Merge pull request #63 from pndineshwaran/main

Lesson1 Added

content.en/nature-of-physical-world-and-measurement/MEASUREMENT/index.md

@@ -1,8 +1,11 @@
 ---
 title: 'MEASUREMENT'
 weight: 7
+extensions: -katex
 ---
 
+{{< katex diaplay >}}  {{< /katex >}}
+
 # Measurement
 
 ---
@@ -21,7 +24,7 @@ Quantities that can be measured, and in terms of which, laws of physics are desc
 
 Physical quantities are classified into two types. They are fundamental and derived quantities.
 
-_Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These_ are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
+_Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities._ These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
 
 _Quantities that can be expressed in terms of fundamental quantities are called derived quantities._ For example, area, volume, velocity, acceleration, force, etc.
 
@@ -40,9 +43,7 @@ A complete set of units which is used to measure all kinds of fundamental and de
 
 (a) **the f.p.s. system** is the British Engineering system of units, which uses **_foot_**, **_pound_** and **_second_** as the three basic units for measuring length, mass and time respectively.
 
-(b) The c.g.s system is the Gaussian system, which uses **_centimeter, gram_**  
-
-and **_second_** as the three basic units for measuring length, mass and time respectively.
+(b) The c.g.s system is the Gaussian system, which uses **_centimeter, gram_** and **_second_** as the three basic units for measuring length, mass and time respectively.
 
 (c) The m.k.s system is based on metre, **_kilogram_** and **_second_** as the three basic units for measuring length, mass and time respectively.
 
@@ -72,9 +73,9 @@ In SI, there are seven fundamental units as given in Table 1.2
 |Mass | kilogram | kg | One kilogram is the mass of the prototype cylinder of platinum iridium alloy (whose height is equal to its diameter), preserved at the International Bureau of Weights and Measures at Serves, near Paris, France. (1901) |
 |Time | second | s | One second is the duration of 9,192,631,770 periods of radiation corresponding to the transition between the two hyperfine levels of the ground state of Cesium-133 atom.(1967) |
 |Electric current | ampere | A | One ampere is the constant current, which when maintained in each of the two straight parallel conductors of infinite length and negligible cross section, held one metre apart in vacuum shall produce a force per unit length of 2 × 10 −7 N/m between them. (1948) |
-|Temperature | kelvin | K | One kelvin is the fraction  of the thermodynamic temperature of the triple point* of the water. (1967) |
+|Temperature | kelvin | K | One kelvin is the fraction {{< katex diaplay >}} \lparen 1/273.16\rparen {{< /katex >}} of the thermodynamic temperature of the triple point* of the water. (1967) |
 |Amount of substance | mole |mol| One mole is the amount of substance which contains as many elementary entities as there are atoms in 0.012 kg of pure carbon-12. (1971) | 
-| Luminous intensity | candela | cd | One candela is the luminous intensity in a given direction, of a source that emits monochromatic radiation of frequency 5.4 × 10^14 Hz and that has a radiant intensity of (1/683)watt/steradian in that direction. (1979)|
+| Luminous intensity | candela | cd | One candela is the luminous intensity in a given direction, of a source that emits monochromatic radiation of frequency 5.4 × {{< katex diaplay >}} 10^14{{< /katex >}} Hz and that has a radiant intensity of (1/683)watt/steradian in that direction. (1979)|
 
