volume and capacity problem solving

VOLUME AND CAPACITY WORD PROBLEMS

The units for capacity and the units for volume are closely related.

1 mL of fluid will fill a cube 1 cm x 1 cm x 1 cm.

1 cm 3 has capacity 1 mL

1 L of fluid will fill a cube 10 cm x 10 cm x 10 cm.

1000 cm 3 has capacity 1 L

1 kL of fluid will fill a cube 1 m x 1 m x 1 m.

1 m 3 has capacity 1 kL.

1 cm 3  = 1 mL

1000 cm 3  = 1 L

1 m 3  = 1 KL

Example 1 :

Calculate the capacity of the container:

Volume of the container = 30 x 10 x 20 cm 3

= 6000 cm 3 

1000 ml = 1 L

= (6000/1000) L

So, the capacity of the container is 6 L.

Example 2 :

Find the capacity in liters of a fish tank 2 m by 1 m and 50 cm.

Volume of tank = l x w x h

Substitute l = 2, w = 1 and h = 50 cm or 0.5 m.

= 2 x 1 x 0.50

So, capacity of the tank is 1 KL.

Example 3 :

A rectangular petrol tank has dimensions 50 cm by 40 cm by 25 cm. How many liters of petrol are needed to fill it?

Volume of rectangular tank  =  l x w x h.

Substitute l = 50, w = 40 and h = 25.

= 50 x 40 x 25

= 50000 cm 3

= 50000/1000

So, the capacity of the petrol tank is 50 L.

Example 4 :

A water trough has triangular cross-section as shown. Its length is 2 m.

a) the area of the triangle in cm 2

b) the volume of space in the trough in cm 3

c) the capacity of the trough in :

i) litres   ii) kilolitres.

Area of triangle :

= (1/2)  ⋅ base  ⋅ height

Base = 60 cm and height  =  50 cm

= (1/2) ⋅ 60 ⋅ 50

= 1500 cm 3

Volume of space :

=  (1/2)  ⋅   Base area x height

= (1/2)  ⋅ 1500  ⋅ 200

= 300000 cm 3

= 300000/1000

= 300 Liter

So, capacity of triangular prism is 300 liter.

1m 3  = 1 KL

1000 ml = 1 kl

So, capacity of triangular prism is 0.3 kl.

Example 5 :

Find the capacity in megaliters of a reservoir with a surface area of 1  hectare and an average depth of 2.5 meters.

1 hectare = 10000 m 2

Volume = Surface area x height

= 10000 x 2.5

= 25000 m 3

1 m 3  = 1 KL

Capacity = 25000/1000 ML

So, the capacity is 25 ML.

Example 6 :

A kidney-shaped swimming pool has surface area 15 m 2 and a constant depth of 2 meters. Find the capacity of the pool in kiloliters.

Surface area = 15 m 2  and height = 2 m

So, the capacity is  30 KL .

Example 7 :

A lake has an average depth of 6 m and a surface area of 35 ha. Find its capacity in megaliters .

35 hectare = 350000 m 2

Height = 6 m

= 350000 x 6

= 2100000  m 3

= 2100000 Kl

1 ML = 1000 KL

= 2100000/1000 ML

So, the capacity of the tank is  2100 ML.

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Volume and Capacity

What is volume .

Volume is the amount of space that a three-dimensional object occupies. It is measured in cubic units , such as cubic centimeters or cubic meters.

How to Calculate Volume

The formula for finding the volume of a rectangular prism is:

Volume = length × width × height

For example, if the length of a rectangular prism is 5 cm, the width is 3 cm, and the height is 2 cm, then the volume would be:

Volume = 5 cm × 3 cm × 2 cm = 30 cubic centimeters

What is Capacity ?

Capacity is the measure of how much a container can hold. It is also measured in cubic units , such as liters or milliliters.

How to Calculate Capacity

The formula for finding the capacity of a container is the same as finding the volume of a rectangular prism . You multiply the length , width , and height of the container to find its capacity .

Conversion between Units

1 liter = 1000 milliliters

1 cubic meter = 1000 liters

Study Guide

• Understand the difference between volume and capacity .
• Learn to calculate volume and capacity using the appropriate formulas .
• Practice converting between different units of volume and capacity .
• Work on real-world problems involving volume and capacity , such as filling containers with liquid or comparing the sizes of different objects.

Read More...

◂ Math Worksheets and Study Guides Fifth Grade. Volume and Capacity

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5th Grade Volume Worksheets

Welcome to our 5th Grade Volume Worksheets page.

Here you will find our collection of worksheets to introduce and help you learn about volume.

These worksheets will help you to understand and practice how to find the volume of rectangular prisms and other simple shapes.

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Volume of Rectangular Prisms

On this webpage you will find our range of worksheets to help you work out the volume of simple 3d shapes such as rectangular prisms.

They are also very useful for introducing the concept of volume being the number of cubes that fill up a space.

These sheets are graded from easiest to hardest, and each sheet comes complete with answers.

Using these sheets will help your child to:

• learn how to find the volume of simple 3d shapes by counting cubes;
• learn how to find the volume of rectangular prisms by multiplying length x width x height
• practice using their knowledge to solve basic volume problems.

What is Volume?

• Volume is the amount of space that is inside a 3 dimensional shape.
• Because it is an amount of space, it has to be measured in cubes.
• If the shape is measured in cm, then the volume would be measured in cubic cm or cm 3
• If the shape is measured in inches, then the volume would be measured in cubic inches or in 3

Volume of a Rectangular Prism

• The volume of a rectangular prism is the number of cubes it is made from.
• To find the number of cubes, we need to multiply the length by the width by the height.
• So Volume = length x width x height or l x w x h.
• We could also multiply the area of the base (which is the length x width) by the height.
• So Volume = l x w x h or b x h (where b is the area of the base)

In the example above, the length is 3, the width is 6 and the height is 2.

So the volume is 3 x 6 x 2 = 36cm 3 or 36 cubic cm.

This tells us that there are 36 cm cubes that make up the shape.

