the line joining the sources.
On the perpendicular bisector of the line joining the sources?

Answer

Given $\triangle\text{x}=10\text{cm},\ \lambda=5.0\text{cm}$

$\Rightarrow\phi=\frac{2\pi}{\lambda}\times\triangle\eta=\frac{2\pi}{5}\times10=4\pi.$

So phase difference is zero.

Zero, as the particle is in same phase because of having same path.

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Question 52 Marks

Two sources of sound S₁ and S₂ vibrate at same frequency and are in phase figure. The intensity of sound detected at a point P as shown in the figure is I₀.

If 0 equals 45^°, what will be the intensity of sound detected at this point if one of the sources is switched off?
What will be the answer of the previous part if = 60^°?

Answer

The two sources of sound S₁ and S₂ vibrate at same phase and frequency. Resultant intensity at P = I₀

Let the amplitude of the waves at S₁ and S₂ be ‘r’

When $\theta=45^\circ,$ path difference = S₁P - S₂P = 0 (because S₁P = S₂P)

So, when source is switched off, intensity of sound at P is $\frac{\text{I}_0}{4}.$

When $\theta=60^\circ,$ path difference is also 0.

Similarly it can be proved that, the intensity at P is $\frac{\text{I}_0}{4}$ when one is switched off.

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Question 62 Marks

Two trains are travelling towards each other both at a speed of 90km/h. If one of the trains sounds a whistle at 500Hz, what will be the apparent frequency heard in the other train? Speed of sound in air = 350m/s.

Answer

According to the given data V_s = 90km/hour = 25m/sec. v₀ = 25 m/sec

So, apparent frequency heard by the observer in train B or

Observer in $=\Big(\frac{350+25}{350-25}\Big)\times500=577\text{Hz}.$

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Question 72 Marks

An open organ pipe has a length of 5cm.

Find the fundamental frequency of vibration of this pipe.
What is the highest harmonic of such a tube that is in the audible range? Speed of sound in air is 340m/s and the audible range is 20-20,000Hz.

Answer

Here given $\text{l}=5\text{cm}=5\times10^{-2}\text{m},\ \text{v}=340\text{m/s}$

$\Rightarrow\text{n}=\frac{\text{V}}{2\text{l}}=\frac{340}{2\times5\times10^{-2}}=3.4\text{Khz}$

If the fundamental frequency $=3.4\text{Khz}$

⇒ Then the highest harmonic in the audible range (20Hz - 20KHz)

$=\frac{20000}{3400}=5.8=5$ (integral multiple of 3.4KHz).

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Question 82 Marks

A small source of sound vibrating at frequency 500Hz is rotated in a circle of radius $\frac{100}{\pi}\text{cm}$ at a constant angular speed of 5.0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air = 332m/s.

Answer

According to the given data

Radius of the circle $\frac{100}{\pi}\times10^{-2}\text{m}=\Big(\frac{1}{\pi}\Big)\ \text{meteres};\ \omega=5\text{rev/sec}.$

So the linear speed $\text{v}=\omega\text{r}=\frac{5}{\pi}=1.59$

So, velocity of the source $\text{V}_\text{s}=1.59\text{m/s}$

As shown in the figure at the position A the observer will listen maximum and at the position B it will listen minimum frequency.

So, apparent frequency at $\text{A}=\frac{332}{332-1.59}\times500=515\text{Hz}$

Apparent frequency at $\text{B}=\frac{332}{332+1.59}\times500=485\text{Hz}.$

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Question 92 Marks

Figure, shows a person standing somewhere in between two identical tuning forks, each vibrating at 512Hz. If both the tuning fbrks move towards right at a speed of 5.5m/s, find the number of beats heard by the listener. Speed of sound in air = 330m/s.

Answer

According to the data, V_s = 5.5m/s for each turning fork.

