DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT

1. **Distance**: Distance refers to the total length traveled by an object regardless of its direction. It is a scalar quantity, meaning it only has magnitude. For example, if you walk around a park in a circular path, the distance traveled is the total length of the path you took.

2. **Displacement**: Displacement, on the other hand, refers to the change in position of an object from its initial position to its final position. It includes both the distance between the initial and final positions and the direction from the initial position to the final position. Displacement is a vector quantity because it has both magnitude and direction. For example, if you walk around a park in a circular path and end up at your starting point, your displacement is zero because you've returned to your initial position, even though the distance you traveled may be nonzero.

MOST IMPORTANT DEFINITION :

Explanations of each concept:

1.**Distance**: Distance is the total length of the path traveled by an object, regardless of direction. It's a scalar quantity and doesn't consider direction.

2. **Displacement**: It refers to the change in position of an object, typically measured in a straight line from the initial position to the final position. It has both magnitude (how far an object has moved) and direction.

3. **Velocity**: Velocity is the rate of change of displacement. It's a vector quantity, meaning it has both magnitude (speed) and direction. Mathematically, velocity is expressed as the change in displacement divided by the time taken.

4. **Acceleration**: Acceleration is the rate of change of velocity. Like velocity, it's also a vector quantity and can be positive (speeding up), negative (slowing down), or changing direction. Acceleration can be calculated by dividing the change in velocity by the time taken.

5. **Force**: Force is a push or pull that can cause an object with mass to change its velocity (accelerate). It's also a vector quantity, meaning it has both magnitude and direction. The standard unit of force is the Newton (N). According to Newton's second law of motion, force equals mass times acceleration (F = ma).

Sure, here are explanations for each term:

6. **Linear Momentum**: Linear momentum is a measure of the motion of an object. It is defined as the product of an object's mass and its velocity. Mathematically, linear momentum (p) is expressed as ( p = mv ), where  m is the mass of the object and  v is its velocity. Like velocity, linear momentum is a vector quantity, meaning it has both magnitude and direction. According to the principle of conservation of momentum, the total momentum of a closed system remains constant if no external forces act on it.

7. **Kinetic Energy**: Kinetic energy is the energy possessed by an object due to its motion. It depends on the mass of the object and its velocity. Mathematically, kinetic energy (KE) is expressed as ( KE = 1/2mv^2 ), where m  is the mass of the object and  v is its velocity. Kinetic energy is a scalar quantity and is always non-negative.

8. **Potential Energy**: Potential energy is the energy possessed by an object due to its position or configuration. It depends on the position of an object within a force field or a configuration of interacting objects. There are different types of potential energy, such as gravitational potential energy, elastic potential energy, and electric potential energy. Gravitational potential energy (PE) near the surface of the Earth is commonly expressed as ( PE = mgh ), where  m  is the mass of the object,  g  is the acceleration due to gravity, and h  is the height above a reference point. Potential energy is also a scalar quantity.

LOGARITHMIC COSINES

Method of using the table of sines : this table is also known as the table of logarithmic cosines . At intervals of 60'(or 1°), we can find the values of cosines angles ranging from 0° to 89°.

         The table of logarithmic cosines is essentially divided into the following (parts) sections, as shown by the following tables. 

        

       (i) the angles in the table's extreme left vertical column are from 0° to 89° at intervals of 1°. 

 

       (ii) at intervals of 6' , the horizontal row at the top of the chart is from 0' to 54'.

                        OR

 at intervals of 0.1° , the horizontal row at the top of the chart is from 0.0° to 0.9° .

       (iii) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5'.

               This portion of the chart is referred to as the mean difference column ( Subtract Mean Difference).

* Note: 

           (i) we obtain the cosines value of a given angle from the chart, which is expressed in five decimal places. 

           (ii)  The table is also drawn in such a way that we can determine the logarithmic cosine values of any given angle between 0° and 89°.

             (iii) sine this portion of the chart is referred to as the subtract mean difference column.