We have split our worksheets up into different sections, to make it easier for you to select the right sheets for your needs.

• Section 1 - Find the Volume by Counting Cubes
• Section 2 - Finding the Volume by multiplication
• Section 3 - Match the Volume (multiplication)
• Section 3 - Volume Problem Solving Riddles

5th Grade Volume Worksheets - Counting Cubes

• Volume - Count the Cubes Sheet 1
• PDF version
• Volume - Count the Cubes Sheet 2

5th Grade Volume Worksheets - Find the Volume by Multiplication

The first sheet is supported, the other two sheets are more independent.

You can choose between the standard or metric versions of sheets 2 and 3 (the measurements are the same)

• Find the Volume Sheet 1 (supported)
• Find the Volume Sheet 2 (standard)
• Find the Volume Sheet 2 (metric)
• Find the Volume Sheet 3 (standard)
• Find the Volume Sheet 3 (metric)

5th Grade Volume Worksheets - Match the Volume

• Match the Volume Sheet 1
• Match the Volume Sheet 2

5th Grade Volume Worksheets - Volume Riddles

• Volume Riddles Sheet 5A
• Volume Riddles Sheet 5B

Volume of Rectangular Prisms Walkthrough Video

This short video walkthrough shows several problems from our Find the Volume Sheet 2 being solved and has been produced by the West Explains Best math channel.

If you would like some support in solving the problems on these sheets, check out the video below!

More Recommended Math Worksheets

Take a look at some more of our worksheets similar to these.

Volume of a Cube/Cuboid/Box Calculators

Each of the pages below includes a handy calculator to help you find the volume of cubes, cuboids and boxes.

• Volume of a Cube Calculator

• Volume of a Box Calculator

Converting Measures Worksheets

Here is our selection of converting units of measure for 3rd to 5th graders.

These sheets involve converting between customary units of measure and also metric units.

• Converting Customary Units Worksheets
• Metric Conversion Worksheets

5th Grade Geometry Worksheets

Here is our selection of 5th grade Geometry worksheets about angles.

The focus on these sheets is angles on a straight line, angles around a point and angles in a triangle.

• 5th Grade Geometry Missing Angles

Area Worksheets

Here is our selection of free printable area worksheets for 3rd and 4th grade.

The sheets are all graded in order from easiest to hardest.

• work out the areas of a range of rectangles;
• find the area of rectilinear shapes.
• Area Worksheets - Rectangles and Rectilinear Shapes
• Area of Right Triangles
• Area of Quadrilaterals Worksheets
• Perimeter Worksheets

Here is our selection of free printable perimeter worksheets for 3rd and 4th grade.

• work out the perimeter of a range of rectangles;
• find the perimeter of rectilinear shapes.

All the math practice worksheets in this section support Elementary Math Benchmarks.

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Volume Problem Solving

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To solve problems on this page, you should be familiar with the following: Volume - Cuboid Volume - Sphere Volume - Cylinder Volume - Pyramid

This wiki includes several problems motivated to enhance problem-solving skills. Before getting started, recall the following formulas:

• Volume of sphere with radius \(r:\) \( \frac43 \pi r^3 \)
• Volume of cube with side length \(L:\) \( L^3 \)
• Volume of cone with radius \(r\) and height \(h:\) \( \frac13\pi r^2h \)
• Volume of cylinder with radius \(r\) and height \(h:\) \( \pi r^2h\)
• Volume of a cuboid with length \(l\), breadth \(b\), and height \(h:\) \(lbh\)

Volume Problem Solving - Basic

Volume - problem solving - intermediate, volume problem solving - advanced.

This section revolves around the basic understanding of volume and using the formulas for finding the volume. A couple of examples are followed by several problems to try.

Find the volume of a cube of side length \(10\text{ cm}\). \[\begin{align} (\text {Volume of a cube}) & = {(\text {Side length}})^{3}\\ & = {10}^{3}\\ & = 1000 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

Find the volume of a cuboid of length \(10\text{ cm}\), breadth \(8\text{ cm}\). and height \(6\text{ cm}\). \[\begin{align} (\text {Area of a cuboid}) & = l × b × h\\ & = 10 × 8 × 6\\ & = 480 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

I made a large ice cream cone of a composite shape of a cone and a hemisphere. If the height of the cone is 10 and the diameter of both the cone and the hemisphere is 6, what is the volume of this ice cream cone? The volume of the composite figure is the sum of the volume of the cone and the volume of the hemisphere. Recall the formulas for the following two volumes: \( V_{\text{cone}} = \frac13 \pi r^2 h\) and \( V_{\text{sphere}} =\frac43 \pi r^3 \). Since the volume of a hemisphere is half the volume of a a sphere of the same radius, the total volume for this problem is \[\frac13 \pi r^2 h + \frac12 \cdot \frac43 \pi r^3. \] With height \(h =10\), and diameter \(d = 6\) or radius \(r = \frac d2 = 3 \), the total volume is \(48\pi. \ _\square \)

Find the volume of a cone having slant height \(17\text{ cm}\) and radius of the base \(15\text{ cm}\). Let \(h\) denote the height of the cone, then \[\begin{align} (\text{slant height}) &=\sqrt {h^2 + r^2}\\ 17&= \sqrt {h^2 + 15^2}\\ 289&= h^2 + 225\\ h^2&=64\\ h& = 8. \end{align}\] Since the formula for the volume of a cone is \(\dfrac {1}{3} ×\pi ×r^2×h\), the volume of the cone is \[ \frac {1}{3}×3.14× 225 × 8= 1884 ~\big(\text{cm}^{2}\big). \ _\square\]

Find the volume of the following figure which depicts a cone and an hemisphere, up to \(2\) decimal places. In this figure, the shape of the base of the cone is circular and the whole flat part of the hemisphere exactly coincides with the base of the cone (in other words, the base of the cone and the flat part of the hemisphere are the same). Use \(\pi=\frac{22}{7}.\) \[\begin{align} (\text{Volume of cone}) & = \dfrac {1}{3} \pi r^2 h\\ & = \dfrac {1 × 22 × 36 × 8}{3 × 7}\\ & = \dfrac {6336}{21} = 301.71 \\\\ (\text{Volume of hemisphere}) & = \dfrac {2}{3} \pi r^3\\ & = \dfrac {2 × 22 × 216}{3 × 7}\\ & = \dfrac {9504}{21} = 452.57 \\\\ (\text{Total volume of figure}) & = (301.71 + 452.57) \\ & = 754.28.\ _\square \end{align} \]

Try the following problems.

