WBBSE Solutions For Class 10 Maths Mensuration Chapter 2 Right Circular Cylinder

WBBSE Solutions For Class 10 Maths Mensuration Chapter 2 Right Circular Cylinder

Right circular cylinders:

A right circular cylinder is a solid generated by the revolution of a rectangle round one of its side as its axis.

In our daily life we see several objects like right circular cylinders, such as, a drum, a lead pencil, a heap of round coins placed one upon another, pipe of water, etc are right circular cylinders.

Let ABCD be a rectangle, it is made to turn round the side AB, it generates a right circular cylinder.

So, AB is the axis of the cylinder.

The positions of C and D always remain equidistant from B and A respectively.

Thus two circles of equal areas are generated by the rotation of C and D respectively, one with centre at B and the other with the centre, at A, AB is called the height of the cylinder.

Obviously, AB = CD = C’D’ = h.

WBBSE Solutions for Class 10 Maths

The circle on the lower side, i.e., the circle with centre at A, is called the base of the cylinder since the cylinder stands on this plane surface.

The constant distance between A and D is called the radius (r) of the base of the cylinder, which is also known as the radius of the cylinder.

The distance between A and B, i..e, the length of the axis AB is called the height (h) of the cylinder.

We may also assume the lengths of CD or C’D’ as the height of the cylinder since AB = CD = C’D’ = h

Curved and plane surface:

The round surface of the cylinder is called the curved surface of the cylinder and the planes on the upper and lower sides of the cylinder are known as the plane surfaces of the cylinder.

If we draw a vertical straight line along the curved surface of a cylinder and cut it along that line, the curved surface of the cylinder will turn into a plane surface.

Evidently, this plane surface will be a rectangle whose length and breadth are equal to the circumference of the base circle and height of the cylinder respectively.

Thus, the area of the curved surface is equal to the area of the rectangle whose length is the circumference of the circle denoted as the base and whose breadth is the height of the cylinder.

Hence the area of the curved surface of the right circular cylinder = circumference of the base circle x height of the cylinder.

The cylinders as shown in the following images are not right circular cylinders since their lateral surface of them are not orthogonal to their bases

Formulas related to right circular cylinder:

Let the radius of the base, i.,e., of the cylinder be r and the height of the cylinder be h.

Then 1. The area of the base of the cylinder = πr² sq-units. Since the base is a circle with radius r units.

2. The area of the curved surface of the cylinder = (circumference of the base of the cylinder) X (height of the cylinder) = 2πr x h sq-units = 2πrh sq-units.

3. The area of the two plane surfaces of the cylinder = 2πr², since there are two such plane surfaces, the areas of which are equal and both are circles with radii r units.

4. The total surface area of the cylinder = area of the curved surface + area of the two ends, i.e., the two plane surfaces.

= (2πrh + 2πr²) sq-units = 2πr (h + r) sq-units

5. The volume of the solid cylinder = (area of the base) X (height of the cylinder)

= πr² x h cubic- units = πr²h cubic-units.

6. The total surface area of a hollow cylinder:

Let the external radius of a hollow cylinder be R units and the internal radius of it be r units.

Also let the height of the cylinder be h units.

Then the area of the external curved surface = 2πRh and the area of the internal curved surface = 2πrh.

Hence, the total area of the curved surfaces = area of the external curved surface + area of the internal curved surface + area of the two ends…..(1)

Now, area of the two ends = 2 (πR² – πr²) sq-units

From (1) we get,

The total surface area of the curved surfaces of a hollow cylinder

= {2πRh + 2πrh + 2 (πR² – πr²)} sq-units = 2π (Rh + rh + R² – r²) sq-units.

7. The volume of a hollow cylinder:

The volume of the hollow cylinder

= (Volume of the external cylinder) – (the volume of the internal cylinder)

= (πR²h – πr²h) cubic-units = πh (R² – r²) cubic-units = πh (R + r) (R – r) cubic-units.

8. The area of the two ends of a hollow cylinder:

The area of the two ends of a hollow cylinder

= 2 x (area of the outer circle – area of the inner circle)

= 2 x (πR² – πr²) sq-units = 2π (R² – r²) sq-units = 2 π (R + r) (R – r) sq-units.

Frustum of a cone:

Let us given a cone, if we cut through it with a plane parallel to its base and remove the cone that is formed on one side of that plane, the part which is left over on the other side of the plane is known as a frustum of the cone.

For example, a glass, used for drinking water, a pail (Balti) etc are frustums.

Let the height of the frustum be h, the slant height of it be l and R and r (R > r) be the two radii of the frustum of a cone.

