2. W HAT IS A CYLINDER ?
A SoLID gENERATED bY THE
REvoLuTIoN of A RECTANgLE AbouT
oNE of ITS SIDES IS CALLED A
CYLINDER.
3. CoNSIDER A CYLINDER WITH RADIuS R AND
HEIgHT H.
A CYLINDER HAS ATWo CIRCuLAR ENDS
WHICH ARE PARALLEL .EACH of THESE IS
CALLED bASE of THE CYLINDER.
THE LINE SEgMENT JoININg THE CENTRES of
THE TWo bASES IS CALLED THE AXIS of THE
CYLINDER.
THE PERPENDICuLAR DISTANCE bETWEEN
THE TWo ENDS IS CALLED THE HEIgHT of THE
CYLINDER.
4. AREA of THE LATERAL SuRfACE of THE
CYLINDER=AREA of THE RECTANguLAR
STRIP of PAPER
=AREA of A RECTANgLE of
LENgTH 2*22/7*R
=2*22/7*R*H
=2*22/7*R*H SquARE uNITS
THuS foR A CYLINDER of RADIuS R AND
HEIgHT H WE HAvE,
LATERAL (CuRvED SuRfACE)
AREA=2*22/7*R*H Sq uNITS
ToTAL SuRfACE AREA
=(2*22/7*R*H+2*22/7*R*R)
5. LET R and D BE THE EXTERNAL AND INTERNAL RADII OF
CYLINDER AND
h BE
ITS HEIGHT .
1-SURFACE AREA OF EACH BASE=22/7(R*R-D*D)
2-curved surface area=(external surface area )+(internal
surface
area)
=2*22/7*R*H+2*22/7*D*H
=2*22/7*h(R+D) sq units
3-total surface area =2*22/7*R*h+2*22/7*D*h+2*22/7(R*R
D*D)
6. EX 1-A RECTANguLAR SHEET of PAPER 44CM*18CM IS RoLLED
ALoNg ITS LENgTH AND A CYLINDER IS foRMED. fIND THE
RADIuS of THE CYLINDER.
SoLuTIoN-WHEN THE RECTANguLAR SHEET IS RooLED ALoNg ITS
LENgTH ,WE fIND THAT THE LENgTH of THE SHEET foRMS THE
CIRCuMfERENCE of ITS bASE AND bREADTH of THE SHEET
bECoMES THE HEIgHT of THE CYLINDER.
LET R CM bE THE RADIuS of THE bASE AND HCM bE THE HEIgHT.
THEN H=18CM
CIRCuMfERENCE of THE bASE=L of THE SHEET
oR 2*22/7*R=44
oR R= 7CM
44cm
18 cm
18 cm
7. EX -2
A CoMPANY PACkAgES ITS MILk PoWDER IN CYLINDRICAL CoNTAINERS
WHoSE bASE HAS A DIAMETRE of 16.8CM AND HEIgHT
20.5CM.CoMPANY PLACES A LAbEL ARouND THE CuRvED SuRfACE of
THE CoNTAINER. If THE LAbEL IS PLACED 1.5CM fRoM THE ToP AND
THE boTToM, WHAT IS THE SuRfACE AREA of THE LAbEL?
SoLuTIoN—CLEARLY SuRfACE AREA of THE LAbEL IS EquAL To THE
CuRvED SuRfACE AREA of A CYLINDER of bASE RADIuS
R=16.8/2CM=8.4CMAND HEIgHT H=(20.5-1.5-1.5)CM=17.5CM
THEREfoRE SuRfACE AREA of THE LAbEL=2*22/7*8.4*17.5CM²
=2*22/7*1.2*17.5CM²=924CM²
1.5CM 20.5C
POWDERED MILK M
16.8
CM
8. TAkE CIRCuLAR SHEETS of
RADIuS R AND STuCk THEM uP
vERTICALLY To foRM A CYLINDER
of HEIgHT H.
voLuME of THE CYLINDER=
MEASuRE of THE SPACE oCCuPIED
bY THE CYLINDER= r
THE AREA of EACH CIRCuLAR
SHEET
=22/7*R²H
h
9. Volume of a hollow cylinder=
Exterior volume-interior
R
volume
=22/7R²h-22/7r²h
=22/7*h(R²-r²)
h
r
10. A WELL WITH 10M INSIDE DIAMETER IS Dug ouT 14M
DEEP. EARTH TAkEN ouT of IT IS SPREAD ALL
ARouND To A WIDTH of 5MTo foRM AN
EMbANkMENT. fIND ITS HEIgHT of EMbANkMENT.
voLuME of THE EARTH Dug ouT=22/7*R²*H*M³
=22/7*5*14M³=1100M³
AREA of THE EMbANkMENT=22/7(R²-R²)
=22/(10²-5²)M²=22/7*75M²
THEREfoRE THE HEIgHT of THE EMbANkMENT=voLuME
of THE EARTH Dug ouT/AREA of THE EMbANkMENT
=1100/22/7*75=7*1100/22*75=4.66M
15m
10m
11. EX-2:THE RAIN WATER THAT fALLS oN A Roof of AREA
6160M² IS CoLLECTED IN A CYLINDRICAL TANk of
DIAMETER 14M AND HEIgHT 10M AND THuS THE TANk IS
CoMPLETELY fILLED. fIND THE HEIgHT of RAIN WATER oN
THE Roof.
SoLuTIoN-LET THE HEIgHT of RAIN WATER oN THE Roof bE H
METRE. CLEARLY RAIN WATER ACCuMuLATED oN THE Roof
foRMS A CuboID of bASE AREA 6160M² AND HEIgHT H
METRE.THEREfoRE THE voLuME of WATER ACCuMuLATED
oN THE Roof= 6160*H M³
THIS WATER CoMPLETELY fILLS A CYLINDER TANk of RADIuS
7M AND HEIgHT 10M.
THEREfoRE WATER ACCuMuLATED oN THE Roof=voLuME of
THE CYLINDER TANk=
6160H=22/7*7²*10 oR
H=1540/6160M=0.25M=25CM
HENCE, THE HEIgHT of THE RAIN WATER oN THE Roof =25CM.
12. Ex 3-how Many Cubic Metres Of Earth Must
Be Dug Out To Sink A Well 22.5m Deep
And Of Diametre 7m?Also,find The Cost
Of Plastering The Inner Curved Surface At
Rs3/M².
Solution-volume Of Earth To Be Dug Out =
Volume Of The Well =
22/7*7/2*7/2*22.5 M³=866.25 M³
Therefore,area Of Inner Curved
Surface=2*22/7*r*h=2*22/7*7/2*22.5
M²=495m²
Cost Of Platering The Inner Curved Surface
=Rs (495*3)=rs 1485m²