            (iv) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5' in contain not available 86° to 89° column.

SOLVED EXAMPLES USING THE TABLE OF LOGARITHMIC COSINES :

1. Find the value of LOG ( cos 60°) ≈ ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 60°.

 Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.6990

   LOG ( cos 60°) ≈ "bar" 1.6990 ... Ans

2. Find the value of LOG ( cos19°30' ) ≈ ? 

   ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 19°.

 Then we move horizontally to the right at the top of the column headed by 30' (or 0.5°) and read the figure "bar" 1.9743

   LOG ( cos19°30' ) ≈  "bar" 1.9743 ... Ans

3. Find the value of LOG ( cos 58°9’) ≈ ? 

     ⇒SOLUTION : 

LOG ( cos 58°9’) ≈LOG ( cos 58°6' + 3’) ≈

Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 58°.

 Then we move horizontally to the right at the top of the column headed by 6' (or 0.1°) and read the figure "bar" 1.7230

Now we move further right along the horizontal line of angle 58° to the column headed by 3' of mean difference and read the figure 7 there; this figure of the table does not contain decimal sign. In fact, 7 implies 0.0007 .                      

LOG ( cos 58°6' + 3’) ≈ (bar 1.7230 - 0.0007)

                                     ≈ "bar" 1.7223

 LOG ( cos 58°9’) ≈  "bar" 1.7223 ... Ans

4. Find the value of LOG ( cos 67°32’) ≈ ? 

   ⇒SOLUTION : 

LOG ( cos 67°32’) ≈LOG ( cos 67°30' + 2’) 

      Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 67°.

      Then we move horizontally to the right at the top of the column headed by 30' (or 0.5°) and read the figure "bar" 1.5828

       Now we move further right along the horizontal line of angle 67° to the column headed by 2' of mean difference and read the figure 6 there; this figure of the table does not contain decimal sign. In fact, 6 implies 0.0006 .  

  LOG ( cos 67°30' + 2’) ≈ (bar 1.5828 - 0.0006) 

                                        ≈ "bar" 1.5822

LOG ( cos 67°32’) ≈ "bar" 1.5822 ... Ans                  

5. Find the value of LOG ( cos 72°8’) ≈ ? 

     ⇒SOLUTION :

   LOG ( cos 72°8’) ≈ LOG (cos 72°6’+ 2') ≈

         Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 72°.

       Then we move horizontally to the right at the top of the column headed by 6' (or 0.1°) and read the figure "bar" 1.4876 

       Now we move further right along the horizontal line of angle 72° to the column headed by 2' of mean difference and read the figure 8 there; this figure of the table does not contain decimal sign. In fact, 8 implies 0.0008  . 

 LOG (cos 72°6’+ 2') ≈"bar"(1.4876 - 0.0008)

                                   ≈ "bar"1.4868

  LOG ( cos 72°8’) ≈ "bar"1.4868  .... Ans

                            **********

LOGARITHMIC SINES

      Method of using the table of sines  : this table is also known as the table of logarithmic sines . At intervals of 60'(or 1°), we can find the values of sines angles ranging from 0° to 89°.

         The table of logarithmic sines is essentially divided into the following (parts) sections, as shown by the following tables. 

        

       (i) the angles in the table's extreme left vertical column are from 0° to 89° at intervals of 1°. 

 

       (ii) at intervals of 6' , the horizontal row at the top of the chart is from 0' to 54'.

       (iii) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5'.

               This portion of the chart is referred to as the mean difference column ( Add Mean Difference).

* Note: 

           (i) we obtain the sine value of a given angle from the chart, which is expressed in five decimal places. 

           (ii) we know that the sine of any given angle is equal to that of the cosine of its complementary angle, i.e., sin(θ) =cos(90º - θ) . The table is also drawn in such a way that we can determine the logarithmic sin values of any given angle between 0° and 89°.

             (iii) sine this portion of the chart is referred to as the add mean difference column.

            (iv) at intervals of 1', the angles in the horizontal row at the top of the chart are from 1' to 5' in contain not available 0° to 3° column.