 Triple point of water is the temperature at which saturated vapour, pure water and melting
 ice are all in equilibrium. The triple point temperature of water is 273.16K
@@ -84,29 +85,29 @@ ice are all in equilibrium. The triple point temperature of water is 273.16K
 |----|----|---|
 |Plane angle | arc / radius | rad |
 |Solid angle | surface area/radius* | sr| 
-|Area | length × breadth | m^2|
-|Volume | area × height | m^3|
-|Velocity | displacement / time | m s−1|
-|Acceleration | velocity / time | m s−2|
-|Angular velocity | angular displacement |rad s−1|
-|Angular acceleration | angular velocity / time | rad s−2|
-|Density | mass / volume | kg m−3|
-|Linear momentum | mass × velocity | kg m s−1 |
-|Moment of inertia | mass × (distance) | kg m2|
-|Force | mass × acceleration | kg m s−2 or N |
-|Pressure |force / area | N m−2 or Pa|
+|Area | length × breadth | {{< katex diaplay >}}  m^2{{< /katex >}}|
+|Volume | area × height | {{< katex diaplay >}}m^3  {{< /katex >}}|
+|Velocity | displacement / time | m \\(s^{−1}\\)|
+|Acceleration | velocity / time | m \\(s^{−1}\\)|
+|Angular velocity | angular displacement |rad \\(s^{−1}\\)|
+|Angular acceleration | angular velocity / time | rad \\(s^{−2}\\)|
+|Density | mass / volume | kg \\(m^{−3}\\)|
+|Linear momentum | mass × velocity | kg m \\(s^{−1}\\) |
+|Moment of inertia | mass × (distance) | kg \\(m^{−1}\\)|
+|Force | mass × acceleration | kg m \\(s^{−2}\\) or N |
+|Pressure |force / area | N \\(m^{−2}\\) or Pa|
 |Energy (work) | force × distance | N m or J|
-|Power | Work / time | J s−1 or watt (W)|
+|Power | Work / time | J \\(s^{−1}\\) or watt (W)|
 |Impulse | force × time | N s|
-|Surface tension | force / length | N m−1|
+|Surface tension | force / length | N \\(m^{−1}\\)|
 |Moment of force (torque) | force × distance | N m |
 |Electric charge | current × time | A s or C|
-|Current density | current / area | A m−2 |
-|Magnetic induction | force / (current × length) | N A−1 m−1 or tesla |
-|Force constant | force / displacement | N m–1 |
+|Current density | current / area | A \\(m^{−2}\\) |
+|Magnetic induction | force / (current × length) | N \\(A^{−1} \\) \\(m^{−1}\\) or tesla |
+|Force constant | force / displacement | N \\(m^{−1}\\) |
 |Plank’s constant | energy of photon|J s|
-|Specific heat (S) | heat energy / (mass * temperature) |J kg–1 K–1 |
-|Bolizmann constant (_k_) | energy/temperature | J K–1 |
+|Specific heat (S) | heat energy / (mass * temperature) |J \\(kg^{–1}\\) \\(K^{−1}\\) |
+|Bolizmann constant (_k_) | energy/temperature | J \\(K^{−1}\\) |
 
 Table 1.3 lists some of the derived quantities and their units.
 
@@ -130,3 +131,5 @@ minutes:
 \end{aligned}
 {{< /katex >}}
 
+
+

content.en/nature-of-physical-world-and-measurement/_index.md

@@ -2,5 +2,19 @@
 title: 'NATURE OF PHYSICAL WORLD  AND MEASUREMENT'
 weight: 1
 ---
+# UNIT 1
 
----
+
+**Learning Objectives**
+
+**In this unit, the student is exposed to** 
+
+- excitement generated by the discoveries in Physics 
+- an understanding of physical quantities of importance 
+- different system of units 
+- an understanding of errors and corrections in physics measurements
+- the importance of significant figures 
+- usage of dimensions to check the homogeneity of physical quantities
+
+
+---

content.en/nature-of-physical-world-and-measurement/accuracy-and-precision/index.md

@@ -11,7 +11,7 @@ If a measurement is precise, that does not necessarily mean that it is accurate.
 
 For example, if the temperature outside a building is 40oC as measured by a weather thermometer and if the real outside
 ![Alt text](<fig 1.8.png>)
-**Figure 1.9** Visual example of accuracy and pr
+**Figure 1.9** Visual example of accuracy and preecision
 
 temperature is 40oC, the thermometer is accurate. If the thermometer consistently registers this exact temperature in a row, the thermometer is precise.
 

content.en/nature-of-physical-world-and-measurement/application-and-Limitations-of-the-Method-of-dimensional-analysis/index.md

@@ -1,28 +1,29 @@
 ---
 title: ' Application and Limitations of the Method of Dimensional Analysis.'
 weight: 22
+extensions:
+    - katex
 ---
+{{< katex diaplay >}}  {{< /katex >}}
 
 # Application and Limitations of the Method of Dimensional Analysis.
 
 
 **This method is used to** 
 
 (i) Convert a physical quantity from one system of units to another. 
-(ii) Check the dimensional correctness of a given physical equation. 
-(iii) Establish relations among various physical quantities.
 
-**(i) To convert a physical quantity from one system of units to another**
+(ii) Check the dimensional correctness of a given physical equation.
 
-This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant. i.e, n [u] = constant (or) n1[u1] = n2[u2].
+(iii) Establish relations among various physical quantities.
 
-Consider a physical quantity which has dimension ‘_a’_ in mass, ‘_b’_ in length and ‘_c’_ in time. If the fundamental units in one system are M1, L1 and T1 and the other system are M2, L2 and T2 respectively, then we can write,
+**(i) To convert a physical quantity from one system of units to another**
 
-FORMULA
+This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant. i.e, n [u] = constant (or) \\(n_1\\)[\\(u_2\\)] = \\(n_2\\)[\\(u_2\\)].
 
