9.6-Volume and Capacity

9.6-Volume and Capacity Important Formulae

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9.6 - Volume and Capacity
  • Volume is the amount of space occupied by a 3D object.
  • Capacity is the amount of liquid a container can hold.
  • Formula for volume of a cube: $V = a^3$ where $a$ is the side length.
  • Formula for volume of a cuboid: $V = l \times b \times h$ where $l$, $b$, and $h$ are the length, breadth, and height.
  • Formula for volume of a cylinder: $V = \pi r^2 h$ where $r$ is radius and $h$ is height.
  • Capacity is typically measured in liters (L) or milliliters (mL).

9.6 - Volume and Capacity

Volume and capacity are important concepts in mensuration. Volume refers to the amount of space occupied by a three-dimensional object, while capacity is the amount of liquid a container can hold. Both are measured in cubic units and are essential for understanding geometric shapes in the real world.

1. Volume of a Cuboid

The volume of a cuboid (rectangular box) is calculated by multiplying its length, width, and height. The formula is:

Formula: $V = l \times w \times h$

Where:

  • $V$ is the volume of the cuboid
  • $l$ is the length of the cuboid
  • $w$ is the width of the cuboid
  • $h$ is the height of the cuboid

2. Volume of a Cube

A cube is a special case of a cuboid where all sides are of equal length. The volume of a cube is calculated by cubing the length of one side. The formula is:

Formula: $V = a^3$

Where:

  • $V$ is the volume of the cube
  • $a$ is the length of a side of the cube

3. Volume of a Cylinder

The volume of a cylinder is determined by the area of its circular base multiplied by its height. The formula is:

Formula: $V = \pi r^2 h$

Where:

  • $V$ is the volume of the cylinder
  • $r$ is the radius of the circular base
  • $h$ is the height of the cylinder
  • $\pi$ is approximately 3.1416

4. Volume of a Cone

The volume of a cone is one-third of the product of the area of the base and the height. The formula is:

Formula: $V = \frac{1}{3} \pi r^2 h$

Where:

  • $V$ is the volume of the cone
  • $r$ is the radius of the base
  • $h$ is the height of the cone
  • $\pi$ is approximately 3.1416

5. Volume of a Sphere

The volume of a sphere is determined by the formula:

Formula: $V = \frac{4}{3} \pi r^3$

Where:

  • $V$ is the volume of the sphere
  • $r$ is the radius of the sphere
  • $\pi$ is approximately 3.1416

6. Capacity

Capacity refers to the amount of liquid a container can hold. It is commonly measured in liters (L) or milliliters (mL) for smaller volumes. One liter is equal to 1000 milliliters.

When converting between volume and capacity, it's important to remember that the capacity of a container is generally expressed in terms of its internal volume, and liquids occupy space in the same way as solid objects. Therefore, the formulae for volume can also be used for calculating the capacity of containers.

7. Relation between Volume and Capacity

The volume of an object is often calculated in cubic units (like cubic meters or cubic centimeters), whereas capacity is usually measured in liters or milliliters. The conversion between the two depends on the context, such as the shape of the object and the material being measured (solid or liquid).

8. Units of Measurement

The standard units for measuring volume and capacity are:

  • Cubic meter (m³)
  • Cubic centimeter (cm³ or cc)
  • Liter (L)
  • Milliliter (mL)

To convert between these units, it is important to remember:

  • 1 liter = 1000 milliliters
  • 1 cubic meter = 1000 liters
  • 1 cubic centimeter = 1 milliliter

9.6-Volume and Capacity

इस खंड में हम आयतन (Volume) और क्षमता (Capacity) के बारे में चर्चा करेंगे। आयतन और क्षमता, दोनों ही 3D (त्रैतीयक) आकृतियों की माप को दर्शाते हैं।

आयतन उस स्थान का माप होता है, जो एक वस्तु या आकृति अपने भीतर घेरती है। वहीं, क्षमता किसी कंटेनर में जितनी सामग्री (जैसे पानी) समा सकती है, उसे हम क्षमता कहते हैं।

आयतन और क्षमता के बीच अंतर यह है कि आयतन वस्तु के भौतिक रूप को दर्शाता है, जबकि क्षमता उस वस्तु के अंदर समाने वाली सामग्री को मापता है।

आयतन के लिए सूत्र (Volume Formulas)