SOLVED EXAMPLES USING THE TABLE OF LOGARITHMIC SINES :

1. Find the value of  LOG ( sin 60°) ≈  ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 60°. 

           Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.9375 ,Therefore LOG ( sin 60°) ≈ "bar" 1.9375   ... Ans

2. Find the value of LOG ( sin 19.4°) ≈ ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 19°.

 Then we move horizontally to the right at the top of the column headed by 24' (or 0.4°) and read the figure "bar" 1.5213

       LOG ( sin 19.4°) or ≈ LOG ( sin 19°24') 

                                 ≈ "bar" 1.5213 ... Ans

3. Find the value of  LOG ( sin 32°5’) ≈ ? 

    ⇒SOLUTION : Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 32°.

 Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.7242 , 

Now we move further right along the horizontal line of angle 32° to the column headed by 5' of mean difference and read the figure 10 there; this figure of the table does not contain decimal sign. In fact, 10 implies 0.0010. 

        LOG ( sin 32°5’) ≈ LOG ( sin 32°0’+5')

                          ≈  ( bar 1.7242 + 0.0010 ) 

                           ≈  "bar" 1.7252 ... Ans

4. Find the value of  LOG ( sin 72°60’) ≈  ? 

    ⇒SOLUTION :  LOG ( sin 72°60’) 

          or ≈ LOG ( sin 73°)      ( •.• 60' = 1°) 

 Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 73°.

 Then we move horizontally to the right at the top of the column headed by 0' (or 0.0°) and read the figure "bar" 1.9806 

.•.  LOG ( sin 72°60’)  ≈ "bar" 1.9806 .. Ans

5. Find the value of  LOG ( sin 67°8’) ≈ ? 

    ⇒SOLUTION : 

    LOG ( sin 67°8’) ≈  LOG ( sin 67°6’ + 2')

   Using the table of logarithmic sines we need to go through the extreme left vertical column 0° to 89° and move downwards till we reach the angle 67°.

 Then we move horizontally to the right at the top of the column headed by 6' (or 0.1°) and read the figure "bar" 1.9643 , 

Now we move further right along the horizontal line of angle 67° to the column headed by 2' of mean difference and read the figure 1 there; this figure of the table does not contain decimal sign. In fact, 1 implies 0.0001 .                      

               ≈ ( bar 1.9643 + 0.0001) 

               ≈ "bar" 1.9644

    

                         ***********

RECIPROCALS OF FOUR-FIGURE NUMBERS

Introduction : 

          Reciprocal is simply defined as the inverse of a number Or a value i.e. the reciprocal of a number is 1 divided by that number. If n is a real number, then its inverse will be 1/n or n⁻¹. A reciprocal is also a number taken to the power of -1.

For Example , the reciprocal (or inverse) of 5 is 1 divided by 5 , which is 1/5 or 5⁻¹.

1/5 or 5⁻¹ = 0.2 ( from chart 0.2000 four- figure number )

         The table of reciprocal numbers is also known as the table of reciprocal four-figure numbers. We can find reciprocal with values ranging from 1.0 to 9.9 at intervals of 0.1. The table of reciprocal four- figure numbers is essentially divided into three sections, as shown in the following tables. 

(i) at intervals of 0.1, the number in the table's extreme left vertical column ranges from 1.1 to 9.9. 

(ii) the horizontal row at the top of the chart is from 0 to 9 at intervals 1. 

(iii) at intervals of 1, the number in the horizontal row at the top of the chart are from 1 to 9 .This portion of the chart is referred to as the mean differences column.

Find the reciprocal of following numbers, using the reciprocal four-figure numbers chart.  

  If x  ≥ 1

(i) 1/1.722 = ? 

⇒ SOLUTION :  The number in the reciprocal four-figure numbers table extreme left vertical column in 1.7 ,the horizontal row at the top of the chart in 2 consist of containing four digits number 0.5814 and horizontal row at the top of the chart in number 2 consist of containing one digit number 7 [= 0.0007 mean difference subtract ]

    1/1.722 = 0.5814 - 0.0007 

                   = 0.5821   .... Ans

  

(ii) 1/17.22 =  ? 

