* सेलों का संयोजन :-* प्रत्येक विशिष्ट विभव के मान की आवश्यकता के लिए विभव स्रोत (बैटरी) बनाना संभव नहीं है। जब एक विशिष्ट विभवांतर की आवश्यक होता है, तो विभव और विद्युत धारा के वांछित मान को प्राप्त करने के लिए दो या दो से अधिक सेलों/बैटरी का उपयोग विभिन्न संयोजनों में किया जाता है। इन सेलों को मुख्यतः दो प्रकार के संयोजनों में जोड़ा जा सकता है। ये दो संयोजन हैं :-

- श्रेणी क्रम संयोजन और
- समानांतर क्रम संयोजन

जब लोड (जैसे विद्युत मोटर, विद्युत उपकरण या इलेक्ट्रॉनिक उपकरण) के लिए आवश्यक विभवांतर, एकल सेल या बैटरी के विभव से अधिक हो जाता है, तो कुल विभवांतर बढ़ाने के लिए सेलों को श्रेणी में जोड़ा जाता है। सेलों के श्रेणी क्रम संयोजन में कई सेलों को एक दुसरे से इस प्रकार जोड़ा जाता है कि एक सेल का धनात्मक सिरा अगली सेल के ऋणात्मक सिरे से जुड़ा हो, और इसी प्रकार आगे की सेलों को जोडा जाता है। इस व्यवस्था में एकल पथ होने से सभी सेलों से सामान विद्युत धारा प्रवाहित होती है।

मान लीजिए ε_{1}, ε_{1} दो सेलों के विद्युत वाहक बल हैं और r_{1}, r_{2} क्रमशः इनके आंतरिक प्रतिरोध हैं। माना कि तीन बिंदुओं A, B और C के विद्युत विभव V_{A}, V_{B}, और V_{C} हैं और इनमें से प्रवाहित विद्युत धारा I है।

उपरोक्त सेलों के लिए विभवांतर :-

…..(1)

…..(2)

सेलों के श्रेणी क्रम संयोजन के लिए विभवांतर :-

…..(3)

अब यदि श्रेणी क्रम संयोजन को विद्युत वाहक बल *ε _{eq}* और आंतरिक प्रतिरोध

…..(4)

समीकरण (3) और (4) की तुलना करने पर,

…..(5)

…..(6)

सामान्यतः *यदि n* *सेलें श्रेणी क्रम में जुड़ी हों** तो*,

**तुल्य विद्युत वाहक बल (ε _{eq}) = ε_{1} + ε_{2} + ε_{3} + …..**

**तुल्य आंतरिक प्रतिरोध (r _{eq}) = r_{1} + r_{2} + r_{3} + …..**

**नोट :-**

श्रेणी क्रम संयोजन में यदि पहले सेल का ऋणात्मक सिरा दूसरे सेल के धानात्मक सिरा से जुड़ा है, तो

…..(7)

तुल्य सेल के लिए भी,

…..(8)

समीकरण (7) और (8) की तुलना करना,

*तुल्य विद्युत वाहक बल* *(ε _{eq}) = ε_{1} – ε_{2}*

**तुल्य आंतरिक प्रतिरोध (r _{eq}) = r_{1} + r_{2}**

निम्न चित्र में बिंदु A और D के मध्य सेलों के समानांतर क्रम संयोजन को दर्शाया गया है जिसमें प्रत्येक सेल का धनात्मक सिरा एक बिंदु से जुड़ा है और प्रत्येक सेल का ऋणात्मक सिरा दूसरे बिंदु से जुड़ा है।

माना ε_{1} , ε_{1} सेलों के विद्युत वाहक बल हैं और r_{1}, r_{2} क्रमशः इनके आंतरिक प्रतिरोध हैं तथा सेलों द्वारा परिपथ को प्रदान विद्युत धारा I_{1 }और I_{2} हैं।

सेलों द्वारा आपूर्ति की गई कुल विद्युत धारा,

I = I_{1} + I_{2} …..(9)

चूँकि सेलें समानांतर क्रम में जुड़ी हुई हैं इसलिए सेलों के मध्य विभवांतर (*ΔV*) समान है। प्रथम सेल के लिए,

दूसरी सेल के लिए,

I_{1} और I_{2} के मानों को समीकरण (9) में रखने पर, हमें प्राप्त होता है

…..(10)

अब यदि सेलों के समानांतर क्रम संयोजन को विद्युत वाहक बल *ε _{eq}* और आंतरिक प्रतिरोध

