**Chapter 12 - Sound**

**1. Fill in the blanks and explain.**

**1. Fill in the blanks and explain.**

**a. Sound does not travel through …………**

**Ans:** vacuum

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**b. The velocity of sound in steel is …………… than the velocity of sound in water. **

**Ans:** greater

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**c. The incidence of**** ……… in daily life shows that the velocity of sound is less than the velocity of light.**

**Ans:** lightning

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**d. To discover a sunken ship or objects deep inside the sea, ………………….. technology is used.**

**Ans:** sonar

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**2. Explain giving scientific reasons.**

**2. Explain giving scientific reasons.**

**a. The roof of a movie theatre and a conference hall is curved.**

**Ans: **

**(i) Sound waves get reflected from the walls and roof of a room multiple times. This causes a single sound to be heard not once but continuously. This is called reverberation.**

**(ii) Due to reverberation, some auditoriums or some particular seats in an auditorium have inferior sound reception. This can be compensated with curtains.**

**(iii) The ceilings of these halls are curved so that sound reflects from the ceiling and reaches all parts of the hall, improving sound quality. **

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**b. The intensity of reverberation is higher in a closed and empty house.**

**Ans:**

**(i) Reverberation occurs due to multiple reflections of sound.**

**(ii) The furniture in the house acts as a sound-absorbing material.**

**(iii) So if the house is closed and empty, a reflection of sound will be at its maximum, and hence, the intensity of reverberation is higher.**

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**c. We cannot hear the echo produced in a classroom.**

**Ans:**

**(i) The classrooms are designed in such a way that the distance between the walls is less than 17.2 m.**

**(ii) Due to this, the reflections of sound from walls or echoes reach our ears within 0.1 s.**

**(iii) Because of this, we are not able to distinguish between the original sound and the echo produced in the classroom.**

**3. Answer the following questions in your own words.**

**3. Answer the following questions in your own words.**

**a. What is an echo? What factors are important to get a distinct echo?**

**Ans: **

**(i) An echo is the repetition of the original sound because of reflection from some surface.**

**(ii) At 220 °C, the velocity of sound in air is 344 m/s.**

**(iii) Our brain retains a sound for 0.1 s. Thus, for us to be able to hear a distinct echo, the sound should take more than 0.1 s after starting from the source to get reflected and come back to us.**

**(iv) We know that,**

**Distance = time x speed**

**∴ Distance = 344 m/s × 0.1 s**

**∴ Distance = 34.4 m**

**(v) Thus, to be able to hear a distinct echo, the reflecting surface should be at a minimum distance of half of the above, i.e., 17.2 m.**

**(vi) Because the velocity of sound varies with air temperature, so does this distance. **

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**b. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.**

**Ans:**

**(i) Golghumat, with a height of 51 m and diameter of 37 m and 3 m thick walls, is spread over approximately 1700 sq. m.**

**(ii) This meets the conditions for an echo, i.e., 17.2 m minimum.**

**(iii) The dome of the golghumat is curved, and hence, sound reflects multiple times before reaching the observer.**

**(iv) This is the reason for the multiple echoes being heard.**

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**c. What should be the dimensions and the shape of classrooms so that no echo can be produced there?**

**Ans:**

**(i) Dimensions: The distance between opposite walls in a classroom must be less than 17.2 m so that the reflected sound returns to the observer within 0.1 s.**

**(ii) Shape: The classrooms should have curved ceilings and walls so that the reflected sound is directed towards the observer instantly within 0.1 s.**

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**4. Where and why are sound absorbing materials used?**

**Ans:**

**(i) The sound-absorbing materials are used in schools, cinemas, concert halls, houses, or other places where sound quality is important.**

**(ii) In the absence of sound-absorbing material, the sound will undergo multiple reflections, causing reverberation.**

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**5. Solve the following examples.**

**a. The speed of sound in air at 0⁰ C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?**

**Given: **

**Initial speed of sound at 0°C = 332 m/s. **

**Final speed of sound = 344 m/s. **

**Rate of increase per degree rise in temperature = 0.6 m/s **

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**To find: **

**Temperature when speed is 344 m/s **

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**Formulae: **

**Increase in temperature = \(\large \frac {Increase\, in\, speed\, of\, sound}{Rate\, of\, increase\, per\, degree\, rise\, in\, temperature}\)**

