Waves Class 11 Important Extra Questions Physics Chapter 15

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 15 Waves. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 15 Important Extra Questions Waves

Waves Important Extra Questions Very Short Answer Type

Harmonic sequence formula is a sequence of real numbers formed by taking the reciprocals of an arithmetic progression.

Question 1.
In a resonance tube, the second resonance does not occur exactly at three times the length at the first resonance. Why?
Answer:
This is due to the end correction.

Question 2.
What is the nature of ultrasonic waves and what is their frequency?
Answer:
Ultrasonic waves are longitudinal waves in nature and have frequencies greater than 20 k Hz.

Question 3.
Is the principle of superposition wave valid in the case of electromagnetic (e.m.) waves?
Answer:
Yes.

Question 4.
A rod is clamped at one end and it is hit by a hammer at the other end
(a) at a right angle to its length
(b) along the length.
What types of waves are produced in each case?
Answer:
(a) Transverse waves
(b) Longitudinal waves.

Question 5.
Why do not we hear beats if the frequency of ìwo sounds is widely different?
Answer:
The beats can’t be heard as separate due to the persistence of hearing if the difference in frequencies is more than 10 Hz.

Question 6.
What causes the rolling sound of thunder?
Answer:
The multiple reflections of the sound of lighting results in the rolling ’ sound of thunder.

Question 7.
A tuning fork produces resonance in a closed pipe. But the ‘ same tuning fork is unable to, produce resonance in an open organ pipe of equal length. Why?
Answer:
It is because the fundamental frequencies of open and closed organ pipes of the same length are different,

Question 8.
Thick and big curtains are preferred in a big hall. Why?
Answer:
To increase the absorption and decrease the reverberation time and hence making the sound uniform,

Question 9.
Why female voice is sweeter than that of a man?
Answer:
This is because the frequency of a lady’s voice is greater than that of a man’s voice.

Question 10.
The frequency of the fundamental note of a tube closed at one end is 200 Hz. What will be the frequency of the fundamental note of a similar tube of the same length but open at both ends?
Answer:
400 Hz.

Question 11.
A wave transmits energy. Can it transmit momentum?
Answer:
Yes.

Question 12.
By how much the wave velocity increases for 1°C rise of temperature?
Answer:
Wave velocity increases by 0.61 ms for 1°C rise of temperature.

Question 13.
Why the sound heard is more in carbon dioxide than in air?
Answer:
This is because the intensity of sound increases with the increase in the density of the medium.

Question 14.
What is the relation between path difference and phase difference?
Answer:
Phase difference = \(\frac{2π}{λ}\) × path difference.

Question 15.
Is it possible to have interference between the waves produced by two violins? Why?
Answer:
No. This is because the sounds produced will not have a constant, phase difference.

Question 16.
The windowpanes of houses sometimes get cracked due to some explosion at a large distance. Which waves are responsible for this?
Answer:
Shockwaves.

Question 17.
Why the velocity of sound is generally greater in solids than in gases?
Answer:
This is because \(\frac{E}{ρ}\) for solids is much greater than for gases.

Question 18.
An observer places his ear at the end of a long steel pipe. He can hear two sounds when a workman hammers the other end of the pipe. Why?
Answer:
This is because the sound is transmitted both through air and medium.

Question 19.
Why a stationary wave is so named?
Answer:
A stationary wave is so named because there is no net propagation of energy. .

Question 20.
How do we identify our friends from his voice while sitting in a dark room?
Answer:
The quality of sound helps us to identify the sound.

Question 21.
Why bells are made of metal and not of wood?
Answer:
This is because the wood has high damping.

Question 22.
What is the nature of light waves?
Answer:
Electromagnetic waves.

Question 23.
When a vibrating tuning fork is moved speedily towards a wall, beats are heard. Why?
Answer:
This is due to the difference in the frequency of the incident wave and the apparent frequency of the reflected wave.

Question 24.
The weight suspended from a sonometer wire is increased by a factor of 4. Will the frequency of the wire be increased exactly by a factor of 2? Justify your answer.
Answer:
No. There will be a slight increase in the length of the wire. So the frequency shall become slightly less than double.

Question 25.
Are sound waves in air longitudinal or transverse?
Answer:
Longitudinal.

Question 26.
Can you notice the Doppler effect if both the listener and the source of sound are moving with the same velocity in the same
direction? Why?
Answer:
No. This is because there is no relative motion between the source of sound and the listener.

Question 27.
If oil of density higher than that of water is used in place of water in a resonance tube, how does the frequency change?
Answer:
The frequency is governed by the air column and does not depend, upon the nature of the liquid. So frequency would not change.

Question 28.
Which property of the medium enables the transverse waves to pass through it?
Answer:
Modulus of rigidity.

Question 29.
In a thunderstorm why flash of light is seen earlier than the sound of thunder?
Answer:
The speed of light is much higher than the speed of sound waves.

Question 30.
What is the main difference between the brazing of a honeybee and the roaring of a lion?
Answer:
Brazing of honeybee is of high pitch and lion’s roar has high intensity.

Question 31.
If you set your watch by the sound of a distant siren, will it go fast or slow?
Answer:
It will go slow due to the low value of the speed of sound in the air.

Question 32.
What is a dispersive medium?
Answer:
A medium in which the wave velocity depends on the frequency of the wave is called a dispersive medium.

Question 33.
Will sound be louder at node or antinode in a stationary wave?
Answer:
It will be louder at node due to large pressure variation there as
ΔP = – strain × elasticity.

Question 34.
Even if a powerful thermonuclear explosion takes place on a planet, the sound is not heard on Earth. Why?
Answer:
The interplanetary space is devoid of continuous material. In the absence of a material, the medium sound is not heard on the earth.

Question 35.
In older days messages were conveyed to distant villages by beating of big drums. Why?
Answer:
It has a larger area and the intensity of sound (I) is directly proportional to the area of an oscillator (A) i.e. I ∝ A.

Question 36.
Which is the most basic property of the wave?
Answer:
Frequency is the basic property of the wave.

Question 37.
You can make waves in a pond by throwing a stone in it. What is the source of energy of the wave?
Answer:
The kinetic energy of the stone is the source of wave energy.

Question 38.
Why echoes are not heard in a small room?
Answer:
Because the minimum distance between the obstacle reflecting sound waves and the source of sound is less than 17 m in a small room, so echoes are not heard.

Question 39.
Name two properties that are common to all types of mechanical waves?
Answer:

  1. They require a material medium for their propagation.
  2. The medium itself does not move with the wave.

Question 40.
In sound, beats are heard when two independent sources are sounded together. Is it possible in the case of sources of light?
Answer:
No. This is because the phase difference between two independent sources of light is random.

Question 41.
Mention a condition when Doppler’s effect in sound is not applicable.
Answer:
When the velocity of the source or listener exceeds the velocity of sound.

Question 42.
What will be the velocity of sound in a perfectly rigid rod and why?
Answer:
Infinite because the value of Young’s modulus of elasticity is infinite for a perfectly rigid rod.

Question 43.
Sound is simultaneously produced at one end of two strings of the same length, one of rubber and the other of steel. In which string will the sound reach the other end earlier and why?
Answer:
In the case of steel-string as \(\frac{Y}{d}\) ratio is larger for steel than rubber.

Question 44.
Two persons cannot talk on the moon just as they do on the earth. Why?
Answer:
Due to the absence of air on the moon.

Question 45.
Graphs between the pressure P and the speed v in a gas are shown below. Explain with the reason which one is correct?
Class 11 Physics Important Questions Chapter 15 Waves 1
Answer:
(c) As the speed of sound (y) is independent of pressure, so y remains constant, hence (e) is correct.

Question 46.
Why the bells of colleges and temples are of large size?
Answer:
The larger the area of the source of sound more is the energy transmitted into the medium. Consequently, the intensity of sound is more and loud sound is heard.

Question 47.
State the limitations of Doppler’s effect.
Answer:
Doppler’s effect is applicable only when there is the relative velocity between the source and the listener and is less than the velocity of sound. The effect is not applicable if the relative velocity is greater than the velocity of sound i.e. if the source or observer moves with supersonic velocity.

Question 48.
Animals and human beings have two ears. What help do they render in listening sound from a distant source?
Answer:
The two ears receive sound in different phases, thus they help in locating the direction of the source of the sound. Turning of head and receiving sound in phase turns the direction.

Question 49.
An observer at a sea-coast observes waves reaching the coast. What type of waves does he observe? Why?
Answer:
Elliptical waves while the waves on the surface of the water are transverse, the waves just below the surface of the water are longitudinal. So the resultant waves are elliptical.

Question 50.
Why the reverberation time is larger for an empty hall than a crowded hall?
Answer:
It is due to the fact that the energy absorption in an empty hall is very small as compared to that of the crowded one.

Waves Important Extra Questions Short Answer Type

Question 1.
Here are the equations of three waves:
(a) y (x, t) = 2 sin (4x – 2t)
(b) y (x, t) = sin (3x – 4t)
(c) y (x, t) = 2 sin (3x – 3t).
Rank the waves according to their (A) wave speed and (B) maximum transverse speed, greatest first.
Answer:
(A) b, c, a.
Standard wave equation is y (x, t) = A sin (ωt – kx)
∴ (a) va = \(\frac{\omega}{k}=\frac{2}{4}=\frac{1}{2}\) unit = 0.5 unit.

(b) vb = \(\frac{\omega}{k}=\frac{4}{3}\)unit = 1.33 unit.

(C) Vc = \(\frac{\omega}{k}=\frac{3}{3}\) = 1 unit.
clearly vb > vc > va, so order is b, c, a.

(b) Transverse speed (vt) = \(\frac{\mathrm{d} y}{\mathrm{dt}}\)
∴ |(vt)a| = 2 × 2 = 4
|(vt)b| = 4
|(vt)c| = 2 × 3 = 6

∴ clearly (c), (a) and (b) tie.

Question 2.
Which physical quantity is represented by the ratio of the intensity of wave and energy density? Why?
Answer:
Velocity.
Class 11 Physics Important Questions Chapter 15 Waves 2

Question 3.
When are the tones called harmonics?
Answer:
The tones are called harmonics if the frequencies of the fundamental tone and other overtones produced by a source of sound are in the harmonic series.

Question 4.
What will be the effect on the frequency of the sonometer wire if the load stretching the sonometer wire is immersed in water?
Answer:
Due to the upthrust due to buoyancy experienced by the load, the effective weight will decrease, so tension and hence frequency will decrease as v ∝ \(\sqrt{T}\).

Question 5.
An organ pipe is in resonance with a tuning fork. If the pressure of air in the pipe is increased by a factor of 139, then how should the length be changed for resonance?
Answer:
We know that the velocity of sound is independent of pressure, so there is no change in frequency and hence there is no need to change the length of the pipe.

Question 6.
Sound waves travel through longer distances during the night than during the day. Why?
Answer:
Earth’s atmosphere is warmer as compared to the surface of the earth at night. The temperature increases with altitude and thus the velocity of sound increases. It is a case of reflection from denser to rarer medium.

The sound waves get totally internally reflected.

Question 7.
Water is being continuously poured into a vessel. Can you estimate the height of the water level reached in the vessel simply by listening to the sound produced?
Answer:
Yes, the frequency of the sound produced by an air column is inversely proportional to the length of the air column. As the level of water in the vessel rises, the length of the air column in the vessel decreases, so the frequency of sound increases, and hence shrillness of sound increases.

From the shrillness of sound, we can have a rough estimate of the level of water in the vessel.

Question 8.
A sonometer wire resonates with a tuning fork. If the length of the wire between the bridges is made twice even then it can resonate with the same fork. Why?
Answer:
When the length of the wire is doubled, the fundamental frequency is halved and the wire vibrates in two segments so the sonometer wire will still resonate with the given tuning fork.

Question 9.
Doppler’s effect in sound is asymmetric. Explain.
Answer:
Sound waves require a material medium for their propagation.

The apparent frequency is different whether the source moves towards the stationary observer or an observer moves towards the stationary source. Thus the Doppler’s effect is said to be asymmetric. No such asymmetry occurs in light because apparent frequency remains the same in either the case whether the source or the listener moves.

Hence Doppler’s effect is said to be symmetric in light.

Question 10.
What is redshift?
Answer:
It is due to Doppler’s effect in the case of light waves. It is known that all stars are moving away from each other. So apparent frequency of light from a star as received by an observer on earth is less than the actual frequency. Since wavelength is inversely proportional to the frequency, the apparent wavelength of light from stars is more than the actual wavelength.

In other words, due to the Doppler effect, the wavelength of light shifts towards a longer end i.e. towards red color and so it is called redshift.

Question 11.
A sitar wife and a tabla when sounded together give 4 beats/ sec. What do we conclude from this? As the tabla membrane is tightened, the beat rate increases or decreases, explain.
Answer:
When sitar and tabla are sounded together, they give 4 seats/ sec. From this, we conclude that the frequencies of the two sounds differ by 4. If the frequency of tabla is greater than that of a sitar, then on tightening the tabla membrane, the frequency of tabla will further increase and hence the difference in frequencies will increase.

Thus beat rate will increase. If the frequency of tabla is less than that of sitar, then on tightening the tabla membrane, the frequency of tabla will increase and the difference in frequencies will decrease. So beat rate will decrease.

Question 12.
Explain why frequency is the most fundamental property of a wave.
Answer:
When a wave passes from one medium to another, its velocity and wavelength change but the frequency remains the same. Hence frequency is said to be the most fundamental property of a wave.

Question 13.
Sound is produced by vibratory motion, explain why then a vibrating pendulum does not produce sound?
Answer:
The sound which we can hear has a frequency from 20 Hz to 20,000 Hz.

The frequency of the vibrating pendulum does not lie within the audible range and hence it does not produce sound.

Question 14.
Explain why stringed instruments are provided with hollow boxes.
Answer:
The hollow boxes are set into forced vibrations along with the strings. The loudness is higher if the area of the vibrating body is more. So hollow boxes attached to increase the loudness of sound.

Question 15.
When we start filling an empty bucket with water, the pitch of sound produced goes on changing. Why?
Answer:
An empty bucket behaves as a closed organ pipe. The frequency of fundamental note produced by it is given by
v = \(\frac{v}{4l}\).

As the bucket starts filling, the length (l) of the resonating air column decreases, and hence frequency increases. Since the pitch of a sound depends upon the frequency. So it changes with the change in frequency.