1. घन (Cube): यदि घन की भुजा $a$ हो, तो घन का आयतन $V$ होगा:

$$ V = a^3 $$

2. द्रव्यमान (Cuboid): यदि आयताकार घन के आयाम $l$, $b$, और $h$ (लंबाई, चौड़ाई, और ऊंचाई) हों, तो इसका आयतन $V$ होगा:

$$ V = l \times b \times h $$

3. गोलार्द्ध (Cylinder): गोलार्द्ध का आयतन $V$ निम्नलिखित सूत्र से दिया जाता है, जहां $r$ गोलार्द्ध का त्रिज्या और $h$ उसकी ऊंचाई है:

$$ V = \pi r^2 h $$

4. गोलक (Sphere): गोलक का आयतन $V$ निम्नलिखित सूत्र से दिया जाता है, जहां $r$ गोलक का त्रिज्या है:

$$ V = \frac{4}{3} \pi r^3 $$

5. शंकु (Cone): शंकु का आयतन $V$ निम्नलिखित सूत्र से दिया जाता है, जहां $r$ शंकु का त्रिज्या और $h$ उसकी ऊंचाई है:

$$ V = \frac{1}{3} \pi r^2 h $$
क्षमता (Capacity)

क्षमता, किसी वस्तु के अंदर समाने वाली सामग्री को मापने के लिए होती है। क्षमता की माप में सामान्यत: लीटर या मिलीलिटर का उपयोग किया जाता है। 1 क्यूबिक सेंटीमीटर (cm³) = 1 मिलिलीटर (ml) और 1 क्यूबिक मीटर (m³) = 1000 लीटर के बराबर होता है।

आयतन और क्षमता के बीच संबंध

किसी वस्तु का आयतन और उसकी क्षमता लगभग समान होते हैं, क्योंकि दोनों ही वस्तु के अंदर समाने वाले स्थान का माप करते हैं। यदि हमें किसी घन के आयतन को लीटर में बदलना हो, तो हमें आयतन को 1000 से विभाजित करना होता है, क्योंकि 1 क्यूबिक मीटर में 1000 लीटर होते हैं।

उदाहरण:

1. यदि एक घन का आयतन 5000 क्यूबिक सेंटीमीटर है, तो उसकी क्षमता कितनी होगी?

यहां, 1 क्यूबिक सेंटीमीटर = 1 मिलिलीटर, तो 5000 क्यूबिक सेंटीमीटर = 5000 मिलिलीटर = 5 लीटर।

2. यदि एक गोलार्द्ध का त्रिज्या 7 मीटर है और उसकी ऊंचाई 10 मीटर है, तो उसका आयतन क्या होगा?

यहां, गोलार्द्ध का आयतन $V = \pi r^2 h$ होता है।

$$ V = \pi \times 7^2 \times 10 = 1540 \, \text{cubic meters} $$

इस प्रकार, इस प्रकार के गणना से हम किसी भी आकृति का आयतन और क्षमता निर्धारित कर सकते हैं।

Given a cylindrical tank, in which situation will you find surface area and in which situation volume.

(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.

Solution:

Given a cylindrical tank, in which situation will you find surface area and in which situation volume?

(a) To find how much it can hold: This involves finding the volume of the cylindrical tank. The volume of a cylinder is given by the formula $V = \pi r^2 h$, where $r$ is the radius of the base and $h$ is the height.

(b) Number of cement bags required to plaster it: This involves finding the surface area of the cylindrical tank. The surface area of a cylinder is given by the formula $A = 2 \pi r (r + h)$, where $r$ is the radius of the base and $h$ is the height.

(c) To find the number of smaller tanks that can be filled with water from it: This involves finding the volume of the cylindrical tank. The volume of a cylinder is given by the formula $V = \pi r^2 h$, where $r$ is the radius of the base and $h$ is the height.

Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Solution:

Diameter of Cylinder A is 7 cm, and the height is 14 cm. Diameter of Cylinder B is 14 cm and height is 7 cm. Without doing any calculations, can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

To compare the volumes and surface areas of Cylinder A and Cylinder B, we use the formulas for volume and surface area of a cylinder:

Volume of a Cylinder: $V = \pi r^2 h$

Surface Area of a Cylinder: $A = 2\pi r(h + r)$

Cylinder A:

  • Diameter = 7 cm, so radius $r_A = \frac{7}{2} = 3.5$ cm
  • Height $h_A = 14$ cm
  • Volume of Cylinder A: $V_A = \pi (3.5)^2 \times 14$
  • Surface Area of Cylinder A: $A_A = 2\pi \times 3.5 \times (14 + 3.5)$

Cylinder B:

  • Diameter = 14 cm, so radius $r_B = \frac{14}{2} = 7$ cm
  • Height $h_B = 7$ cm
  • Volume of Cylinder B: $V_B = \pi (7)^2 \times 7$
  • Surface Area of Cylinder B: $A_B = 2\pi \times 7 \times (7 + 7)$

Find the height of a cuboid whose base area is 180 cm$^2$ and volume AB is 900 cm$^3$?