Find the volume (in \(\text{cm}^3\)) of a cube of side length \(5\text{ cm} \).

A spherical balloon is inflated until its volume becomes 27 times its original volume. Which of the following is true?

Bob has a pipe with a diameter of \(\frac { 6 }{ \sqrt { \pi } }\text{ cm} \) and a length of \(3\text{ m}\). How much water could be in this pipe at any one time, in \(\text{cm}^3?\)

What is the volume of the octahedron inside this \(8 \text{ in}^3\) cube?

A sector with radius \(10\text{ cm}\) and central angle \(45^\circ\) is to be made into a right circular cone. Find the volume of the cone.

\[\] Details and Assumptions:

• The arc length of the sector is equal to the circumference of the base of the cone.

Three identical tanks are shown above. The spheres in a given tank are the same size and packed wall-to-wall. If the tanks are filled to the top with water, then which tank would contain the most water?

A chocolate shop sells its products in 3 different shapes: a cylindrical bar, a spherical ball, and a cone. These 3 shapes are of the same height and radius, as shown in the picture. Which of these choices would give you the most chocolate?

\[\text{ I. A full cylindrical bar } \hspace{.4cm} \text{ or } \hspace{.45cm} \text{ II. A ball plus a cone }\]

How many cubes measuring 2 units on one side must be added to a cube measuring 8 units on one side to form a cube measuring 12 units on one side?

This section involves a deeper understanding of volume and the formulas to find the volume. Here are a couple of worked out examples followed by several "Try It Yourself" problems:

\(12\) spheres of the same size are made from melting a solid cylinder of \(16\text{ cm}\) diameter and \(2\text{ cm}\) height. Find the diameter of each sphere. Use \(\pi=\frac{22}{7}.\) The volume of the cylinder is \[\pi× r^2 × h = \frac {22×8^2×2}{7}= \frac {2816}{7}.\] Let the radius of each sphere be \(r\text{ cm}.\) Then the volume of each sphere in \(\text{cm}^3\) is \[\dfrac {4×22×r^3}{3×7} = \dfrac{88×r^3}{21}.\] Since the number of spheres is \(\frac {\text{Volume of cylinder}}{\text {Volume of 1 sphere}},\) \[\begin{align} 12 &= \dfrac{2816×21}{7×88×r^3}\\ &= \dfrac {96}{r^3}\\ r^3 &= \dfrac {96}{12}\\ &= 8\\ \Rightarrow r &= 2. \end{align}\] Therefore, the diameter of each sphere is \[2\times r = 2\times 2 = 4 ~(\text{cm}). \ _\square\]

Find the volume of a hemispherical shell whose outer radius is \(7\text{ cm}\) and inner radius is \(3\text{ cm}\), up to \(2\) decimal places. We have \[\begin{align} (\text {Volume of inner hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × R^3\\ & = \dfrac {1 × 4 × 22 × 27}{2 × 3 × 7}\\ & = \dfrac {396}{7}\\ & = 56.57 ~\big(\text{cm}^{3}\big) \\\\ (\text {Volume of outer hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × r^3\\ & = \dfrac {1 × 4 × 22 × 343}{2 × 3 × 7}\\ & = \dfrac {2156}{7}\\ & = 718.66 ~\big(\text{cm}^{3}\big) \\\\ (\text{Volume of hemispherical shell}) & = (\text{V. of outer hemisphere}) - (\text{V. of inner hemisphere})\\ & = 718.66 - 56.57 \\ & = 662.09 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

A student did an experiment using a cone, a sphere, and a cylinder each having the same radius and height. He started with the cylinder full of liquid and then poured it into the cone until the cone was full. Then, he began pouring the remaining liquid from the cylinder into the sphere. What was the result which he observed?

There are two identical right circular cones each of height \(2\text{ cm}.\) They are placed vertically, with their apex pointing downwards, and one cone is vertically above the other. At the start, the upper cone is full of water and the lower cone is empty.

Water drips down through a hole in the apex of the upper cone into the lower cone. When the height of water in the upper cone is \(1\text{ cm},\) what is the height of water in the lower cone (in \(\text{cm}\))?

On each face of a cuboid, the sum of its perimeter and its area is written. The numbers recorded this way are 16, 24, and 31, each written on a pair of opposite sides of the cuboid. The volume of the cuboid lies between \(\text{__________}.\)

A cube rests inside a sphere such that each vertex touches the sphere. The radius of the sphere is \(6 \text{ cm}.\) Determine the volume of the cube.

If the volume of the cube can be expressed in the form of \(a\sqrt{3} \text{ cm}^{3}\), find the value of \(a\).

A sphere has volume \(x \text{ m}^3 \) and surface area \(x \text{ m}^2 \). Keeping its diameter as body diagonal, a cube is made which has volume \(a \text{ m}^3 \) and surface area \(b \text{ m}^2 \). What is the ratio \(a:b?\)

Consider a glass in the shape of an inverted truncated right cone (i.e. frustrum). The radius of the base is 4, the radius of the top is 9, and the height is 7. There is enough water in the glass such that when it is tilted the water reaches from the tip of the base to the edge of the top. The proportion of the water in the cup as a ratio of the cup's volume can be expressed as the fraction \( \frac{m}{n} \), for relatively prime integers \(m\) and \(n\). Compute \(m+n\).

The square-based pyramid A is inscribed within a cube while the tetrahedral pyramid B has its sides equal to the square's diagonal (red) as shown.

Which pyramid has more volume?