Then,

1. The curved surface area of the frustum of the cone = π(R + r) l sq-units, where l= \(\sqrt{h^2+(\mathrm{R}-r)^2}\)
2. the total surface area of the frustum of the cone = {πl (R + r) + πR² + πr²} sq-units, where l= \(\sqrt{h^2+(\mathrm{R}-r)^2}\)
3. Volume of the frustum of the cone = \(\frac{1}{3}\) πh (r² +r² + Rr) cubic-units.

Mensuration Chapter 2 Right Circular Cylinder Multiple Choice Questions

“WBBSE Class 10 right circular cylinder solved examples”

Example 1. If the length of the radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:3, then the ratio of their lateral surfaces is

1. 2: 5
2. 8: 7
3. 10:9
4. 16: 9

Solution:

Given

If the length of the radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:3,

Let the radii of two solid right circular cylinders are 2x units and 3x emits, such that their ratio is 2x: 3x = 2 : 3, which satisfies the given condition.

Also, let their heights be 5 h units and 3 h units respectively.

∴ the area of the lateral surface of the first = 2π X 2x X 5h sq-units = 20πxh sq-units.

and the area of the lateral surface of the second = 2π X 3x X 3h sq-units = 18πxh sq-units.

Hence the required ratio = 20πxh : 18πxh =10:9

∴ 3. 10:9 is correct.

The ratio of their lateral surfaces is 10:9.

Example 2. If the length of the radii of two solid right circular cylinders are in the ratio 2 : 3 and their height are in the ratio 5:3, then the ratio of their volumes is

1. 27: 20
2. 20: 27
3. 4: 9
4. 9: 4

Solution:

Given

If the length of the radii of two solid right circular cylinders are in the ratio 2 : 3 and their height are in the ratio 5:3

Let the radii are 2r units and 3r units and their heights are 5h units and 3h units respectively.

Then the volume of the first = π (2r)² x 5h cubic-units = 20 πr²h cubic-units

and the volume of the second = π (3r)² x 3h cubic-units = 27πr²h cubic-units.

Hence the required ratio = 20πr²h : 27πr²h = 20 : 27

∴ 2. 20: 27 is correct.

The ratio of their volumes is 20: 27

Example 3. If volumes of two solid right circular cylinder are same and their heights are in the ratio 1: 2, then the ratio of lengths of their radii is

1. 1: √2
2. √2 : 1
3. 1: 2
4. 2: 1

Solution:

Given

If volumes of two solid right circular cylinder are same and their heights are in the ratio 1: 2,

Let the heights of the cylinders arc h units and 2h units (since the ratio is 1: 2) and let the radii of the cylinders arc R and r units respectively.

Since the volume of the cylinders are same,

∴ πR²h = πr².2h

⇒ R² = 2r²

⇒ \(\sqrt{\mathrm{R}^2}=\sqrt{2 r^2}\) [Taking square root of both the sides]

⇒ R = √2r

⇒ \(\frac{R}{r}\) = √2

⇒ R : r = √2 : 1

Hence the required ratio = √2 : 1

∴ 2. is correct.

The ratio of lengths of their radii is √2 : 1

“Mensuration problems on right circular cylinder for Class 10”

Example 4. In a right circular cylinder, if the length of radius is halved and height is doubled, then the volume of the cylinder will be

1. Equal
2. Double
3. Half
4. 4 times

Solution:

Given

In a right circular cylinder, if the length of radius is halved and height is doubled

Let the radius of the cylinder be r units and height of it be h units.

Then the volume of the cylinder = πr²h cubic-units when the radius is halved,

i.e., \(\frac{r}{2}\) units and the height is doubled,

i.e., 2h units, then the volume becomes  π \(\left(\frac{r}{2}\right)^2\) x 2h cubic – units

= \(\pi \times \frac{r^2}{4}\) x 2h cubic – units

= \(\frac{\pi r^2 h}{2}\) cubic – units

So, the volume will be half.

∴ 3. Half is correct.

The volume of the cylinder will be Half

Example 5. If the length of radius of a right circular cylinder is doubled and height is halved, then its lateral surface area will be

1. Equal
2. Double
3. Half
4. 3 times

Solution:

Given

If the length of radius of a right circular cylinder is doubled and height is halved

Let the radius of the cylinder be r units and height of it be h units,

∴ the lateral surface area = 2πrh sq-units

If the radius is doubled, i.e., 2r units and the height is halved, i.e., \(\frac{h}{2}\) units, then the lateral surface area becomes 2π x 2r x \(\frac{h}{2}\) sq-units = 2πrh sq-units.