Question 16.
Two loud-speakers have been installed in an open space to listen to a speech. When both are operational, a listener sitting at a .particular- place receives a very faint sound. Why? What will happen if one loud-speaker is kept off?
Answer:
When the distance between two loud-speakers from the position of listener is an odd multiple of \(\frac{λ}{2}\), then due to destructive interference between sound waves from two loud-speakers, a feeble sound is heard by the listener.

When one loud-speaker is kept off, no interference will take place and the listener will hear the full sound of the operating loud-speaker.

Question 17.
Why is the sound produced in the air not heard by a person deep inside the water?
Answer:
The velocity of sound in water is much lesser than the velocity of sound in air. So the sound waves are mostly reflected from the surface of the water. Only little refraction of sound from air to water takes place. Moreover, the refracted sound waves die off after traveling a small distance in the water. Hence no sound waves reach deep inside the water.

Question 18.
The reverberation time is larger for an empty hall than a crowded hall. Explain why?
Answer:
The sound prolongs for a longer time in an empty hall as it does not get absorbed. In the case of a crowded hall, the sound is absorbed by the audience. Hence reverberation time is less in the crowded hall.

Question 19.
If a balloon is filled with C02 gas, then how will it behave for sound as a lens? On filling it with hydrogen gas, what will happen?
Answer:
The speed of sound in CO2 is less than in air (∴ v ∝ \(\frac{1}{\sqrt{\rho}}\)). So the balloon filled with CO2 gas behaves as a concave lens.

When it is filled with hydrogen, the speed of sound in H2 gas becomes greater than in air. So it will behave as a convex lens.

Question 20.
Why on reflection from the rigid surface, the wave does not change type, and only phase change occurs, while on reflection from a smooth surface, type changes while phase does not change?
Answer:
When a compression strikes the rigid surface, the surface does not move but pushes the compression back as such. On the other hand, the direction of vibration of the particles of the medium is reversed. Thus the type of wave does not change while phase change occurs.

When a compression strikes the smooth surface (i.e. yielding surface), the compression continues to move forward with the surface. But due to refraction, a part of the wave moves backward. Thus type changes while phase does not change.

Question 21.
Explain why transverse elastic waves can’t propagate through a fluid?
Answer:
When a transverse elastic wave travels through a solid, a shearing strain develops which is supported by the elastic solid because of the development of a restoring force. Thus elastic waves can propagate through solids.

But a liquid or a gas (i.e. fluid) can’t support a shearing strain. So there will be no restoring forces when there are transverse displacements and so transverse vibrations are not possible.

Question 22.
Two strings of the same material and length under the same tension may vibrate with different fundamental frequencies. Why?
Answer:
The frequency of vibration of the string is given by
Class 11 Physics Important Questions Chapter 15 Waves 3
If ρ = density of the material of string.
l, r = its length and radius.
D = diameter = 2r, then
Class 11 Physics Important Questions Chapter 15 Waves 4
Hence the two strings may vibrate with different frequencies when they have different diameters.

Question 23.
Distinguish between transverse waves and longitudinal waves.
Answer:
Longitudinal waves:

  1. Particles of the medium vibrate along the direction of propagation of the wave.
  2. They travel in the form of alternate compressions and rare¬factions.
  3. They can be formed in any medium i.e. solid, liquid, or gas.
  4. When these waves propagate, there are pressure changes in ‘ the medium.
  5. They can’t be polarised.

Transverse waves:

  1. Particles of the medium vibrate, to the direction of propagation of the wave.
  2. They travel in the form of alternate crests and troughs.
  3. These can be formed in solids and on the surfaces of liquids only.
  4. There are no pressure changes due to the propagation of these waves in the medium.
  5. They can be polarised.

Question 24.
Distinguish between progressive waves and stationary waves.
Answer:
Progressive waves:

  1. The disturbance travels onward. It is1 handed over from one particle to the next.
  2. Energy is transported in the medium along with the propagation of waves.
  3. Each particle of the medium executes S.H.M. with the same amplitude.
  4. No particle of the medium is permanently at rest.
  5. Changes in pressure and density are the same at all points of the medium.

Stationary waves:

  1. The disturbance is confined to a particular region and there is no onward motion.
  2. No energy is transported in the medium.
  3. All the particles of the medium except at nodes execute S.H.M. with different amplitude.
  4. The particles of the medium at nodes are at rest.
  5. The changes of pressure and density are maximum at nodes and minimum at antinodes.

Question 25.
Distinguish between musical sound and noise.
Answer:
Musical sound:

  1. It produces a pleasant effect on the ear.
  2. It has a high frequency.
  3. There are no sudden changes in the amplitude of the musical sound waves.
  4. It is a desirable sound.

Noise:

  1. It produces an unpleasant effect on the ear.
  2. It has a low frequency.
  3. There are sudden changes in the amplitude of noise waves.
  4. It is an undesirable sound.

Question 26.
What are the characteristics of wave motion?
Answer:

  1. Wave motion is a form of disturbance that travels in a medium due to repeated periodic motion of the particles of the medium.
  2. The wave velocity is different from the particle velocity.
  3. The vibrating particles of the medium possess both K.E. and P.E.
  4. The particle velocity is different at different positions of its vibrations whereas wave velocity is constant throughout a given medium.
  5. Waves can undergo reflection, refraction, diffraction, dispersion, and interference.

Question 27.
Show that for 1°C change in temperature, the velocity of sound changes by 0.61 ms-1.
Answer:
We know that v ∝ \(\sqrt{T}\) .
If vt and vo be the velocity of sound at T°C and 0°C respectively,
Class 11 Physics Important Questions Chapter 15 Waves 5
where α = \(\frac{\mathbf{v}_{\mathrm{t}}-\mathbf{v}_{0}}{\mathrm{t}}\) is called temp. coefficient of the velocity of sound.

Putting vo = 332 ms-1 at T0 i.e. 0°C, we get
α = \(\frac{332}{546}\) = 0.61 ms-1 °C-1

Question 28.
An electric bell is put in an evacuated room (a) near the center (b) close to the glass window, in which case the sound is heard (i) inside the room, (ii) out of the room.
Answer:

  1. Sound is not heard in cases (a) and (b) inside the room as the medium is not there for the propagation of sound.
  2. In case (a) sound cannot be heard outside for the reason given in (i) above.

In case (b) since the bell is very close to the window, the glass pane picks up its vibrations which are conveyed to the eardrum through the air outside the room. So, the sound can be heard in condition (b).

Question 29.
One of the primitive musical instrument is flute, yet produces good musical sound, how?
Answer:
The flute is an open organ pipe instrument having some holes which determine the wavelength and hence the frequency of sound. produced. By closing one or more holes, the length of the vibrating air column is changed and thus different harmonics are produced. The harmonic rich sound is a good musical sound.

Question 30.
Write basic conditions for the formation of stationary waves.
Answer:
The basic conditions for the formation of stationary waves are listed below:

  1. The direct and reflected waves must be traveling along the same line.
  2. For stationary wave formation, the superposing waves should either be longitudinal or transverse. A longitudinal and a transverse wave cannot superpose.
  3. For the formation of stationary waves, there should not be any relative motion between the medium and oppositely traveling waves.
  4. The amplitude and period of the superposing waves should be the same.

Question 31.
What is the difference between interference and stationary waves? In which phenomenon, out of the two, energy is not propagated? Why there is no energy at interference minimum?
Answer:
The superposition of two waves close to each other traveling in the same direction produces interference. The energy gets redistributed. It is minimum or zero at points of destructive interference and maximum at points of constructive interference. It is to be noted that the interference minima may not be points of zero energy unless •, the frequencies and amplitudes of the superposing waves are exactly equal.

The superposition of two similar waves (waves having the same amplitude and period) traveling in opposite direction produce stationary waves The nodes have no vibration of particles but antinodes have a maximum amplitude of vibration.

Question 32.
Write the applications of beats.
Answer:
Beats are used to:

  1. Determine an unknown frequency by listening to the best frequency Δv. Then unknown frequency V’ = v ± Δv where v is known and it is close to the unknown frequency. The exact value of v’ is found by loading and filling the tuning fork of unknown frequency from which + or – sign is chosen.
  2. Tune musical instruments by sounding them together and reducing beats number to zero.
  3. Make a sound rich in musical effect by the deliberate introduction of beats between different musical instruments.
  4. To produce very low-frequency pulses which otherwise cannot be produced. The beat frequency is the low-frequency sound.
  5. Receive radio program by the superheterodyne method.
  6. Detect harmful gases in mines.

Waves Important Extra Questions Long Answer Type

Question 1.
Derive expressions for apparent frequency when
(i) source is moving towards an observer at rest.
(ii) the observer is moving towards the source at rest.
(iii) both source and observer are in motion.
Answer:
Let S and O be the positions of. source and observer respectively.
υ = frequency of sound waves emitted by the source.
v = velocity of sound waves.
Class 11 Physics Important Questions Chapter 15 Waves 6
Case (i) Source (S) in motion and observer at rest: When S is at rest, it will emit waves in one second and these will occupy a space of length v in one second.
If λ = wavelength of these waves, then
λ = \(\frac{v}{υ}\) …(i)

Let vs = velocity of a source moving towards O at rest and let S reaches to S’ in one second. Thus the sound waves will be crowded in length (v – vs). So if λ’ be the new wavelength,
Then λ’ = \(\frac{v-v_{s}}{v}\)

if υ’ be the apparent frequency, then
υ’ = \(\frac{v}{\lambda^{\prime}}=\frac{v}{v-v_{s}}\)υ

∴ υ’ > v i.e. when S moves towards O, the apparent frequency of sound waves is greater than the actual frequency.

(ii) If the observer moves towards the source at rest:
Class 11 Physics Important Questions Chapter 15 Waves 7

Let vo = velocity of an observer moving towards S at rest.
As the observer moves towards S at rest, so the velocity of sound waves w.r.t. the observer is v + vo.

If υ’ = apparent frequency, then
υ’ = \(\frac{\mathbf{v}+\mathbf{v}_{0}}{\lambda}=\frac{\mathbf{v}+\mathbf{v}_{0}}{\mathbf{v}}\)υ
clearly v’ > υ.

(iii) If both S and O are moving
(a) towards each other: We know that when S moves towards stationary observer,
Class 11 Physics Important Questions Chapter 15 Waves 8
then υ’ = \(\frac{\mathbf{v}}{\mathbf{v}-\mathbf{v}_{\mathrm{s}}}\)

When O moves towards S, then
υ” = \(\left(\frac{v+v_{0}}{v}\right) v^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) v\)

(b) If both S and O move in the direction of sound waves:
Then the apparent frequency is given by
Class 11 Physics Important Questions Chapter 15 Waves 9
υ” = \(\left(\frac{\mathbf{v}-\mathbf{v}_{0}}{\mathbf{v}-\mathbf{v}_{\mathrm{s}}}\right)\)υ

(c) When both S and O are moving away from each other:
When source moves away from O at rest, then apparent frequency is given by
υ’ = \(\frac{v}{v+v_{s}}\)υ
Class 11 Physics Important Questions Chapter 15 Waves 10

When the observer is also moving away from the source, the frequency v’ will change to v” and is given by
Class 11 Physics Important Questions Chapter 15 Waves 11

Question 2.
Give the analytical treatment of beats.
Answer:
Consider two simple harmonic progressive waves traveling simultaneously in the same direction and in the same medium. Let

  • A be the amplitude of each wave.
  • There is no initial phase difference between them.
  • Let v1 and v2 be their frequencies.

If y1 and y2 be displacements of the two waves, then
y1 = A sin 2π v1 t and
y2 = A sin 2π v2 t

If y be the result and displacement at any instant, then
y = y1 + y2
= A (sin 2π v1 t) + sin (2π v2 t)
Class 11 Physics Important Questions Chapter 15 Waves 12
is the amplitude of the resultant displacement and depends upon t. The following cases arise.
(a) If R is maximum, then
cos π (v1 – v2) t = max. = ±1 = cos nπ
∴ π (v1 – v2) t = nπ
or
t = \(\frac{\mathrm{n}}{v_{1}-v_{2}}\) …(3)
where n = 0, 1, 2, …..

∴ Amplitude becomes maximum at times given by
t = \(0, \frac{1}{v_{1}-v_{2}}, \frac{2}{v_{1}-v_{2}}, \frac{3}{v_{1}-v_{2}}\)…..

∴ Time interval between two consecutive maxima is
= \(\frac{1}{v_{1}-v_{2}}\)

∴ Beat period = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat frequency = v1 – v2
∴ no. of beasts formed per sec. = v1 – v2.

(b) If R is minimum, then
Class 11 Physics Important Questions Chapter 15 Waves 13
∴ Time interval between two consecutive minima is = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat period = \(\frac{1}{v_{1}-v_{2}}\)
∴ Beat frequency = v1 – v2
∴ no. of beasts formed per sec. = v1 – v2.
Hence the number of beats formed per second is equal to the difference between the frequencies of two component waves.

Question 3.
What conditions are necessary for the good acoustical properties of the building? How are they met?
Answer:
An acoustically good building is one in which the sound is heard clearly in every nook and corner, some conditions must be fulfilled.

These are:
(a) the building should have proper reverberation time, the reverberation time is given by Sabine’s formula which for a hall is
T = \(\frac{0.166 \mathrm{~V}}{\sum \alpha \mathrm{A}}\)
The reverberation time is adjusted by:

  1. changing the volume (V): This can be changed little due to the size of the hall already fixed.
  2. changing effective absorbing area: This can be done artificially by putting heavy curtains, paintings, providing open windows, wall coverings, etc.
  3. providing sufficiently energetic sound: The sound should be sufficiently loud and intelligible at every point.
  4. eliminating echo: Except for the desired one, echoes must be eliminated.
  5. properly focussing the sound: The sound has to be properly focussed to avoid the source of silence and also unreliable focussing.
  6. avoiding unique reinforcement: No single overtone should uniquely be reinforced then the total quality of the note will be affected.
  7. avoiding extra noise: Extra noises including resonance within the building has to be avoided.
  8. eliminating smooth curved surfaces: Smooth surfaces reflect sound sharply which may cause several problems.

To achieve these goals following steps need to be taken:
(a) Properly design the building to optimize its acoustic condition.
(b) Decorate selectively the building with paintings, floral designs, perforations.
(c) Use false perforated or cardboard ceilings, and perforated structures at the curved walls.
(d) Use carpets, upholster seats with holes in the bottom.
(e) Use carpets or mats etc. on floors.
(f) Heavy curtains, wall hangings, etc. should be used.
(g) Properly place the mike and loud-speakers in the hall.
(h) Avoid sharp corners in the hall and make the stage back parabolic with mike at its focus.