Please remember this section contains highly advanced problems of volume. Here it goes:

Cube \(ABCDEFGH\), labeled as shown above, has edge length \(1\) and is cut by a plane passing through vertex \(D\) and the midpoints \(M\) and \(N\) of \(\overline{AB}\) and \(\overline{CG}\) respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).

If the American NFL regulation football

has a tip-to-tip length of \(11\) inches and a largest round circumference of \(22\) in the middle, then the volume of the American football is \(\text{____________}.\)

Note: The American NFL regulation football is not an ellipsoid. The long cross-section consists of two circular arcs meeting at the tips. Don't use the volume formula for an ellipsoid.

Answer is in cubic inches.

Consider a solid formed by the intersection of three orthogonal cylinders, each of diameter \( D = 10 \).

What is the volume of this solid?

Consider a tetrahedron with side lengths \(2, 3, 3, 4, 5, 5\). The largest possible volume of this tetrahedron has the form \( \frac {a \sqrt{b}}{c}\), where \(b\) is an integer that's not divisible by the square of any prime, \(a\) and \(c\) are positive, coprime integers. What is the value of \(a+b+c\)?

Let there be a solid characterized by the equation \[{ \left( \frac { x }{ a } \right) }^{ 2.5 }+{ \left( \frac { y }{ b } \right) }^{ 2.5 } + { \left( \frac { z }{ c } \right) }^{ 2.5 }<1.\]

Calculate the volume of this solid if \(a = b =2\) and \(c = 3\).

• Surface Area

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Volume Word Problems

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Course: 3rd grade   >   Unit 13

Understanding volume (liters).

• Estimate volume (milliliters and liters)
• Word problems with volume
• Measurement: FAQ

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• Volume And Capacity Worksheets And Resources For Ks1 And Ks2 Maths

7 of the best volume and capacity worksheets and resources for KS1 and KS2 maths

It doesn't matter whether you're a 'glass half full' or 'glass half empty' kind of person, these primary maths resources will help your students get their fill of capacity and volume…

1 | Introduce, measure and compare capacity

These activity sheets have been created to match the small steps on the White Rose maths schemes of work, with questions that include varied fluency with reasoning with problem solving, and an additional sheet with extension activities.

Children are given a variety of pictorial examples to work with and questions to provoke deeper thinking to help them with the curriculum requirements of Year 1 Measurement to “Compare, describe and solve practical problems for: capacity and volume (for example, full/empty, more than, less than, quarter, half full, half).”

There are three separate worksheets for this, so click the links for introducing capacity , measuring capacity and comparing capacity .

2 | George’s Marvellous Medicine KS1 capacity lesson plan

Can children estimate, decant and measure their way to discovering the elusive formula for George’s Marvellous Medicine , asks Jonathan Lear?

That’s the subject of this KS1 lesson plan that lets them explore measurements of capacity using standard metric units.

Download it here.

3 | Capacity and volume factsheets and worksheets

On the BBC Skillswise page for measuring capacity you’ll find a brief introductory video on the topic, plus a collection of free printable factsheets and worksheets.

These cover everything from labels, instruments for measuring capacity and non-standard measures of capacity to a matching exercise, reading scales and choosing litres or millilitres.

Check this all out here.

4 | Cover volume and capacity without pouring and filling

Looking for something a bit different? Mike Askew has ideas for teaching this topic without all the usual tricks.

Check them out here.

5 | Estimating volume for Year 5

These volume worksheets provide extra challenge for Year 5 children, with a variety of volume problems spread across three sections, enabling you to use the whole sheet during a lesson or to select specific problems for different teaching sessions.

Plus, a separate answer sheet for all sections is included.

Get this resource here.

6 | Volume of cuboids worksheet

This three-page, 12-question worksheet for upper KS2 is a quick and easy way to check pupils’ knowledge on the topic.

Get the worksheet here and the answer sheet here .

7 | Nrich capacity problems

What happens when you pour the water from one of these glasses into the other? Obvious right? Or is it? Watch the video from Nrich and explore with your class what you’ve seen.

This one is called Pouring Problem , but there are loads more volume and capacity problems to try. Here are a few of our favourites:

• Multilink cubes
• Next size up
• Cuboid-in-a-box
• Double Your Popcorn, Double Your Pleasure
• The Big Cheese

Or check out the full selection here.

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Volume word problems

Thinking about volume and capacity.

These measurement word problems focus on volumes and capacities; students add / subtract / multiply / divide amounts measured in customary or metric units of volume. No mixing or converting of units is required.

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Worksheet on Word Problem on Measuring Capacity

Practice the questions given in the worksheet on word problem on measuring capacity (i.e. addition and subtraction). Addition and subtraction of word problems in liters and milliliters is done in the similar way as in the case of ordinary numbers.

Solve the following word problems on addition and subtraction of Capacity:

1. A milkman sold 46 l 200 ml of milk on 3 days of a week and 53 l 195 ml of milk in next 2 days. What quantity of milk did he sell in 5 days?

2. Gary the milkman sells 35 l 800 ml of milk in the morning and 24 l 500 ml of milk in the evening. How much milk is he able to sell in a day?

3. Sara bought 500 ml of mustard oil, 250 ml of coconut oil and 2 l of refined oil. What is the total quantity of the 3 oils together?

4.  Ruby bought a 5 l Sunflower oil can from the market on Tuesday. By Friday only 2 l 440 ml of oil was left in it. How much quantity of oil was consumed in 3 days?

5. Alex consumed 15 l of water in 2 days and 27 litres of water in the rest of the days of the week. How much water was consumed in a week by him?

6. Gary the milkman delivers 1 l 250 ml of milk everyday at Mr. Jones house. What quantity of milk is delivered by the milkman to Mr. Jones house in a week?

7. Mary purchased 500 ml bottle of Pepsi, 300 ml can of Frooti and 2 litres of Limca. What quantity of cold drinks did Mary purchases altogether?

8. Shelly has 2 l of oil. She wants to pour it equally into 250 ml bottles. How many bottles is she able to fill with the oil?