…..(11)

समीकरण (10) और (11) की तुलना करने पर,

…..(12)

और

…..(13)

अथवा

…..(14)

समीकरण (12) को समीकरण (13) से विभाजित करने पर हमें प्राप्त होता है,

…..(15)

सामान्यतः ** यदि n सेलें समानांतर क्रम संयोजन में जुड़ी हों** तब,

और

**नोट :-**

* यदि समान विद्युत वाहक बल ε और समान आंतरिक प्रतिरोध r की दो सेलें समानांतर क्रम में जुड़ी हुई हों*, तब

और

]]>

* Mixed grouping of cells | Mixed grouping of identical cells :-* There are three types of grouping of identical cells :-

- Series Grouping of cells
- Parallel Grouping of cells
- Mixed Grouping of cells

In series grouping of identical cells, several identical cells are connected end-to-end in an electrical circuit where the positive terminal of one cell is connected to the negative terminal of the next cell, and so on. This setup is typically used to increase the total voltage output of the battery pack.

Consider ‘n’ cells each of emf *ε* and internal resistance *r* are connected in series to an external resistor *R* as shown in figure below :

Equivalent emf of n cells in series,

Equivalent internal resistance of n cells in series,

Total resistance of the circuit = (nr + R)

Current flowing in the circuit,

…..(1)

**Special cases :-**

Case (i). If ** R << nr**, then

Thus same current flows through the external resistor as the current due to a single cell.

Case (ii). If ** R >> nr**, then

Thus, the current in the external resistor is n times the current due to a single cell.

Hence, we conclude that

- Series combination of cells is advantageous when
, then*R >> nr* - Series combination of cells is not advantageous when
, then .*R << nr*

In short * using series combination of identical cells, the current drawn is maximum if the value of external resistor is very high as compared to the total internal resistance of the cells*.

Parallel grouping of cells involves connecting the positive terminals of multiple cells together and the negative terminals together to get more current. Consider ‘m’ cells each of emf *ε* and internal resistance* r* are connected in parallel and this combination is connected to an external resistor *R* as shown in figure below :

In parallel grouping of cells the equivalent emf is same as that of a single cell, hence

Equivalent emf of m cells in parallel = emf of a single cell = ε

Equivalent internal resistance of m cells in parallel,

As *R* and *r _{p}* are in series so,

Total resistance of the circuit = *R* + *r*/*m*

∴ Current in the external resistance *R* is given by,

…..(2)

**Special cases :-**

Case (i). If ** R << r/m**, then

Thus, the current in the external resistor is m times the current due to a single cell.

Case (ii). If ** R >> r/m**, then

Thus same current flows through the external resistor as the current due to a single cell.

Hence, we conclude that

- Parallel combination of cells is advantageous when
, then*R << r/m (or r >> mR)* - Parallel combination of cells is not advantageous when
, then .*R >> nr (or r << mR)*

In short * using parallel combination of identical cells, the current drawn is maximum if the value of external resistor is very low as compared to the total internal resistance of the cells*.

If some cells are connected in series and to form a row and a number of such rows of cells are connected in parallel, then this type of grouping of cells is called mixed grouping. This approach allows for customization of the voltage * to get maximum power*.

Let n identical cells each of emf *ε* and internal resistance *r* in a row are connected in series and m such rows are connected in parallel. This arrangement is connected to an external resistor *R* as shown in figure below :-

Total emf of each row (of whole combination) = *nε*

Total internal resistance of each row = *nr*

As m identical rows of cells are connected in parallel, therefore,

Total internal resistance (*r _{p}*) of all the cells is given by,

Total resistance of the circuit = *R* + *nr*/*m*

The current in the external resistor *R* is given by,

**Note :-**

(1). The current in the mixed grouping of cells will be maximum when (*mR + nr*) is minimum, i.e.,

Here cannot be zero, so I will be maximum when :-

Hence **current drawn by mixed grouping of cells will be maximum if the value of external R is equal to the total internal resistance of all the cells.**

(2). * When one cell is wrongly connected in series combination of n identical cells*, each emf ε, then total emf is reduced by 2ε.

so the **effective emf = (nε – 2ε)**

Here,

(3). When n cells each of internal resistance r are connected wrongly in series, then * total internal resistance of the combination = nr*. This is because there is no effect of order of cells on total resistance in series combination.