Numerical Problems:

Question 1.
How far does the sound travel in the air when a tuning fork of frequency 280 Hz makes 15 vibrations? Given the velocity of sound is 336 ms-1.
Answer:
Velocity of sound, v = 336 ms-1
frequency, v = 280 Hz

∴ Time of one complete vibration,
T = \(\frac{1}{v}=\frac{1}{280}\) s.
Time to complete 15 vibrations,
t = 15 T
= \(\frac{1}{280}\) × 15

If x be the distance covered in time t, then
x = v × t
= 336 × \(\frac{15}{280}\) = 18 m
x = 18 m.

Question 2.
Audible frequencies have a range of 20 Hz to 20 × 103 Hz.
Express this range in terms of
(i) period T,
(ii) wavelength λ,
(iii) angular frequency CD.
Given the velocity of sound = 340 ms-1.
Answer:
V = 340 ms-1
Using v1 = 20 Hz
v2 = 20 × 103 Hz
(i) T = \(\frac{1}{v}\), we get
T1 = \(\frac{1}{20}\) = 0.005 s
and T2 = \(\frac{1}{20}\) × 103
= 0.00005 s = 5 × 10-5 s.

(ii) we know that
Class 11 Physics Important Questions Chapter 15 Waves 14
∴ The wavelength range is 17 m to 0.0 17 m.

(iii) Angular frequency, ω = 2πv
∴ ω1 = 2πv1 = 2π × 20= 40π rads-1
ω2 = 2πv2 = 2π × 20 × 103
= 40π × 103 rad s-1
∴ angular frequency range is 40π rad s-1 to 40π × 103 rad s-1

Question 3.
A copper wire is held at the two ends by rigid supports. At 30°C, the wire is just taut with negligible tension. Find the speed of transverse waves in the wire at 10°C. α = 1.7 × 10-5 °C-1, λ = 1.4 × 1011 Nm-1 and ρ = 9 × 103 kg m-3
Answer:
When the temperature changes from 30°C to 10°C, then the change in length of the wire is
Δl = l ∝ Δt = l × 1.7 × 10-5 × 20
= 3.4 × l × 10-4 m

Here, ρ = 9 × 103 kg m-3
Y = 1.4 × 1011 Nm-2
Δt = 30 – 10 = 20°C
α = 1.7 × 10-5 °C-1

Now Y = \(\frac{\mathrm{F} / \mathrm{a}}{\mathrm{\Delta}l / \mathrm{l}}\)
or
F = Ya \(\frac{Δl}{l}\)
where a = area of cross section of the wire.
Class 11 Physics Important Questions Chapter 15 Waves 15

Question 4.
The speed of sound in hydrogen is 1270 ms-1. What will be the speed in a mixture of oxygen and hydrogen mixed in a volume ratio of 1:4?
Answer:
Let V1 and V2 be the volumes of O2 and H2 respectively in the mixture.
ρ1 and ρ2 be their respective densities.

If m1 and m2 be the mass of oxygen and hydrogen respectively, then
m1 = V1ρ1
and m2 = V2ρ2

If ρ be the density of the mixture, then
Class 11 Physics Important Questions Chapter 15 Waves 16
Class 11 Physics Important Questions Chapter 15 Waves 17
Class 11 Physics Important Questions Chapter 15 Waves 18
Also, let v1 = velocity of sound in the mixture
and v2 = velocity of sound in the hydrogen
= 1270 ms-1.
Class 11 Physics Important Questions Chapter 15 Waves 19

Question 5.
A tuning fork A of unknown frequency is sounded with a tuning fork B of frequency 288. Four beats are heard in one second. Tuning fork A is then loaded with a little wax and again sounded with fork B. Again 4 beats are heard in one second. What is the frequency of A?
Answer:
Since tuning fork A of known frequency say v gives 4 beats/ sec with B of frequency 288, so frequency of tuning fork A is given by
∴ v = 288 ± 4 = 284 or 292.
On loading A with wax, it again produces 4 beats/s with B of frequency 288.

If the frequency of A is 292, it means on loading A with wax, its frequency falls to 284 which can produce 4 beats/s with a B frequency of 288. So the frequency of A maybe 292.

If the frequency of A is 284, on loading it with wax, its frequency falls below 284 so that when it is sounded with B of frequency 288, it forms more than 4 beats/s which is not consistent with the given condition.

So the frequency of A can’t be 284. Hence the frequency of A = 292.

Question 6.
The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of a closed organ pipe is 110 Hz. Find the lengths of the pipes. The velocity of sound in air = 330 ms-1.
Answer:
Fundamental frequency of closed organ pipe is
vc = \(\frac{\mathrm{v}}{4 \mathrm{~L}_{\mathrm{c}}}\)
Lc = length of closed pipe v
or
Lc = \(\frac{\mathrm{v}}{4 \mathrm{~v}_{\mathrm{c}}}\)

Here, νc = 110 Hz
v = 330 ms-1

∴ Lc = \(\frac{330}{4 \times 110}\) = 0.75m = 75 cm. 0.4 × 110
Now frequency of first overtone of open pipe is
v1 = \(\frac{\mathrm{v}}{\mathrm{L}_{0}}\)
where Lo = length of open pipe Frequency of first overtone of closed pipe is

Frequency of first overtone of closed pipe is
ν2 = \(\frac{3 v}{4 L_{c}}=3\left(\frac{v}{4 L_{c}}\right)\)
= 3νc = 3 × 110 = 330 Hz
Class 11 Physics Important Questions Chapter 15 Waves 20

Question 7.
An open pipe is suddenly closed at one end with the result that the frequency of the 3rd harmonic of the closed pipe is found to be higher by loo Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of open pipe?
Answer:
νo = \(\frac{v}{2L}\)
where ν0 = fundamental frequency of open pipe. The frequency of the third harmonic of closed pipe is
Class 11 Physics Important Questions Chapter 15 Waves 21

Question 8.
A set of 24 tuning forks is so arranged that each gives 4 beats per second with the previous one and the last sounds the octave of the first. Find the frequency of the first and last.
Answer:
As the last is the octave (double) of the first, so the frequencies are ¡n the increasing order.
Let frequency of the first = x
frequencyofthe2nd = x + 4
frequency olthe3rd = x + 2(4)
frequency of the 24th = x + 23 (4)
But this is the octave of the first
∴ frequency of the 24th = 2(x) (given)
∴ x + 23(4) = 2x
or
x = 92
∴ frequency of last tuning fork 2 × 92 = 184 Hz.

Question 9.
A drop of water 2 mm in diameter falling from a height of 50 cm in a bucket generates a sound that can be heard from a 5 m distance. Take all gravitational energy difference equal to sound energy, the transformation being spread in time over 0.2s, deduce the average intensity. Take g = 10 ms2.
Answer:
Here, ρ of water = 103 kg m-3
D = 2mm

∴ radius = r = \(\frac{2}{2}\) = 1 mm = 10-3 m
h = 50cm = 0.50 m
g = 10 ms-2
d = 5m
∴ a = area in which sound is heard
=4πd2
I = intensity of sound =?

Now I = \(\frac{\text { energy transference }}{\text { time } \times \text { area }}\)
= \(\frac{\text { P.E. of drop }}{\text { time } \times \text { area }}=\frac{m g h}{t \times a}\) …(1)
Class 11 Physics Important Questions Chapter 15 Waves 22

Question 10.
A train moves towards a stationary observer with a velocity of \(\frac{1}{30}\)th of the velocity of sound. The whistle of the engine regularly blows after one second. What ¡s the interval between successive sounds of the whistle as heard by the observer?
Answer:
Actual time interval between successive blows of whistle = 1 sec.

∴ Actual frequency of whistle, v = 1 Hz
Let v = velocity of sound
∴ velocity of source, vs = \(\frac{v}{30}\)
∴ apparent frequency of the whistle is
ν’ = \(\frac{v}{v-v_{s}}\) × ν = \(\frac{v}{v-\frac{v}{30}}\) × 1
or
ν’ = \(\frac{30}{29}\) Hz

∴ Apparent time interval between two successive sounds of whistle is
t = \(\frac{1}{v^{\prime}}=\frac{29}{30}\) s.

Question 11.
The splash is heard 4.2,3 s after a stone is dropped into a well which is 78.4 m deep. Find the velocity of sound in the air.
Answer:
Here, Depth of well, h = 78.4 m
v = velocity of sound = ?
t = total time after which splash is heard = 4.23 s.
t1 = time taken by stone to hit water surface in the well.
t2 = time the sound takes in reaching top of the well.
∴ t1 +t2 = 4.23 s
u = 0, g = 9.8 ms-2, h = 78.4m, t = t1

using the relation, s = ut + \(\frac{1}{2}\) at2, we get
Class 11 Physics Important Questions Chapter 15 Waves 23

Question 12.
What is the intensity level in dB of sound whose intensity is 10-6 watt m-2. Take zero levels of intensity = 10-2 watt m-2.
Answer:
Here,
I = 10-6 w m-2
I0 = 10-12 w m-2

L =?
L = log10 \(\left(\frac{I}{I_{0}}\right)\) = log10 \(\left(\frac{10^{-6}}{10^{-12}}\right)\)
= log10 106
= 6 log10 10 = 6 × 1 = 6B
As 1B = 10 dB
∴ L = 6 × 10 = 60 dB

Question 13.
Two closed organ pipes when sounded together produce 12 beats in 4s at a temperature of 27°C. Find the temperature at which the number of beats produced is 16 during the same period.
Answer:
Let v1 and v2 be the frequencies produced by the two organ pipes at 27°C. Let v27 be the velocity of sound at 27°C.
and vt = velocity of sound at t°C
l1, l2 be their lengths respectively.
Class 11 Physics Important Questions Chapter 15 Waves 24
Class 11 Physics Important Questions Chapter 15 Waves 25

Question 14.
A police-man on duty detects a drop of 15% in the pitch of the horn of a motor car as it crosses him. If the velocity of sound is 330 ms-1, calculate the speed of a car.
Answer:
vs = speed of car = ?
v = velocity of sound = 330 ms-1
v = frequency of sound emitted by horn
If v’ be the apparent frequency’of sound when car approaches the police man, then
ν’ = \(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\) × ν …(i)

Also let ν” be the apparent frequency of sound when car recedes away from the police man, then
ν” = \(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\) × ν …(ii)
Class 11 Physics Important Questions Chapter 15 Waves 26

Question 15.
A wire of length 80 cm has a frequency of 250 Hz. If the length of the wire is increased to 100 cm and tension is reduced to \(\frac{1}{4}\)th of its original value, then calculate the new frequency.
Answer:
l1 = initial length = 80 cm
ν1 = 250 Hz

Let T1 = T be the tension
l2 = 100 cm
T2 = \(\frac{1}{4}\)T1 = \(\frac{T}{4}\).
ν = ?

Using the relation
Class 11 Physics Important Questions Chapter 15 Waves 27
Class 11 Physics Important Questions Chapter 15 Waves 28

Question 16.
A tuning fork A of unknown frequency is sounded with a tuning fork B of frequency 512 Hz. 8 beats are heard in 1 second. Tuning fork A is then loaded with a little wax and sounded together with B. Again 8 beats are heard in 1 second. Find the frequency of
Answer:
Let ν = frequency of tuning fork A
No. of beats formed/sec. =8
∴ ν = 512 ± 8 = 520, 504

On loading A with wax, it again produces 8 beats/second with B of frequency 512.
Now if νA = 520, then it means its frequency decreases on loading. If v. falls to 504 then it will produce 8 beats/sec. with B.
∴ νA = 520

If νA is 504, then on loading its frequency falls below 504, so it forms more than 8 beats/sec. when sounded together with B which is not consistent with the given condition.

So the frequency of A cannot be 504.
∴ νA = 520 Hz.

Question 17.
Two engines pass each other in opposite directions with a velocity of 60 km ph. One of them is emitting a note of frequency 540 Hz. Calculate the frequencies heard in the other engine before and after they have passed each other. The velocity of sound = 316.67 ms-1.
Answer:
Here, vs = vo = 60 kmph = 60 × \(\frac{5}{18}\) = 16.67 ms-1
ν = 540 Hz
ν’ = ?
v = velocity of sound = 316.67 ms-1

Case A: Before passing each other:
v = 316.67 ms-1
vs = vo = 16.67 ms-1
∴ ν’ = ?

Using the relation
Class 11 Physics Important Questions Chapter 15 Waves 29
Case B: After passing each other:
vs = – 16.67 ms-1
vo = – 16.67 ms-1
v = 316.67 ms-1
Class 11 Physics Important Questions Chapter 15 Waves 30

Question 18.
The audible frequency range of a human ear is 20 Hz to 20 kHz. Convert this into the corresponding wavelength range. Take the speed of sound in the air to be 340 ms-1.
Answer:
Here, ν1 = 20 Hz
ν2 = 20000 Hz
v = 340 ms-1
λ1 = ?
λ2 = ?

∴ λ1 = \(\frac{v}{v_{1}}=\frac{340}{20}\) = 17 m
and λ2 = \(\frac{v}{v_{2}}=\frac{340}{20000}\) = 0.017 m

∴ Corresponding wavelength range is 0.017 to 17 m.

Question 19.
For aluminum, the bulk modulus and the modulus of rigidity are 7.5 × 1010 Nm-2 and 2.1 × 1010 Nm-2. Find the velocity of longitudinal and transverse waves in the medium. The density of aluminum is 2.7 × 103 kg m-3.
Answer:
Here, for aluminium, ρ = 2.7 × 103 kg m-3 .
η = 2.1 × 1010 Nm-2
k = 7.5 × 1010 Nm-2
vl = velocity of longitudinal waves = ?
vt = velocity of transverse waves = ?

Using the relation,
Class 11 Physics Important Questions Chapter 15 Waves 31
Class 11 Physics Important Questions Chapter 15 Waves 32
Also using the relation, for transverse waves,
Class 11 Physics Important Questions Chapter 15 Waves 33

Question 20.
The velocity of sound at 27°C is 350 ms-1. If the density of air at N.T.P. is 1.293 kg m-3, calculate the ratio of the specific heats of air.
Answer:
Here, T = 27 + 273 = 300k
v27 =. velocity of sound at 27°C
= 350 ms-1
ρ = 1.293 kg m-3
To = 273 k
vo = velocity at N.T.P. = ?
P = 76 cm of Hg
= 0.76 × 13600 × 9.8 Nm-2

Using the relation,
Class 11 Physics Important Questions Chapter 15 Waves 34
Using the relation,
Class 11 Physics Important Questions Chapter 15 Waves 35

Question 21.
A tuning fork B produces 6 beats per second with another tuning fork of frequency 288 Hz. The tuning fork of unknown frequency is filed and the number of beats now produced is 4 per second. Calculate the frequency of the tuning fork.
Answer:
Known frequency of the fork =288 Hz
No. of beats/second = 6
∴ Unknown frequency = 288 ± 6 = 294 or 282.