9. Harry had 3 buckets of different capacity which could hold 9 l 257 ml, 12 l 420 ml and 30 l 100 ml of water. How much water could be stored in all?

10. Petrol costs $ 75 per liter. Tina gets her car tank filled with 46 liters of petrol. How much amount does she pays at the petrol pump.

11. There is 478 liters and 360 milliliters of water in the tank. 239 liters and 125 milliliters of water is consumed. How much water is left in the tank?

12. A swimming pool with a capacity of 200000 l. If there is 115730 l of water in the pool, how much more water can be put in the pool?

13. The capacity of the milk boiler is 2 l 500 ml of milk. If 1 l 200 ml of milk is put into the vessel then how much more quantity of milk can be filled in the vessel?

14. A cow gives 22 l 350 ml of milk each day. If the milk man has 20 cows then:

(i) How much milk will he get in a day?

(ii) If all the milk is filled in the bottles of capacity 500 ml, how many bottles will be required?

15. An oil can holds 5 l of oil. How much oil is left in the can if 2 l 750 ml of oil is used?

16. Neil has 2 cars, one with a capacity of 33 l 770 ml and other with a capacity of 42 l 550 ml. If petrol tanks of both the cars are empty, how much petrol can he fill in the tanks?

17. 35 l 450 ml of petrol was filled in the car and 29 l 561 ml of petrol was used in a month. How much petrol is left in the car?

18. Adrian has 44 l of liquid soap and wants to fill it in 25 Cans of capacity 550 ml each.

(i) Will he be able to fill all the Cans completely?

(ii) How much quantity of liquid soap will be left out?

19. In 1 ½ liter of cold drink bottle only 320 ml of cold drink is left. How much cold drink is consumed?

20. A water purifier cleans 100 l 150 ml of water each day. How much water will be cleaned by the cleaner in a week?

21. The consumption of diesel in a truck. A on one day is 102 l 208 ml and the consumption of truck B is 105 l 196 ml. Whose consumption is less and by how much?

22. Clara made 20 l 500 ml of lime squash on Saturday and 18 l 255 ml of lime squash on Sunday. On which day lime squash was prepared in more quantity and by how much?

23. Jug A is filled with 45 l 670 ml of orange juice and Jug B is filled with 67 l 890 ml of apple juice. Calculate the total litres of juice filled in both the jugs.

24. Sam filled 34 l 487 ml petrol in a truck, 23 l 790 ml in a car and 19 l 560 ml in another car. How much petrol he filled in all the three vehicles?

25. An oil tank has 162 l 345 ml oil. 78l 589 ml oil is removed from the tank. How much oil is left in the tank now?

26. There are two water tanks on a terrace of a building. Tank A has the capacity to hold 57 l 900 ml and tank B can hold 63 l 500 ml.

(i)Which tank can hold more and how much more it can hold?

(ii) What is the total capacity of both the tanks?

27. Factory A produces 146 l of juices and factory B produces 234l 467 ml of juices. Calculate the total litres of juices produced in both the factories?

Answers for the worksheet on word problem on measuring capacity (i.e. addition and subtraction in liters and milliliters) are given below.

1. 99 l 395 ml

2. 60 l 300 ml

3. 2 l 750 ml

4. 2 l 560 ml

6. 8 l 750 ml

7. 2 l 800 ml

9. 51 l 777 ml

11. 239 l 235 ml

12. 84270 l

13. 1 l 300 ml

14. (i) 447 l

15. 2 l 250 ml

16. 76 l 320 ml

17. 5 l 889 ml

18. (i) Yes

(ii) 30 l 250 ml

19. 1 l 180 ml

20. 701 l 50 ml

21. Truck A by 2 l 988 ml

22. Saturday by 2 l 245 ml

23. 113 ℓ 560 mℓ

24. 77 ℓ 837 mℓ

25. 83 ℓ 756 mℓ

26.  (i) Tank B, 5 ℓ 600 mℓ

(ii) 121 ℓ 400 mℓ

27.  380 ℓ 467 mℓ

Standard Unit of Capacity

Conversion of Standard Unit of Capacity

Addition of Capacity

Subtraction of Capacity

Addition and Subtraction of Measuring Capacity

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• Maths Questions

Volume Questions

Volume questions and answers are available in an easily understandable format along with required formulas. Students can practise questions on finding the volume of various solids provided here, and verify their solutions with the available answers. This is the best way to learn about various problem-solving techniques of solids in geometry.

What is volume?

Volume is the amount of space occupied by a solid shape in a three-dimensional plane or region. Some examples of solids include cubes, cuboids, spheres, cones, cylinders, etc.

• Volume of cube = a 3 , where a is the edge of the cube.
• Volume of cuboid = lbh, where l = length, b = breadth and h = height.
• Volume of sphere = (4/3) πr 3 , where r is the radius of the sphere.
• Volume of cylinder = πr 2 h, where r is the radius of the circular bases and h is the height.
• Volume of cone = (⅓) πr 2 h, where r is the radius of the circular base and h is the height.
• Volume of hemisphere = (2/3) πr 3 , where r is the radius of the sphere.
• Volume of frustum = πh/3 (R 2 + r 2 + Rr), where ‘R’ and ‘r’ are the radii of the base and top of the frustum.
• Volume of prism = Base Area x Height.
• Volume of pyramid = ⅓ (Area of base) (Height).

Click here to learn more about volume .

Volume Questions and Answers

1. Find the volume of a cuboidal box with dimensions 11 cm × 8 cm × 13 cm.

Given, the dimensions of a cuboidal box: 11 cm × 8 cm × 13 cm

Length = l = 11 cm

Breadth = b = 8 cm

Height = h = 13 cm

As we know, volume of a cuboid = lbh

= 11 × 8 × 13

Thus, the volume of the cuboidal box is 1144 cm 3 .

2. What is the volume of a sphere of diameter 21 units?

Diameter of the sphere = 21 units

Radius of the sphere = (21/2) units

Volume of a sphere = (4/3) πr 3

= (4/3) × (22/7) × (21/2) × (21/2) × (21/2)

Therefore, the volume of the sphere is 4851 cubic units.