The lateral surface area remains the same, i.e., in both the cases the lateral surface areas are equal.

∴ 1. Equal is correct.

Lateral surface area will be

Mensuration Chapter 2 Right Circular Cylinder True Or False

“Chapter 2 right circular cylinder exercises WBBSE solutions”

Example 1. The length of right circular drum is r cm and height is h cm. If half part of the drum is filled with water, then the volume of water will be πr² h cubic cm.

Solution: False

Since the volume of water will be \(\frac{1}{2}\) πr²h cu – cm = \(\frac{\pi r^2 h}{2} \mathrm{cc}\) cc.

Example 2. If the length of radius of a right circular cylinder is 2 units, then the numerical value of volume and surface area of cylinder will be equal for any height.

Solution: True

since volume = π2²h, h = height of the cylinder; surface area = 2π.2h = 4πh sq-cm.

But the numerical value of π.2²h cc = 4πh and of 2.π.2.h sq-cm = 4πh are equal.

Hence the statement is true.

Mensuration Chapter 2 Right Circular Cylinder Fill In The Blanks

Example 1. The length of a rectangular paper is l units and the breadth is b units. The rectangular paper is rolled and a cylinder is formed of which perimeter is equal to the length of the paper. The lateral surface area of the cylinder is ______ sq-unit.

Solution: lb

since we know that lateral surface area = circumference of the base x height = l x b sq-units = lb sq-units.

Example 2. The longest rod that can be kept in a right circular cylinder having a diameter of 3 cm and height of 4 cm, then the length of rod is _______ cm.

Solution: 5

since the length of rod = \(\sqrt{3^2+4^2}\) cm = √25 cm = 5cm.

 

 

Example 3. If the numerical values of volume and lateral surface area of a right circular cylinder are equal, then the length of diameter of cylinder is _______ unit.

Solution: 4

since let the diameter be d units and height of the cylinder is h units.

As per question, π x \(\left(\frac{d}{2}\right)^2\) x h = 2π x \(\frac{d}{2}\) x h ⇒ d = 4.

Mensuration Chapter 2 Right Circular Cylinder Short Answer Type Questions

“Class 10 Maths volume of right circular cylinder problems”

Example 1. If the lateral surface area of a right circular cylindrical pillar is 264 sq-metres and volume is 924 cubic-metres. Find the length of radius of the base of the cylinder.

Solution:

Given:

If the lateral surface area of a right circular cylindrical pillar is 264 sq-metres and volume is 924 cubic-metres.

Let the radius of the base of the cylinder be r metres and height be h metres.

∴ Its lateral surface = 2 πrh sq-metres and volume = πr²h cubic-metres.

As per questions, 2πrh = 264 or, πrh =132……(1)

and πr²h = 924 or, πrh x r = 924 or, 132 x r = 924 by (1) or, r = \(\frac{924}{132}\) = 7

Hence the required radius = 7 metres.

Example 2. If the lateral surface area of a right circular cylinder is c square unit, length of radius of base is r unit and volume is V cubic-unit. Find the value of \(\frac{cr}{V}\)

Solution:

Given:

If the lateral surface area of a right circular cylinder is c square unit, length of radius of base is r unit and volume is V cubic-unit.

Let the height of the cylinder be h units.

As per question, c = 2πrh……(1)

V = πr²h……(2)

∴ \(\frac{cr}{V}\) =\(\frac{2 \pi r h \times r}{\pi r^2 h}\) =\(\frac{2 \pi r^2 h}{\pi r^2 h}=2\)

Hence the required value of \(\frac{cr}{V}\)  = 2.

Example 3. If the height of a right circular cylinder is 14 cm and lateral surface area is 264 sq-cm, find the volume of the cylinder.

Solution:

Given:

If the height of a right circular cylinder is 14 cm and lateral surface area is 264 sq-cm

Let the radius of the base of the cylinder be r cm.

As per question, 2πr x 14 = 264 or, 2 x \(\frac{22}{7}\) x r x l4 = 264 or, r = \(=\frac{264}{2 \times 22 \times 2}\) = 3

∴ the volume of the cylinder = \(\frac{22}{7}\) x (3)² x 14 cc = 22 x 9 x 2 cc = 396 cc

Hence the required volume = 396 cc.

“Understanding right circular cylinder in Class 10 Maths”

Example 4. If the height of two right circular cylinder are in the ratio of 1 : 2 and perimeters are in the ratio of 3 : 4. Find the ratio of their volumes.