(4). The * series combination of cells is used to get more voltage*,

**Example.**

8 cells each of internal resistance 0.5 Ω and emf 1.5V are used to send a current through an external resistor of (a) 200 Ω (b) 0.002 Ω (c ) 1.0 Ω. How would you arrange them to get the maximum current in each case? Find the value of current in each case.

**Solution :**

Here, total number of cells= 8,

r = 0.5 Ω and ε = 1.5 V

(a). When external resistor R = 200 Ω, then R >> r, so for maximum current, the cells are to be connected in series in the circuit.

Total internal resistance of 8 cells = 8r

Current in circuit,

(b) When R = 0.002 Ω, then R << r, so for maximum current, the cells are to be connected in parallel in the circuit.

Total internal resistance of 8 cells = r / 8

Total resistance of circuit = R + r / 8 = 0.002 + 0.5 / 8 = 0.0645 Ω

Effective emf of all cells = emf of each cell 1.5 V

Current in circuit,

(c ) When R = 1.0 Ω, then R is comparable to r. For maximum current, the cells are to be connected in mixed grouping.

Let there be m rows of cells in parallel with n cells in series, in each row. Then,

Total number of cells, mn = 8 …..(1)

Total internal resistance of combination = nr/m

∴ Total resistance in the circuit , here R = 1Ω

The emf each row = nE, here E = 1.5V

∴ Current in external circuit,

As mnRr = constant, so I will be maximum if,

Hence for maximum current,

So now

From equation (1),

And

n = 2 × 2 = 4

Thus, 4 cells in series in a row and 2 such rows of cells in parallel.

Max. current,

]]>

* Grouping of Cells :-* It isn’t possible to make voltage sources and batteries for each specific voltage requirement. When a different voltage is necessary, two or more voltage sources(cells/batteries) are used in different combinations to produce the desired value of voltage and current. These cells can be connected in two basic types of combinations. These two combinations are :

- Series Combination
- Parallel Combination.

When the voltage required by a load (such as a motor, appliance or electronic device) exceeds the voltage output of a single cell or battery, cells are connected in series to increase the total voltage. A series combination of cells refers to connecting multiple cells end-to-end so that the * positive terminal of one cell is connected to the negative terminal of the next cell, and so on*. This arrangement creates a single path for current flow through all the cells.

Let ε_{1} , ε_{1} be the emfs of two cells and r_{1} , r_{2} be their internal resistances respectively. Let the potentials of the three points A, B and C is denoted by V_{A}, V_{B}, and V_{C }and the current flowing trough them is I.

Potential difference for the above cells :-

…..(1)

…..(2)

Potential difference for the series combination of cells :-

…..(3)

Now if the series combination is replaced by a single cell of emf *ε _{eq}* and internal resistance

…..(4)

Comparing equations (3) and (4), we get

…..(5)

…..(6)

In general * if n cells are connected in series* then,

**Equivalent emf (ε _{eq}) = ε_{1} + ε_{2} + ε_{3} + …..**

**Equivalent internal resistance (r _{eq}) = r_{1} + r_{2} + r_{3} + …..**

**Note :-**

In series combination if negative terminal of first cell is connected to the negative terminal of the second cell, then

…..(7)

Also for equivalent cell,

…..(8)

Comparing equations (7) and (8),

*Equivalent emf* *(ε _{eq}) = ε_{1} – ε_{2}*

**Equivalent internal resistance (r _{eq}) = r_{1} + r_{2}**

The figure below shows a parallel combination of cells between points A and D in which positive terminal of each cell is connected to one point and negative terminal of each cell is connected to the other point.

Let ε_{1} , ε_{1} be the emfs of two cells and r_{1} , r_{2} be their internal resistances respectively and the cells are supplying currents I_{1} and I_{2} to the circuit.

Total current supplied by the cells,

I = I_{1} + I_{2} …..(9)

As the cells are connected in parallel so the potential difference (*ΔV*) across the cell is equal. For the first cell,

For the second cell,

Putting the values of I_{1} and I_{2} in equation (9), we get

…..(10)

Now if the parallel combination of cells is replaced by a single equivalent cell of emf *ε _{eq}* and internal resistance

…..(11)

Comparing equations (10) and (11), we get

…..(12)

and

…..(13)

Or

…..(14)

Dividing equation (12) by equation (13) we get,

…..(15)

In general * if n cells are connected in parallel* then,

and

**Note :-**

If ** two cells of same emf ε and same internal resistance r are connected in parallel**, then

and

]]>

* No of Images Formed by Two Plane Mirrors *depends on the position of mirrors with respect to each others and also on the position of object between the mirrors. Let us discuses different positions of mirrors:-

If two plane mirrors are placed parallel to each other and an object is placed anywhere between them, then:-

- The number of images formed is infinite.
- The images are created by the successive reflections of light between the two mirrors.
- The images are formed at regular intervals, and their brightness decreases with each reflection.