3. A cuboidal block of wood was cut into eight equal cubes of edges 4 cm. Find the volume of the initial block of wood.

Edge of a cubical wood = 4 cm

Volume of the cube = a 3

= 4 × 4 × 4

Volume of 8 such cubes = 8 volume of one cube

Hence, the volume of the initial cuboidal block of wood is 512 cm 3 .

4. If three solid spherical beads of radii 3 cm, 4 cm, and 5 cm, respectively, are melted into one spherical bead, then find its radius in cm.

Radii of three solid spherical beads: 3 cm, 4 cm and 5 cm

As we know, the volume of a sphere = (4/3) πr 3

Let R be the radius of the new spherical bead, which is made by melting three spherical beads.

Volume of a new spherical bead = Sum of volumes of three spherical beads

(4/3)πR 3 = (4/3) × 3 × 3 × 3 + (4/3) × 4 × 4 × 4 + (4/3) × 5 × 5 × 5

(4/3)πR 3 = (4/3)π (27 + 64 + 125)

Hence, the radius of the spherical bead is 6 cm.

5. How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?

Dimensions of a brick = 25 cm x 11.25 cm x 6 cm

Dimensions of a wall = 8 m x 6 m x 22.5 cm

= 800 cm x 600 cm x 22.5 cm

Number of bricks = volume of the wall/ volume of a brick

= (800 x 600 x 22.5) / (25 x 11.25 x 6)

6. Find the depth of the cylindrical tank if its capacity is 1848 m 3 and the diameter of the base is 14 m.

Let h be the depth of the cylindrical tank.

Diameter of the base = 14 m

Radius of the circular base = r = 14/2 = 7 m

Volume of the cylindrical tank = πr2h

So, πr 2 h = 1848 m 3

(22/7) × 7 × 7 × h = 1848

h = 1848/(22 × 7)

Therefore, the depth of the cylindrical tank is 12 m.

7. What is the ratio of the volume of a cone, a sphere and a cylinder if each has the same radius and height?

Let r be the radius of the cone, sphere and cylinder.

Let h be the height of the cone and cylinder.

Also, r = h

We know that,

Volume of cone = (1/3)πr 2 h = (1/3)πr 3

Volume of sphere = (4/3)πr 3

Volume of cylinder = πr 2 h = πr 3

Volume of cone : Volume of sphere : Volume of cylinder

= (1/3)πr 2 h : (4/3)πr 3 : πr 2 h

= (1/3)πr 3 : (4/3)πr 3 : πr 3

= (1/3) : (4/3) : 1

= 1 : 4 : 3

Therefore, the required ratio is 1 : 4 : 3.

8. Find the water weight in a conical vessel that is 21 cm deep and 16 cm in diameter.

Depth of height of a conical vessel = h = 21 cm

Diameter of the circular base = 16 cm

Radius = r = 16/2 = 8 cm

Volume of a cone = (1/3)πr 2 h

= (1/3) × (22/7) × 8 × 8 × 21

= 1408 cm 3

= (1408/1000) kg

Therefore, the weight of the water in a conical vessel is 1.408 kg.

9. A vessel is in the form of a frustum of a cone. Its radius at one end and the heights are 8 cm and 14 cm, respectively. If its volume is 5676/3 cm 3 , find the radius at the other end.

Radius of one circular end = r 1 = 8 cm

Height of the frustum = h = 14 cm

Let r2 be the radius of the other circular end.

Volume of the frustum = πh/3 (r 1 2 + r 2 2 + r 1 r 2 )

= (1/3) × (22/7) × 14 × [(8) 2 + (r 2 ) 2 + 8r 2 ]

According to the given,

(1/3) × (22/7) × 14 × [(8) 2 + (r 2 ) 2 + 8r 2 ] = 5676/3 cm3

64 + r 2 2 + 8r 2 = 5676/44

r 2 2 + 8r 2 = 129 – 64

r 2 2 + 8r 2 – 65 = 0

r 2 2 + 13r 2 – 5r 2 – 65 = 0

r 2 (r 2 + 13) – 5(r 2 + 13) = 0

(r 2 – 5)(r 2 + 13) = 0

r 2 = 5, r 2 = -13

Hence, the radius of the other circular end is 5 cm.

10. If the curved surface area of a right circular cone is 10010 cm 2 and its radius is 35 cm, find the volume of the cone.

Radius of the circular base of a cone = r = 35 cm

Let h and l be the height and slant height of the right circular cone.

Also, given that the curved surface area = 10010 cm 2

πrl = 10010

(22/7) × 35 × l = 10010

l = (10010 × 7)/ (22 × 35)

As we know, l 2 = r 2 + h 2

h 2 = l 2 – r 2

= (91) 2 – (35) 2

= 8281 – 1225

Volume of the cone = (1/3) πr 2 h

= (1/3) × (22/7) × 35 × 35 × 84

Hence, the volume of the right circular cone is 107800 cm 3 .

Practice Questions on Volume

• The height of the wall is six times its width, and the length of the wall is seven times its height. If the volume of the wall is 16128 cm 3 , what is the width of the wall?
• A fish tank has a length of 45 cm, a width of 25 cm, and a depth of 10 cm. Calculate the volume of the fish tank.
• Find the volume of the cylinder with a radius of 5.4 units and a height of 16 units.
• Find the number of small cubes with edges of 10 cm that can be accommodated in a cubical box with a 1 m edge.
• The radius of a sphere is increased by 10%. Prove that the volume will be increased by 33.1% approximately.