Solution:

Given:

If the height of two right circular cylinder are in the ratio of 1 : 2 and perimeters are in the ratio of 3 : 4.

Let the heights of the two cylinders be h unit and 2h units [ratio =1:2]

Also, let the radii of the cylinders be r₁ units and r₂ units.

As per questions, 2πr₁h : 2πr₂.2h = 3:4

Now, ratio of volumes = \(=\frac{\pi r_1^2 h}{\pi r_2^2 \cdot 2 h}=\left(\frac{r_1}{r_2}\right)^2\) x \(\frac{1}{2}\)

Hence the required ratio = 9 : 16.

Example 5. The length of radius of a right circular cylinder is decreased by 50% and height is increased by 50%. Calculate how much percent of the volume will be changed.

Solution:

Given:

The length of radius of a right circular cylinder is decreased by 50% and height is increased by 50%

Let the radius of the cylinder be r units and height be h units,

∴  the volume of the cylinder = πr²h cubic-units

If radius is decreased by 50% the radius becomes \(\left(r-r \times \frac{50}{100}\right)\) units = \(\left(r-\frac{r}{2}\right)\) units = \(\frac{r}{2}\) units.

Also, if height is increased by 50%, then it becomes \(\left(h+h \times \frac{50}{100}\right)\) units = \(\left(h+\frac{h}{2}\right)\) units = \(\frac{3h}{2}\) units.

Then the volume of the cylinder = π x \(\left(\frac{r}{2}\right)^2\) x \(\frac{3h}{2}\) cubic-units = \(\frac{3 \pi r^2 h}{8}\) cubic units.

So, the volume decreases by \(\left(\pi r^2 h-\frac{3 \pi r^2 h}{8}\right)\) cubic-units.

= \(\frac{8 \pi r^2 h-3 \pi r^2 h}{8}\) cubic-units = \(=\frac{5 \pi r^2 h}{8}\) cubic-units

So, the required percentage = \(\frac{\frac{5 \pi r^2 h}{8}}{\pi r^2 h} \times 100 \%=\frac{5}{8} \times 100 \%=\frac{125}{2} \%=62 \frac{1}{2} \%\)

Hence the volume will be decreased by 62 \(\frac{1}{2}\) %

Mensuration Chapter 2 Right Circular Cylinder Long Answer Type Questions

“Step-by-step solutions for right circular cylinder Class 10”

Example 1. Calculate how many cubic decimetre of concrete materials will be needed to construct two cylindrical pillars each of whose diameter is 5.6 decimetres and height is 2.5 metres. Calculate the cost of plastering the curved surface area of the two pillars at ₹125 per sq-metres.

Solution:

The radius of the pillar = \(\frac{5 \cdot 6}{2}\) dcm = 2.8 dcm = 0.28 m

and height = 2.5 metres = 2.5 x 10 dcm = 25 dcm

Total volume of the two pillars = 2 x \(\frac{22}{7}\) x(2.8)² x 25 cubic-dcm

= 2 x \(\frac{22}{7}\) x 2.8 x 2.8 x 25 cubic – dcm = 1232 cubic – decm

Again, the total curved surface area of two pillars = 2 x 2 x \(\frac{22}{7}\) x 0.28 x 2.5 sq- metres = 8.8 sq-metres.

Hence the required cost = ₹ 8.8 x 125 = ₹1100.

Example 2. Out of three jars of equal diameter and height, 2/3 part of the first, 5/6 part of the second and 7/9 part of the third were filled with dilute sulphuric acid. Whole of acid in the three jars were poured into a large jar of 2.1 dcm diameter, as a result the height of acid in the jar becomes 4.1 dcm. If the length of diameter of each of the three equal jars is 1.4 dem. Calculate the height of three jars.

Solution:

Given:

Out of three jars of equal diameter and height, 2/3 part of the first, 5/6 part of the second and 7/9 part of the third were filled with dilute sulphuric acid. Whole of acid in the three jars were poured into a large jar of 2.1 dcm diameter, as a result the height of acid in the jar becomes 4.1 dcm. If the length of diameter of each of the three equal jars is 1.4 dem.

Let the height of three jars be h dcm.

