Mensuration Chapter 2 Right Circular Cylinder
⇒ A right circular cylinder is a solid generated by the revolution of a rectangle round of its side as its axis.
⇒ In a right circular cylinder, we have a curved surface or lateral surface and two circular plane surfaces with same radius.
⇒ The lateral surface area of a right circular cylinder = 2πrh units.
⇒ [Where r = length of the radius of the base of the right circular and h = height of the cylinder]
Read and Learn More WBBSE Solutions for Class 10 Maths
= 2πr x h units
= perimeter of the circular plane surface x height
⇒ Total surface area = Area of the lateral surface + area of two circular plane surface
= [2πrh + 2πr2] square unit
= 2 πr (h + r) square unit
⇒ Total surface area of one surface is open = 2 πrh + πr2
⇒ Volume = area of the base x height
= πr2 x h square unit
= πr2h square units
⇒ Volume of the material of a hollow cylinder = \(\left(\pi r_1{ }^2 h-\pi r_2{ }^2 h\right)\) cubic units.
= \(\pi\left(r_1^2-r_2^2\right) h\) cubic units
[where length of the outer radius is r1 units, the inner radius is r2 units and height is h units]
Total surface area of the hellowy cylinder
= 2 π (r1 + r2) h + \(2\pi\left(r_1^2-r_2^2\right)\) sq units
Class 10 Maths Mensuration Chapter 2 Solutions
Mensuration Chapter 2 Right Circular Cylinder True Or False
Example 1. The length of right circular drum is r cm and height is h cm. If half part of the drum is tilled with water then the volume of water will be πr2h cubic cm.
Solution: False
Example 2. If the length of radius of a right circular cylinder is 20 unit, the numerical value of volume and lateral surface area of cylinder will be equal for any height.
Solution: True
Example 3. Volume of a cylinder = Perimeter of the base x height.
Solution: False
Example 4. If length of outer radius and inner radius of a hollow right circular cylinder r1 units and r2 units respectively and height is h units, sum of outer and inner curved surface area = 2π (r1 + r2) h sq. units.
Solution: True
Example 5. Number of curved surfaces is 2 of a cylinder.
Solution: False
Class 10 Mensuration Chapter 2 Solved Examples
Example 6. In a solid cylinder no. of plane surfaces are 2.
Solution: True
Example 7. Total surface area of a cylinder with one end open = area of the base + area of the lateral surface.
Solution: True
Example 8. Radius of the right circular cylinder means radius of the base.
Solution: True
Example 9. If the volume of a cylinder is V cubic cm and height is h cm the radius of the base will be \(\sqrt{\frac{\mathrm{V}}{\pi h}}\)cm.
Solution: True
Example 10. Base of a cylinder is a square.
Solution: False
Wbbse Class 10 Mensuration Notes
Mensuration Chapter 2 Right Circular Cylinder Fill In The Blanks
Example 1. The length of a rectangular paper is b units and the breadth is b units. The rectangular paper is rolled and a cylinder is formed of which the perimeter is equal to the length of the paper. The lateral surface area of the cylinder is _______ sq. units.
Solution: lb
Example 2. The largest rod that can be kept in a right circular cylinder having the diameter of 3 cm and height 4 cm, then the length of rod is ______ cm.
Solution: 5
Surface Area And Volume Of Frustum Class 10
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
Example 4. Volume of a cylinder = ______ x height.
Solution: Area of the base
Example 5. Total surface area of cylinder = 2π x _______ x ______
Solution: radius, radius + height
Example 6. The ratio of radii of two cylinders having equal height is 3. : 4. The ratio of volumes is ______
Solution: 9: 16
Example 7. A _______ is formed when a cylinder is cut off according to its generating line.
Solution: rectangle
Example 8. If a rectangular paper is rolled with respect to its breadth as an axis then a _____ is formed.
Solution: cylinder
Surface Area And Volume Of Frustum Class 10
Mensuration Chapter 2 Right Circular Cylinder Short Answer Type Questions
Example 1. If the lateral surface area of a right circular cylindrical piller is 264 sq. metres and volume is 924 cubic metres then write the length of radius of the base of the cylinder.
Solution: 2πrh = 264, πr2h = 924
∴ \(\frac{\pi r^2 h}{2 \pi r h}=\frac{924}{264}\)
or, \(r=\frac{924 \times 2}{264}=7 \mathrm{mt}\)
Example 2. If the lateral surface area of a right circular cylinder is C sq. unit, length of radius of base is r unit and volume is V cubic unit, then write the value of \(\frac{Cr}{V}\).
Solution: 2πrh = C
\(\frac{\mathrm{Cr}}{\mathrm{V}}=\frac{2 \pi r h \cdot r}{\pi r^2 h}=2\)Example 3. If the height of a right circular cylinder is 14 cm and lateral surface area is 264 sq. cm, then write the volume of the cylinder.
Solution: 2πrh = 264
or, \(r=\frac{264 \times 7}{2 \times 22 \times 14}\) = 3
∴ Volume = πr2h
= \(\frac{22}{7}\) x 9 x 14 cubic cm = 396 cubic cm
Example 4. If the height of two right circular cylinder are in the ratio of 1 : 2 and perimeter are in the ratio of 3 : 4, then write two ratio of their volumes.
Solution: \(\frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} \quad\left[\frac{2 \pi r_1}{2 \pi r_2}=\frac{3}{4} \Rightarrow \frac{r_1}{r_2}=\frac{3}{4}\right]\)
= \(\frac{9}{16}\) x \(\frac{1}{2}\) = 9: 32
Example 5. The length of the radius of a right circular cylinder is decreased by 50% and height is increased by 50%, then write how much percent of the volume will be changed.
Solution: New volume = \(\pi\left(\frac{r}{2}\right)^2 \cdot\left(3 \frac{h}{2}\right)\) cubic unit
= \(\frac{3}{8} \pi r^2 h\) cubic unit
% change = (1 – \(\frac{3}{8}\)) x 100 cubic unit
= \(\frac{5}{8}\) x 100 cubic unit
= \(\frac{125}{2}\) cubic unit = 62 \(\frac{1}{2}\) cubic unit
Example 6. The volume of a right circular cylinder is 352 c.c. If the height of the cylinder is 7 cm, what is the length of the radius?
Solution: πr2 x 7 = 352
⇒ \(r^2=\frac{352 \times 7}{7 \times 22}=16\)
∴ r = 4 cm.
Class 10 Maths Mensuration Important Questions
Example 7. The area of the curved surface area of a cylinder is 528 sq. mt and its volume is 792 cu. mt. What is its radius?
Solution: 2πrh = 528, πr2h = 792
∴ \(\frac{\pi r^2 h}{2 \pi r h}=\frac{792}{528} \quad r=\frac{3}{2} \times 2=3 \mathrm{mt} .\)
Example 8. Find the radius of a solid cylinder of height 6 cm, if the magnitude of the volume of the cylinder be equal to magnitude of its whole surface area.
Solution: 2πr (h + r) = πr2h
or, 2 (h + r) = rh
∴ 2 (6 + r) = 6r
∴ r = 3 cm
Example 9. The diameter and height of a cylindrical drum are 1 cm and 14 cm respectively, then find the volume.
Solution: Volume = πr2h = \(\frac{22}{7} \times\left(\frac{1}{2}\right)^2\) x 14 cu. cm = 11 cu. cm
Example 10. Volume and lateral surface area of a solid cylinder are equal by numerical value. find its radius of the base.
Solution: πr2h = 2πrh
or, r = 2 units