When the prongs of tuning fork B are filed, its frequency increases. If the frequency of B is originally 294 Hz, on filing it will increase. So the difference between the frequencies will increase and hence no. of beats will also increase.

If the frequency of B is originally 282 Hz, then on filing it, its frequency will increase, so the difference between the frequencies will decrease, and hence the no. of beats will decrease.

In this case, when the prongs of B are filed, the no. of beats decreases from 6 to 4. Thus the frequency of the given tuning fork is 282 Hz.

Question 22.
Find the change in the frequency observed by a listener when a source approaching him with a velocity of 54 km h-1 starts going away from him with the same velocity. Take velocity of sound = 330 ms-1 and v = 300 Hz.
Answer:
Here, vs = 54 km h-1 = 54 × \(\frac{5}{18}\) = 15 ms-1
v = 330 ms-1
ν = 300 Hz
Case I: When the source is approaching, then
ν’1 = \(\frac{\mathbf{v}}{\mathbf{v}-\mathbf{v}_{\mathbf{s}}}\)ν = \(\frac{330}{330-15}\) × 300
= 314 Hz

Case II: When the source reduces then
ν’2 = \(\frac{\mathbf{v}}{\mathbf{v}+\mathbf{v}_{\mathbf{s}}}\)ν = \(\frac{330}{330+15}\) × 300
= \(\frac{330}{345}\) × 300
= 287 Hz

∴ Change in frequency,
ν’1 – ν’2 = 314 – 287 = 27 Hz.

Question 23.
A progressive wave is given by
y = 12 sin (5t – 4x)
On this wave, how far away are the two points having a phase difference of \(\frac{π}{2}\)?
Answer:
Here, ΔΦ = phase difference = \(\frac{π}{2}\)

Let Δx be the corresponding path difference,
Class 11 Physics Important Questions Chapter 15 Waves 36

Question 24.
The figure here shows the wave, y = A sin (ωt – kx) at any instant traveling in the +x direction. What is the slope of the curve at B?
Class 11 Physics Important Questions Chapter 15 Waves 37
Answer:
Here, the particle velocity is maximum at B and is given by
vo = ωA.
Also, wave velocity is given by
C = \(\frac{ω}{k}\)
∴ So the slope \(\frac{v_{0}}{C}=\frac{\omega A}{\omega / k}\) = kA

Question 25.
Two sound waves
y1 = A1 sin 1000 π(t – \(\frac{x}{220}\) )
and y2 = A2 sin 1010 π(t – \(\frac{x}{220}\) ) are superposed. What is the frequency with which the amplitude varies?
Answer:
Rate of variation of amplitude is equal to the beat frequency.
Here, 2πν1 = 1000 π
or
ν1 = 500
and 2πν2 = 1010 π
or
ν2 = 505

∴ beat frequency = ν2 – ν1
= 505 – 500
= 5.

Value-Based Type:

Question 1.
A group of students went to a place on an excursion. While boating on seawater, the students identified a submerged Torpedo-shaped structure. The boys debated among themselves on what they saw. A student by the name of Sharath considering it as a threat informed the police. The police took necessary steps to protect the country from the enemy submarine. Sharath was rewarded.
(a) What can you say about the qualities exhibited by Sharath?
Answer:
Navigator is a responsible citizen, he is duty-minded, having a presence of mind.

(b) A SONAR system fixed in a submarine operates at a frequency of 40 kHz. An enemy submarine moves towards the
SONAR with a speed of 360 km/hr. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to bel450 m/s.
Answer:
Here, frequency of SONAR (Source) = 40 k Hz
= 4 × 103 Hz
Speed of the sound wave, V = 1450 m/s
Speed of the observer, Vo = 360 kM/h = 100 m/s

∴ Apparent frequency received by an enemy submarine ;
V’ = \(\left(\frac{V+V_{0}}{V}\right) \cdot V=\left(\frac{1450+100}{1450}\right)\) × 40 × 103
=4.276 × 104 Hz.

This frequency is reflected by the enemy submarine (Source) and observed by SONAR (Now observer).
In this case, Apparent frequency is given by
V” = \(\left(\frac{V}{V-V_{s}}\right) \times V^{\prime}=\left(\frac{1450}{1450-100}\right)\) × 4.276 × 104
=45.9k Hz

Question 2.
Jagat and Ram are working in the same company. Jagat has noticed that Ram is suffering from Cancer. Ram is not aware of this. When Jagat asks him to go for a checkup, Ram refused. He gets convinced how even when he realizes normal, it is very important to get the checkup done once a year.
(a) What according to you, are the values displayed by Jagat in helping Ram.
Answer:
He has concern for hi^ friend, also he has the knowledge of medical facilities available

(b) A hospital uses an ultrasonic scanner to locate tumors in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km/s? The operating fre¬quency of the scanner is 4.2 MHz.
Answer:
Class 11 Physics Important Questions Chapter 15 Waves 38
that is λ = 4.05 × 10-4 m

Question 3.
‘Preeti a student of class XI was reading the newspaper, The Headlines in the Newspaper were about the earthquake that had taken place in Assam on the previous day. She was very depressed seeing the loss of life and property. She approached her physics teacher and got the information about how an earthquake occurs.
(a) What can you say about the inquisitiveness of Preeti?
Answer:
She has concern for society and is sympathetic towards others.

(b) Earthquake generates sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about 4 km/s, and that of the P wave is 8 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, how far away does the earthquake occur?
Answer:
LetV1 and V2 be the velocities of S waves and P waves and t1 , t2 be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earthquake from the seismograph, .
∴ l = V1 t1 = V2t2
⇒ 4t1 = 8 t2
or
t1 = 2 t2 [∴ V1 = 4 kmh-1 ,V2 = 8 kmh-1]

Also, t1 – t2 = 4 min = 240 S.
2t2 – t2 = 240 [Using (i)]
or
t2 = 240 S and t1 = 2 × 240 = 480 S.
l = v1 t1 = 4 × 480 = 1920 km

Question 4.
Rajesh was waiting for the train on the platform with his parents. They were going to Mumbai and the train arrived half an hour late. He felt that when the train was coining towards the station, the intensity of the sound of the whistle was gradually increasing and on the platform the sound was maximum and when the train passes away the intensity of the whistle was decreasing.
Rajesh got confused and asked his father the reason behind it. His father could not give the correct answer. So, he decided to ask his physics teacher.
(i) What are the values displayed here by Rajesh?
Answer:
Values displayed by Rajesh are:
Awareness, curiosity, creativity, and intelligence,

(ii) Write the answer given by your teacher.
Answer:
The teacher explained that it is due to “Doppler’s effect”. Whenever there is a relative motion between the source and observer O. There is a change in observed frequency. So, the apparent frequencies of sound heard by the listener are different from the actual frequency of the sound of the source, which is given by.
V = Vo\(\left(\frac{V+V_{0}}{V+V_{s}}\right)\)

Here, V is the speed of sound through the j medium, Vo is the velocity of the observer relative to the medium, and Vs is the source velocity relative to the medium.

Binomial Theorem Class 11 Notes Maths Chapter 8

By going through these CBSE Class 11 Maths Notes Chapter 8 Binomial Theorem Class 11 Notes, students can recall all the concepts quickly.

Binomial Theorem Notes Class 11 Maths Chapter 8

Expanding binomials calculator. This calculators lets you calculate expansion of a binomial in your website.

Binomial theorem for any positive integer
1. (a + b)n = nC0an + nC1an-1 + nC2an-2 + ………………..nCran-rbr + ………..+ nCnbn

2. Tr+1 = general term = nCran-rbr … (1)
(a + b)n = \(\sum_{r=0}^{n}\)nCran-rbr

3. Some observations : (i) Number of terms in binomial expansion = Index of the binomial + 1 = n + 1.

(ii) In the successive terms of the expansion, the index of the first term is n and it goes on decreasing by unity. Thus, the indices of first term and n, n – 1, n – 2, …, 1, 0 while the index of second term starts from 0 and goes on increasing by 1. Therefore, the indices of second term are 0, 1, 2, … , n.

(iii) In any term of the expansion, the sum of indices of first term and second term is constant and it is equal to n.

(iv) Coefficients of these terms in the expansion are nC0, nC1, nCr. ……, nC2….., nCn

Note : These coefficients may also be denoted as
C (n, 0), C(n, 1), C (n, 2),…, C (n, r), …, C (n,n).

To find their values, we apply the formula
Binomial Theorem Class 11 Notes Maths Chapter 8 1
Whenever r > \(\frac{n}{2}\) , apply the formula
nCr = nCn-r
These coefficients may also be obtained with the help of Pascal’s triangles.

Binomial Theorem Class 11 Notes Maths Chapter 8 2

Now look at this pattern :
(a) First row; index 0; coefficient = 1.
(b) Second row; index 1; coefficients are 1, 1. A triangle is inserted between 1 and 1 as shown above.
(c) Third row; index 2; coefficients are 1, 2, 1.

Method : Two triangles, are drawn below each number of 1 of the second row. Adding 1 and 1 of second row we get 2 which is put below the third vertex of triangle. Write 1 each at the beginning arid end.

(d) Fourth row, index 3; coefficients are 1, 3, 3, 1. Method : Insert three triangles below 1, 2 and 1. 1 + 2 = 3 is written below the third vertex of first triangle, 2 + 1 = 3 is written below the second triangle.

Further, we proceed in the same manner.
(v) Coefficients nC0, nC1, nC2…, nCr, nCn in Binomial Theorem are known as Binomial Coefficients.

(vi) Replacing b by – b, in (1), we obtain
(a-b)n = nC0annC1an-1b + nC2an-2b2 + ……….(-1)ⁿⁿCran-rbr + ……….nCn(-b)n

(vii) Putting a = 1, b = x in (1), we obtain
(1+x)n = nC0 + nC1x + nC2x2 + ……nCrxr + …….. + nCnxn

(viii) Putting a = 1, b = – x in (1), we obtain
(1-x)n = nC0nC1x + nC2x2 – ….+nCr(-x)r + …… + nCn(-x)n

(ix) Putting a = 1, b = 1 in (1), we obtain
(1+1)n = 2n = nC0 + nC1 + nC2 + ………..+nCn

(x) Putting a = 1, b = – 1, we obtain
0 = nC0nC1 + nC2nC3 + …..
or nC0 + nC2 + nC4 + ….. = nC1 + nC3 + …………+ nCr + ………

4. Middle Term : The expansion (a + b)n has n + 1 terms.
(i) If n + 1 is odd, the middle term is th = \(\frac{(n+1)+1}{2}\) th \(\frac{(n+2)}{2}\)th term.
(ii) If n + 1 is even, then
1st middle term = \(\frac{n+1}{2}\)th term
2nd middle term = \(\left(\frac{n+1}{2}+1\right)\)th term

Online Education NCERT Solutions for Class 11 English Hornbill Chapter 6 The Browning Version

Here we are providing Online Education NCERT Solutions for Class 11 English Hornbill Chapter 6 The Browning Version. Students can get Class 11 English The Browning Version NCERT Solutions, Questions and Answers designed by subject expert teachers.

Online Education The Browning Version NCERT Solutions for Class 11 English Hornbill Chapter 6

The Browning Version NCERT Text Book Questions and Answers

The Browning Version Understanding the text

Question 1.
Comment on the attitude shown by Taplow towards Crocker-Harris.
Answer:
Taplow has very more bitter feelings about his teacher Crocker-Harris. He is a student in the lower fifth grade and feels that he would specialise next term if he got his remove, of which he is uncertain as Mr Crocker-Harris doesn’t tell the students the results like the other teachers. As a rule, the class results should only be announced by the headmaster on the last day of term but Taplow feels that none other than Mr Crocker-Harris waits to inform students of their result. He is not interested in the Classical literature that is taught by Mr Crocker-Harris.

He feels science is more interesting than studying Classics such as The Agamemnon, which he calls “muck”. Moreover, he does not like the way it is taught to them. The Agamemnon had a lot of Greek words and Mr Crocker-Harris punished them for not getting them right.

Taplow feels more bitter as he had been given extra work to do for missing a day of school the previous week when he was ill. It was the last day of school and he wished to play golf instead. It was just on the previous day that Mr Crocker- Harris had told Taplow that he had got what he deserved. Taplow feels that Mr Harris might have given him lesser marks to make him do extra work. He adds that Mr Harris is “hardly human”. He also imitates his teacher.

When Frank suggests that Taplow could go and play golf, Taplow is shocked as nobody takes that kind of liberty with Mr Crocker-Harris. Taplow calls Mr Crocker-Harris, “the Crock”, and says that he is worse than a sadist. If he were a sadist, he wouldn’t be as frightening because he would then show he had some feelings. His inside, feels Taplow, is like a “shrivelled nut” and he seems to hate people who like him.

However, Taplow admits that despite everything Mr Crocker-Harris does, he still likes him. Although, he says that Mr Crocker-Harris feels uncomfortable about people liking him. He says once in class Mr Crocker-Harris made one of his classical jokes, and nobody laughed because nobody understood it. However Taplow knew that it was meant to be funny, so he laughed. Mr Crocker-Harris said that he was pleased with Taplow’s knowledge of Latin and wanted him to explain the joke to the rest of the class.

Question 2.
Does Frank seem to encourage Taplow’s comments on Crocker-Harris?
Answer:
Taplow comes to meet Mr Crocker-Harris when he meets Frank. From his conversation with Taplow, Frank realises that the boy does not like Mr Crocker-Harris. Frank then confirms with Taplow, “You sound a little bitter, Taplow.” He then pretends to console him by reasoning that he would get his remove the next day for taking on extra work. Taplow vents his dislike for Mr Harris and says that he is “hardly human”.

But after saying so, he apologises to Frank for talking too much. Frank pretends to be unhappy but asks Taplow to “repeat” what Mr Harris had said to him. Taplow imitates him. Frank pretends to look strict and asks him to be.quiet. He then asks Taplow at what time he was supposed to meet Mr Crocker-Harris. He then tells Taplow that Mr Crocker-Harris was already ten minutes late and suggests that Taplow could go and play golf.