-c\] We have thus converted the numerical
+Consider a physical quantity which has dimension ‘_a’_ in mass, ‘_b’_ in length and ‘_c’_ in time. If the fundamental units in one system are \\(M_1\\), \\(L_1\\) and \\(T_1\\) and the other system are \\(M_2\\), \\(L_2\\) and \\(T_2\\) respectively, then we can write,\\(n_1\\)  \\([M_1^aL_1^bT_1^C]\\)\\(n_{2}=[M_{2}^{a}L_{2}^{b}T_{2}^{c}]\\)
 
-value of physical quantity from one system of units into the other system.
+We have thus converted the numerical value of physical quantity from one system of units into the other system.
 
 **EXAMPLE 1.12**
 
@@ -32,17 +33,56 @@ Convert 76 cm of mercury pressure into Nm−2 using the method of dimensions.
 
 In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm−2
 
-The dimensional formula of pressure P is \[ML−1T−2\]
+The dimensional formula of pressure P is \\([ML^{−1}T^{−2}\]\\)
+
+\\(P_{1}[M_{1}^{a}L_{1}^{b}T_{1}^{c}]=P_{2}[M_{2}^{a}L_{2}^{b}T_{2}^{c}]\\)
+
+We have \\(P_{2}=P_{1}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c}\\)
+
+\\(M₁ = 1g, M₂ = 1kg,
+L₁ = 1 cm, L₂ = 1m,
+T₁ = 1 s, T₂ = 1s\\)
+
+As a =1, b= -1, and c = -2
+
+Then
+
+\\(P_{2}=76\times13.6\times980\left[\frac{1g}{1kg}\right]\left[\frac{1cm}{1m}\right]^{-1}\left[\frac{1s}{1s}\right]^{-1}\\)
+
+\\(=76\times13.6\times980\left[\frac{10^{-3}kg}{1kg}\right]\left[\frac{10^{-2}m}{1m}\right]^{-1}\left[\frac{1s}{1s}\right]^{-2}\\)
+
+\\(=76\times13.6\times980\times[10^{-3}]\times10^{2}\\)
+
+\\(P_{2}=1.01\times10^{5}Nm^{-2}\\)
+
 
 **EXAMPLE 1.13**
 
-If the value of universal gravitational constant in SI is 6.6x10−11 Nm2 kg−2, then find its value in CGS System?
+If the value of universal gravitational constant in SI is 6.6 × 10⁻¹¹ Nm² kg⁻² = 6.6, then find its value in CGS System?
 
 **_Solution_**
 
-Let GSI be the gravitational constant in the SI system and Gcgs in the cgs system. Then
+Let \\(G_{SI}\\) be the gravitational constant in the SI system and \\(G_{cgs}\\) in the cgs system. Then
+
+\\(G_{SI}=6.6\times10^{-11}Nm^{2}kg^{-2}\\)
+
+\\(G_{cgs}\\)
+
+\\(n_{2}=n_{1}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c}\\)
+
+\\(G_{cgs}=G_{SI}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c}\\)
 
-GSI = 6.6 × -10 11Nm2 kg−2
+M₁ = 1 kg
+
+L₁ = 1 m
+
+T₁ = 1 s
+
+M₂ = 1 g
+
+L₂ = 1 cm
+
+T₂ = 1 s
 
 **(ii) To check the dimensional correctness of a given physical equation** Let us take the equation of motion v = u + at Apply dimensional formula on both sides [LT−1] = [LT−1] + [LT−2] [T]  [LT−1] = [LT−1] + [LT−1]
 
@@ -52,83 +92,61 @@ We see that the dimensions of both sides are same. Hence the equation is dimensi
 
 **EXAMPLE 1.14**
 
-Check the correctness of the equation 1 2
-
-2_mv mgh_\= using dimensional analysis
+Check the correctness of the equation \\(\frac{1}{2}mv^2 = mgh\\) using dimensional analysis
 
 method.
 
 **_Solution_**
 
 Dimensional formula for
 
-1 2
-
-2 1 2 2 2_mv_ \= =- -\[M\]\[LT \] \[ML T \]
+\\(\frac{1}{2}mv^2 = [M][LT^{-1}]^{2} = [ML^{2}T^{-2}]\\)
 
 Dimensional formula for
 
-_mgh_ \= =- -\[M\]\[LT \]\[L\] \[ML T \]2 2 2
+\\(mgh = [M][LT^{-2}][L] = [ML^{2}T^{-2}]\\)
 
-\[ML2T−2\] = \[ML2T−2\]
+\\([ML^2T^{-2}]\\) = \\(ML^2T^{-2}\\)
 
 Both sides are dimensionally the
 
-same, hence the equations 1 2
+same, hence the equations \\(\frac{1}{2}mv^2 = mgh\\) is dimensionally correct.
 
-2_mv mgh_\= is dimensionally correct.
-
-**(iii) To establish the relation among various physical quantities** If the physical quantity Q depends upon the quantities Q1, Q2 and Q3 ie. Q is proportional to Q1, Q2 and Q3.
+**(iii) To establish the relation among various physical quantities** If the physical quantity Q depends upon the quantities \\(Q_1\\), \\(Q_2\\) and \\(Q_3\\) ie. Q is proportional to \\(Q_1\\), \\(Q_2\\) and \\(Q_3\\).
 