So, the apparent frequency heard from the tuning fork on the left,

$\text{f}'=\Big(\frac{330}{330-5.5}\Big)\times512=527.36\text{Hz}=527.5\text{Hz}$

Similarly, apparent frequency from the tunning fork on the right,

$\text{f}'=\Big(\frac{330}{330-5.5}\Big)\times512=510\text{Hz}$

So, beats produced $527.5-510=17.5\text{Hz}.$

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Question 102 Marks

A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in figure, It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of 20cm. Find the frequency of the sound emitted. Velocity of sound in air is 336m/s.

Answer

According to the given data

$\text{V}=336\text{m/s},$

$\frac{\lambda}{4}$ distance between maximum and minimum intensity

$=(20\text{cm})\Rightarrow\lambda=80\text{cm}$

⇒ n = frequency = $\frac{\text{V}}{\lambda}=\frac{336}{80\times10^{-2}}$

$=420\text{Hz}.$

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Question 112 Marks

At what temperature will the speed of sound be double of its value at 0^°C?

Answer

$\text{T}_1=273,\ \text{V}_2=2\text{v}_1$

$\text{V}_1=\text{v},\ \text{T}_2=?$

We know that $\text{V}\propto\sqrt{\text{T}}$

$\Rightarrow\frac{\text{T}_2}{\text{T}_1}=\frac{\text{V}^2_2}{\text{V}^2_1}$

$\Rightarrow\text{T}_2=273\times2^2=4\times273\text{K}$

So temperature will be $(4\times273)-273=819^\circ\text{c}.$

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Question 122 Marks

A vertical rod is hit at one end. What kind of wave propagates in the rod if:

The hit is made vertically.
The hit is made horizontally?

Answer

If hit is made vertically the wave is longitudinal.
If hit is made horizontal the wave is transverse.

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Question 132 Marks

The separation between a node and the next antinode in a vibrating air column is 25cm. If the speed of sound in air is 340m/s, find the frequency of vibration of the air column.

Answer

$\text{v}=340\text{m/s}$

Distances between two nodes or antinodes

$\Rightarrow\frac{\lambda}{4}=25\text{cm}$

$\Rightarrow\lambda=100\text{cm}=1\text{m}$

$\Rightarrow\text{n}=\frac{\text{v}}{\lambda}=340\text{Hz}.$

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Question 142 Marks

Figure, shows two coherent sources S₁ and S₂ which emit sound of wavelength $\lambda$ in phase. The separation between the sources is $3\lambda.$ A circular wire of large radius is placed in such a way that S₁S₂ lies in its plane and the middle point of S₁S₂ is at the centre of the wire. Find the angular positions $\theta$ on the wire for which constructive interference takes place.

Answer

Let the sound waves from the two coherent sources S₁ and S₂ reach the point P.

$\text{OQ}=\text{R}\cos\theta$

$\text{OP}=\text{R}\cos\theta$

$\text{OS}_2=\text{OS}_1=1.5\lambda$

From the figure, we find that:

$\text{PS}_1^2=\text{PQ}^2+\text{QS}62=(\text{R}\sin\theta)^2+(\text{R}\cos\theta-1.5\lambda)^2$

$\text{PS}_1^2=\text{PQ}^2+\text{QS}^2_1=(\text{R}\sin\theta)^2+(\text{R}\cos\theta+1.5\lambda)^2$

Path difference between the sound waves reaching point P is given by:

$(\text{S}_1\text{P})^2-(\text{S}_2\text{P})^2=\Big[(\text{R}\sin\theta)^2+(\text{R}\cos\theta+1.5)^2\Big]\\-\Big[(\text{R}\sin\theta)^2+(\text{R}\cos\theta-1.5\lambda)^2\Big]$

$=(1.5\lambda+\text{R}\cos\theta)^2-(\text{R}\cos\theta-1.5\lambda)^2$

$=6\lambda\text{R}\cos\theta$

$(\text{S}_1\text{P}-\text{S}_2\text{P})=\frac{6\lambda\text{R}\cos\theta}{2\text{R}}$

Suppose, for construction interference, the path difference be made equal to the integral multiple of $\lambda.$

Hence,

$(\text{S}_1\text{P}-\text{S}_2\text{P})=3\lambda\cos\theta=\text{n}\lambda$

$\Rightarrow\cos\theta=\frac{\text{n}}{3}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{\text{n}}{3}\Big)$

Where, n = 0, 1, 2, ...

