Comprehensive Notes on Physical Quantities and Measurement: Volume, Area, Density, and Speed

Measurement of Volume of a Liquid

  • Instruments Used for Volume Measurement:

    • Measuring Cylinder: Used in laboratories to measure the volume of liquids and irregular objects using the displacement method. It is typically marked with a scale in mLmL. Common capacities include 100 mL100\,mL, 200 mL200\,mL, and 500 mL500\,mL.
    • Note: 100 cm3=100 mL100\,cm^3 = 100\,mL.
    • A specific form of the measuring cylinder is used by pharmacists to measure liquid medicines.
    • Measuring Beakers: Used generally for measuring a fixed volume of liquids like milk, oil, or lubricating oil.
    • Available capacities: 50 mL50\,mL, 100 mL100\,mL, 200 mL200\,mL, 500 mL500\,mL, and 1000 mL1000\,mL.
    • Design feature: Capacity is marked on the beaker, and it is provided with a handle for holding.

  • Procedure for Using a Measuring Cylinder:

    • Select a clean, dry measuring cylinder.
    • Place it on a flat, horizontal surface.
    • Pour the liquid gently to avoid splashing.
    • Wait for the liquid to become stationary. The upper surface (meniscus) will appear curved.
    • Reading Technique: Keep the eye horizontally in line with the lower surface of the liquid (meniscus).
    • In the provided example (Fig. 1.6), the recorded volume is 70 mL70\,mL.

  • Procedure for Using a Measuring Beaker:

    • Select a beaker with the required capacity (e.g., 500 mL500\,mL for measuring milk from a bucket).
    • Wash and dry the beaker.
    • Immerse it fully into the liquid so it is completely filled.
    • Remove it gently to avoid splashing and pour it into the destination vessel.

Volume Measurement of Regular and Irregular Objects

  • Volume of Regular Objects: Volume is determined by measuring dimensions (length, breadth, height, or radius) and applying specific formulas:

    • Cube: Volume=(side)3\text{Volume} = (side)^3
    • Cuboid: Volume=length×breadth×height\text{Volume} = length \times breadth \times height
    • Sphere: Volume=43π×(radius)3\text{Volume} = \frac{4}{3} \pi \times (radius)^3
    • Cylinder: Volume=π×(radius)2×height\text{Volume} = \pi \times (radius)^2 \times height
    • Cone: Volume=π3×(radius)2×height\text{Volume} = \frac{\pi}{3} \times (radius)^2 \times height
    • Constant: π=3.14\pi = 3.14 or 227\frac{22}{7}

  • Volume of an Irregular Body (Displacement Method):

    • Every body occupies space equal to its own volume. When immersed in a liquid, it displaces a volume of liquid equal to its own volume.
    • Formula: Volume of body=Final level (V2)−Initial level (V1)\text{Volume of body} = \text{Final level } (V_2) - \text{Initial level } (V_1)
    • Activity 2 (Measuring a Stone):
    • Tie the stone with a fine thread.
    • Initial water level (V1V_1) = 60 mL60\,mL.
    • Level after immersion (V2V_2) = 80 mL80\,mL.
    • Volume displaced = 80 mL−60 mL=20 mL80\,mL - 60\,mL = 20\,mL.
    • Volume of stone = 20 cm320\,cm^3 (since 1 mL=1 cm31\,mL = 1\,cm^3).

  • Determining Volume of Empty Vessels: Fill the vessel (like a bottle) completely with water, then pour that water into a measuring cylinder to find the volume.

Measurement of Area

  • Concept of Area: The surface occupied by an object is its area.

    • S.I. Unit: Square metre (m2m^2). It is the area of a square with sides of 1 m1\,m.

  • Units of Area and Relationships:

    • Large Units:
    • Are: Area of a square with 10 m10\,m sides. 1 are=10 m×10 m=100 m21\,are = 10\,m \times 10\,m = 100\,m^2.
    • Hectare: Area of a square with 100 m100\,m sides. 1 hectare=100 m×100 m=10,000 m2=104 m2=100 ares1\,hectare = 100\,m \times 100\,m = 10,000\,m^2 = 10^4\,m^2 = 100\,ares.
    • Square Kilometre (km2km^2): Area of a square with 1 km1\,km sides. 1 km2=1000 m×1000 m=106 m21\,km^2 = 1000\,m \times 1000\,m = 10^6\,m^2.
    • Small Units:
    • Square Centimetre (cm2cm^2): 1 cm2=1100 m×1100 m=10−4 m21\,cm^2 = \frac{1}{100}\,m \times \frac{1}{100}\,m = 10^{-4}\,m^2.
    • Square Millimetre (mm2mm^2): 1 mm2=11000 m×11000 m=10−6 m21\,mm^2 = \frac{1}{1000}\,m \times \frac{1}{1000}\,m = 10^{-6}\,m^2.