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Volume and Capacity KS2

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19.4: Heat Capacity and Equipartition of Energy

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Learning Objectives

By the end of this section, you will be able to:

• Solve problems involving heat transfer to and from ideal monatomic gases whose volumes are held constant
• Solve similar problems for non-monatomic ideal gases based on the number of degrees of freedom of a molecule
• Estimate the heat capacities of metals using a model based on degrees of freedom

In the chapter on temperature and heat, we defined the specific heat capacity with the equation \(Q = mc\Delta T\), or \(c = (1/m)Q/\Delta T\). However, the properties of an ideal gas depend directly on the number of moles in a sample, so here we define specific heat capacity in terms of the number of moles, not the mass. Furthermore, when talking about solids and liquids, we ignored any changes in volume and pressure with changes in temperature—a good approximation for solids and liquids, but for gases, we have to make some condition on volume or pressure changes. Here, we focus on the heat capacity with the volume held constant. We can calculate it for an ideal gas.

Heat Capacity of an Ideal Monatomic Gas at Constant Volume

We define the molar heat capacity at constant volume \(C_V\) as

\[\underbrace{C_V = \dfrac{1}{n} \dfrac{Q}{\Delta T}}_{\text{with constant V}}\nonumber \]

This is often expressed in the form

\[Q = nC_V\Delta T\nonumber \]

If the volume does not change, there is no overall displacement, so no work is done, and the only change in internal energy is due to the heat flow \(\Delta E_{int} = Q\). (This statement is discussed further in the next chapter.) We use the equation \(E_{int} = 3nRT/2\) to write \(\Delta E_{int} = 3nR\Delta T/2\) and substitute \(\Delta E\) for Q to find \(Q = 3nR\Delta T/2\), which gives the following simple result for an ideal monatomic gas:

\[C_V = \dfrac{3}{2}R.\nonumber \]

It is independent of temperature, which justifies our use of finite differences instead of a derivative. This formula agrees well with experimental results.

In the next chapter we discuss the molar specific heat at constant pressure \(C_p\), which is always greater than \(C_V\).

Example \(\PageIndex{1}\): Calculating Temperature

A sample of 0.125 kg of xenon is contained in a rigid metal cylinder, big enough that the xenon can be modeled as an ideal gas, at a temperature of \(20.0^oC\). The cylinder is moved outside on a hot summer day. As the xenon comes into equilibrium by reaching the temperature of its surroundings, 180 J of heat are conducted to it through the cylinder walls. What is the equilibrium temperature? Ignore the expansion of the metal cylinder.

• Identify the knowns: We know the initial temperature \(T_1\) is \(20.0^oC\), the heat Q is 180 J, and the mass m of the xenon is 0.125 kg.
• Identify the unknown. We need the final temperature, so we’ll need \(\Delta T\).
• Determine which equations are needed. Because xenon gas is monatomic, we can use \(Q = 3nR\Delta T/2\). Then we need the number of moles \(n = m/M\).
• Substitute the known values into the equations and solve for the unknowns.

The molar mass of xenon is 131.3 g, so we obtain

\[n = \dfrac{125 \, g}{131.3 \, g/mol} = 0.952 \, mol, \nonumber\nonumber \]

\[\Delta T = \dfrac{2Q}{3nR} = \dfrac{2(180 \, J)}{3(0.952 \, mol)(8.31 \, J/mol \cdot \, ^oC)} = 15.2^oC. \nonumber\nonumber \]

Therefore, the final temperature is \(35.2^oC \). The problem could equally well be solved in kelvin; as a kelvin is the same size as a degree Celsius of temperature change, you would get \(\Delta T = 15.2 \, K \).

Significance

The heating of an ideal or almost ideal gas at constant volume is important in car engines and many other practical systems.

Exercise \(\PageIndex{1}\)

Suppose 2 moles of helium gas at 200 K are mixed with 2 moles of krypton gas at 400 K in a calorimeter. What is the final temperature?

As the number of moles is equal and we know the molar heat capacities of the two gases are equal, the temperature is halfway between the initial temperatures, 300 K.

We would like to generalize our results to ideal gases with more than one atom per molecule. In such systems, the molecules can have other forms of energy beside translational kinetic energy, such as rotational kinetic energy and vibrational kinetic and potential energies. We will see that a simple rule lets us determine the average energies present in these forms and solve problems in much the same way as we have for monatomic gases.

Degrees of Freedom

In the previous section, we found that \(\frac{1}{2}mv^2 = \frac{3}{2}k_BT\) and \(v^2 = 3v_x^2\), from which it follows that \(\frac{1}{2}mv_x^2 = \frac{1}{2}k_BT\). The same equation holds for \(\frac{3}{2}k_BT\) as the sum of contributions of \(\frac{1}{2}k_BT\) from each of the three dimensions of translational motion. Shifting to the gas as a whole, we see that the 3 in the formula \(C_V = \frac{3}{2}R\) also reflects those three dimensions. We define a degree of freedom as an independent possible motion of a molecule, such as each of the three dimensions of translation. Then, letting d represent the number of degrees of freedom, the molar heat capacity at constant volume of a monatomic ideal gas is \(C_V = \frac{d}{2}R\), where \(d = 3\).

The branch of physics called statistical mechanics tells us, and experiment confirms, that \(C_V\) of any ideal gas is given by this equation, regardless of the number of degrees of freedom. This fact follows from a more general result, the equipartition theorem , which holds in classical (non-quantum) thermodynamics for systems in thermal equilibrium under technical conditions that are beyond our scope. Here, we mention only that in a system, the energy is shared among the degrees of freedom by collisions.

Equipartition Theorem

The energy of a thermodynamic system in equilibrium is partitioned equally among its degrees of freedom. Accordingly, the molar heat capacity of an ideal gas is proportional to its number of degrees of freedom, d : \[C_V = \dfrac{d}{2}R.\nonumber \]

This result is due to the Scottish physicist James Clerk Maxwell (1831−1871), whose name will appear several more times in this book.

For example, consider a diatomic ideal gas (a good model for nitrogen, \(N_2\), and oxygen, \(O_2\)). Such a gas has more degrees of freedom than a monatomic gas. In addition to the three degrees of freedom for translation, it has two degrees of freedom for rotation perpendicular to its axis. Furthermore, the molecule can vibrate along its axis. This motion is often modeled by imagining a spring connecting the two atoms, and we know from simple harmonic motion that such motion has both kinetic and potential energy. Each of these forms of energy corresponds to a degree of freedom, giving two more.