$\Rightarrow\theta=0^\circ,48.2^\circ,70.5^\circ$ and $90^\circ$ are similar point in other quadrants.

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Question 152 Marks

Show that if the room temperature changes by a small amount from T to $\text{T}+\triangle\text{T},$ the fundamental frequency of an organ pipe changes from v to $\text{v}+\triangle\text{v},$ where-

$\frac{\triangle\text{v}}{\text{v}}=\frac{1}{2}\frac{\triangle\text{T}}{\text{T}}.$

Answer

We know that $\text{f}\propto\sqrt{\text{T}}$

According to the question $\text{f}+\triangle\text{f}\propto\sqrt{\triangle\text{T}}+\text{T}$

$\Rightarrow\frac{\text{f}+\triangle\text{f}}{\text{f}}=\sqrt{\frac{\triangle\text{t}+\text{T}}{\text{T}}}$

$\Rightarrow1+\frac{\triangle\text{f}}{\text{f}}=\Big(1+\frac{\triangle\text{T}}{\text{T}}\Big)^{\frac{1}{2}}=1+\frac{1}{2}\frac{\triangle\text{T}}{\text{T}}+\ \dots$ (neglecting other terms)

$\Rightarrow\frac{\triangle\text{f}}{\text{f}}=\Big(\frac{1}{2}\Big)\frac{\triangle\text{T}}{\text{T}}$

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Question 162 Marks

The fundamental frequency of a closed pipe is 293Hz when the air in it is at a temperature of 20^°C. What will be its fundamental frequency when the temperature changes to 22^°C?

Answer

We know that the frequency = f, T = temperatures

$\text{f}\propto\sqrt{\text{T}}$

So, $\frac{\text{f}_1}{\text{f}_2}=\frac{\sqrt{\text{T}_1}}{\sqrt{\text{T}}}$

$\Rightarrow\frac{293}{\text{f}_2}=\frac{\sqrt{293}}{\sqrt{295}}$

$\Rightarrow\text{f}_2=\frac{293\times\sqrt{295}}{\sqrt{293}}=294$

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Question 172 Marks

Three sources of sound S₁, S₂ and S₃ of equal intensity are placed in a straight line with S₁S₂ = S₂S₃ figure. At a point P, far away from the sources, the wave coming from S is 120^° ahead in phase of that from S₁. Also, the wave coming from S₃ is 120^° ahead of that from S₂. What would be the resultant intensity of sound at P?

Answer

Because the 3 sources have equal intensity, amplitude are equal
So, A₁ = A₂ = A₃
As shown in the figure, amplitude of the resultant = 0 (vector method)
So, the resultant, intensity at B is zero.

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Question 182 Marks

Two loudspeakers are arranged facing each other at some distance. Will a person standing behind one of the loudspeakers clearly hear the sound of the other loudspeaker or the clarity will be seriously damaged because of the 'collision' of the two sounds in between?

Answer

The sound waves of both speakers would collide with each other and the sound waves of the speaker facing towards man would be damaged up to some extent.

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Question 192 Marks

Draw a diagram to show the standing pressure wave and standing displacement wave for the 3rd overtone mode of vibration of an open organ pipe.

Answer

The displacement node is a pressure anti-node and via-versa.

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Question 202 Marks

Two tuning forks vibrate with the same amplitude but the frequency of the first is double the frequency of the second. Which fork produces more intense sound in air?

Answer

If both forks have same amplitude then the fork having high frequency will produce more intense sound in air.

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Question 212 Marks

In discussing Doppler effect, we use the word "apparent frequency". Does it mean that the frequency of the sound is still that of the source and it is some physiological phenomenon in the listener's ear that gives rise to Doppler effect? Think for the observer approaching the source and for the source approaching the observer.

Answer

In both cases if observer approaches source or source approaches observer the apparent frequency is physiological phenomenon as the frequency of the sound is still the same but due to motion w.r.t. each other the apparent frequency changes.

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Question 222 Marks

Can you hear your own words if you are standing in a perfect vacuum? Can you hear your friend in the same conditions?