  • Area of Regular Objects:

    • Square: side2side^2
    • Rectangle: length×breadthlength \times breadth
    • Circle: π×(radius)2\pi \times (radius)^2
    • Surface Area of Cylinder: 2π×(radius)×length2\pi \times (radius) \times length
    • Surface Area of Sphere: 4π×(radius)24\pi \times (radius)^2

  • Area of an Irregular Object (Lamina):

    • Measurements are estimated using graph paper (ruling at 1 mm1\,mm intervals, thick lines at 1 cm1\,cm intervals).
    • Procedure:
    • Draw the boundary of the lamina on the graph paper.
    • Count nn, where n=Number of complete squares+Number of squares half or more than half fulln = \text{Number of complete squares} + \text{Number of squares half or more than half full}.
    • Ignore squares less than half full.
    • Area=n×1 cm2Area = n \times 1\,cm^2.
    • Activity 3 Example: 4 complete squares + 4 half/more-than-half squares = 8 squares. Area = 8 cm28\,cm^2.

  • Volume of an Irregular Sheet: If the sheet has a finite thickness (tt), volume is calculated as: Volume=Area×thickness\text{Volume} = \text{Area} \times \text{thickness}.

Density: Concepts and Principles

  • Key Observations:

    • Equal masses of different substances have different volumes (e.g., 1 kg1\,kg iron vs. 1 kg1\,kg sugar/cotton).
    • Equal volumes of different substances have different masses (e.g., milk is heavier than an equal volume of water).

  • Definitions and Formulas:

    • Density: The mass of a unit volume of a substance.
    • Formula: d=MVd = \frac{M}{V}
    • Units:
    • S.I. Unit: kg m−3kg\,m^{-3} (kilogram per cubic metre).
    • C.G.S. Unit: g cm−3g\,cm^{-3} (gram per cubic centimetre).
    • Relationship: 1 g cm−3=1000 kg m−31\,g\,cm^{-3} = 1000\,kg\,m^{-3}.

  • Density Characteristics:

    • Density does not change with shape or size.
    • Density usually decreases as temperature increases (expansion), with the exception of water.
    • Water Anomaly: Water contracts when heated from 0∘C0^{\circ}C to 4∘C4^{\circ}C and expands above 4∘C4^{\circ}C. Density is maximum at 4∘C4^{\circ}C (1.0 g cm−31.0\,g\,cm^{-3} or 1000 kg m−31000\,kg\,m^{-3}).

  • Substance Comparisons (Denser vs. Less Dense):

    • Iron is denser than sugar because iron particles are closely packed while sugar particles are loosely packed.
    • Iron is denser than aluminium; aluminium is denser than glass; glass is denser than wood.

Determination of Density for Different States

  • Regular Solids:

    • Measure mass (MM) using a beam balance.
    • Find volume (VV) using dimensions and regular formulas (l×b×hl \times b \times h).
    • Calculate d=MVd = \frac{M}{V}.

  • Irregular Solids:

    • Measure mass (MM) with a beam balance.
    • Measure volume (VV) using the displacement method in a measuring cylinder (V=V2−V1V = V_2 - V_1).
    • Calculate density in g cm−3g\,cm^{-3}.

  • Liquids (Example: Milk):

    • Measure mass of empty beaker (M1M_1).
    • Pour a known volume of milk (VV, e.g., 50 mL50\,mL) into the beaker using a measuring cylinder.
    • Measure mass of beaker + milk (M2M_2).
    • Mass of milk (MM) = M2−M1M_2 - M_1.
    • Calculate density (dd). Example calculation: 51.5 g50 cm3=1.03 g cm−3\frac{51.5\,g}{50\,cm^3} = 1.03\,g\,cm^{-3}.

  • Table: Densities of Common Substances (at 4∘C4^{\circ}C for water):

    • Cork: 0.25 g cm−30.25\,g\,cm^{-3} (250 kg m−3250\,kg\,m^{-3})
    • Wood: 0.7 g cm−30.7\,g\,cm^{-3} (700 kg m−3700\,kg\,m^{-3})
    • Ice: 0.92 g cm−30.92\,g\,cm^{-3} (920 kg m−3920\,kg\,m^{-3})
    • Glass: 1.97 g cm−31.97\,g\,cm^{-3} (1,970 kg m−31,970\,kg\,m^{-3})
    • Aluminium: 2.7 g cm−32.7\,g\,cm^{-3} (2,700 kg m−32,700\,kg\,m^{-3})
    • Iron: 7.8 g cm−37.8\,g\,cm^{-3} (7,800 kg m−37,800\,kg\,m^{-3})
    • Brass: 8.4 g cm−38.4\,g\,cm^{-3} (8,400 kg m−38,400\,kg\,m^{-3})
    • Silver: 10.3 g cm−310.3\,g\,cm^{-3} (10,300 kg m−310,300\,kg\,m^{-3})
    • Lead: 11.5 g cm−311.5\,g\,cm^{-3} (11,500 kg m−311,500\,kg\,m^{-3})
    • Alcohol: 0.8 g cm−30.8\,g\,cm^{-3} (800 kg m−3800\,kg\,m^{-3})
    • Mercury: 13.6 g cm−313.6\,g\,cm^{-3} (13,600 kg m−313,600\,kg\,m^{-3}) (used in barometers due to high density)

Speed and Motion

  • Definition of Speed: The distance covered or travelled by a body in unit time.