⇒SOLUTION : 1/1.722 x 10¹(decimal place change  right to left and multiply  10¹ )

      = ( 1/1.722 ) x 10⁻¹  (·.· 1/10¹ = 10⁻¹ )

      = 0.5821 x 10⁻¹ 

      = 0.05821     .....  Ans

 

(iii) 1/172.2 =  ? 

⇒SOLUTION : 1/1.722 x 10² (decimal place change from right to left and multiply  10² )

     = ( 1/1.722 ) x 10⁻²   ( ·.· 1/10² = 10⁻² )

     = 0.5821 x 10⁻² 

     = 0.005821    ..... Ans

 

 

(iv) 1/1722 =  ? 

⇒SOLUTION : 1/1.722 x 10³ (decimal place change from  right to left and multiply x 10³ )

    = ( 1/1.722 ) x 10⁻³ ( ·.· 1/10³ = 10⁻³ )

    = 0.5821 x 10⁻³ 

    = 0.0005821  ..... Ans

  

Find the reciprocal of following numbers, using the reciprocal four-figure numbers chart.  

 If x < 1

(i) 1/0.1722 =  ? 

⇒SOLUTION : 1/1.722 x 10⁻¹ (decimal place change from left to right and multiply x 10⁻¹ )

      = ( 1/1.722 ) x 10¹    ( ·.· 1/10⁻¹ = 10¹ )

      = 0.5821 x 10¹ 

      = 5.821   ...... Ans

(ii) 1/0.01722 =  ? 

⇒SOLUTION : 1/1.722 x 10⁻² (decimal place change from  left to right  and multiply x 10⁻² )

     = ( 1/1.722 ) x 10²   ( ·.· 1/10⁻² = 10² )

     = 0.5821 x 10² 

     = 58.21      ..... Ans

(iii) 1/0.001722 =  ? 

⇒SOLUTION : 1/1.722 x 10⁻³(decimal place change from left to right and multiply x 10⁻³ )

     = ( 1/1.722 ) x 10³  ( ·.· 1/10⁻³ = 10³ )

     = 0.5821 x 10³

     = 582.1     .....  Ans

(iii) 1/0.001722 =  ? 

⇒ SOLUTION : 1/1.722 x 10⁻⁴ (decimal place changes from left to right and multiply x 10⁻⁴  )

    = ( 1/1.722 ) x 10⁴    ( ·.· 1/10⁻⁴ = 10⁴ )

    = 0.5821 x 10⁴ 

    = 5821     ......  Ans

  

                      *********

SQUARE ROOTS(FROM 1 to 10 AND 10 TO 99)

By MR. DEVIDAS MADHAO MISTARI

                            M.Sc. (physics) 

                          Material science

 Introduction :      

           A number is a square root that, when multiplied by itself, corresponds to the desired value. So, for example, the square root of x = 64 is 8 ( 8x8 = 64 ). Squaring is the act of multiplying a number by itself.

        [ Using square root symbol :  √(x)  ] where, x is number. 

  

           Method of using the table of square roots (from 1 to10 ) : this table is also known as the table of square roots . At intervals of 0.1, we can find the values of square roots  ranging from 1.0 to 9.9.

         The table of square roots is essentially divided into the following parts (three parts) , as shown in the following tables. 

        

       (i) the number in the table's extreme left vertical column are from 1.1 to 9.9 at intervals of 0.1 . 

 

        (ii) at intervals of 1 , the horizontal row at the top of the chart is from 0 to 9.

    

       (iii) at intervals of 1, the number in the horizontal row at the top of the chart are from 1 to 9 .This portion of the chart is referred to as the mean difference column.

      * NOTE :  The power of 10 extracted under the root sign must always be even number. 10⁻²,10⁻⁴,10² and 10⁴ ⇒ Even power (-2),(-4),(2) and (4)

1. Square roots (from 1 to 10) 

      For Example :

   

     ( i )   √(6.479) = ? 