So the volume of the acid in the large jar

= \(\frac{22}{7}\) x \(\left(\frac{2 \cdot 1}{2}\right)^2\) 4 . 1 cubic.dcm = \(\frac{22}{7}\) x 1.05 x 1.05 x 4.1 cubic – dccm = 14.2065 cubic – dcm

As per question,

\(\frac{2}{3} \times \frac{22}{7} \times\left(\frac{1 \cdot 4}{2}\right)^2 \times h+\frac{5}{6} \times \frac{22}{7} \times\left(\frac{1 \cdot 4}{2}\right)^2 h+\frac{7}{9} \times \frac{22}{7} \times\left(\frac{1 \cdot 4}{2}\right)^2 \times h=14 \cdot 2065\)

or, \(\frac{22}{7} \times(0 \cdot 7)^2 \times h\left(\frac{2}{3}+\frac{5}{6}+\frac{7}{9}\right)\) = 14.2065

or, 22 x 0.1 x 0.7 x h x \(\frac{12+15+14}{18}\) = 14.2065

or, 1.54 h x \(\frac{41}{18}\) = 14.2065

or, h = \(\frac{14 \cdot 2065 \times 18}{1 \cdot 54 \times 41}\)

Hence the required height of the jar = 4 05 dcm.

Example 3. If a pump set with a pipe of 14 cm diameter can drain 2500 metres water per minute, then calculate how much kilolitres water will that pump drain per hour. [1 litre = 1 cubic.dcm]

Solution:

Given:

If a pump set with a pipe of 14 cm diameter can drain 2500 metres water per minute

14cm = \(\frac{14}{10}\) dcm = 1.4 dcm

2500 metres = 2500 x 10 dcm = 25000 dcm

So, the volume of water poured by the pipe per minute

= \(\frac{22}{7} \times\left(\frac{1 \cdot 4}{2}\right)^2\) x 2500 cubic.dcm = 38500 cubic.dcm = 38500 litres

= \(\frac{38500}{1000}\) kilolitres = 38.5 kilolitres

∴ the volume of water poured by the pipe per hour = 38.5 x 60 kilolitres = 2310 kilolitres.

Hence the required water = 2310 kilolitres.

Example 4. There are some water in a long gas jar of 7 cm diameter. If a solid right circular cylindrical pipe of iron having 5 cm length and 5.6 cm diameter be immersed completely in that water, calculate how much the level of water will rise.

Solution:

Given:

There are some water in a long gas jar of 7 cm diameter. If a solid right circular cylindrical pipe of iron having 5 cm length and 5.6 cm diameter be immersed completely in that water

The volume of the solid cylindrical iron pipe

= \(\frac{22}{7} \times\left(\frac{5 \cdot 6}{2}\right)^2\) x 5cc = 123.2cc

Let the level of water in the gas jar be h cm.

Then the volume of water raised in the jar = \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2\) x hcc = 38.5 hcc

As per question, 38.5 h = 123.2

or, h = \(\frac{123 \cdot 2}{38 \cdot 5}\) = 3.2

Hence the level of water will rise 3.2 cm.

Example 5. If the surface area of a right circular cylindrical pillar is 264 sq-metres and volume is 396 cubic-metres, then calculate height and length of diameter of this pillar.

Solution:

Given:

If the surface area of a right circular cylindrical pillar is 264 sq-metres and volume is 396 cubic-metres

Let the height and radius of the pillar be h metres and r metres respectively.

∴ the surface area of the pillar = 2 x \(\frac{22}{7}\) x r x h sq.metres = \(\frac{44}{7}\) rh sq.metres

As per question, \(\frac{44}{7}\) rh = 264 or, rh = \(\frac{264 \times 7}{44}\) = 42……..(1)

Again, volume of the pillar = \(\frac{22}{7}\) x r² x h cubic.metres

As per question, \(\frac{22}{7}\) r²h = 396

or, \(\frac{22}{7}\) x rh x r = 396 or, \(\frac{22}{7}\) x 42 x r = 396 [from (1)]

or, r = \(=\frac{396}{22 \times 6}\) = 3

∴ 2r = 2 x 3 = 6

∴ h = \(\frac{42}{3}\) = 14

Hence the required height = 14 cm and length of diameter = 6 cm.

Example 6. A right circular cylindrical tank of 9 metres height is filled with water. Water comes out from there through a pipe having length of 6 cm diameter with a speed of 225 metre per minute and the tank becomes empty after 2 hours 24 minutes. Calculate the length of radius of the tank.

Solution:

Given:

A right circular cylindrical tank of 9 metres height is filled with water. Water comes out from there through a pipe having length of 6 cm diameter with a speed of 225 metre per minute and the tank becomes empty after 2 hours 24 minutes.

Let the radius of the base of the tank be r metres,

∴ the volume of the tank (i.e., of the water)

= \(\frac{22}{7}\) x r² x 9 cubic.metres = \(=\frac{198 r^2}{7}\) cubic metres.

Again, 6 cm = \(\frac{6}{100}\) m = 0.06 metre.

