physics.

GENERAL PHYSICS

1. Quantity and Unit
Physical Quantities
  • Types of Quantities:

    • Base Quantity: Represented by a single SI unit.

    • Examples:

    • Mass

    • Length

    • Time

    • Current

    • Temperature

    • Speed

    • Volume

    • Area

    • Force

    • Derived Quantity: Composed of two or more base quantities.

    • Examples:

    • Force (in terms of mass and acceleration)

    • Speed (in terms of distance and time)

SI Units
  • Definition: The value of a physical quantity expressed in terms of a number and unit.

  • Common SI Units Table:

  • Definition: Used for very large or very small quantities in physical measurements.

  • Examples of Prefixes Table:

  • Scalar Definition: A quantity with magnitude only.

    • Examples: Mass, length, area, volume, density, time, distance, speed, energy, temperature, current, voltage.

  • Vector Definition: A quantity with both magnitude and direction.

    • Examples: Weight, displacement, velocity, acceleration, force, moment.

1. Length and Time
Length
  • Definition: Length is the measurement from one end to another.

  • SI Unit: Metre (m)

  • Instruments for Measuring Length:

    • Measuring Tape: For long length (Accuracy: 1mm)

    • Ruler: For medium length (Accuracy: 1mm)

    • Vernier Calipers: For short length (Accuracy: 0.1mm)

    • Micrometer Screw Gauge: For very short length (Accuracy: 0.01mm)

How to Use Ruler or Measuring Tape
  1. Align the 0 mark with one end of the object.

  2. Read the scale at the other end of the object.

  3. Ensure the line of sight is directly above the scale.

Vernier Calipers Usage
  1. Position the object between the jaws.

  2. Take the main scale reading before the zero mark on the vernier.

  3. Identify the closest vernier mark to the main scale.

  4. Add the two readings for the complete measurement.

Micrometer Screw Gauge Usage
  1. Rotate the thimble until the object is held gently between the anvil and spindle.

  2. Record the main scale reading on the sleeve just before the thimble edge.

  3. Note the smallest circular scale marking aligned with the main scale.

  4. Combine main scale and circular scale readings for final measurement.

Time
  • SI Unit: Second (s); Other units include minute, hour, day, etc.

  • Conversion Tables:

    • 1 year = 365 days = 8760 hours = 31536000 seconds

    • 1 day = 24 hours = 86400 seconds

    • 1 hour = 60 minutes = 3600 seconds

    • 1 minute = 60 seconds

Instruments for Measuring Time
  • Clock, Watch, Stopwatch, Pendulum

Simple Pendulum
  • A simple pendulum oscillates with length defined from ceiling to centre of the bob.

  • Period (T): Time taken for one full oscillation.

  • Formula for Period: T=tnT = \frac{t}{n}

    • Where:

    • TT = period [s]

    • tt = total time for nn oscillations

    • nn = number of oscillations

Factors Affecting Pendulum Period
  1. Length of pendulum (l).

  2. Acceleration due to gravity (g).

Sample Problem: Find the Period of a Pendulum
  • Data: t = 3s for n = 3 (three-quarter oscillations)

Solution

T=tn=334=4sT = \frac{t}{n} = \frac{3}{\frac{3}{4}} = 4s

Exercises
  1. Use Vernier calipers for measuring wooden cubes.

  2. Measure with micrometer screw gauge.

  3. Calculate the period of a pendulum if it oscillates 15 times for 45 seconds.

  4. A pendulum's period is calculated with examples.

1. Speed, Velocity, and Acceleration
Distance and Displacement
  • Distance Definition: The total length between two points (scalar).

  • Displacement Definition: The change of position in a specific direction (vector).

Example 1
  • Car movement:

    • Moves 5 km East and 3 km North.

    • Distance = 8 km; Displacement = 5 km East + 3 km North.

Example 2
  • Roundabout passage:

    • Circumference = 10 m.

    • Distance = 10 m; Displacement = 0 m (because returned to starting point).

Example 3
  • Walking forward and backward:

    • Forward = 15 m; Backward = 5 m.

    • Distance = 20 m; Displacement = 10 m forward.

Speed
  • Definition: The rate of change of distance with time (scalar).

  • Formula:
    Speed=Distance traveledTime taken\text{Speed} = \frac{\text{Distance traveled}}{\text{Time taken}}

  • Average Speed Formula:
    Average Speed=Total distanceTotal time\text{Average Speed} = \frac{\text{Total distance}}{\text{Total time}}

Example
  • A car covers 540 km in 10 hours.

Calculate Average Speed
  • Average Speed in km/hr=54010=54 km/hr\text{Average Speed in km/hr} = \frac{540}{10} = 54 \text{ km/hr}

  • Average Speed in m/s=540,00036,000=15 m/s\text{Average Speed in m/s} = \frac{540,000}{36,000} = 15 \text{ m/s}

Velocity
  • Definition: Rate of change of displacement with time (vector).

  • Example: Car1 has 10 m/s East; Car2 has 10 m/s North (same speed, different velocities).

Acceleration
  • Definition: Rate of change of velocity with time (vector).

  • Formula: a=vuta = \frac{v-u}{t}

    • Where vv = final velocity, uu = initial velocity, tt = time taken.

Example 1
  • A car accelerates from rest to 15 m/s in 5 seconds.

Calculate acceleration
  • a=vut=1505=3 m/s2a = \frac{v-u}{t} = \frac{15-0}{5} = 3 \text{ m/s}^2

Example 2
  • A car slows from 72 km/h to a stop in 10 seconds.

Calculate acceleration
  • Convert 72 km/h to 20 m/s:
    a=02010=2 m/s2a = \frac{0-20}{10} = -2 \text{ m/s}^2

  • Negative acceleration indicates deceleration.

Motion with Uniform Acceleration
  • Key Equations:

    • v=u+atv = u + at

    • x=ut+12at2x = ut + \frac{1}{2}at^2

    • v2=u2+2axv^2 = u^2 + 2ax

Example 1
  • A car traveling at 10 m/s accelerates at 2 m/s² for 3 seconds.

Calculate final velocity
  • v=u+at=10+2×3=16 m/sv = u + at = 10 + 2 \times 3 = 16 \text{ m/s}

Example 2
  • A motorcycle starts from rest acquiring 72 km/h in 5s.

Calculate (a) acceleration; (b) distance traveled.
Solution
  • Convert 72 km/h = 20 m/s.

  1. (a) a=2005=4 m/s2a = \frac{20-0}{5} = 4 \text{ m/s}^2

  2. (b) Use x=ut+12at2=0×5+12×4×52=50 mx = ut + \frac{1}{2}at^2 = 0 \times 5 + \frac{1}{2} \times 4 \times 5^2 = 50 \text{ m}

Example 3
  • For a uniformly decelerating object from 50 m/s to 30 m/s with deceleration of -4 m/s², find distance covered.

Solution
  • Use v2=u2+2axv^2 = u^2 + 2ax:
    \begin{align} 30^2 &= 50^2 + 2 \times (-4) \times x \ 900 &= 2500 - 8x \ 8x &= 1600 \ x &= 200 \text{ m} \end{align}

  • Acceleration due to gravity (g): 9.8 \text{ m/s}^2 (approximately 10 \text{ m/s}^2)

  • Free-fall condition: u = 0, a = g (down).