It’s important to note that the images formed by parallel mirrors are virtual images, meaning they cannot be projected onto a screen. These virtual images exist only in the apparent location created by the reflection of light rays.

* Special case :-* If t

Here 3 images are formed :

- Virtual image
*I*of real object O_{1} - Virtual image
*I*of real object O_{2} - Virtual image
*I*of image_{3}*I*(image_{1}*I*acts as virtual object for_{1}*I*) or image_{3}*I*(image_{2}*I*acts as virtual object for_{2}*I*)_{3}

**Special case :-** If two plane mirrors are placed perpendicular to each other but are of insufficient length and the object is placed vertically above the ends of both of the mirrors as shown in figure below then number of images formed will be 2.

Consider two plane mirrors M_{1} and M_{2} inclined at an angle (θ_{1}+θ_{2}) and an object O placed between them making an angle θ_{1} with M_{1} and θ_{2} with M_{2}.

Here in triangles AI_{1}B and AOB :-

AB = AB (common)

OB (distance of object) = I_{1}B (distance of image)

∠B = ∠B = 90°

Hence from side-angle-side theorem, triangles AI_{1}B and AOB are congruent. Hence

**AO = AI _{1}**

Similarly, in triangles AI_{2}C and AOC :-

AC = AC (common)

OC (distance of object) = I_{2}C (distance of image)

∠C = ∠C = 90°

So from side-angle-side theorem, triangles AI_{2}C and AOC are congruent. Hence

**AO = AI _{2}**

In the same way,

**AO = AI _{1} = AI_{2} = AI_{3} = AI_{4} = AI_{5} = AI_{6}**

Here I_{1} , I_{2} , I_{3} , I_{4} , I_{5} , I_{6} are nothing but the multiple images of object O formed by the mirrors M_{1} and M_{2 }and we can see that * they all lie on a circle of radius AO*.

I_{1} = image of object O formed by mirror M_{1} at an angle θ_{1}, anticlockwise from mirror M_{1}

I_{2} = image of object O formed by mirror M_{2} at an angle θ_{2}, clockwise from mirror M_{2}

I_{3} = image of image I_{1} formed by mirror M_{2} at an angle (2θ_{1 }+ θ_{2}), clockwise from mirror M_{2}

I_{4} = image of image I_{2} formed by mirror M_{1} at an angle (2θ_{2 }+ θ_{1}), anticlockwise from mirror M_{1}

I_{5} = image of image I_{4} formed by mirror M_{2} at an angle (2θ_{1 }+ 3θ_{2}), clockwise from mirror M_{2}

I_{6} = image of image I_{3} formed by mirror M_{1} at an angle (2θ_{2 }+ 3θ_{1}), anticlockwise from mirror M_{1}

Now here is a * circular method* to find out total number of images formed by two inclined plane mirrors :-

**Example 1.**

Two plane mirrors are inclined at an angle of 72º. Find the number of images of a point object placed between.

**Solution.**

Here

Now

If the object is placed **symmetrically** between the mirrors, then no. of images = 5 – 1 = **4 **and

If the object is placed **asymmetrically** between the mirrors, then no. of images = **5**

**Example 2.**

Two plane mirrors are at 45º to each other. If an object is placed between them, then the number of images will be

(a) 5 (b) 9 (c) 7 (d) 8

**Solution.**

As 8 is even number, hence **no. of images** = 8 – 1 = **7**

**Example 3.**

Two plane mirrors are inclined at an angle of (a) θ = 112.5º (b) θ = 75º . Find the number of images of a point object placed between.

**Solution.**

(a)

Now 3 is an odd number, so :-

If object is symmetrically placed between the mirrors then number of images = 3.2 – 1 = 2.2

Number of **complete images** = **2**

If object is asymmetrically placed between the mirrors then number of images = 3.2

Number of **complete images** = **3**

(b)

Now 4 is an even number so no matter how the object is placed(symmetrically or asymmetrically) between the mirrors,

Number of images = 4.8 – 1 = 3.8

Number of **complete images** = **3**

]]>

* Refraction of Sound Waves :-* Refraction of sound waves refers to the bending of sound waves as they pass from one medium into another, where the speed of sound differs. This bending occurs due to the change in the speed of sound between the two mediums, causing the direction of the sound wave to change. The degree of bending depends on the angle of incidence, the difference in the speed of sound between the two mediums, and the density or temperature gradients present at the boundary.