∴ the volume of water comes out per minute

= \(\frac{22}{7}\) x \(\left(\frac{0.06}{2}\right)^2\) x 225 cubic.metres

= \(\frac{22}{7}\) x 0.03 x 0.03 x 225 cubic.metres = \(\frac{4 \cdot 455}{7}\) cubic.metres

2 hours 24 minutes = (2 x 60 + 24) minutes = 144 minutes

∴ the volume of water comes out in 2 hours 24 minutes = \(\frac{4 \cdot 455}{7}\) x l44 cubic. metres

As per condition given, \(\frac{198 r^2}{7}\) = \(\frac{4 \cdot 455}{7}\) x 144

or, r² = \(\frac{4 \cdot 455}{198}\) x 144 or, r² = 0.0225 x 144

or, r = V0.0225x 144 = 0.15×12 = 1.8

Hence the length of radius = 1.8 m = 1.8 x 100 cm = 180 cm.

Example 7. Curved surface area of a right circular cylindrical log of wood of uniform density is 440 sq-dcm. If 1 cubic dcm of wood weighs 1.5 kg and weight of the log is 9.24 quintals, then calculate the length of diameter of log and its height.

Solution:

Given:

Curved surface area of a right circular cylindrical log of wood of uniform density is 440 sq-dcm. If 1 cubic dcm of wood weighs 1.5 kg and weight of the log is 9.24 quintals

1 quintal = 100 kg

9.24 quintal = 100 x 9.24 kg = 924 kg

Also 1.5 kg is the weight of 1 cu.dcm wood

∴ 1 kg is the weight of \(\frac{1}{1 \cdot 5}\) cu.dcm wood

∴ 924 kg is the weight of \(\frac{1 \times 924}{1 \cdot 5}\) cu.dcm wood = 616 cubic.dcm

Let the radius of the log of wood be r dcm and its height be h dcm.

∴ curved surface area =2 x \(\frac{22}{7}\) x r x h sq.dcm

As per question, 2 x \(\frac{22}{7}\) x rh = 440

⇒ rh= \(\frac{440 \times 7}{2 \times 22}\) ⇒ rh = 70…….(1)

Also, volume of the log = \(\frac{22}{7}\) x r x h cu – dcm.

As per question, \(\frac{22}{7}\) x r² x h = 616

⇒ r x rh = \(=\frac{616 \times 7}{22}\)

⇒ r x 70 = 19l [rh = 70]

⇒ r = \(\frac{196}{20}\) = 2.8

∴ 2r = 2 x 2.8 = 5.6. From (1) we get, h = \(\frac{70}{2 \cdot 8}\) = 25

Hence the required diameter = 56 dcm and height = 25 dcm.

“WBBSE Mensuration Chapter 2 practice questions on cylinders”

Example 8. The length of inner and outer diameter of a right circular cylindrical pipe open at two ends are 30 cm and 26 cm respectively and length of pipe is 14.7 metres. Calculate the cost of painting its all surfaces with coaltar at ₹ 2.25 per sq-dcm.

Solution:

Given:

The length of inner and outer diameter of a right circular cylindrical pipe open at two ends are 30 cm and 26 cm respectively and length of pipe is 14.7 metres.

Inner diameter of the cylinder = 30 cm 30

∴ Inner radius = \(\frac{30}{2}\) cm = 15cm = 1.5dcm

∴ Inner curved surface =2 x \(\frac{22}{7}\) x l.5 x l47 sq.dcm [v 14.7 m = 147 dcm] = 1386 sq.dcm.

Outer diameter of the cyclinder = 26 cm

∴ Outer radius = \(\frac{26}{2}\) cm = 13cm = 1.3 dcm

∴ Outer curved surface area =2 x \(\frac{22}{7}\) x 1.3 x 147 sq. dcm = 1201.2 sq. dcm

∴ Total curved surface of the pipe = (1386 + 1201.2) sq. dcm = 2587.2 sq. dcm

∴ the cost of painting = ₹ 2587.2 x 2.25 = ₹ 5821.2

Hence the required cost of painting = ₹ 5821.2.

Example 9. The height of a hollow right circular cylinder, open at both ends, is 2.8 metres. If length of inner diameter of the cylinder is 4.6 dcm and the cylinder is made up of 84.48 cubic-dcm of iron, then calculate the length of outer diameter of the cylinder.

Solution:

Given:

The height of a hollow right circular cylinder, open at both ends, is 2.8 metres. If length of inner diameter of the cylinder is 4.6 dcm and the cylinder is made up of 84.48 cubic-dcm of iron,

Let the outer diameter be d dcm.