Solution:

Find the height of a cuboid whose base area is 180 cm2 and volume AB is 900 cm3

Given:

  • Base area of the cuboid = 180 cm2
  • Volume of the cuboid = 900 cm3

The formula for the volume of a cuboid is:

Volume = Base area × Height

Using the formula, we have:

900 = 180 × Height

To find the height, divide both sides of the equation by 180:

Height = $\frac{900}{180}$

Height = 5 cm

A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

Solution:

1. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

To find the number of small cubes that can be placed in the given cuboid, we first need to calculate the volume of the cuboid and the volume of a small cube.

The volume of the cuboid is given by:

Volume of cuboid = length × width × height = $60 \, \text{cm} \times 54 \, \text{cm} \times 30 \, \text{cm}$

Volume of cuboid = $60 \times 54 \times 30 = 108000 \, \text{cm}^3$

The volume of a small cube with side 6 cm is:

Volume of small cube = side³ = $6 \, \text{cm} \times 6 \, \text{cm} \times 6 \, \text{cm}$

Volume of small cube = $6^3 = 216 \, \text{cm}^3$

Now, the number of small cubes that can be placed in the cuboid is:

Number of small cubes = $\frac{\text{Volume of cuboid}}{\text{Volume of small cube}} = \frac{108000}{216}$

Number of small cubes = $500$

Find the height of the cylinder whose volume is 1.54 m$^3$ and diameter of the base is 140 cm?

Solution:

1. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm.

The formula for the volume of a cylinder is given by:

$ V = \pi r^2 h $

Where:

  • $V$ = Volume of the cylinder = 1.54 m³
  • $r$ = Radius of the base of the cylinder
  • $h$ = Height of the cylinder
  • Diameter of the base = 140 cm = 1.4 m (since 100 cm = 1 m)

Radius, $r = \frac{Diameter}{2} = \frac{1.4}{2} = 0.7$ m

Now, substitute the values in the formula:

$ 1.54 = \pi (0.7)^2 h $

We know that $ \pi \approx 3.1416 $

$ 1.54 = 3.1416 \times 0.49 \times h $

$ 1.54 = 1.539 \times h $

Now, solve for $h$:

$ h = \frac{1.54}{1.539} \approx 1$ m

The height of the cylinder is approximately 1 meter.

A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?

Solution:

Question:

A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank.

Solution:

The formula for the volume of a cylinder is given by:

Volume of cylinder $V = \pi r^2 h$,

where $r$ is the radius and $h$ is the height (length) of the cylinder.

Given: $r = 1.5 \, \text{m}$ and $h = 7 \, \text{m}$

Substitute the values into the formula:

$V = \pi (1.5)^2 \times 7$

$V = \pi \times 2.25 \times 7$

$V = \pi \times 15.75$

Using $\pi \approx 3.14$:

$V = 3.14 \times 15.75$

$V \approx 49.395 \, \text{m}^3$

Now, to convert cubic meters to litres, we multiply by 1000:

Quantity of milk = $49.395 \times 1000 = 49395 \, \text{litres}$

Therefore, the quantity of milk that can be stored in the tank is approximately 49395 litres.

If each edge of a cube is doubled,

(i) how many times will its surface area increase?
(ii) how many times will its volume increase?

Solution:

1. (i) How many times will its surface area increase?

The surface area of a cube is given by the formula $A = 6a^2$, where $a$ is the length of the edge of the cube. If the edge is doubled, the new edge length becomes $2a$. The new surface area will be:

$A_{new} = 6(2a)^2 = 6 \times 4a^2 = 24a^2$

So, the surface area increases by a factor of $24a^2 / 6a^2 = 4$ times.

1. (ii) How many times will its volume increase?

The volume of a cube is given by the formula $V = a^3$, where $a$ is the length of the edge of the cube. If the edge is doubled, the new edge length becomes $2a$. The new volume will be:

$V_{new} = (2a)^3 = 8a^3$

So, the volume increases by a factor of $8a^3 / a^3 = 8$ times.

Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m$^3$, find the number of hours it will take to fill the reservoir.

Solution:

Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m³, find the number of hours it will take to fill the reservoir.

Given:

  • Rate of water pouring = 60 litres per minute
  • Volume of the reservoir = 108 m³

1 m³ = 1000 litres

So, the volume of the reservoir in litres = $108 \times 1000 = 108000$ litres

Now, the time taken to fill the reservoir can be calculated as:

Time = $\frac{\text{Volume of the reservoir}}{\text{Rate of water pouring}}$

Time = $\frac{108000 \text{ litres}}{60 \text{ litres per minute}} = 1800 \text{ minutes}$

Convert time into hours:

Time in hours = $\frac{1800}{60} = 30$ hours