SOLUTION : The number in the square roots (from 1 to 10) table extreme left vertical column in 6.4 ,the horizontal row at the top of the chart in 7 consist of containing four digits number 2.544 and horizontal row at the top of the chart in number 9 consist of containing one digit number 2 = 0.002 (mean difference add)·

  √(6.479) = ( 2.544 + 0.002 ) = 2.546 (·.·1 < 5 < 9 )

Therefore √(6.479) = 2.546  ... Answer

     

     ( ii )  √(647.9) =  ? 

     SOLUTION : √(6.479 x 10²)   (decimal place changes from 3ʳᵈ number to 2ⁿᵈ number digit right to left and multiply x 10² )

 = √(6.479)  x 10¹   [ ·.· √(10²) = 10¹ ]

 = ( 2.544 + 0.002 ) x 10¹

 = 2.546 x 10¹ = 25.46

Therefore √(6.479) = 25.46 ... Answer

   ( iii )  √(0.06479) =  ? 

   SOLUTION : √(6.479 x 10⁻² )   (decimal place changes from 1ˢᵗ to 2ⁿᵈ left to right and multiply x 10⁻² )

   = ( 2.544 + 0.002 ) x 10⁻¹

            [·.· √(10⁻²)=10⁻¹] 

   = 2.546 x 10⁻¹ = 0.2546

   ( iv )  √(0.0006479) = ? 

SOLUTION : √(6.479 x 10⁻⁴)   (decimal place changes from 1ˢᵗ to 4ᵗʰ left to right and multiply x10⁻⁴)

 = ( 2.544 + 0.002 ) x 10⁻²  

         [·.· √(10⁻⁴)=10⁻²]

 = 2.546 x 10⁻² = 0.02546

        Method of using the table of square  (from 10 to 99 ): this table is also known as the table of square roots . At intervals of 1, we can find the values of square roots  ranging from 10 to 99.

         The table of square roots is essentially divided into the following parts, as shown in the following tables. 

        

       (i) the number in the table's extreme left vertical column are from 10 to 99 at intervals of 1 . 

 

        (ii) at intervals of 1 , the horizontal row at the top of the chart is from 0 to 9.

    

       (iii) at intervals of 1, the number in the horizontal row at the top of the chart are from 1 to 9 .This portion of the chart is referred to as the mean difference column.

2. Square roots (from 1 to 99) 

      For Example :

 ( i ) √(64.79) = ? 

SOLUTION  : √(64.79) = (8.044 + 0.006) 

  = 8.050   The number in the square roots (from 10 to 99) table extreme left vertical column in 64 ,the horizontal row at the top of the chart in 7 consist of containing four digits number 8.044 and horizontal row at the top of the chart in number 9 consist of containing one digit number 6 = 0.006 (mean difference add)·

   Therefore √(64.79) = 8.050

  

( ii )  √(6479) = ? 

  SOLUTION :  √(64.79 x 10²) (decimal place changes from 4ᵗʰ number to 2ⁿᵈ number digit right to left and multiply x 10² )

    = (8.044 + 0.006) x 10¹

            [·.· √(10²)=10¹] 

    = 8.050 x 10¹ = 80.50

   

 ( iii )  √(0.6479) = ? 

SOLUTION : √(64·79 x 10⁻²) (decimal place changes from 1ˢᵗ  to 2ⁿᵈ left to right and multiply x 10⁻² )

= (8.044 + 0.006) x 10⁻¹     

     [·.· √(10⁻²)=10⁻¹] 

= 8.050x 10⁻¹ = 0.8050

   

  ( iv )  √(0.006479) = ? 