Answer

We can hear our own words if we are in vacuum as sound wave would travel through our body.
But we cannot hear friend’s voice as it does not has any material to propagate.

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Question 232 Marks

The voice of a person, who has inhaled helium, has a remarkably high pitch. Explain on the basis of resonant vibration of vocal cord filled with air and with helium.

Answer

As helium is light gas it is very less dense so sound wave propagates faster in helium gas. Thus frequency of the speech is high thus making high pitch voice.

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Question 242 Marks

The speed of sound as measured by a student in the laboratory on a winter day is 340m/s when the room temperature is 17^°C. What speed will be measured by another student repeating the experiment on a day when the room temperature is 32^°C?

Answer

$\text{V}_1=330\text{m/s},\ \text{V}_2=?$

$\text{T}_1=273+17=290\text{K},\ \text{T}_2=272+32=305\text{K}$

We know $\text{v}\propto\sqrt{\text{T}}$

$\frac{\sqrt{\text{V}_1}}{\sqrt{\text{V}_2}}=\frac{\sqrt{\text{T}_1}}{\sqrt{\text{T}}_2}$

$\Rightarrow\text{v}_2=\frac{\text{V}_1\times\sqrt{\text{T}_2}}{\sqrt{\text{T}_1}}$

$=340\times\sqrt{\frac{305}{290}}=349\text{m/s}$

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Question 252 Marks

Sound with intensity larger than 120dB appears painful to a person. A small speaker delivers 2.0W of audio output. How close can the person get to the speaker without hurting his ears?

Answer

If sound level = 120dB, then I = intensity = 1W/m²

Given that, audio output = 2W

Let the closest distance be x.

So, intensity $=\Big(\frac{2}{4\pi\text{x}^2}\Big)=1$

$\Rightarrow\text{x}^2=\Big(\frac{2}{2\pi}\Big)$

$\Rightarrow\text{x}=0.4\text{m}$

$=40\text{cm}.$

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Question 262 Marks

Find the fundamental, first overtone and second overtone frequencies of an open organ pipe of length 20cm. Speed of sound in air is 340m/s.

Answer

If $\text{V}=340\text{m/s},\text{I}=20\text{cm}=20\times10^{-2}\text{m}$
Fundamental frequency $=\frac{\text{V}}{21}=\frac{340}{2\times20\times10^{-2}}=850\text{Hz}$
We know first over tone $=\frac{2\text{V}}{21}=\frac{2\times340}{2\times20\times10^{-2}}$ (for open pipe) = 1750Hz
Second over tone $=3\Big(\frac{\text{V}}{21}\Big)=3\times850=2550\text{Hz}.$

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Question 272 Marks

In a Quincke's experiment the sound detected is changed from a maximum to a minimum when the sliding tube is moved through a distance of 2.50cm. Find the frequency of sound if the speed of sound in air is 340m/s.

Answer

Distance between tow maximum to a minimum is given by, $\frac{\lambda}{4}=2.50\text{cm}$

$\Rightarrow\lambda=10\text{cm}=10^{-1}\text{m}$

We know, V = nx

$\Rightarrow\text{n}=\frac{\text{V}}{\lambda}=\frac{340}{10^{-1}}$

$=3400\text{Hz}$

$=3.4\text{KHz}$

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Question 282 Marks

A tuning fork produces 4 beats per second with another tuning fork of frequency 256Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?

Answer

Frequency of tuning fork A,

n₁ = 256 Hz

No. of beats/ second m = 4

Frequency of second fork B: n₂ = ?

$\text{n}_2=\text{n}_1\pm\text{m}$

$\Rightarrow\text{n}_2=256\pm4$

$\Rightarrow\text{n}_2=260\text{Hz or 252 Hz}$

Now, as it is loaded with wax, its frequency will decrease.

As it produced 6 beats per second, the original frequency, the beats second should decreas, wich is not possible.

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Question 292 Marks

A cylindrical metal tube has a length of 50cm and is open at both ends. Find the frequencies between 1000Hz and 2000Hz at which the air column in the tube can resonate. Speed of sound in air is 340m/s.