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**Solution:**

**Increase in temperature = \(\large \frac {Increase\, in\, speed\, of\, sound}{Rate\, of\, increase\, per\, degree\, rise\, in\, temperature}\)**

**Increase in temperature = \(\large \frac{344\;-\;332}{0.6}\)**

**Increase in temperature = \(\large \frac{12}{0.6}\)**

**Increase in temperature = 20° C**

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**∴ The temperature is 20° C when the speed of sound is 344 m/s. **

**b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her? **

**(The velocity of sound in air is 340 m/s)**

**Given: **

**Speed of sound (v) = 340 m/s **

**Time taken (t) = 4 sec**

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**To find: **

**Distance (s) = ?**

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**Formulae: **

**$$ {\style{font-family:stix}{\style{font-size:12px}{\mathrm{Velocity}\;=\;\frac{\mathrm{Distance}}{\mathrm{Time}}\\\\\\}}} $$**

**Solution:**

**$$ {\style{font-family:stix}{\style{font-size:12px}{\mathrm{Velocity}\;=\;\frac{\mathrm{Distance}}{\mathrm{Time}}\\\\\mathrm{Distance}\;=\;\mathrm{Velocity}\;\times\;\mathrm{Time}\\\\\mathrm{Distance}\;=\;340\;\times\;4\\\\\mathrm{Distance}\;=\;1360\;m\\\\}}} $$**

**∴ The lightning has struck at a distance of 1360 m from the observer.**

**c. Sunil is standing between two parallel walls. The wall closest to him is at a distance of 660 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.**

**1. What is the velocity of sound in air?**

**2. What is the distance between the two walls?**

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**Given:**

**Distance of the closer wall (S₁) = 660 m **

**Time of first echo (t₁) = 4 sec**

**Time of second echo (t₂) = 2 sec**

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**To find:**

**Velocity of sound in air (v) = ? **

**Distance between two walls (S₁ + S₂) = ?**

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**Formulae :**

**$$ {\style{font-family:stix}{\style{font-size:12px}{\mathrm{Velocity}\;=\;\frac{\mathrm{Distance}}{\mathrm{Time}}\\\\}}} $$**

**Solution: **

**1. **For** Velocity of sound in air (v),**

**$$ {\style{font-family:stix}{\style{font-size:12px}{\mathrm{Velocity}\;=\;\frac{\mathrm{Distance}}{\mathrm{Time}}\\\mathrm{Velocity}\;=\;\frac{660}2\\\\\mathrm{Velocity}\;=\;330\;m/s}}} $$**

**2.** For distance between two walls (S₁ + S₂),

**For the first wall,**

**$$ {\style{font-family:stix}{\;V_1\;=\;\frac{2\;S_1}{t_1}\\330\;\;=\;\frac{2\;S_1}4\\330\;\;=\;\frac{S_1}2\\330\;\times\;2\;=\;S_1\\S_1\;=\;660\;m\;\\}} $$**

**For the second wall,**

**$$ {\style{font-family:stix}{\;V_2\;=\;\frac{2\;S_2}{t_2}\\330\;\;=\;\frac{2\;S_2}6\\330\;\;=\;\frac{S_2}3\\330\;\times\;3\;\;=\;S_2\\S_2\;=\;990\;m\;\\}} $$**

**∴ Distance between two wall = S₁ + S₂ = 660 + 990 = 1650 m**

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**∴ The velocity of sound in air is 330 m/s and the distance between two walls is 1650 m.**

**d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How many times as fast as the other? **

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**Given:**

**Mass of hydrogen in bottle A (mA) = 12gm **

**Mass of hydrogen in bottle B(mB) = 48gm**

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**To find:**

**In which bottle sound travels faster**

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**Formulae:**

**$$ {\style{font-family:stix}{\style{font-size:16px}{\mathrm\rho=\;\frac mV\\v\;\;\propto\frac1{\sqrt{\mathrm\rho}}\\؞\;v\;\;\propto\frac1{\sqrt{\frac{\mathrm m}{\mathrm V}}}\\؞\;v\;\;\propto\frac{\sqrt V}{\sqrt m}\\}}} $$**