Taplow is shocked and expresses his apprehension if Mr Crocker-Harris should know. Frank envies the effect Mr Crocker-Harris seems to have on boys in the class; they seem to be scared to death of him. Taplow confesses that Mr Crocker-Harris, unlike any other person, does not care for being liked. Frank attempts to instigate Taplow by deriding students for using the teacher’s need to be liked to their own advantage. Taplow remarks that a few teachers were sadists, and Mr Crocker-Harris was worse because he had no feelings.

When Taplow recounts the episode when he had laughed at Mr Crocker-Harris’s jokes, and Mr Crocker-Harris wanted ‘ him to explain it to the rest of the class, but Frank just laughs at that. He, thus, seems to enjoy the low opinion Taplow has of Mr Crocker-Harris.

Question 3.
What do you gather about Crocker-Harris from the play?
Answer:
Mr Crocker-Harris is an old Classics teacher at a British public school, where he’s been teaching for many years. He apparently wants the children to work hard at their lessons and it is for this reason that he has called Taplow to his office. Unfortunately, students do not like him and neither do they like his teaching methods. Taplow feels science is more interesting than studying Classics such as The Agamemnon, which he calls “muck”.

Moreover, he does not like the way it is taught to them. It has a lot of Greek words and Mr Crocker-Harris punishes them for not getting them right. Taplow feels Mr Harris might have given him lesser marks to make him do extra work. He adds that Mr Harris is “hardly human”. Thus underlining that Mr Harris has lost the student’s trust and respect.

He is a fastidious and a rule-bound person who is the only one who follows the rule of letting the headmaster announce the results on the last day of term.Taplow imitates Mr Crocker-Harris but all the same is frightened of letting Mr Harris know. The students seem to be scared to death of him. Calling Mr Crocker-Harris, the Crock, Taplow says that he is worse than a sadist as he shows no feelings.

He feels uncomfortable about people liking him. Mr Harris does not seem to respond to students who try to warm up to him. When Taplow laughed at his joke, Mr Crocker-Harris had wanted him to explain it to the rest of the class. The poor man is an unfortunate teacher.

The Browning Version Talking about the text

Discuss with your partners

Question 1.
Talking about teachers among friends.
(Answers will vary)

Question 2.
The manner you adopt when you talk about a teacher to other teachers.
(Answers will vary)

Question 3.
Reading plays is more interesting than studying science.
(Answers will vary)

The Browning Version Working with words

Question 1.
A sadist is a person who gets pleasure out of giving pain to others. Given below are some dictionary definitions of certain kinds of persons. Find out the words that fit these descriptions.
Answer:

  • A person who considers it very important that things should be correct or genuine, for example, in the use of language or in the arts: Perfectionist/Purist
  • A person who believes that war and violence are wrong and will not fight in a war: Pacifist
  • A person who believes that nothing really exists: Nihilist
  • A person who is always hopeful and expects the best in all things: Optimist
  • A person who follows generally accepted norms of behaviour: Conformist
  • A person who believes that material possessions are all that matter in life: Materialist
We’re Not Afraid to Die… If We Can All Be Together Important Extra Questions and Answers Class 11 English Hornbill

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We’re Not Afraid to Die… If We Can All Be Together Extra Questions and Answers Short Answer Type

We Are Not Afraid To Die If We Are Together Extra Questions Question 1.
Who was the narrator? What adventurous task did he take on?
Answer:
The narrator was a thirty-seven-year-old businessman, who along with his family, set from Plymouth, England, on a round-the-world voyage like Captain James Cook had done 200 years earlier in a 30-ton wooden-hulled boat.

We Are Not Afraid To Die Extra Questions Question 2.
How did they prepare for this onerous task?
Answer:
For sixteen years, they spent all their leisure time improving their seafaring skills in British waters. They bought a boat, Wavewalker, a 23-metre, 30-ton wooden-hulled vessel that had been professionally built. They spent months fitting it out and testing it in the roughest weather that they could find.

We Are Not Afraid To Die Question Answer Question 3.
How many people were there in the boat?
Answer:
The four of them the narrator, his wife Mary, son Jonathan, and daughter Suzanne sailed for 105,000 kilometres to the west coast of Africa to Cape Town. They took on two crewmen with them an American, Larry Vigil, and a Swiss, Herb Seigler, before settling sail on the southern Indian Ocean.

We’re Not Afraid To Die Extra Questions Question 4.
What was the first indicator of rough weather?
Answer:
On their second day out of Cape Town, they encountered strong winds. For the next few weeks, the gales blew continuously. The gales did not worry the narrator but the sizes of the waves were disturbing.

Class 11 English Hornbill Chapter 2 Extra Questions And Answers Question 5.
What ordeal awaited them on 2 January?
Answer:
After they celebrated Christmas, the weather changed for the worse. On the early morning of 2 January, the waves became huge. As the ship rose to the top of each wave, they could see the vast sea rolling towards them. The wind seemed to be howling.

Class 11 English Hornbill Chapter 2 Extra Questions Question 6.
What measures did they take to counter this ordeal?
Answer:
They dropped the storm jib and lashed a heavy mooring rope in a loop across the stem to slow the boat, and then double-lashed everything, went through their life-raft drill, attached lifelines, put on oilskins and life jackets.

We Are Not Afraid To Die Question And Answers In Short Question 7.
What happened on the evening of 2 January?
Answer:
On the evening of 2 January there was a lull before the storm. As the sky grew dark, they heard a growing roar, and saw a massive cloud rising at the rear of the ship. To their dismay, it was a huge wave, almost twice the height of other waves, with a fearsome breaking top.

We Are Not Afraid To Die Important Questions Question 8.
What happened when they tried to ride over the wave?
Answer:
When they tried to ride over the wave, there was a loud blast that shook the deck. Water gushed over the ship, the narrator’s head hit the wheel and he was thrown overboard into the water. He accepted his impending death, and while he was losing consciousness, he felt peaceful.

Important Questions Of We’re Not Afraid To Die Question 9.
How did the narrator get back to the ship after having been thrown into the sea?
Answer:
After the narrator felt he was losing consciousness, his head suddenly popped out of the water. A few metres away, he saw Wavewalker, nearly overturned. Then, a wave threw it upright. He grabbed the guardrails and sailed through the air into Wavewalker’s main boom. The waves tossed him onto the deck like a rag doll.

We Are Not Afraid To Die Class 11 Important Questions Question 10.
How did they manage to throw out water from the ship?
Answer:
With the narrator’s wife, Mary, at the wheel, the narrator half-swam, half-crawled into the children’s cabin, where he found a hammer, screws and canvas, and struggled back on deck. He secured waterproof hatch covers across the wide-open holes. With Herb and Larry’s assistance, he managed to throw out the water.

We Are Not Afraid To Die Extract Based Questions Question 11.
What were the difficulties that they faced that night?
Answer:
The night was bitterly cold, and they were pumping water out of the ship, steering the ship and working the radio. Moreover, they were getting no replies to their calls for help, as they were in a remote comer of the world.

We’re Not Afraid To Die Class 11 Extra Questions Question 12.
What injuries did Sue sustain? What does it reveal about her?
Answer:
Sue had bumped her head and there was a big bump above her eyes. She had two black eyes, and a deep cut on her arm. She showed remarkable maturity for a seven-year-old when she said that she didn’t want to worry them when her father was trying to save all of them.

We Are Not Afraid To Die Short Question Answer Question 13.
After the water level receded, what was their next concern? What did they decide to do?
Answer:
Having survived fifteen hours since the wave hit, the narrator checked the charts and calculated that there were two small islands a few hundred kilometres to the east. One of them was lie Amsterdam. Knowing Wavewalker would not hold for much longer, they aimed to reach the island.

We Are Not Afraid To Die Class 11 Extra Questions Question 14.
“But our respite was short-lived.” Why does the narrator say so?
Answer:
By 4 January, they ate their first meal in almost two days after pumping out most of the water. But their breather was short-lived. Soon after, black clouds gathered and the wind rose to 40 knots; the sea kept getting higher. The weather deteriorated and by dawn on 5 January, the situation turned hopeless, again.

We Are Not Afraid To Die Question Answer Extra Question 15.
What did Jon say that left the narrator speechless?
Answer:
When the narrator tried to comfort and reassure the children, Jon said that they were not afraid of dying if all four of them could be together. The narrator could find no words to respond, but he left the children’s cabin determined to fight the sea with everything he had.

Question 16.
What action did the narrator take, after having decided to fight the sea?
Answer:
To protect the weakened starboard side, he decided to heave to with the undamaged port hull facing the oncoming waves, using an improvised sea anchor of heavy nylon rope and two 22-litre plastic barrels of paraffin.

Question 17.
How did the narrator make his calculations to find out their position on 6 January?
Answer:
The Wavewalker rode out the storm and by the morning of 6 January, the narrator worked on wind speeds, changes of course, drift and current in an effort to calculate their position.

Question 18.
What instruction did the narrator give Larry? What did he expect?
Answer:
At about 2 p.m., the narrator asked Larry to steer a course of 185 degrees and said that if they were lucky, they would see the island at about 5 p.m. He was not optimistic himself so he went below, climbed on his bunk and slept off.

Question 19.
Why did the narrator feel that he was not the best captain? What was the surprise in store for him?
Answer:
When Jon called him the best daddy in the whole world and the best captain, the narrator was dejected for not being able to locate the island, so he refuted the statement. The truth was that the island was just in front of them.

Question 20.
Why did the narrator feel that it was the most beautiful island?
Answer:
The narrator saw lie Amsterdam. It was an unwelcoming piece of volcanic rock, with little vegetation, but to them it was the most beautiful island in the world because it held for them the hope of their survival.

We’re Not Afraid to Die… If We Can All Be Together Extra Questions and Answers Long Answer Type

Question 1.
The narrator and his wife had longed to sail. What did they do to accomplish their dream?
Answer:
The narrator and his wife had always dreamt of sailing. They wanted to do a round-the-world voyage like Captain James Cook had done 200 years earlier. For sixteen years they spent all their leisure time improving their seafaring skills in the British waters. They took a boat, Wavewalker, that was 23 metres, and weighed 30 ton. It had been professionally built and they spent months fitting it out and testing it in the roughest weather that they could find. Finally, in July 1976, the family set out to sail from Plymouth, England.

Question 2.
What were the troubles that they faced on the morning of 2 January? How did they counter nature’s wrath?
Answer:
When they reached the southern Indian Ocean, one of the world’s roughest seas, they began to encounter strong winds. Apart from the gales, the size of the waves was alarming. It was as high as the main mast. Things became worse on 2 January when the waves became huge. The ship rose to the top of each wave and they could see endless waves approaching them, and the screaming of the wind seemed horrifying to them. To slow the boat down, they dropped the storm jib and lashed a heavy mooring rope in a loop across the stem. Then they double-lashed everything, went through their life-raft drill, attached lifelines, donned oilskins and life jackets.

Question 3.
“The first indication of impending disaster came at about 6 p.m.” What was the warning? What was the disaster that followed?
Answer:
The first warning of the approaching disaster was the threatening stillness. The wind dropped, and the sky grew dark. Then with a roar, an enormous cloud seemed to come after the ship. It turned out to be a vertical wave, almost twice the height of the other waves, and had fearsome breaking crests. When they tried to move over it, a monstrous explosion shook the deck. Water broke over the ship, and the narrator’s head hit against the wheel and he was thrown into the sea. The narrator accepted his impending death, and felt he was losing consciousness. But soon, he was tossed back into the ship like a ‘rag doll’.

Question 4.
How did they deal with the water that had gushed into the ship?
Answer:
As Mary took control of the wheel, the narrator made his way towards the hatch. Larry and Herb were pumping out water frantically. He saw broken timbers hanging, the starboard side bulged inwards; clothes, crockery, charts, tin and toys sloshed about in deep water. So he struggled into the children’s cabin, found a hammer, screws and canvas, and laboured back on deck. He managed to stretch the canvas and secure waterproof hatch covers across the gaping holes.

Some water continued to stream below, but most of it was now being deflected over the side. The problems cropped up when the hand pumps started to block up with the fragments floating around the cabins and the electric pump short-circuited. The water level rose ominously. On the deck he missed the two spare hand pumps, forestay sail, jib, dinghies and the main anchor, which were pitched overboard. He found another electric pump and connected it to an out-pipe, and this worked.

Question 5.
Why were they desperate to look for an island? How did they manage?
Answer:
After having survived for fifteen hours since the wave hit, they knew that Wavewalker could not hold together long enough for them to reach Australia. The narrator checked the charts and calculated that there were two small islands a few hundred kilometres to the east. One of them, lie Amsterdam, was a French scientific base. But the waves had put the auxiliary engine out of action. To make matters worse, the weather continued to worsen. The wind finally eased, and the ship rode out the storm by the morning of 6 January.

The narrator deliberated on wind speeds, changes of course, drift and current in an effort to calculate their position. What he could determine was that they were somewhere in 150,000 kilometres of ocean looking for a 65 kilometre-wide island. About 2 p.m., he asked Larry to steer a course of 185 degrees. He expected to see the island at about 5 p.m., and eventually reached it by 6 p.m.

Question 6.
The children braved the situation more maturely than their years. Discuss.
Answer:
The children, certainly braved the situation more maturely than their years. Sue had her head hit and swollen, worryingly. She had two huge black eyes, and a deep cut on her arm. She did not make much of her injuries because she did not want to worry her father when he was trying to save them. Jon, the narrator’s six-year- old son, assured him that they were not afraid of dying if the family could all be together.

When Sue’s head injury worsened with her blackened eyes narrowed to slits, she held on to her spirit and gave the narrator a card with drawn caricatures of Mary and him with the words: ‘Here are some funny people. Did they make you laugh? I laughed a lot as well. ’ The underlying message of love and positive hope overwhelmed the narrator. He was touched with the thoughtfulness of a seven-year-old girl, who did not want her parents to worry about a head injury, and that of the boy who was not afraid to die.

The Tale of Melon City Important Extra Questions and Answers Class 11 English Snapshots

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The Tale of Melon City Extra Questions and Answers Short Answer Type

The Tale Of Melon City Extra Questions Class 11 Question 1.
What do the words ‘just and placid’ imply?
Answer
The phrase implied that the king was fair and mild. The king, ‘a great believer injustice’ ensured justice was meted out to his subjects. He was also mild mannered and rarely showed any displeasure and even if he did frown, he quickly wiped the frown off his face.