Answer

Given:

Length of cylindrical metal tube L = 50cm

Speed of sound in air v = 340ms^-1

Fundamental frequency (f₁) of an open pipe:

$\text{f}_1=\Big(\frac{\text{v}}{2\text{L}}\Big)$

$=\frac{340}{2\times 50\times 10^{-2}}=340\text{Hertz}$

So, the required hamonics will be in the range of 1000Hz to 2000Hz.

$\text{f}_2=2\times340=380\text{Hz}$

$\text{f}_3=3\times340=1020\text{Hz}$

$\text{f}_4=4\times 340=1360\text{Hz}$

$\text{f}_5=5\times 340=1700\text{Hz}$

$\text{f}_6=6\times 340=2040\text{Hz}$

f₂, f₃, f₄... are the second, third , fourth overtone, and so on.

The possible frequecies between 1000Hz and 2000Hz are 1020Hz and 1700Hz.

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Question 302 Marks

In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0cm. If the speed of sound in air is 328m/s, what is the frequency of 73 the source?

Answer

Here given distance between two nodes is = 4.0cm,

$\Rightarrow\lambda=2\times4.0=8\text{cm}$

We know that $\text{V}=\text{n}\lambda$

$\therefore \text{f}=\frac{\text{v}}{\lambda}=\frac{328}{8\times 10^{-2}}=4.1\text{KHz}$

$\eta=\frac{328}{8\times10^{-2}}=4.1\text{Hz}.$

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Question 312 Marks

Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range (20Hz-20,000Hz). Speed of sound in air = 340m/s.

Answer

Let the length will be l.
Here given that $\text{V}=340\text{m/s}$ and $\text{n}=20\text{Hz}$
Here $\frac{\lambda}{2}=\text{l}\Rightarrow\lambda=2\text{l}$
We know $\text{V}=\text{n}\lambda\Rightarrow\text{l}=\frac{\text{V}}{\text{n}}=\frac{340}{2\times20}=\frac{34}{4}=8.5\text{cm}$ (for maximum wavelength, the frequency is minimum).

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Question 322 Marks

Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker.

Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20cm.
Sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of the speaker described above. Take the speed of sound to be 340m/s.

Answer

According to the question,

$\lambda=20\text{cm}\times10=200\text{cm}=2\text{m}$

$\text{v}=340\text{m/s}$

So, $\text{n}=\frac{\text{v}}{\lambda}=\frac{340}{2}=170\text{Hz}.$

$\text{N}=\frac{\text{v}}{\lambda}=\frac{340}{2\times10^{-2}}=17.000\text{Hz}$

$=17\text{KH}_2$ $\big($because $\lambda=2\text{cm}=2\times10^{-2}\text{m}\big)$

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Question 332 Marks

Two electric trains run at the same speed of 72km/h along the same track and in the same direction with a separation of 2.4km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500Hz and the speed of sound in air is 340m/s, find the frequencies heard by the person.

Answer

Given:

Speed of sound in air v = 340 ms⁻¹

Frequency of whistles f₀= 500Hz

Speed of train v_s= 72 km/ h

$=72\times \frac{5}{18}=20\text{m}/\text{s}$

The person will receive the sound in a direction that makes an angle θ with the track. The angle θ is given by,

$\theta =\tan^{-1}\bigg(\frac{0.5}{\frac{2.4}{2}}\bigg)=22.62^\circ$

The velocity of the source will be $'\text{v}\cos \theta'$ when heard by the observer.