∴ outer radius = \(\frac{d}{2}\) dcm.

As per question, π \(\pi\left\{\left(\frac{d}{2}\right)^2-\left(\frac{4 \cdot 6}{2}\right)^2\right\}\) x 28

= 84.48 [2.8 m = 28 dcm]

or, \(\frac{22}{7}\left\{\left(\frac{d}{2}\right)^2-(2 \cdot 3)^2\right\}\) x 28 = 84.48

or, \(\left\{\left(\frac{d}{2}\right)^2-5 \cdot 29\right\}\) x 28 = 84.48 x \(\frac{7}{22}\)

or, \(\left(\frac{d}{2}\right)^2\) – 5.29 = \(\frac{84 \cdot 48 \times 7}{28 \times 22}\)

or, \(\left(\frac{d}{2}\right)^2\) -5.29 = 0.96

or, \(\left(\frac{d}{2}\right)^2\) =0.96+5.29

or, \(\left(\frac{d}{2}\right)^2\) = 6.25= (2.5)^2

\(\frac{d}{2}\) =2.5

or, d = 2.5 x 2 = 5

Hence the length of the diameter of the cylinder = 5 dcm.

Example 10. The height of a right circular cylinder is twice of its radius. If the height would be 6 times of its radius, then the volume of the cylinder would be greater by 4312 cubic. dcm. Calculate the radius of the cylinder.

Solution:

Given:

The height of a right circular cylinder is twice of its radius. If the height would be 6 times of its radius, then the volume of the cylinder would be greater by 4312 cubic. dcm

Let the radius of the cylinder be r dcm.

∴ height = 2r dcm

∴ volume = \(\frac{22}{7}\) x r² x 2r cubic.dcm = \(\frac{44}{7}\) r³ cubic.dcm.

If the height be 6 times of its radius, i.e if h = 6r, then volume

= \(\frac{22}{7}\) x r² = 6r cubic.dcm

= \(\frac{132 r^3}{7}\) cubic.dcm

As per question, \(\frac{132 r^3}{7}\) – \(\frac{44}{7} r^3\) = 4312

or, \(\frac{88 r^3}{7}\) = 4312

or, r³ = \(\frac{4312 \times 7}{88}\)

or, r³ = 49 x 7 = 7³ or, r = 7

Hence the required radius = 7 dcm.

Example 11. A group of fire brigade personnel carried a right circular cylindrical tank filled with water and pumped out water at a speed of 420 metres per minute to put out the fire in 40 minutes by three pipes of 2 cm diameter each. If the diameter of the tank is 2.8 metres and its length is 6 metre, then calculate (1) what volume of water has been spent in putting out the fire and (2) the volume of water that still remains in the tank.

Solution:

Given:

A group of fire brigade personnel carried a right circular cylindrical tank filled with water and pumped out water at a speed of 420 metres per minute to put out the fire in 40 minutes by three pipes of 2 cm diameter each. If the diameter of the tank is 2.8 metres and its length is 6 metre

2 cm = \(\frac{2}{100}\) metre = 0.02 metre.

The volume of water pumped out in 40 minutes by one pipe

= \(\frac{22}{7} \times\left(\frac{0 \cdot 02}{2}\right)^2\) x 420 x 40 cubic.metres

= \(=\frac{22 \times 0.01 \times 0.01 \times 420 \times 40}{7}\) cubic.metres.

= 22 x 0.01 x 0.01 x 60 x 40 cubic.metres.

∴ The volume of water pumped out in 40 minutes by three pipes = 3 x 22 x 0.01 x 0.01 x 60 x 40 cubic.metres

= 3 x 22 x 0.01 x 0.01 x 60 x 40 x 1000 cubic.dcm = 15840 cubic.dcm = 15840 litres

The volume of the tank = \(\frac{22}{7} \times\left(\frac{2.8}{2}\right)^2\) x 6 cubic.metres

= \(\frac{22 \times 1 \cdot 4 \times 1 \cdot 4 \times 6}{7}\) = 36960 cubic.dcm = 36960 litres

∴ the water still remains = (36960 – 15840) litres = 21120 litres.

Example 12. It is required to make a plastering of sand and cement with 3.5 cm thick, surrounding four cylindrical pillars, each of whose diameter is 17.5 cm.

1. If each pillar is of 3 metre height, calculate how many cubic dcm of plaster materials will be needed?
2. If the ratio of sand and cement in the plaster material be 4 : 1, then how many cubic- dcm of cement will be needed?