A denser medium refers to a material or substance in which sound travels at a slower speed, while a rarer medium refers to a material or substance in which sound travels at a faster speed. It is the speed of wave which decides whether the medium is denser or rarer for that particular wave, i.e.,

Although the concepts of denser and rarer mediums apply to both sound waves and electromagnetic waves but remember that if a medium is denser for one type of wave then at the same time the same medium can be rarer for the other type of wave. For example * air is considered a rarer medium compared to water for light (electromagnetic wave)* because light travels faster in air than water. At the same time

Notice that as the wavefronts cross the boundary the wavelength changes, but the frequency remains constant and also note that no phase change occur during refraction.

The laws of refraction and reflection remains the same, i.e., the wave bends towards the normal if it travels from rarer medium to denser medium and vice – versa. Snell’s law relates the directions of the wave before and after it crosses the boundary between the two media. If *v _{1}* and

]]>

* Reflection of Sound | Reflection of Waves :-* In this article, we are going to study about the reflection of waves (transverse and longitudinal) at the boundary of the denser medium and rarer medium. When a wave travelling in a homogeneous medium strikes an interface separating two media, a part of incident wave is reflected back into the first medium while the remainder is partly absorbed and partly refracted(or transmitted) into the second medium. Both light waves (transverse waves) and sound waves (longitudinal waves) exhibit this phenomenon. As the wavelength of a light wave is very very small of order 10

When the pulse reaches the right end which is clamped at the wall [Fig (b)], the leading edge of the pulse exerts a force on the wall and the wall exerts equal and opposite force on the element(According to Newton’s Third Law). Now as the wall is rigid, it cannot move but the reaction force on the rope by the wall produces a pulse that is inverted but identical to the original pulse [Fig (d)]. The result of this change is that the pulse undergoes reflection— that is, the pulse moves back along the string in the opposite direction [Fig (e)].

Thus after reflection at a denser medium, a crest returns as a trough i.e. there is a phase change of π radian or 180° between the incident wave and the reflected wave.

Incident wave (Travelling along + x-axis) :-

Reflected wave (Travelling along – x-axis) :-

Here A’ < A, because of the energy loss and a phase change of π – radian, because of the reflection from rigid boundary.

The right end of the string is attached to a light frictionless ring which can freely move on a vertical rod. A wave pulse is sent on the string from left. When the wave pulse reaches the right end, the ring at this end is acted on by the force to go up. As the ring is free to slide along the rod, it is displaced in upward direction more than the height of the pulse i.e., it overshoots the normal maximum displacement [Fig (b)]. At this position the ring and string come momentary to rest. But since the string is stretched in this position, so the free end of the string is pulled back down and again an extra force acts from right which sends a wave from right to left with its shape identical to the original one [Fig (c)]. Thus, a wave is reflected by the free end (rarer medium) without inversion.

Thus after reflection at a rarer medium, a crest returns as a crest i.e. there is a no phase difference between the incident wave and the reflected wave.

Incident wave (Travelling along + x-axis) :-

Reflected wave (Travelling along – x-axis) :-

Here A’ < A, because of the energy loss.

Assume that longitudinal waves in air (sound waves) are incident normally on a rigid wall. As the compression strikes the wall, it applies a force on the wall. But since the wall is rigid, it exerts an equal and opposite force on the layer of air in compression and thus pushes the compression in the backward direction. There is a * phase difference of π radian or 180°* between the incident wave and the reflected wave. This is because

Thus a compression travelling towards the right is reflected as a compression travelling towards the left. Similarly, incident rarefaction is reflected as a rarefaction and hence nature of the wave remains the same i.e., incident and reflected waves looks similar.

Assume that a longitudinal wave travelling in a denser medium is incident at the Interface (boundary) of a rarer medium. As the compression in the incident wave strikes the interface (e.g. air at the open end of a pipe) then due to the high pressure of compression, the interface is pushed back towards the rarer medium. This creates a low pressure region because the surrounding air goes away quickly and compression is converted into a rarefaction before the wave is reflected.