The Tale Of Melon City Question Answer Class 11 Question 2.
Where did the king want the arch constructed? Why?
Answer
The king wanted an arch to be erected which extended over the major main road. He felt, the road would edify the spectators it would improve the morals and knowledge of the onlookers there.

The Tale Of Melon City Class 11 Extra Questions Question 3.
What happened to the king as he rode down the road?
Answer
After the arch was built, the king rode through the street. He wanted to edify the spectators there. But as he was crossing below the arch, his crown fell off as the arch was built too low. This angered the king.

The Tale Of Melon City Class 11 Question Answer Question 4.
What order did the king give when his crown was knocked off his head?
Answer
The king was angry because his crown was knocked off his head as he tried to ride under the arch. He ordered the chief of the builders, responsible for building the arch, to be hanged.

Tale Of Melon City Questions And Answers Class 11 Question 5.
How did the chief of the builders escape hanging?
Answer
When the chief of the builders was led away to be hanged, he pleaded innocence. He claimed that it was the fault of the workers that the arch was built so low. He escaped hanging as the ‘just and placid’ king could not bear to punish an innocent man.

Tale Of Melon City Extra Questions Class 11 Question 6.
Why were the workmen to be hanged? How did they escape hanging?
Answer
The king ordered the workmen to be put to death as they were painted responsible, for building the low arch, by the chief of the builders. The workmen protested that they were not the ones at fault and blamed the masons who had made bricks of the wrong size. They, too, escaped death by hanging.

Tale Of Melon City Question Answer Class 11 Question 7.
Whom did the architect lay the blame on?
Answer
The masons blamed the architect for the poor design of the arch. The architect, in turn, passed on the blame to the king who had made certain changes in the architectural plans of the arch.

The Tale Of Melon City Questions Class 11 Question 8.
How did the king react to the architect’s accusation? Why did he react that way?
Answer
When the king heard the architect’s accusation, he was so angry that he almost lost his ability to reason. Since, he was righteous and tolerant, he admitted that this was a difficult situation. The king solicited advice and called for the wisest man in the country for counsel.

The Tale Of Melon City Class 11 Questions And Answers Question 9.
How was the wise man brought to court? What advice did he offer?
Answer
The wisest man was found and carried to the royal court, as he could neither walk nor see. He was an old and experienced man. He said in a trembling, feeble voice that the offender must be penalized. He condemned the arch, guilty, for throwing the crown off the king’s head.

The Tale Of The Melon City Question Answer Class 11 Question 10.
The arch was not punished in the end. Why?
Answer
The wise man declared that it was the arch that had thrown the crown off, and it must be hanged. A councillor objected to the arch being hanged; he called it a disgrace to hang something that had touched the honourable head of the king. The king agreed with the councillor and the arch was spared.

The Tale Of A Melon City Question Answer Class 11 Question 11.
What circumstances led to the execution of the king?
Answer
The crowd grew restless, tired to see the offenders escape death, by hanging. The king grew fearful of their agitation and decided that someone must be hanged. All the people were measured, one by one, along the noose, to see who fit it. They found that only the king reached the noose, leading to his execution.

The Tale Of Melon City Question Answers Class 11 Question 12.
What was the result of the king’s execution? How was the problem resolved?
Answer
After the king’s execution, the ministers realized they had to find a new king. They perplexed over the problem and sent out messengers to make known that the next person to cross the city gate would decide the ruler of the kingdom.

Tale Of Melon City Class 11 Question 13.
How did the melon become king?
Answer
The next man who crossed the city gate, entrusted to choose a ruler for the kingdom, was a fool. He liked melons and named a melon as the king. The ministers crowned a melon and accepted it as their king. They carried the melon to the throne and respectfully placed it on it.

Question 14.
How did the people of the kingdom react to their melon king?
Answer
The people of the kingdom were content with their melon king. They found no reason to criticize him as long as he left them in ‘Peace and Liberty’. In that kingdom, the philosophy of Taissez faire’ (refusal to interfere) . seemed to be well established.

The Tale of Melon City Extra Questions and Answers Long Answer Type

Question 1.
How did the ‘just and placid king’ get executed?
Answer
The arch, commissioned by the king, was built low that knocked off the king’s crown as he was crossing below the arch. He sentenced the chief of builders to death for causing him the dishonour, who passed on the blame to the workers. The workmen blamed the masons and the masons, in turn, blamed the architect. The architect reminded the king that he had made certain changes in the plans himself when they were shown to him.

The king was infuriated to hear that. Being righteous and tolerant, he called for the wisest man in the country, for counsel, who advised the king to hang the arch as it was the real offender who had thrown the crown off. A councillor objected to this ruling and declared it a disgrace to hang something that touched the honourable head.
The crowd, gathered for the hanging, became restless. The king apprehended their mood and ruled for someone to be hanged immediately. The noose, hung high, fitted the king alone and he was hung as per the royal ruling.

The Address Important Extra Questions and Answers Class 11 English Snapshots

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The Address Extra Questions and Answers Short Answer Type

The Address Class 11 Extra Questions And Answers Question 1.
Where had the narrator come? Why was she back?
Answer:
The narrator is a Dutch Jew, who had to leave Holland during the Second World War. She had left along with her mother for safety. Now she was back to where her past ‘things’ lay. She wanted to see and touch her belongings in order to relive those memories.

The Address Extra Questions Class 11 Question 2.
Whom did the narrator desire to meet in Holland? Why?
Answer:
The narrator was told by her mother to remember ‘Number 46 Marconi Street’, where Mrs Dorling lived; she had insisted on keeping their things safely till the war was over. After the war, the narrator was curious about their possessions that were still at that address and she went to meet Mrs Dorling.

The Address Class 11 Questions And Answers Question 3.
What kind of a welcome did the narrator get from Mrs Dorling?
Answer:
Mrs Dorling was cold and indifferent and evidently displeased to see the author. In fact, she tried to prevent her from entering by blocking her entrance. Later, she said it was not convenient for her to talk to the narrator at that point of time and refused to meet her.

Extra Questions Of The Address Class 11 Question 4.
When did the narrator first learn about the existence of Mrs Dorling?
Answer:
The narrator recalled the time when she was home during the first half of the War. She had noticed that various things were missing. Her mother then told her about Mrs Dorling, an old acquaintance who renewed their contact, and came regularly, each time, carrying away some of their things.

The Address Important Questions Class 11 Question 5.
What was the narrator’s mother’s opinion about Mrs Dorling?
Answer:
The narrator’s mother considered Mrs Dorling a very benevolent lady, who strived to ‘save’ their ‘nice things’ by carrying some of them away, each time she visited. The narrator’s mother was unable to see through the lady who wished to cheat her out of her valuables, instead she felt grateful to Mrs Dorling.

The Address Class 11 Important Questions Question 6.
What did the narrator recall about her first meeting with Mrs Dorling?
Answer:
The narrator saw Mrs Dorling for the first time on the morning after the day she came to know about her. Coming downstairs, the narrator saw her mother about to see someone out. It was a woman, dressed in a brown coat and a shapeless hat, with a broad back; she nodded and picked up the suitcase.

The Address Short Questions And Answers Question 7.
Why did the narrator return to Marconi Street after a long time?
Answer:
The narrator returned to Marconi Street after a long time because in the beginning, after the Liberation, she was not interested in all that stored stuff. She had lost her mother and was also afraid of being confronted with things that remained as a painful reminder to their past.

The Address Question Answer Class 11 Question 8.
How did the narrator decide to go back to the ‘things’?
Answer:
Gradually, when everything became normal again the bread was of a lighter colour and she had a bed to sleep in, securely, and the surroundings became familiar again the narrator was curious about all the possessions that must still be at that address that her mother had talked about and went there to relive her memories.

The Address Class 11 Short Questions And Answers Question 9.
Explain: “I stopped, horrified. I was in a room I knew and did not know.”
Answer:
When the narrator went to Mrs Dorling’s house the second time, a girl of about fifteen let her in. She saw familiar things but arranged differently that lent unfamiliarity to the surroundings. She found herself surrounded by things that she had wanted to see again but which really oppressed her in that strange atmosphere.

Class 11 English Snapshot Chapter 2 Extra Question Answer Question 10.
Why did the narrator not want to remember the place?
Answer:
The narrator had primarily returned for the sake of memories that were linked to the things that had once belonged to her mother. However, she realized, the objects linked in her memory with the familiar life that she had once lived lost their value as they had been removed and put in strange surroundings.

The Summer of the Beautiful White Horse Extra Questions and Answers Long Answer Type

The Address Question Answers Class 11 Question 1.
Describe the narrator’s first post-War meeting with Mrs Dorling.
Answer:
When the narrator knocked at Mrs Dorling’s door and introduced herself as Mrs S’s daughter, Mrs Dorling showed no sign of recognition. She held the door in a way making clear that the narrator was not welcome. For sometime, she stared quietly at the narrator at which she felt that it was not the person that she had been looking for. When Mrs Dorling let her in, the narrator noticed her wearing her mother’s green knitted cardigan.

The lady saw her looking at the cardigan and hid herself partially behind the door. When the narrator mentioned her mother, she said that she had thought that none of the people who had left had come back. The lady expressed regret at her inability to do anything for her but the narrator insisted on talking to her having come all the way for it. However, the lady refused to talk to her, claiming it was not a convenient time; the narrator had no option but to leave.

Extra Questions Of The Address Class 11 Question 2.
Contrast the character of the narrator’s mother and Mrs Dorling.
Answer:
The narrator’s mother was a trusting woman. She told her daughter about Mrs Dorling, an old acquaintance, who had suddenly turned up and renewed their contact and since then had been a regular visitor. The mother did not doubt her kindness and was obliged that she insisted on taking all her nice things with her to save them. The mother was worried about Mrs Dorling getting a crick in her back from carrying the crockery and lugging the large vases. When the narrator showed her scepticism, she was annoyed.

On the other hand, Mrs Dorling was an opportunist. She renewed her contact primarily to take the antique things the narrator’s mother owned. When the narrator came back after years, she made her feel unwelcome as she did not wish to part with the things that had belonged to the narrator’s mother. She was rude and brusque with the narrator and did not allow her to enter the house.

Address Extra Questions Class 11 Question 3.
Describe the narrator’s second visit to Mrs Dorling’s place.
Answer:
On her second visit, a girl of about fifteen led the narrator in and she noticed an old-fashioned iron Hanukkah candle-holder that belonged to them. In the living room, the sight was dismal. The room had a strange, stressful effect the atmosphere, the tasteless way everything was arranged, the ugly furniture or the muggy smell that hung there. She noticed the woollen tablecloth and recalled the bum mark on it that had never been repaired.

When the girl put cups on the tea table and poured tea from a white pot with a gold border on the lid and the pewter plate these things clouded her mind with strange emotions. The narrator noticed various things that brought back memories of the past. The narrator rushed out for her train as the girl went to get their cutlery. As the narrator walked out, she heard jingling of spoons and forks.

Online Education NCERT Solutions for Class 11 English Snapshots Chapter 4 Albert Einstein at School

Here we are providing Online Education NCERT Solutions for Class 11 English Snapshots Chapter 4 Albert Einstein at School. Students can get Class 11 English Albert Einstein at School NCERT Solutions, Questions and Answers designed by subject expert teachers.

Online Education Albert Einstein at School NCERT Solutions for Class 11 English Snapshots Chapter 4

Albert Einstein at School NCERT Text Book Questions and Answers

Question 1.
What do you understand of Einstein’s nature from his conversations with his history teacher, his mathematics teacher and the head teacher?
Answer:
Albert Einstein was an intelligent student but was not good at rote learning of dates and facts in history. However, only his mathematics teacher acknowledged his brilliance. Young Albert Einstein hated learning dates and facts by heart. This forever displeased Mr Braun, his history teacher.

Einstein explained his desire to know the reason a battle is fought, rather than remember its date. Mr Braun taunted him by calling his views the “Einstein theory of education” and punished him by detaining him for an extra period at school.

Albert’s brilliance was recognised by his mathematics teacher, Mr Koch, who accepted Einstein’s superiority of knowledge over his own. He gave Einstein a “glowing reference” to aid him continue his higher education. However, his principal clearly did not recognise his merit and expelled him from school. He accused Einstein of disrupting the class and of not making an effort to learn.

Question 2.
The school system often curbs individual talents. Discuss.
Answer:
Hint
1. If no

  • It often advances by giving opportunities in co-curricular and extra-curricular activities
  • Different options for different aptitudes
  • Offers subject choices/options that help develop the talent a student may not know she/he possesses
  • Encourages interaction

2. If yes

  • Examination system thrust on learning
  • Teacher student ratio often does not permit individual attention
  • Time bound classes/specific syllabi permits adherence to standard procedures
  • Facilities to cater to individual choices may not be permissible due to lack of resources

Question 3.
How do you distinguish between information gathering and insight formation?
Answer:
Gathering information: Gathering information, generally means, sorting out facts that are relevant to your work. It is usually a passive process and involves a lower level of understanding. It is more of a mechanical process based on the memorisation of facts.

Insight formation: This is a more complex process by which the actual learning occurs. The learner assimilates facts, and based on the acquired knowledge, is able to develop a better understanding of herself/himself, her/his world, and the people in her/his life. Insight formation comprises three ‘primary processes’:

  • information reception or perception,
  • encoding or interpretation, and
  • recall and use.

 

Online Education Clauses Exercises for Class 11 CBSE With Answers

Clauses Class 11

This grammar section explains Online Education English Grammar in a clear and simple way. There are example sentences to show how the language is used. Students can also read NCERT Solutions for Class 11 English to get good marks in CBSE Board Exams. https://ncertmcq.com/clauses-class-11/

Online Education Clauses Exercises for Class 11 CBSE With Answers

Clause Exercise For Class 11

♦ Structure of Sentences:

According to their grammatical structure, sentences can be classified into three types:

  1. Simple sentence
  2. Compound sentence
  3. Complex sentence

1. A Simple Sentence has only one clause, i.e., one subject and one predicate, e.g.
The students are learning grammar.
2. A Compound Sentence has two or more main clauses joined together by coordinating conjunctions like and, but, yet, still, so and so, therefore, for, now, otherwise, either ………….. or, neither ………….., nor, not only …………….. but also, e.g.
(i) The child ran fast and won the race.
(ii) Hari works hard and therefore stands first.
3. A Complex Sentence has one main clause and one or more subordinate clauses, e.g.
I read the book which I had bought from the Book Fair.

♦ Subordinate Clauses are of three kinds:

  • Noun Clause
  • Adjective Clause
  • Adverb Clause

1. Noun Clause:

The Noun Clause does the work of a noun in the sentence. It is introduced by the following connectives:

Pronouns: who, what, which, whom, whose
Example:

  • I know who he is.