So, the apparent frequency received by the man from train A is:

$\text{f}_1=\Big(\frac{\text{v}}{\text{v}-\text{v}_\text{s}\cos\theta}\Big)\times \text{f}_0$

$\Rightarrow \text{f}_1=\Big(\frac{340}{340-\text{v}_\text{s}\cos22.62^\circ}\Big)\times500$

$\Rightarrow \text{f}_1=\Big(\frac={340}{340-20\times \cos 22.62^\circ}\Big)\times 500$

$\Rightarrow \text{f}_1=528.70\text{Hz}\approx529\text{Hz}$

The apparent frequency heard by the man from train B is:

$\text{f}_2=\Big(\frac{\text{v}}{\text{v}+\text{v}\cos\theta}\Big)\times \text{f}_0$

$\Rightarrow \text{f}_2=\Big(\frac{340}{340+20\times \cos22.62^\circ}\Big)\times 500$

$\Rightarrow \text{f}_2=474.24\text{Hz}\approx 474\text{hz}$

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Question 342 Marks

A Kundt's tube apparatus has a steel rod of length 1.0m clamped at the centre. It is vibrated in its fundamental mode at a frequency of 2600Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5cm. Calculate the speed of sound in steel and in air.

Answer

Here given,

$\text{L}_\text{r}=\frac{1.0}{2}=0.5\text{m},\ \text{d}_\text{a}=6.5\text{cm}\times10^{-2}\text{m}$

As Kundt’s tube apparatus is a closed organ pipe, its fundamental frequency

$\text{n}=\frac{\text{V}_\text{r}}{4\text{L}_\text{r}}$

$\Rightarrow\text{V}_\text{r}=2600\times4\times0.5=5200\text{m/s}$

$\frac{\text{V}_\text{r}}{\text{V}_\text{a}}=\frac{2\text{L}_\text{r}}{\text{d}_\text{a}}$

$\Rightarrow\text{V}_\text{a}=\frac{5200\times6.5\times10^{-2}}{2\times0.5}=338\text{m/s}$

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Question 352 Marks

A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either 476Hz or 480Hz. What is the frequency of the tuning fork?

Answer

Given,

First Frequency f₁ = 476Hz

Second frequency f₂ = 480Hz

Number of beats produced per second by the tuning fork m = 2

As the tuning fork produces 2 beats, its frequency should be an average of two. This is given by,

$\text{f}=\frac{(\text{f}_1+\text{f}_2)}{2}$

$\text{f}=\frac{(476+480)}{2}=478\text{Hz}$

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Question 362 Marks

The first overtone frequency of a closed organ pipe P₁ is equal to the fundamental frequency of an open organ pipe P₂. If the length of the pipe P₁ is 30cm, what will be the length of P₂?

Answer

According to the questions

f₁ first overtone of a closed organ pipe $\text{P}_1=\frac{3\text{v}}{4\text{I}}=\frac{3\times\text{V}}{4\times30}$

f₂ fundamental frequency of a open organ pipe $\text{P}_2=\frac{\text{V}}{2\text{I}_2}$

Here given $\frac{3\text{V}}{4\times30}=\frac{\text{V}}{2\text{I}_2}\Rightarrow\text{I}_2=20\text{cm}$

$\therefore\ $length of the pipe P₂ will be 20cm.

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Question 372 Marks

Ultrasonic waves of frequency 4.5MHz are used to detect tumour in soft tissues. The speed of sound in tissue is 1.5km/s and that in air is 340m/s. Find the wavelength of this ultrasonic wave in air and in tissue.

Answer

Given $\text{V}_\text{air}=340\text{m/s},\ \text{n}=4.5\times10^6\text{Hz}$

$\Rightarrow\lambda_\text{air}=\Big(\frac{340}{4.5}\Big)\times10^{-6}=7.36\times10^{-5}\text{m}$

$\text{V}_\text{tissue}=1500\text{m/s}$

$\Rightarrow\lambda_\text{t}=\Big(\frac{1500}{4.5}\Big)\times10^{-6}$

$=3.3\times10^{-4}\text{m}.$

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Question 382 Marks

A 30.0cm long wire having a mass of 10.0g is fixed at the two ends and is vibrated in its fundamental mode. A 50.0cm long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air = 340m/s.