We might expect that for a diatomic gas, we should use 7 as the number of degrees of freedom; classically, if the molecules of a gas had only translational kinetic energy, collisions between molecules would soon make them rotate and vibrate. However, as explained in the previous module, quantum mechanics controls which degrees of freedom are active. The result is shown in Figure \(\PageIndex{1}\). Both rotational and vibrational energies are limited to discrete values. For temperatures below about 60 K, the energies of hydrogen molecules are too low for a collision to bring the rotational state or vibrational state of a molecule from the lowest energy to the second lowest, so the only form of energy is translational kinetic energy, and \(d = 3\) or \(C_V = 3R/2\) as in a monatomic gas. Above that temperature, the two rotational degrees of freedom begin to contribute, that is, some molecules are excited to the rotational state with the second-lowest energy. (This temperature is much lower than that where rotations of monatomic gases contribute, because diatomic molecules have much higher rotational inertias and hence much lower rotational energies.) From about room temperature (a bit less than 300 K) to about 600 K, the rotational degrees of freedom are fully active, but the vibrational ones are not, and \(d = 5\). Then, finally, above about 3000 K, the vibrational degrees of freedom are fully active, and \(d = 7\) as the classical theory predicted.

Polyatomic molecules typically have one additional rotational degree of freedom at room temperature, since they have comparable moments of inertia around any axis. Thus, at room temperature, they have \(d = 6\) and at high temperature, \(d = 8\). We usually assume that gases have the theoretical room-temperature values of d .

As shown in Table \(\PageIndex{1}\), the results agree well with experiments for many monatomic and diatomic gases, but the agreement for triatomic gases is only fair. The differences arise from interactions that we have ignored between and within molecules.

What about internal energy for diatomic and polyatomic gases? For such gases, \(C_V\) is a function of temperature (Figure \(\PageIndex{1}\)), so we do not have the kind of simple result we have for monatomic ideal gases.

Molar Heat Capacity of Solid Elements

The idea of equipartition leads to an estimate of the molar heat capacity of solid elements at ordinary temperatures. We can model the atoms of a solid as attached to neighboring atoms by springs (Figure \(\PageIndex{2}\)).

Analogously to the discussion of vibration in the previous module, each atom has six degrees of freedom: one kinetic and one potential for each of the x -, y -, and z -directions. Accordingly, the molar specific heat of a metal should be 3 R . This result, known as the Law of Dulong and Petit , works fairly well experimentally at room temperature. (For every element, it fails at low temperatures for quantum-mechanical reasons. Since quantum effects are particularly important for low-mass particles, the Law of Dulong and Petit already fails at room temperature for some light elements, such as beryllium and carbon. It also fails for some heavier elements for various reasons beyond what we can cover.)

Problem-Solving Strategy: Heat Capacity and Equipartition

The strategy for solving these problems is the same as the one in Phase Changes for the effects of heat transfer. The only new feature is that you should determine whether the case just presented—ideal gases at constant volume—applies to the problem. (For solid elements, looking up the specific heat capacity is generally better than estimating it from the Law of Dulong and Petit.) In the case of an ideal gas, determine the number d of degrees of freedom from the number of atoms in the gas molecule and use it to calculate \(C_V\) (or use \(C_V\) to solve for d ).

Example \(\PageIndex{2}\): Calculating Temperature: Calorimetry with an Ideal Gas

A 300-g piece of solid gallium (a metal used in semiconductor devices) at its melting point of only \(30.0^oC\) is in contact with 12.0 moles of air (assumed diatomic) at \(95.0^oC\) in an insulated container. When the air reaches equilibrium with the gallium, 202 g of the gallium have melted. Based on those data, what is the heat of fusion of gallium? Assume the volume of the air does not change and there are no other heat transfers.

We’ll use the equation \(Q_{hot} + Q_{cold} = 0\). As some of the gallium doesn’t melt, we know the final temperature is still the melting point. Then the only \(Q_{hot}\) is the heat lost as the air cools, \(Q_{hot} = n_{air}C_V\Delta T\), where \(C_V = 5R/2\). The only \(Q_{cold}\) is the latent heat of fusion of the gallium, \(Q_{cold} = m_{Ga}L_f\). It is positive because heat flows into the gallium.

• Set up the equation:\[n_{air}C_V\Delta T + m_{Ga}L_f = 0.\nonumber \]
• Substitute the known values and solve: \[(12.0 \, mol) \left(\dfrac{5}{2}\right) \left(8.31 \dfrac{J}{mol \cdot \, ^oC}\right)(30.0^oC - 95.0^oC) + (0.202 \, kg)L_f = 0.\nonumber \]

We solve to find that the heat of fusion of gallium is 80.2 kJ/kg.

Fixed-point algorithms for solving the critical value and upper tail quantile of Kuiper's statistics

• Zhang, Hong-Yan
• Lin, Rui-Jia

Kuiper's statistic is a good measure for the difference of ideal distribution and empirical distribution in the goodness-of-fit test. However, it is a challenging problem to solve the critical value and upper tail quantile, or simply Kuiper pair, of Kuiper's statistics due to the difficulties of solving the nonlinear equation and reasonable approximation of infinite series. In this work, the contributions lie in three perspectives: firstly, the second order approximation for the infinite series of the cumulative distribution of the critical value is used to achieve higher precision; secondly, the principles and fixed-point algorithms for solving the Kuiper pair are presented with details; finally, finally, a mistake about the critical value c n α for (α , n) = (0.01 , 30) in Kuiper's distribution table has been labeled and corrected where n is the sample capacity and α is the upper tail quantile. The algorithms are verified and validated by comparing with the table provided by Kuiper. The methods and algorithms proposed are enlightening and worth of introducing to the college students, computer programmers, engineers, experimental psychologists and so on.

• Kuiper's statistic;
• Upper tail quantile;
• Fixed-point algorithm;
• Numerical approximation;
• Algorithm design;
• STEM education;
• Statistics - Computation