 Then,
 
-Q ∝ Q1 a Q2
-
-b Q3 c
+\\(Q \propto Q_{1}^{a}Q_{2}^{b}Q_{3}^{c}\\)
 
-Q = k Q1 a Q2
+\\(Q = K Q_{1}^{a}Q_{2}^{b}Q_{3}^{c}\\)
 
-b Q3 c
-
-where k is a dimensionless constant. When the dimensional formula of Q, Q1, Q2 and Q3
+where k is a dimensionless constant. When the dimensional formula of Q, \\(Q_1\\), \\(Q_2\\) and \\(Q_3\\)
 
 are substituted, then according to the principle of homogeneity, the powers of M, L, T are made equal on both sides of the equation. From this, we get the values of a, b, c
 
 **EXAMPLE 1.15**
 
 Obtain an expression for the time period T of a simple pendulum. The time period T depends on (i) mass ‘_m’_ of the bob (ii) length ‘_l’_ of the pendulum and (iii) acceleration due to gravity _g_ at the place where the pendulum is suspended. (Constant k = 2π) i.e **_Solution_**
 
-_T_ ∝ _m_a _l_b _g_c
+\\(T \propto m^{a}l^{b}g^{c}\\)
 
-_T = km_a _l_b _g_c  
+\\(T = km^{a}l^{b}g^{c}\\)  
 
 Here k is the dimensionless constant. Rewriting the above equation with dimensions
 
-\[T1\] = \[Ma\] \[Lb\] \[LT−2\]c
+\\([T^{1}]=[M^{a}][L^{b}][LT^{-2}]^{c}\\)
 
-\[M0L0T1\] = \[Ma Lb+c T−2c\]
+\\([M^{0}L^{0}T^{1}]=[M^{a}L^{b+c}T^{-2c}]\\)
 
 Comparing the powers of M, L and T on both sides, a=0, b+c=0, -2c=1
 
 Solving for a,b and c a = 0, b = 1/2, and c = −1/2
 
-From the above equation T = km0 l
-
-1/2 g−1/2
-
-_T k g_
-
-_k g_\=  
+From the above equation T = km0 l \\(C^{1/2}g^{-1/2}\\)
 
- 
+\\(T = k\left(\frac{l}{g}\right)^{1/2} = k\sqrt{\frac{l}{g}}\\)
 
-\=l
-
-l
-
-1 2
-
-Experimentally k = 2π, hence _T g_\= 2π l
+Experimentally k = 2π, hence \\(T = 2\pi\sqrt{\frac{l}{g}}\\)
 
 **Limitations of Dimensional analysis**
 
@@ -141,5 +159,5 @@ Experimentally k = 2π, hence _T g_\= 2π l
 4. It cannot be applied to an equation involving more than three physical quantities.
 
 5. It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation. For example using
-
-dimensional analysis, s = ut + 1 3
+dimensional analysis, \\(s = ut + \frac{1}{3}\\)
+is dimensionally correct whereas the correct relation is \\(s = ut + \frac{1}{3}\\)

content.en/nature-of-physical-world-and-measurement/arithmetical-operations-with-significant-figures/index.md

@@ -12,14 +12,22 @@ result should retain as many decimal places as there are in the number with the
 
 **Example:**
 
-1\. 3.1 + 1.780 + 2.046 = 6.926 Here the least number of significant digits after the decimal is one. Hence the result will be 6.9.
+1\. 3.1 + 1.780 + 2.046 = 6.926 
 
-2\. 12.637 – 2.42 = 10.217 Here the least number of significant digits after the decimal is two. Hence the result will be 10.22
+Here the least number of significant digits after the decimal is one. Hence the result will be 6.9.
+
+2\. 12.637 – 2.42 = 10.217 
+
+Here the least number of significant digits after the decimal is two. Hence the result will be 10.22
 
 **(ii) Multiplication and Division** In multiplication or division, the final result should retain as many significant figures as there are in the original number with smallest number of significant figures.
 
-**Example:** 1\. 1.21 × 36.72 = 44.4312 = 44.4
+**Example:** 
+
+1\. 1.21 × 36.72 = 44.4312 = 44.4
 
 Here the least number of significant digits in the measured values is three. Hence the result when rounded off to three significant digits is 44.4
 
-2\. 36.72 ÷ 1.2 = 30.6 = 31 Here the least number of significant digits in the measured values is two. Hence the result when rounded off to significant digit becomes 31.  
+2\. 36.72 ÷ 1.2 = 30.6 = 31 
+
+Here the least number of significant digits in the measured values is two. Hence the result when rounded off to significant digit becomes 31.