Adverbs: when, where, why, how
Examples:

  • I asked him if he knew where the post office was.
  • I don’t know why he does not study.

Conjunctions: that, whether, if
Examples:

  • I think that Raghu is a liar.
  • I asked her whether she wanted a book to read.

Clauses Mcq Class 11

2. Adjective Clause:

Adjective Clauses describe a noun or a pronoun in the main clause or in another subordinate clause.
Adjective clauses are also known as relative clauses as they are usually introduced by relative pronouns like ‘who’, ‘whom’, ‘whose’, ‘which’, ‘that’, e.g.
The house which has large glass windows was burgled yesterday.
The adjective clause describes the noun ‘house’ in the main clause.
I know the boy who won the national championship.
The adjective clause describes the noun ‘boy’ in the main clause.
He who laughs last laughs the best.
The adjective clause describes the pronoun ‘He’ in the main clause.

Clause Class 11

3. Adverb Clause:

The adverb clause functions as an adverb, i.e. it modifies verbs. Therefore, an adverb clause may appear anywhere in a sentence. It tells us why, where, under what conditions, or to what degree the action occurred or the situation existed. Unlike adjective clauses, they frequently change their position within the sentence. Example: When the timer rings, we know the cake is done.

OR

We know the cake is done when the timer rings.
Adverb Clauses may be classified as Adverb Clauses of Condition, Time, Place, Reason, Manner, Purpose, etc.

Question 1.
Vipul and Mahesh are discussing their plans for the summer vacation. Complete their dialogue in an appropriate manner. Write your answers in the space provided. The first one has been done for you as an example.
Answer:

How are you going to spend your summer vacation?
I haven’t thought about it yet.
How do you like the idea (a) ………………… (Shimla)?
That’s nice, (b) ………………. (but).
I’ll come to your house today (c) ………………. (and).

  • Mahesh: Oh! that’ll be fine.
  • Vipul: (d) in the evening?
  • Mahesh: Yes, I don’t think he is going out tonight.
  • Vipul: OK. I’ll certainly come and meet him.

Answer:

(a) that we should go to Simla?
(b) but I will have to take my father’s permission.
(c) and request him to allow you to go to Simla.
(d) Should I come

Question 2.
Complete the sentences given below using subordinate clauses.

  1. My eyes have become red
  2. You should use sunglasses
  3. I always wash my eyes
  4. One should not use eyecare products
  5. I hope

Answer:

  1. because I have caught some eye infection.
  2. if you cannot tolerate glare.
  3. when they become red.
  4. which have not been produced by a reputable company.
  5. your eyes are free of infection soon.

Question 3.
Complete the dialogue using subordinate clauses.

  1. Customer: Will you tell me …………..?
  2. Shopkeeper: The price of this watch ……………….. is 900.
  3. Customer: I want to know ………………..
  4. Shopkeeper: Yes sir, the guarantee is of one year ……………..
  5. Customer: Of course, no one …………….. will damage it purposely.

Answer:

  1. what this watch costs?
  2. which/that you are looking at
  3. if it carries a guarantee.
  4. if it is not damaged carelessly.
  5. who buys it

Question 4.
Complete the letter given below with suitable clauses.
Dear son,
I am well. Hope you are taking good care of yourself. You must lock the doors properly at night (a) ……………..
You are alone (b) ……………… There is no need to be afraid (c) ………………. I shall come back (d) …………….. I miss you (e) ………………..

Yours affectionately
Dad

Answer:
(a) because there have been some thefts in the neighbourhood
(b) and so you must be careful
(c) because you can depend on our neighbours for help
(d) when the work in hand is finished.
(e) and am looking forward to being with you.

Question 5.
Chiki and Miks are planning to visit Kurukshetra during Vacation. Complete the dialogue between them using the given clauses.
(i) how are we going
(ii) what are we carrying?
(til) when I was a child
Chiki: I’m really excited about the trip.
Miki: Let’s decide what we are going to require for the journey.

(a) Miki: I have been to this place long back
(b) Chiki: Tell me by car or by train?
(c) Miki: We’ll be going by train. Have you decided
(d) Chiki: I’m not very good at packing. Do you know how to pack the rucksacks?
Answer:
(a) When I was a child
(b) how are we going
(c) what we are carrying?

Question 6.
Fill in the blanks in the following conversation with meaningful clauses.

Sheela: Tomorrow is my birthday. I want to prepare a cake myself. Please tell me (a) …………………………………….
Radha: I can make many types of cakes. Tell me (b) ……………………..
Sheela: I am very fond of chocolates and want to prepare a chocolate cake.
Radha: Please let me know (c) …………….. so that I can come and help you.
Sheela: Oh! Thank you.
Answer:

(a) how I should start
(b) what you want to make
(c) when you wish to make the cake

Question 7.
Fill in the blanks with suitable clauses.

  1. He ran so fast …………………
  2. You should act ………………
  3. He is the boy …………………

Answer:

  1. that he became breathless
  2. as you feel right
  3. who deserves admiration.

Question 8.
Complete the dialogue using the clauses given in the box.

if you want; what you require for packing; how I should pack it nicely

Chiki: Have you bought a birthday gift for Pooja?
Miki: Yes, but I don’t know (a) ……………………
Chiki: I can help you (b) ………………….
Miki: That’s really nice of you. Tell me (c) ………………..

(a) how I should pack it nicely
(b) if you want
(c) what you require for packing

Question 9.
Complete the following using clauses given in the box.

which make housing difficult for the poor; that the poor suffer the most; how it should cut down the cost of housing,

The finding of a housing study shows (a) ………………… The lack of resources, rising cost of material, and of land are the problems (b) ………………. The government’s worry is (c) ……………… so that the poor can buy houses.
Answer:

(a) that the poor suffer the most
(b) which make housing difficult for the poor
(c) how it should cut down the cost of housing.

Question 10.
Complete the following letter using appropriate clauses to fill in the blanks.
Answer:

Dear Sonia
How are you? I haven’t heard from you (a) ………………… I wanted to send you the books (b) ………………. We are all eagerly waiting to catch a glimpse of the man (c) ………………… Do send a quick reply.

Yours affectionately
Raina

Answer:
(a) or received any communication
(b) that you wanted
(c) who wrote them

Question 11.
Combine the following sets of sentences using clauses.

  1. You have met Ashok. He is my friend’s brother.
  2. A laser is a new device. It makes the light shine in a red beam.

Answer:

  1. You have met Ashok who is my friend’s brother.
  2. A laser is a new device which makes the light shine in a red beam.

Question 12.
Complete the sentences using a noun clause with the words given in brackets.

  1. The Principal announced that …………….. (holiday after the Annual day).
  2. Everybody cheered and felt that …………….. (good idea).
  3. ………… the day after the Annual Day would be a holiday.
  4. ……….. it was a good idea.

Question 13.
Given below are some sets of simple sentences. Combine each set into a complex sentence by using one of the sentences in each set into an adjective clause.

  1. The dog has been shot dead. It went mad.
  2. The lady brought to the party the little box. She was very fond of it.
  3. She is a girl. I wanted to meet her.

Answer:

  1. The dog that went mad has been shot dead.
  2. The lady brought to the party the little box that she was very fond of.
  3. She is the girl whom I wanted to meet.

Online Education Important Questions for Class 11 Biology Chapter Wise with Answers

Online Education Important Questions for Class 11 Biology Chapter Wise with Answers Pdf Download 2020-21: Here we are providing CBSE Important Extra Questions for Class 11 Biology Chapter Wise State Board Pdf download in Hindi and English Medium. Students can get Class 11 Biology NCERT Solutions, Biology Class 11 Important Extra Questions and Answers designed by subject expert teachers.

Online Education for CBSE Class 11th Biology Important Extra Questions and Answers Chapter Wise Pdf

  1. The Living World Biology Class 11 Important Questions
  2. Biological Classification Class 11 Biology Important Questions
  3. Plant Kingdom Class 11 Biology Chapter Wise Questions
  4. Animal Kingdom Question Bank Biology Class XI
  5. Morphology of Flowering Plants Biology Important Questions
  6. Anatomy of Flowering Plants Class 11 Important Questions
  7. Structural Organisation in Animals Important Questions
  8. Cell: The Unit of Life Class 11 Important Questions
  9. Biomolecules 11th Std Biology Important Questions
  10. Cell Cycle and Cell Division Class 11 Important Questions
  11. Transport in Plants Class 11 Important Questions
  12. Mineral Nutrition Class 11 NCERT Biology Important Questions
  13. Photosynthesis in Higher Plants Class 11 Important Questions
  14. Respiration in Plants Biology Important Questions for Class 11
  15. Plant Growth and Development Class 11 Important Questions
  16. Digestion and Absorption Biology Questions and Answers for Class 11
  17. Breathing and Exchange of Gases Class 11 Important Questions
  18. Body Fluids and Circulation Class 11 Important Questions
  19. Excretory Products and their Elimination Class 11 Important Questions
  20. Locomotion and Movement Class 11 Important Questions
  21. Neural Control and Coordination Class 11 Important Questions
  22. Chemical Coordination and Integration Class 11 Important Questions

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Discovering Tut: The Saga Continues Important Extra Questions and Answers Class 11 English Hornbill

Online Education for Discovering Tut: The Saga Continues Important Extra Questions and Answers Class 11 English Hornbill

Here we are providing Online Education for The Portrait of a Lady Important Extra Questions and Answers Class 11 English Hornbill, Extra Questions for Class 11 English was designed by subject expert teachers.

Online Education for The Portrait of a Lady Important Extra Questions and Answers Class 11 English Hornbill

The Portrait of a Lady Extra Questions and Answers Short Answer Type

Discovering Tut Important Questions Class 11 Question  1.
What generated the interest of the world in King Tut?
Answer:
King Tut was just a teenager when he died. He was the last heir of a powerful family that had ruled Egypt and its empire for centuries. Since the discovery of his tomb in 1922, the modem world wondered about what happened to him and wondered if he could have been murdered.

Discovering Tut Extra Questions Class 11 Question 2.
How did nature seem to echo the unnatural happening?
Answer:
As King Tut was taken from his resting place in the ancient Egyptian cemetery, dark-bellied clouds that had scudded across the desert sky all day, veiled the stars in grey. It seemed that the wind was angry and had roused the dust devils.

Discovering Tut The Saga Continues Extra Questions Class 11 Question 3.
Why did the tourists throng to see Tut’s tomb? What was their reaction?
Answer:
The tourists came to pay their respects to King Tut. They admired the murals and Tut’s gilded face on his mummy-shaped outer coffin. They read from the guidebooks in whisper, or stood silently, pondering over Tut’s untimely death, dreading, lest the pharaoh’s curse befall those who disturbed him.

Discovering Tut: The Saga Continues Extra Questions Class 11 Question 4.
Who was Howard Carter? What did he find?
Answer:
Howard Carter was the British archaeologist who in 1922 discovered Tut’s tomb after years of unsuccessful search. He discovered the richest royal collection ever found that included stunning artifacts in gold that caused a sensation.

Discovering Tut Question Answers Class 11 Question 5.
Tut was buried in March-April. How did Carter conclude this?
Answer:
On opening a coffin, Carter found a shroud decorated with garlands of willow and olive leaves, wild celery, lotus petals and cornflowers. Since these flowers grow in March or April, Carter concluded that the burial was in these months.

Discovering Tut The Saga Continues Question And Answer Class 11 Question 6.
“When he finally reached the mummy, though, he ran into trouble.” Why was it so?
Answer:
When Carter tried to raise the mummy out of the coffin, he could not. The ritual resins had hardened, cementing Tut’s body to the bottom of his solid gold coffin. No amount of force could pull it out.

Discovering Tut The Saga Continues Question Answer Class 11 Question 7.
How did he decide to detach the mummy? Why?
Answer:
First Carter tried to loosen the resins with the heat of the sun. For several hours, he put the mummy outside in blazing sunshine that heated it to 149 degrees Fahrenheit but it was in vain. Then he decided to carve it out from beneath the limbs and trunk as there was no other way of raising the king’s remains.

Discovering Tut The Saga Continues Important Questions Class 11 Question 8.
What were the treasures found in the coffin? Why were they put there?
Answer:
King Tut’s coffin contained precious collars, inlaid necklaces and bracelets, rings, amulets, a ceremonial apron, sandals, sheaths for his fingers and toes, and his inner coffin and mask, all of which were made of pure gold. The royals, in King Tut’s time, hoped to take their riches along with them for their next life.

Important Question Of Discovering Tut Class 11 Question 9.
How has the viewpoint of archaeologists changed with the passage of time?
Answer:
The archaeologists, earlier, focussed on the treasures that the tomb would yield. The centre of attention, now, is more on the fascinating details of life and intriguing mysteries of death. Moreover, now they use more sophisticated tools, including medical technology.

Question 10.
What was the interesting fact about Tut that was brought to light in the late sixties?
Answer:
In 1968, more than forty years after Carter’s discovery, an anatomy professor X-rayed the mummy and revealed a startling fact: beneath the resin that caked his chest, his breast-bone and front ribs were missing.

Question 11.
Why was King Tut’s death a big event?
Answer:
King Tut’s demise was a big event as he was the last of his lineage and his funeral sounded the death rattle of a dynasty. Moreover, he died at the very young age of about eighteen.

Question 12.
What is known about Tut’s predecessor Amenhotep IV?
Answer:
Amenhotep IV, during his reign, promoted the worship of the Aten, the sun disk, and changed his own name to Akhenaten, or ‘servant of the Aten’, and moved the religious capital to the new city of Akhetaten. He outraged the country by attacking Amun, a major god, smashing his images and closing his temples.

Question 13.
What made a guard remark, ‘curse of the pharaoh’?
Answer:
When Tut’s body was taken out to be scanned and the million-dollar scanner had stopped functioning because of sand in a cooler fan, the guard jokingly remarked that the king had expressed his annoyance at being disturbed.

Question 14.
With King Tut was being finally laid to rest, nature was at rest too. Explain.
Answer:
When King Tut was finally laid to rest, the wind stopped blowing and was still, like death itself. Orion, the constellation that the ancient Egyptians knew as the soul of Osiris, the god of the afterlife, was sparkling. It seemed to be watching over the boy king.