Answer

Given, $\text{m}=10\text{g}=10\times10^{-3}\text{kg},\ \text{l}=30\text{cm}=0.3\text{m}$

Let the tension in the string will be = T

$\mu=\frac{\text{Mass}}{\text{Unit length}}=33\times10^{-3}\text{kg}$

The fundamental frequency $\Rightarrow\text{n}_0=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}\ \dots(1)$

The fundamental frequency of closed pipe

$\Rightarrow\text{n}_0=\Big(\frac{\text{v}}{4\text{l}}\Big)=\frac{340}{4\times50\times10^{-2}}=170\text{Hz}\ \dots(2)$

According equations (1) × (2) we get

$170=\frac{1}{2\times30\times10^{-2}}\times\sqrt{\frac{\text{T}}{33\times10^{-3}}}$

$\Rightarrow\text{T}=347\ \text{Newton}$

Hence, then tansion in the wire is 347 N.

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Question 392 Marks

In a resonance column experiment, a tuning fork of frequency 400Hz is used. The first resonance is observed when the air column has a length of 20.0cm and the second resonance is observed when the air column has a length of 62.0cm.

Find the speed of sound in air.
How much distance above the open end does the pressure node form?

Answer

Here given $\text{I}_2=0.67\text{m},\ \text{I}_1=0.2\text{m},\ \text{f}=400\text{Hz}$

We know that $\lambda=2(\text{I}_2-\text{I}_1)$

$\Rightarrow\lambda=2(62-20)=84\text{cm}=0.84\text{m}$

So, $\text{v}=\text{n}\lambda=0.84\times400=336\text{m/s}$

We know from above that, $\text{I}_1+\text{d}=\frac{\lambda}{4}$

$\Rightarrow\text{d}=\frac{\lambda}{4}-\text{I}_1=21-20=1\text{cm}.$

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Question 402 Marks

Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35cm travelling in air. The pressure at a point varies between $(1.0\times10^5\pm14)$ Pa and the particles of the air vibrate in simple harmonic motion of amplitude 5.5 × 10^-6m.

Answer

We know that,

Bulk modulus $\text{B}=\frac{\triangle\text{p}}{\big(\frac{\triangle\text{V}}{\text{V}}\big)}=\frac{\text{P}_0\lambda}{2\pi\text{S}_0}$

Where P₀ = pressure amplitude $\Rightarrow\text{P}_0=1.0\times10^5$

S₀ = displacement amplitude $\Rightarrow\text{S}_0=5.5\times10^{-6}\text{m}$

$\Rightarrow \text{B}=\frac{\text{P}_0\lambda}{2\pi\text{S}_0}=\frac{\triangle \text{p}}{\frac{\triangle\text{V}}{\text{V}}}$

$\Rightarrow\text{B}=\frac{14\times35\times10^2\text{m}}{2\pi(5.5)\times10^{-6}\text{m}}$

$=1.4\times10^5\text{N/m}^2.$

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Question 412 Marks

A man stands before a large wall at a distance of 50.0m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.

Answer

He has to clap 10 times in 3 seconds.

So time interval between two clap $=\Big(\frac{3}{10}\text{second}\Big).$

So the time taken go the wall $=\Big(\frac{3}{2}\times10\Big)$

$=\frac{3}{20}\text{second}$

$=333\text{m/s}$

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Question 422 Marks

An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is 80cm. The frequency of the loudspeaker can be varied between 20Hz-2kHz. Find the frequencies at which the column will resonate. Speed of sound in air = 320m/s.

Answer

The resonance column apparatus is equivalent to a closed organ pipe.

Here $\text{l}=80\text{cm}=10\times10^{-2}\text{m};\ \text{v}=320\text{m/s}$

$\Rightarrow\text{n}_0=\frac{\text{v}}{4\text{l}}=\frac{320}{4\times50\times10^{-2}}=100\text{Hz}$

So the frequency of the other harmonics are odd multiple of n₀ = (2n + 1)100Hz.

According to the question, the harmonic should be between 20Hz and 2KHz.

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Question 432 Marks

A closed organ pipe can vibrate at a minimum frequency of 500Hz. Find the length of the tube. Speed of sound in air = 340m/s.

Answer

According to the questions $\text{V}=340\text{m/s},\ \text{n}=500\text{Hz}$

$\text{f}=\frac{\text{v}}{4\text{L}}$

$\therefore \text{L}=\frac{\text{v}}{4\text{f}}$

$\therefore \text{L}=\frac{340}{4\times 500}=0.17\text{m}$

$=17\text{cm}$

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Question 442 Marks

A tuning fork of frequency 256Hz produces 4 beats per second with a wire of length 25cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?