    • Formula: Speed (v)=Distance (D)Time (t)\text{Speed } (v) = \frac{\text{Distance } (D)}{\text{Time } (t)}

  • Units and Conversions:

    • S.I. Unit: Metre per second (m s−1m\,s^{-1}).
    • Other Units: Kilometre per hour (km h−1km\,h^{-1}), centimetre per second (cm s−1cm\,s^{-1}), and km min−1km\,min^{-1}.
    • Relationships:
    • 1 km h−1=1000 m3600 s=13.6 m s−11\,km\,h^{-1} = \frac{1000\,m}{3600\,s} = \frac{1}{3.6}\,m\,s^{-1}
    • 3.6 km h−1=1 m s−13.6\,km\,h^{-1} = 1\,m\,s^{-1}
    • 18 km h−1=5 m s−118\,km\,h^{-1} = 5\,m\,s^{-1}
    • 100 cm s−1=1 m s−1100\,cm\,s^{-1} = 1\,m\,s^{-1}
    • 1 km min−1=1000 m60 s=16.67 m s−11\,km\,min^{-1} = \frac{1000\,m}{60\,s} = 16.67\,m\,s^{-1}
    • Speedometer: A device in vehicles (scooters, cars) that shows speed at a specific instant.

  • Approximate Speed of Common Objects:

    • Man walking: 1 m s−11\,m\,s^{-1} (3.6 km h−13.6\,km\,h^{-1})
    • Man running: 5 m s−15\,m\,s^{-1} (18 km h−118\,km\,h^{-1})
    • Bicycle: 7 m s−17\,m\,s^{-1} (25 km h−125\,km\,h^{-1})
    • Scooter: 11 m s−111\,m\,s^{-1} (40 km h−140\,km\,h^{-1})
    • Car: 14 m s−114\,m\,s^{-1} (50 km h−150\,km\,h^{-1})
    • Train: 17 m s−117\,m\,s^{-1} (60 km h−160\,km\,h^{-1})
    • Sound: 330 m s−1330\,m\,s^{-1} (1188 km h−11188\,km\,h^{-1})
    • Light: 3×108 m s−13 \times 10^8\,m\,s^{-1} (1.08×109 km h−11.08 \times 10^9\,km\,h^{-1})

Solved Examples

  • Example 1: Box Volume. Dimensions: 2.4 m×1.0 m×75 cm2.4\,m \times 1.0\,m \times 75\,cm.
    • Convert 75 cm=0.75 m75\,cm = 0.75\,m.
    • V=2.4×1.0×0.75=1.8 m3V = 2.4 \times 1.0 \times 0.75 = 1.8\,m^3.
  • Example 2: Book Volume. Dimensions: 24 cm,15 cm,1 cm24\,cm, 15\,cm, 1\,cm.
    • (a) V=24×15×1=360 cm3V = 24 \times 15 \times 1 = 360\,cm^3.
    • (b) Convert to m3m^3: 360×10−6 m3=3.6×10−4 m3360 \times 10^{-6}\,m^3 = 3.6 \times 10^{-4}\,m^3.
  • Example 3: Copper Piece Volume. V1=22 mL,V2=30 mLV_1 = 22\,mL, V_2 = 30\,mL.
    • (a) V=30−22=8 mL=8 cm3V = 30 - 22 = 8\,mL = 8\,cm^3.
    • (b) V=8×10−6 m3V = 8 \times 10^{-6}\,m^3.
  • Example 4: Circular Park Area. Diameter = 30 m30\,m, Radius = 15 m15\,m.
    • Area=πr2=3.14×(15)2=706.5 m2Area = \pi r^2 = 3.14 \times (15)^2 = 706.5\,m^2.
  • Example 5: Irregular Lamina. 14 complete squares, 11 more-than-half squares.
    • n=14+11=25n = 14 + 11 = 25.
    • Area=25×1 cm2=25 cm2Area = 25 \times 1\,cm^2 = 25\,cm^2.
  • Example 6: Iron Density. Volume = 25 cm325\,cm^3, Mass = 195 g195\,g.
    • (a) d=19525=7.8 g cm−3d = \frac{195}{25} = 7.8\,g\,cm^{-3}.
    • (b) Convert: 195 g=0.195 kg195\,g = 0.195\,kg; V=25×10−6 m3V = 25 \times 10^{-6}\,m^3.

  1. Measurement of Volume:
    a. What instrument is used to measure the volume of irregular objects?
    b. Describe the procedure for using a measuring cylinder to measure liquids.

  2. Volume Calculation of Regular Objects:
    a. What is the formula for the volume of a sphere?
    b. Calculate the volume of a cube with a side length of 4 cm.

  3. Density Concepts:
    a. Define density and provide the formula for calculating it.
    b. Explain why density does not change with the shape of an object.

  4. Density Determination of Different States:
    a. What are the steps to determine the density of an irregular solid using the displacement method?
    b. If a piece of metal has a mass of 150 g and a volume of 50 cm³, what is its density?

  5. Application Questions:
    a. Why do materials like iron sink in water while materials like cork float?
    b. Calculate the area of a rectangle with a length of 10 m and a breadth of 5 m.