Solution: Diameter = 17.5 cm = 1.75 dcm

∴ Radius = \(\frac{1 \cdot 75}{2}\) dcm = 0.875 dcm

Thickness of each pillar = 3.5 cm = 0.35 dcm

∴ Outer radius of the pillars after plastering will be (0.875 + 0.35) dcm = 1.225 dcm

∴ Volume of materials = \(\frac{22}{7}\){(1.225)² -(0.875)²} x 30 cubic.dcm [3m = 30dcm]

= \(\frac{22}{7}\) x (1.225 + 0.875)(1.225-0.875) x 30 cubic -dcm

= \(\frac{22}{7}\) x 2.1 x 0.35 x 30 cubic-dcm= 69.3 cubic.dcm

∴ Volume of materials needed for 4 pillars = 69.3 x 4 cubic.dcm = 277.2 cubic,dcm

1. 277.2 cubic.dcm plaster materials will be needed.
2. The ratio of sand and cement is 4 : 1

∴ part of cement = \(\frac{1}{4+1}=\frac{1}{5}\)

Hence the required cement =277.2 x \(\frac{1}{5}\) cubic.dcm = 55.44 cubic.dcm.

Example 13. The length of outer and inner diameter of a hollow right circular cylinder are 16 cm and 12 cm respectively. Height of cylinder is 36 cin. Calculate how many solid cylinders of 2 cm radius and 6 cm length may be obtained by melting this cylinder.

Solution:

Given:

The length of outer and inner diameter of a hollow right circular cylinder are 16 cm and 12 cm respectively. Height of cylinder is 36 cin.

The outer diameter =16 cm

∴ the outer radius = \(\frac{16}{2}\) cm = 8 cm.

The inner diameter = 12 cm

∴ the inner radius = \(\frac{12}{2}\) cm = 6 cm.

Height of the cylinder = 36 cm.

So, the volume of the materials of the hollow cylinder

= \(\frac{22}{7}\){(8)²-(6)²} x 36cc = \(\frac{22}{7}\) x 28 x 36 cc = 3168 cc.

The radius of small cylinders to be made = 2 cm and their lengths = 6 cm each.

∴ The volume of each small cylinders to be made = \(\frac{22}{7}\) X (2)² x 6cc = \(\frac{22}{7}\) x 4 x 6cc

So the required number of small cylinders

= \(\frac{\frac{22}{7} \times 28 \times 36}{\frac{22}{7} \times 4 \times 6}\) = 42.

Hence the required number of cylinders to be made is equal to 42.

[If the outer and inner radius (not diameter) be 16 cm and 12 cm respectively, then the number of cylinders is 168.]

Example 14. 11 cubic centimetres of iron is drawn into a wire of 56 cm long. Find the radius of the end of the wire.

Solution:

Given:

11 cubic centimetres of iron is drawn into a wire of 56 cm long.

Let the radius of the end of the wire be r cm.

∴ Volume of the wire = \(\frac{22}{7}\) x r² x 56 cc

Also, the volume of iron = 11 cc

∴ \(\frac{22}{7}\) x r² x 56= 11 or r² = \(\frac{11 \times 7}{22 \times 56}\)=\(\frac{1}{2 \times 8}\)=\(\frac{1}{16}\)

or, r = \(\sqrt{\frac{1}{16}}=\frac{1}{4}\) = 0.25

Hence the radius of the end of the wire is 0.25 cm.

Example 15. The upper portion of a cylindrical pillar is a hemisphere. If the radius of its base is 2 metres and its total length is 10 metres. Find out the volume of the pillar.

Solution:

Given:

The upper portion of a cylindrical pillar is a hemisphere. If the radius of its base is 2 metres and its total length is 10 metres.

Since the radius of the base is 2 metres, the height of the hemisphere on top of the pillar will be 2 metres.

∴ the height of the cylindrical portion = (10 – 2) metres = 8 metres

Now the volume of the hemisphere = \(\frac{1}{2}.\frac{4}{3}\) π. (2)³ cu.m = \(\frac{16}{3}\) π cu.m,

and the volume of the cylindrical portion = π.2².8 = 32π cu.m.

∴ The total volume of the pillar = \(\left(\frac{16}{3} \pi+32 \pi\right)\) cu-m. = \(\pi\left(\frac{16}{3}+32\right)\) cu-m

= \(\frac{22}{7} \times \frac{112}{3}\) cu-m.

= \(\frac{352}{3}\) cu-m.

= 117 \(\frac{1}{3}\) cu-m.

Hence the required volume = 117 \(\frac{1}{3}\) cu-m.