Answer

Given that, $\text{l}=25\text{cm}=25\times10^{-2}\text{m}$

By shortening the wire the frequency increases, $\Bigg[\text{f}=\Big(\frac{1}{2\text{l}}\Big)\sqrt{\Big(\frac{\text{TB}}{\text{M}}\Big)}\Bigg]$

As the vibrating wire produces 4 beats with 256Hz, its frequency must be 252Hz or 260Hz.

Its frequency must be 252Hz, because beat frequency decreases by shortening the wire.

So, $252=\frac{1}{2\times25\times10^{-2}}\sqrt{\frac{\text{T}}{\text{M}}}\ \dots(1)$

Let length of the wire will be l, after it is slightly shortened,

$\Rightarrow256=\frac{1}{2\times\text{l}_1}\sqrt{\frac{\text{T}}{\text{M}}}\ \dots(2)$

Dividing (1) by (2) we get

$\frac{252}{256}=\frac{\text{l}_1}{25\times10^{-2}}$

$\Rightarrow\frac{252\times25\times10^{-2}}{256}$

$\Rightarrow 0.24609\text{m}$

So, it should be shorten by (25 - 24.61) = 0.39cm.

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Question 452 Marks

A person can hear sound waves in the frequency range 20Hz to 20kHz. Find the minimum and the maximum wavelengths of sound that is audible to the person. The speed of sound is 360m/s.

Answer

For maximum wavelength $\text{n}=20\text{Hz}.$

As $\Big(\eta\propto\frac{1}{\lambda}\Big)$

For minimum wavelength, $\text{n}=20\text{kHz}$

$\therefore\lambda=\frac{360}{(20\times10^3)}=18\times10^{-3}\text{m}=18\text{mm}$

$\Rightarrow\text{x}=\Big(\frac{\text{v}}{\text{n}}\Big)=\frac{360}{20}=18\text{m}.$

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Question 462 Marks

Two successive resonance frequencies in an open organ pipe are 1944Hz and 2592Hz. Find the length of the tube. The speed of sound in air is 324m/s.

Answer

Let the length of the resonating column will be = 1

Here $\text{V}=320\text{m/s}$

Then the two successive resonance frequencies are $\frac{(\text{n}+1)\text{v}}{4\text{l}}$ and $\frac{\text{nv}}{4\text{l}}$

Here given $\frac{(\text{n}+1)\text{v}}{4\text{l}}=2592;\ \lambda=\frac{\text{nv}}{4\text{l}}=1944$

$\Rightarrow\frac{(\text{n}+2)\text{v}}{4\text{l}}-\frac{\text{nv}}{4\text{I}}=2592-1944=648$

$\Rightarrow\frac{2\text{v}}{\text{I}}=648$

$\Rightarrow\text{l}=\frac{2\times324\times100}{4\times648}\text{cm}=25\text{cm}$

Hence, the length of the tube is 25cm.

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Question 472 Marks

If the sound level in a room is increased from 50dB to 60dB, by what factor is the pressure amplitude increased?

Answer

$\beta_1=50\text{dB},\ \beta_2=60\text{dB}$

$\therefore\text{I}_1=10^{-7}\text{W/m},\ \text{I}_2=10^{-6}\text{W/m}^2$

$\Big($because $\beta=10\log_{10}\big(\frac{\text{I}}{\text{I}_0}\big),$ where $\text{I}_0=10^{-12}\text{W/m}^2\Big)$

Again, $\frac{\text{I}_2}{\text{l}_1}=\Big(\frac{\text{P}_2}{\text{P}_1}\Big)^2$

$=\Big(\frac{10^{-6}}{10^{-7}}\Big)$

$=10$ (where p = pressure amplitude)

$\therefore\Big(\frac{\text{P}_2}{\text{P}_1}\Big)=\sqrt{10}.$

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2 Marks Questions - Physics STD 11 Science Questions - Vidyadip