Thus, the compression after reflection at a rarer medium returns as a rarefaction and similarly, incident rarefaction is reflected as a compression. Thus * no phase change takes place when a longitudinal wave is reflected from the surface of a rarer medium* but the nature of the wave is changed i.e., incident and reflected waves looks different.

]]>

* Factors Affecting Speed of Sound :- *Several factors affect the speed of sound in a medium like air :-

**(1) Effect of temperature**

For a gas γ & M_{W} are constant and hence from , we get

Here *v _{t}* is velocity of sound at t °C and

By applying Binomial theorem,

(i) For any gas medium

(ii) For air

(As *v _{o}* = 332 m/s)

**(2) Effect of Relative Humidity**

Moist air is less dense than dry air so from the expression , we conclude that with rise in humidity velocity of sound increases. This is why sound travels faster in humid air (rainy season) than in dry air (summer) at same temperature. Due to this in rainy season the sound of factories siren and whistle of train can be heard more than summer.

**(3) Effect of Pressure**

Looking at the expression , we conclude that, as long as temperature remains constant, there is no effect of change of pressure on speed of sound. This is because as pressure changes, density also changes proportionately.

**(4) Effect of Wind**

Wind can affect the speed of sound, particularly when it is blowing in the same direction as the sound waves. In such cases, it can increase the speed of sound relative to an observer on the ground, and decrease it when blowing in the opposite direction.

**(5) Effect of Frequency**

There is no effect of frequency on the speed of sound. Sound waves of different frequencies travel with the same speed in air although their wavelength in air are different. That is why we are able to enjoy orchestra producing sound waves of different frequencies.

]]>

* Laplace Correction :-* Laplace modified Newton’s formula assuming that propagation of sound in air is an

As a result, the sound propagates under adiabatic conditions and not under isothermal conditions.

Now in an adiabatic process,

Here

For air γ = 1.4, therefore

Which is in good agreement with the experimental value (332 m/s). This in turn establishes that sound propagates adiabatically through gases.

The velocity of sound in air at NTP is 332 m/s which is much lesser than that of light and radio–waves (= 3 x 10^{8} m/s). This implies that :–

(a) If we set our watch by the sound of a distant siren it will be slow.

(b)If we record the time in a race by hearing sound from starting point it will be lesser than actual.

(c)In a cloud–lightening, though light and sound are produced simultaneously but as c > v, light proceeds thunder.

In case of gases :–

And from kinetic-theory of gases :–

]]>

* Speed of Sound in Air :-* The formula for velocity of sound in air was first obtained by Newton. He assumed that when sound propagates through a gas the temperature remains constant (i.e. the process is isothermal). So a compressed layer of air gradually loses heat to the surroundings and a rarefied layer gains heat from the surroundings, so that temperature of air remains constant.

We know in an isothermal process,

Now we know that speed of longitudinal waves is given by , so

At S.T.P. *P _{atm}* = 1.013 × 10

However this value is about 16 % less than the experimental value of sound in air which is about 332 m/s. In order to remove this large discrepancy between theoretical and experimental values of velocity of sound, Laplace modified Newton’s formula and this modification is called **Laplace’s Correction**.

]]>

* Sound Intensity In Decibels | Sound Level In Decibels :- *Human ear respond to sound intensities from 10

We use logarithmic scale of intensity called the “sound level” which is defined as below :

Here *I* is the intensity of sound wave. Sound intensity level (*SL*) is expressed in **decibels** (*dB*).

For example a sound of intensity I_{0 }has sound level :

Similarly, sound level of threshold of pain :

Hence we can say :

*Threshold of hearing = 0 dB **and*

*Threshold of pain = 120 dB*

Sr. No. |
Source of sound |
Intensity (in Wm |
Sound intensity level ( (in |

1. |
Threshold of hearing |
10^{-12} |
0 |

2. |
Rustle of leaves |
10^{-11} |
10 |

3. |
Whisper |
10^{-10} |
20 |

4. |
Quiet room |
10^{-8} |
40 |

5. |
Normal conversation |
3.2 × 10^{-6} |
65 |

6. |
Street traffic |
10^{-5} |
70 |

7. |
Thunder |
10^{-1} |
110 |

8. |
Threshold of pain |
1 |
120 |

9. |
Military jet aircraft (30 m away) |
10^{2} |
140 |

**Example :**

Calculate the change in intensity level when the intensity of sound increases by 10^{6} times its original intensity.

**Solution :**