Discovering Tut: The Saga Continues Extra Questions and Answers Long Answer Type

Question 1.
Nature echoed the unnatural happenings with King Tut’s body. Comment.
Answer:
To set to rest the modem world’s speculation about King Tut, the body was taken out of its resting place some 3,300 years later. He was required to undergo a CT scan to generate precise data for an accurate forensic reconstruction. As the body was taken out, raging wind began to blow which seemed to arouse the eerie devils of dust. Dark clouds gathered and appeared to shroud the stars in a grey-coloured coffin. When the body was put down for scan, the million-dollar scanner seemed to keep from functioning.

There was sand in a cooler fan. It was when he was finally laid to rest, that the winter air lay cold and still, like death itself, in this valley of the departed. Just above the entrance to Tut’s tomb stood Orion the constellation that the ancient Egyptians knew as the soul of Osiris, the god of the afterlife, supervising the young pharaoh returning to his rightful place.

Question 2.
“The mummy is in a very bad condition because of what Carter did in the 1920s.” What did Carter do and why?
Answer:
Howard Carter was the British archaeologist who in 1922 discovered Tut’s tomb. He searched its contents in haste. The tomb, which had stunning artefacts in gold, caused a sensation at the time of the discovery.

After months of carefully recording the treasures in the pharaoh’s coffin, Carter began investigating the three nested coffins. When he finally reached the mummy, he found that the ritual resins had hardened. Thus, Tut’s body was cemented to the bottom of his solid gold coffin. Carter set the mummy outside in blazing sun that heated it up to 149 degrees Fahrenheit, to no avail.

To prevent the thieves from ransacking, he chiselled the body free. To separate Tut from his embellishments, Carter’s men removed the mummy’s head and severed nearly every major joint.

Question 3.
Describe the changing attitudes of the archaeologists over a span of time.
Answer:
Archaeology has changed substantially in the intervening decades. It now focusses less on treasure and more on the interesting details of life and the intriguing mysteries of death. It also uses more sophisticated tools, including medical technology. In 1968, more than forty years after Carter’s discovery, an anatomy professor X-rayed the mummy and revealed a startling fact: beneath the resin that cakes King Titu’s chest, his breast bone and front ribs were missing.

Today, diagnostic imaging can be done with computed tomography, or CT, by which hundreds of X-rays in cross section are put together like slices of bread to create a three dimensional virtual body. It can even answer questions such as how a person died, and how old he was at the time of his death.

Question 4.
What are the facts that are known about King Tut’s lineage?
Answer:
Amenhotep III, Tut’s father or grandfather, was a powerful pharaoh who ruled for almost four decades at the height of the eighteenth dynasty’s golden age. His son Amenhotep IV succeeded him and initiated one of the strangest periods in the history of ancient Egypt. The new pharaoh promoted the worship of the Aten, the sun disk, changed his name to Akhenaten, or ‘servant of the Aten’, and moved the religious capital

from the old city of Thebes to the new city of Akhetaten, now known as Amama. He further shocked the country by attacking Amun, a major god, smashing his images and closing his temples. After Akhenaten’s death, a mysterious ruler named Smenkhkare appeared briefly and exited with hardly a trace. A very young Tutankhaten took the throne as the king, thereafter.

Online Education for A Photograph Important Extra Questions and Answers Class 11 English Hornbill

Here we are providing Online Education for A Photograph Extra Questions and Answers Class 11 English Hornbill, Extra Questions for Class 11 English was designed by subject expert teachers. Students can also read A Photograph Poem. https://ncertmcq.com/extra-questions-for-class-11-english/

Online Education for A Photograph Important Extra Questions and Answers Class 11 English Hornbill

A Photograph Extra Questions and Answers Short Answer Type

A Photograph Class 11 Extract Questions And Answers Question 1.
The poet talks about a particular cardboard. How is it special to her?
Answer:
The poet talks about a particular cardboard to which is pasted her mother’s photograph taken at the sea beach. The mother seems to have been enjoying her sea holiday. The photograph is special as she has lost her mother sometime back and looking at the photograph makes her happy as well as sad.

A Photograph Class 11 Extract Questions And Answers Pdf Question 2.
What can you say about the childhood of the poet’s mother?
Answer:
The childhood of the poet’s mother must have been filled with fun and happiness. This is clear from the snapshot of the sea holiday. They are enjoying their holiday.

The mother of the poet laughs when looking at the snapshot even after many years have passed since the sea holiday. All this shows us that it was a very pleasant childhood.

A Photograph Class 11 Extra Questions And Answers Question 3.
What moment does the photograph depict?
Answer:
The photograph clicked by. the uncle of the poet’s mother depicts a sea-holiday being enjoyed by the poet’s mother and her two cousins Dolly and Betty. They are full of smiles in their beach dresses, not worrying about their flying hair.

A Photograph Extract Based Questions With Answers Question 4.
Were the three cousins camera friendly? Who was taking their photograph?
Answer:
The three cousins appear to be camera-friendly as they stood at the sea beach without moving when the uncle took the photograph.

The Photograph Class 11 Extra Questions Question 5.
The poet’s mother would laugh looking at the photograph. Why?
Answer:
The poet’s mother was in middle age when she looked at that photograph and used to laugh remembering those golden days of her childhood, enjoying a sea holiday. She would also laugh at the beach dresses which looked weird after many years.

A Photograph Question Answers Question 6.
What impression do you form about the poet’s mother?
Answer:
The poet’s mother was very pretty at the age of twelve. She enjoyed the sea holiday. This is indicated by the happiness that she gets in middle age after looking at her childhood photograph.

Photograph Class 11 Extra Questions Question 7.
The sea ‘appears to have changed less’ in comparison to the three girls who enjoyed the sea holiday. Comment.
Answer:
The poet compares the mortal nature of human beings with the eternal nature of the sea or natural objects. With the passage of time, the poet’s mother died but the vast sea has remained as it was since the photograph was taken.

The Photograph Extra Questions Question 8.
Why does the poet feel nostalgic?
Answer:
The poet sees an old photograph of her mother in which she was standing on the beach with her two cousins — Dolly and Betty. They were enjoying themselves. The photograph captured her mother’s sweet and smiling face. At that time, she was around twelve years old. The poet remembers how her mother used to laugh whenever she looked at that old photograph. But time has passed and now the poet has been left only with the memories of her mother. Thus, she feels nostalgic

A Photograph Class 11 Important Question Question 9.
What does the poet say about her mother’s face?
Answer:
The poet remarks that her mother had a sweet face, smiling and caring for her cousins who were younger to her. The poet also says that her mother used to enjoy these sea holidays, and would laugh heartily, later on when she saw the photograph.

A Photograph Short Question Answer Question 10.
How does the poet react to her past? Why has she not mentioned anything about her mother’s death?
Answer:
The poet remembers with sadness her mother’s laughter which she cannot hear any more. The poet is full of a sense of loss and does not mention about her mother’s death, as it may bring more gloom to her and make her speechless.

A Photograph Class 11 Important Questions And Answers Question 11.
Does the poet appear to be grieving?
Answer:
The poet is certainly filled with a sense of loss. Her mother is long dead and though the poet has adjusted to her absence, she is not able to completely overcome her loss.

She remembers how it used to be when her mother was still with her. The last line is an apt depiction of her state of mind. The loss has filled her life with silence.

A Photograph Question And Answer Question 12.
Comment on the tone of the poem.
Answer:
The tone of the poem is that of sadness. Shirley Toulson looks at an old photograph of her mother and is sadly reminded of her mother who is no more. She mentions about death of her mother indirectly only but this photograph has made her speechless and silent.

A Photograph Extra Questions and Answers Long Answer Type

Class 11 English Poem A Photograph Extra Questions Question 1.
The poet has paid a tribute to her mother. Similar instances can be seen in ‘The Portrait of a Lady’. This made you think that writing about a loved one is much better than building their statues or drawing their portraits. Comment.
Answer:
Many writers have paid tributes to their loved ones through beautiful writing. Khushwant Singh gave an adorable description about his grandmother through his story. Shirley Toulson remembered her mother through her heart-touching poem.

In my opinion, writing about a loved one is much better than building their statues or drawing their portraits. One can never tell the true personality of a person just by looking at their sculptures or portraits. One can never know about the amazing time someone has spent with them. That magic can only be created by words.

Words stand the passage of time, whereas sculptures or portraits may get damaged by it. Hence, words are the best way by which anyone can pay a tribute to one’s loved ones.

A Photograph Important Questions Question 2.
“Its silence silences,” writes Shirley Toulson. The loss of her mother has silenced her. Do you think that this attitude of the poet is the right attitude to live life? Why/ why not?
Answer:
There is no doubt that Shirley Toulson has given a very touching tribute to her mother by remembering her through her verses. It is apparent that she is very much nostalgic and is grieving at the loss of her mother. Though she says that over the years she has adjusted to her mother’s absence, but circumstances have surely filled her with silence and a deep void.

We cannot deny that it hurts very much to lose someone, but the attitude shown by the poet at the end is not the right way to live your life. Life will keep going on even if we stop to lament our loss.

Loss is universal. It is the law of nature. We cannot let ourselves get depressed just because of this. It is also understandable that we will grieve. However, grieving to the point of hampering the normal functioning of our lives is not acceptable.

Extract Based Questions Class 11 English A Photograph Question 3.
Happy moments are short-lived but provide a lifetime memory. They provide a cushion to bear the difficulties which the future has in store for you. Comment in the light of the poem ‘A Photograph’ by Shirley Toulson.
Answer:
Our life is a mixture of happy as well as adverse times. We must learn to move on with the help of those happy memories which provided us with so much enjoyment and happiness. As life is not a bed of roses, everyone at one stage or another is likely to face difficulties.

At the time of difficulties, happy moments can give us solace and fill us with positivity which is required during difficult times. Happy moments will certainly provide us with a hope that, as happy moments are short-lived, so are difficult times. One must learn to cope with the situation. The memories of happy times can provide us a cushion to bear difficulties with patience and peace.

A Photograph Poem Question Answer Question 4.
‘Both wry with the laboured ease of loss.’ The poet is missing her mother. What is the role of the mother in forming the personality of a child?
Answer:
A mother’s role in shaping the personality of a child is of paramount importance. The child watches his/ her mother intently and learns about the world and how to react to it at the initial stages. The mother can

play an important role by making the child to deal constructively with mistakes, forgive others, handle frustration, show kindness and share love.
When a mother is nurturing and caring the child, it will develop a healthy bond with not only the mother but will be willing to form new relationships with others.

Children and adults both want a sense of independence and autonomy. It is very important on a mother’s part to offer choices to the child. This makes the child feel that he/she is smart enough to make choices.

The mother’s thoughts nourish a child’s mind and soul as her personal attention nourishes a child’s body. She is a child’s first teacher. She tries to imbibe such values that may help a child lifelong.

A Photograph Extract based Questions and Answers

I. Read the extract given below and answer any two of the questions that follow.

“The cardboard shows me how it was When the two girl cousins went paddling, Each one
holding one of my mother’s hands,
And she the big girl – some twelve years or so.”

Question 1.
What does the cardboard here refer to?
(a) A thick paper on which the poet’s photograph was pasted
(b) A thick envelope
(c) A thick paper on which the poet’s mother’s photograph was pasted
(d) A paper boat
Answer:
(c) A thick paper on which the poet’s mother’s photograph was pasted

Question 2.
What does the cardboard depict?
(a) It depicts a scenery
(b) It depicts the picture of a house
(c) It depicts the picture of a school
(d) It depicts the picture of three girls
Answer:
(d) It depicts the picture of three girls

Question 3.
Who is the ‘big girl’ mentioned here?
(a) The big girl is the poet herself
(b) The big girl is the poet’s mother
(c) The big girl is the poet’s relative
(d) The big girl is the poet’s friend
Answer:
(b) The big girl is the poet’s mother

II. Read the extract given below and answer any two of the questions that follow.

“All three stood still to smile through their hair At the uncle with the camera. A sweet face,
My mother’s, that was before I was born.
And the sea, which appears to have changed less,
Washed their terribly transient feet.”

Question 1.
What does the poet mean by ‘smile through their hair’?
(a) It means that a smile was painted on the hair of the photographed girls
(b) It means that the photographed girls were wearing a mask
(c) It means that the hair of the photographed girls were covering their face when they were smiling
(d) It means that the hair of the girls in the photograph was smiling too
Answer:
(c) It means that the hair of the photographed girls were covering their face when they were smiling

Question 2.
What has not changed over a period of time?
(a) The photo
(b) The cardboard
(c) The girls
(d) The sea
Answer:
(d) The sea

Question 3.
Find a word from the extract which means “lasting only for a short time”?
(a) Still
(b) Transient
(c) Changed
(d) Less
Answer:
(b) Transient

III. Read the extract given below and answer any two of the questions that follow.

“Some twenty-thirty – years later She’d laugh at the snapshot. “See Betty And Dolly,” she’d
say, “and look how they Dressed us for the beach.” The sea holiday Was her past, mine is her
laughter. Both wry With the laboured ease of loss.”

Question 1.
Why did ‘she’ laugh?
(a) Because of the funny dresses that they were wearing at the sea holiday
(b) Because one of them cracked a joke
(c) Because of the funny dresses they were wearing at the party
(d) Because of the funny man they saw at the sea holiday
Answer:
(a) Because of the funny dresses that they were wearing at the sea holiday

Question 2.
Who are Betty and Dolly?
(a) They are poet’s cousins
(b) They are poet’s friends
(c) They are poet’s mother’s friends
(d) They are poet’s mother’s cousins
Answer:
(d) They are poet’s mother’s cousins

Question 3.
…………. in the extract is the synonym of ‘photograph’.
(a) Snapshot
(b) Picture
(c) Mine
(d) Laboured
Answer:
(a) Snapshot

IV. Read the extract given below and answer any two of the questions that follow.

“Now she’s been dead nearly as many years As that girl lived. And of this circumstance There
is nothing to say at all.
Its silence silences.”

Question 1.
Who does ‘she’ refer to?
(a) The poet’s dead aunt
(b) The poet’s dead mother
(c) The poet’s dead cousin
(d) The poet’s sister
Answer:
(b) The poet’s dead mother

Question 2.
Why is there nothing to say about the death of the poet’s mother?
(a) Because the poet is confused
(b) Because the poet was not in her senses when her mother expired
(c) Because the death of the poet’s mother has left a deep void in the poet’s heart
(d) Because the poet did not have a good relationship with her mother
Answer:
(c) Because the death of the poet’s mother has left

Question 3.
Which word in the extract means the same as “events that change your life, over which you have no control”?
(a) Silences
(b) Circumstances
(c) Situation
(d) Circumstance
Answer:
(d) Circumstance