Right Circular and Hollow Cylinder
✔ A right circular cylinder is a solid generated by the revolution of a rectangle about one of its
sides as axis
✔ If the axis is perpendicular to the radius then the cylinder is called a tight circular cylinder,
✔A solid cylinder is an object bounded by two circular plane surfaces and a curved surface.
✔ C.S.A. of a right circular cylinder-2xth sq. units.
✔TS.A. of a right circular cylinder = 2h+r) sq units
✔ An object bounded by two co-axial cylinders of the same height and different radit is called a "hollow cylinder".
✔ C.S.A of a hollow cylinder 2x(R+r)h sq. units
✔TS.A. of a hollow cylinder - 2x(R +r)(R-r+h) sq. units
✔Right Circular and Hollow Cone
✔A right circular cone is a solid generated by the revolution of a right angled triangle about one of the sides containing the right angle as axis. If the right triangle ABC revolves about AB as axis, the hypotenuse AC generates the curved surface of the cone.
✓ The height of the cone is the length of the axis AB, and the slant height is the length of the hypotenuse AC
✔Curved Surface Area of the cone Area of the Sector=nr/sq units.
✔TSA of a right circular cone nr/+r) sq units, where / √√² +²
CSA of a right circular cylinder = 2🈷️rh sq. units.
TSA of a right circular cylinder = 2pr (htr) sq. units.
EXERCISE 7.1
1. The radius and height of a cylinder are in the ratio 5:7 and its curved surface area is 5500 sq.em. Find its radius and height.
Solution:
Given r: h=5:7
h 7
⇒7r=Sh
7r
e
N
CSA of Cylinder-5500
2arh=5500CSA of Cylinder-5500
2arh=5500
22. 7r
44
5500
5500×5 44
500x5
-125x5
625 P=25
2 h = 75×25=35
Radius - 25 cm, Height -35 cm
Subr=7 in (1)
22
2x x7xh-1540
1540
h-35
Radius=7 m, Height = 35 m.
The external radius and the length of a
hollow wooden log are 16 cm and 13 cm respectively. If its thickness is 4 cm then
2. A solid iron cylinder has total surface area of 1848 sq.m. Its curved surface. area is five-sixth of its total surface area. Find the radius and height of the Given, external radius of iron cylinder.
find its T.S.A.
Solution:
hollow cylinder-16 cm - R
length of log=13cm R
thickness-4 cm-1
t =R-r
Solution:
Given total surface area of cylinder
= 1848 m² &
(TSA) CSA (TSA) =
27th=-x1848
6
= 5x308
27th=1540
2ar (h+r) 1848
⇒ 2mh+2m² = 1848
1540+2m² -1848
-16-4
r = 12
.. TSA of hollow cylinder
-2x (R+ 1) (R-r+h)
-2x - (28) (4+13)
=44x4x17 -2992 cm²
A right angled triangle PQR where Q -90° is rotated about QR and PQ. If QR = 16 cm and PR = 20 cm, compare the curved surface areas of the right circular cones so formed by the triangle.
(1)
A
08
L.K. AL
r=7m
<18-x36×12-3×216] =#36x18
648-144-144]
x360
22
7920
7
=1131.42cm
5. A capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?
Solution:
15
15
12cm
Hemisphere r-1.5mm-3
Cylinder
H-9mm
T=1.5 mm= Volume of the Capsule
Vol. of Cylinder +2 (Vol, of hemisphere).
- mr²H+2
6cm 12cm
60 m
Hemisphere Cylinder
r=6cm
16cm
H=18cm
Solution:
Cone
r=6cm
h=12cm
Volume of water displaced out of cylinder= Volumme of cone + Volume of HS
1 22 x-x36(12+12)
22 =x12x24
905.14cm³
Note: When the conical hemisphere is com pletely submerged in water inside the cylinder,
Volume of water left in the cylinder
-Volume of cylinder -[Volume of cone + Vol of Hemisphere)
A
Stame
08
L.K. AL
22 81+18