SOLUTION  : √(64·79 x 10⁻⁴) (decimal place changes from 1ˢᵗ to 4ᵗʰ right to left and multiply x 10⁻⁴ )

= (8.044 + 0.006) x 10⁻²   

    [·.· √(10⁻⁴)=10⁻²] 

= 8.050 x 10⁻² = 0.08050

   

SQUARES

By MR. DEVIDAS MADHAO MISTARI

                                             M.Sc. (physics) 

                                            Material science

HOW TO USE SQUARES CALCULATIONS WITH ILLUSTRATIONS . 

          

The Squares table consists of two parts:

1. The extreme left column consisting of one to two digit numbers from 0 to 99.

2. Ten right columns containing one to six digit numbers , headed by the digits from 0 to 9.

     * Note : From the identity, exact squares of four figure numbers can be easily calculated.

               ( x + y)² = x² + 2xy +y²

              ( x - y)² = x² - 2xy + y²

A Square table is shown below:

    

Solved examples using the table of  squares : 

 1. Find the square of one digit number   (i) 3²   (ii) 5²   (iii) 9²    (iv) 8²

  Solution :  Easily find the square of one digit number and remember in mind.One digit numbers 1 to 9.

         (i) 3² = 9  →Solution :  Find the value of 3 by the using the squares table . Then we move horizontally to the right at the top of the column headed by 3 and the extreme left column consisting of  number 0  read the figure 9. Therefore 3 squares is 9.

        (ii) 5² = 25  →Solution : Find the value of 5 by the using the squares table . Then we move horizontally to the right at the top of the column headed by 5 and the extreme left column consisting of number 0 read the figure 25. Therefore 5 squares is 25.

       (ii) 8² = 64 → Solution : Find the value of 8by the using the squares table . Then we move horizontally to the right at the top of the column headed by 8 and the extreme left column consisting of number 0 read the figure 64. Therefore 8 squares is 64.

      (iv) 9² = 81 →Solution : The extreme left column consisting of number 0. Right columns containing read two digit number 81, headed by the digits 9 .Therefore 9 squares is 81.

2. Find the square of two digit number Quickly find the square of two digit number using squares table . Two digit numbers 10 to 99. 

            For example  ⇒

           (i) 10² = 100 →Solution :Determine the value of squres easily find the square of 10 using the squares table. The extreme left column consisting of number 1. Then we move horizontally to the right at the top of the column headed by 0 and Right columns containing read three digit number 100.

          (ii) 12² = 144 →Solution : Determine the value of squres easily find the square of 12 using the squares table. The extreme left column consisting of number 1. Then we move horizontally to the right at the top of the column headed by 2 and Right columns containing read three digit number 144.

         (iii) 15² = 225 →Solution : Find the value of squres easily find the square of 15 using the squares table. The extreme left column consisting of number 1. Then we move horizontally to the right at the top of the column headed by 5 and Right columns containing read three digit number 225.

         (iv) 98² = 9604→Solution : Determine the value of squres quickly find the square of 98 using the squares table. The extreme left column consisting of number 9. Then we move horizontally to the right at the top of the column headed by 8 and Right columns containing read four significant figure number 9604.

3. Find the square of three digit number quickly find the square of three digits numbers using squares table . Three digits numbers 100 to 999.  

           For example  ⇒

          (i) 103² = 10609 →Solution : The extreme left column consisting of number 10. Then we move horizontally to the right at the top of the column headed by 3 and Right columns containing read five digit number 10609.

          (ii) 121² = 14641 →Solution : The extreme left column consistingSolution of number 12. Then we move horizontally to the right at the top of the column headed by 1 and Right columns containing read five digit number 14641.

         (iii) 125² = 15625 → Solution : The extreme left column consisting of number 12. Then we move horizontally to the right at the top of the column headed by 5 and Right columns containing read five digit number 15625.

         (iv) 888² = 788544 → Solution : The extreme left side column consisting of number 88. Then we move horizontally to the right at the top of the column headed by 8 and Right columns containing read six significant figures 788544.

         (v)  987² = 974169 → Solution : The extreme left side column consisting of number 98. Then we move horizontally to the right at the top of the column headed by 7 and Right columns containing read six significant digits 974169.

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