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### Measurement of Area:

#### Area of Irregular Shape:

- The area of an irregular object can be obtained by drawing outline of the shape of the object on graph paper of 1 square centimeter marked square.
- The number of complete squares is counted.
- Next squares more than half are also counted as a complete square.
- Squares less than half are left and not counted.
- The sum of all the squares counted gives the area of the shape in square centimetres.
- In above figure, the area of the shape is 51 square units.

#### Area of Regular Shapes:

- Area of a triangle = 1/2 x base x height
- Area of a square = side²
- Area of a rectangle – length x breadth
- Area of a circle = π(radius)² = π/4 x (diameter)²
- Area of a parallelogram = base x height
- Area of a rhombus = 1/2 x product of diagonals
- Area of trapezium = 1/2 x sum of parallel sides x Distance between parallel lines

#### Units of Area:

- S.I. unit of area is m² and read as a square metre.
- c.g.s unit of area is cm² and read as square centimeters.
- Land areas are bigger hence the area of land is measured in a hectare. 1 hectare = 10000 m²

#### Example – 1:

- An artist designed two postage stamps. One was square and measured 2 cm X 2 cm. The Other was rectangular and measured 3 cm X 2 cm. Before the stamp was issued, it was decided to double the length of each side of a square stamp and triple the length of each side of a rectangular stamp. By how many times will the area of the paper required for each stamp increase.
**Solution:****Square Stamp:**

Initial area of square stamp = 2 cm 2 cm = 4 cm².

New area of square stamp = (2 x 2)cm x (2 x 2) cm = 16 cm².

The area of the paper for square stamp will increase by (16 cm²./4 cm²) 2 times

**Rectangular Stamp:**

Initial area of rectangular stamp = 2 cm x 3 cm = 6 cm².

New area of rectangular stamp = (3 x 2)cm (3 x 3) cm = 54 cm².

The area of the paper for rectangular stamp will increase by(54 cm²./6 cm²) 9 times

#### Example – 2:

- The length of a school compound is 500 m and the breadth is 120 m. Find the area of the school in hectares.
**Solution:**

Area = Length Breadth = 500 x 120 = 60000 m²

Now 1 hectare = 10000 m²

The are of school = (60000 m² / 60000 m²) = 6 hectares,

### Measurement of Volume:

- Volume is the space occupied by an object.
- The S.I. unit of volume is m³. Its c.g.s. unit is c.c. i.e cm³. Other practical units are millilitre (ml), litre (l) etc.
- The volume of regular bodies can be obtained by measuring necessary dimension for the calculation of the volume.

#### Measurement of Volume of Regular Solids:

- Volume of rectangular parallelepiped = length x breadth x height
- Volume of cube = side³.
- Volume of cylinder = πr²h
- Volume of cone = 1/3 x πr²h

#### Volume of Liquids:

- Measurement of volume of a liquid is done using following apparatus

#### a) Measuring Jar:

- It is graduated glass jar marked in ml from bottom to top. It is used to measure required amount of liquid.

#### b) Measuring Flask:

- It has one mark etched at the neck and the liquid of only volume of its capacity can be measured.

#### c) Pipette:

- It is filled by sucking the liquid into it. It is used to take fixed amount of liquid.

#### d) Burette:

- It is a tube similar to measuring jar, with a pitch cock provided at the bottom. It is used to take desired amount of liquid.Precaution:
- Read from the bottom of the curved surface of the liquid called meniscus.

### Measurement of volume using Measuring jar by the direct method:

- Pour the liquid in the measuring jar carefully and fill the jar with liquid as per required quantity by reading the meniscus of the liquid in the measuring jar.

### Measurement of Volume of Irregular Solids:

#### Solid Insoluble in Water but Heavier Than Water:

- Take some water in the measuring jar. Note the volume V
_{1}cm³ of this water. - Immerse the solid carefully and completely in water.
- Note the volume V
_{2}cm³ of water with solid in it. - Calculate increase in volume = V
_{2}– V_{1} - Then volume of solid = (V
_{2}– V_{1}) cm³.

#### Solid Insoluble in Water but Lighter Than Water:

- For this sinker (heavy body) has to be used.
- Take some water in the measuring jar. Note the volume V
_{1}cm³ of this water when the sinker is in water. - Tie sinker to the solid and immerse the solid with sinker carefully and completely in water.
- Note the volume V
_{2}cm³ of water with solid and sinker in it. - Calculate increase in volume = V
_{2}– V_{1} - Then volume of solid = (V
_{2}– V_{1}) cm³. **Note:**For solids soluble in water a liquid other than water, in which the solid is insoluble is used.

#### To Find Average Volume of a single Drop of water:

- Take clean burette with water. Clamp it upright. Remove any air bubbles formed.
- Now allow the water to trickle slowly drop by drop in measuring jar.
- count the number of drops collected in the measuring jar.
- Measure the volume of water in the measuring jar and use following formula to find the volume of a drop of water.
- Volume drop = Volume of water collected / No. of drops

#### To Find Average Volume of Lead Shot:

- Take some water in the measuring jar. Note the volume V
_{1}cm³ of this water. - Drop ‘n’ numbers of lead shot in the measuring jar carefully.
- Note the volume V
_{2}cm³ of water with lead shots in it. - Calculate increase in volume = V
_{2}– V_{1} - Volume of each drop = Increase in volume / No. of lead shots (n)

#### Example – 1:

- A stone X is immersed in water taken in a measuring jar. The water level rises from 60 cm³ to 78 cm³. If a second stone Y is immersed in water, the level rises from 60 cm³ to 86 cm³. What is the rise in the level of water when stone X and Y are immersed simultaneously? What is reading in the measuring jar?
**Solution:**

Volume of stone X = 78 cm³ – 60 cm³ = 18 cm³.

Volume of stone Y = 86 cm³ – 60 cm³ = 26 cm³.

Total Volume of stone X and stone Y = 18 cm³ + 26 cm³ = 44 cm³.

Rise in the level of water in measuring flask = 44 cm³.

The level of water in the measuring jar after immersing both the stones = 60 cm³ + 44 cm³= 104 cm³.

Rise in the level of water in measuring flask = 44 cm³. and Reading of measuring jar = 104 cm³

#### Example – 2:

- Two identical lead balls are immersed into a measuring jar containing water, the level of water raises from 16 cm³ to 26 cm³. If one of the above lead ball and another metallic ball are immersed simultaneously, then the level of water in measuring jar raises from 20 cm³ to 40 cm³. Find the volume of the lead ball and metallic ball.
**Solution:**

Volume of two lead balls = 26 cm³ – 16 cm³ = 10 cm³.

Thus, volume of each lead ball = 10 cm³/2 = 5 cm³.

Volume of one lead ball + Volume of metallic ball = 40 cm³ – 20 cm³ = 20 cm³.

Volume of metallic ball = 20 cm³ – Volume of one lead ball = 20 cm³ – 5 cm³ = 15 cm³

Hence the volume of lead ball is 5 cm3 and that of metallic ball is 15 cm³.

#### Example – 3:

- When 20 drops of water are allowed to collect in a measuring jar containing water, the level of water raises from 26 cm³ to 28 cm³. Find the average volume of each drop.
**Solution:**

Volume of 20 water drops = 28 cm³ – 26 cm³ = 2 cm³.

Volume of each drop = Volume of 20 water drops / 20 = 2 cm³ / 20 = 0.1 cm³

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