**Solution:**

**$$ {\style{font-family:stix}{\style{font-size:16px}{v_A\;\;\propto\frac{\sqrt V}{\sqrt{m_A}}\;\;\;\;\;\;\;\;\;\;…\;(i)\\\\v_B\;\;\propto\frac{\sqrt V}{\sqrt{m_B}}{\;\;\;\;\;\;\;\;\;\;…\;(ii)}}}} $$**

**Since both the bottes are identical, hence, the volume is same, i.e. V**

**Dividing (i) by (ii), we get,**

**$$ {\style{font-family:stix}{\style{font-size:16px}{\frac{v_A}{v_B}\;\;=\;\frac{\frac{\sqrt V}{\sqrt{m_A}}}{\frac{\sqrt V}{\sqrt{m_B}}}\\؞\;\frac{v_A}{v_B}\;\;=\;\frac{\sqrt{m_B}}{\sqrt{m_A}}\\؞\;\frac{v_A}{v_B}\;\;=\;\sqrt{\frac{m_B}{m_A}}\\؞\;\frac{v_A}{v_B}\;\;=\;\sqrt{\frac{48}{12}}\\؞\;\frac{v_A}{v_B}\;\;=\;\sqrt4\\؞\;\frac{v_A}{v_B}\;\;=\;2\\؞\;v_A\;=\;2\;v_B}}} $$**

**∴ Velocity of the sound will be more in bottle A, and Velocity of sound in Bottle A is twice of that in bottle B.**

**e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what concl****usions will you draw?**

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**Given:**

**Mass of helium in bottle A (mA) = 10gm **

**Mass of hydrogen in bottle B (mB) = 40gm**

** **

**To find:**

**To draw the conclusions related to the conditions given in the sum.**

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**Formulae:**

**$$ {\style{font-family:stix}{\style{font-size:16px}{v\;\alpha\;\frac1{\sqrt M}\\v\;\alpha\;\sqrt T\\\rho\;=\;\frac mv}}} $$**

**Solution:**

**$$ {v_A\;=\;\frac{\sqrt{T_A}}{\displaystyle\frac{\sqrt{m_A}}V}\;\;\;\;\;\;\;…(i)\\v_B\;=\;\frac{\sqrt{T_B}}{\displaystyle\frac{\sqrt{m_B}}V}{\;\;\;\;\;\;\;…(ii)}\\\\\\} $$**

**Since both the bottes are identical, hence, the volume is same, i.e.**

**$$ {\style{font-family:stix}{\style{font-size:16px}{v_A\;=\;v_B\\i.e.\;\frac{\sqrt{T_A}}{\displaystyle\frac{\sqrt{m_A}}V}\;=\;\frac{\sqrt{T_B}}{\displaystyle\frac{\sqrt{m_B}}V}\\؞\;\frac{\sqrt{T_A}}{\displaystyle\sqrt{m_A}}\;=\;\frac{\sqrt{T_B}}{\displaystyle\sqrt{m_B}}\\؞\;\frac{\sqrt{T_A}}{\displaystyle\sqrt{T_B}}\;=\;\frac{\sqrt{m_A}}{\displaystyle\sqrt{m_B}}\\؞\;\frac{\sqrt{T_A}}{\displaystyle\sqrt{T_B}}\;=\;\frac{\sqrt{10}}{\displaystyle\sqrt{40}}\\؞\;\frac{\sqrt{T_A}}{\displaystyle\sqrt{T_B}}\;=\;\sqrt{\frac{10}{40}}\\؞\;\frac{\sqrt{T_A}}{\displaystyle\sqrt{T_B}}\;=\;\sqrt{\frac14}\\\\\\\\\\\\}}} $$**

**Squaring both sides, we get,**

**$$ {\style{font-family:stix}{\style{font-size:16px}{؞{\;\left(\frac{\sqrt{T_A}}{\displaystyle\sqrt{T_B}}\right)}^2\;={\;\left(\sqrt{\frac14}\right)}^2\\؞{\;\frac{T_A}{\displaystyle T_B}}\;={\;\frac14}\\؞\;4\;T_A\;=\;T_B\\i.e.\;T_B=\;4\;T_A\\\\\\\\\\}}} $$**

**∴ The temperature of Bottle B is 4 times the temperature of Bottle A.**