Ex 13.2 NCERT Solutions Class 9 Chapter 13
The solutions to exercise 13.2 from class 9th's maths NCERT book is given below.
The exercise has 11 questions which have been explained thoroughly with proper reasoning.
Answer: Radius of cylinder = r
Height of the cylinder = 14 cm
CSA of cylinder = 88 cm2
Curved surface area of a cylinder = 2 π rh
88 = 2 x (22/7) x 14 x r
r= 1 cm
Diameter of cylinder = 2 r
= 2 x 1 cm
= 2 cm
Answer: Height of the cylindrical tank = 1 m
Diameter of the cylindrical rank = 140 cm = 1.4 m
Radius = 0.7 m
Since it is to be a closed cylindrical tank we will find TSA
TSA of the cylindrical tank
= 2Ï€ r h ( r + h)
= 2 x (22/7) x 0.7 x 1 x ( 0.7 + 1 )
= 2 x 22 x 0.1 x 1.7
= 4.4 x 1.7
= 7.48 m2
Answer:
(i) Inner diameter of the cylinder = 4 cm
Inner radius of the cylinder = 2 cm
Height of the cylinder = 77 cm
Inner CSA
= 2Ï€rh
= 2 x (22/7) x 2 x 77
= 2 x 22 x 2 x 11
= 44 x 22
= 968 cm2
(ii)Outer diameter of the cylinder = 4.4 cm
Outer radius of the cylinder = 2.2 cm
Height of the cylinder = 77 cm
Inner CSA
= 2Ï€rh
= 2 x (22/7) x 2.2 x 77
= 2 x 22 x 2.2 x 11
= 1064.80 cm2
(iii) The total surface area of the cylinder:
CSA of (inner cylinder + outer cylinder) + 2 x( area of the outer circle - the area of the inner circle)
= 968 + 1064.80 + 2 [π (2.2) - π(2)]
= 2032.80 + 2 π[ (2.2)2 - (2)2]
= 2032.80 + 2 π[ (2.2)2 - (2)2]
= 2032.80 + [2 x (22/7) x 4.2 x 0.2]
= 2032.80 + (2 x 22 x 0.6 x 0.2)
= 2032.80 + (44 x 0.12)
= 2032.80 +5.28
= 2038.08 cm2
A roller is in the form of a cylinder and when it revolves, only its curved area touches the earth, not the cylinders.
So we will only find the CSA.
CSA of the roller
= 2Ï€rh
= 2 x (22/7) x 42 x120
= 2 x 22 x 6 x 120
= 12 x 22 x 120
= 31680 cm2
In 1 revolution the roller covers area
= 31680 cm2 = 3.1680 m2
In 500 revolutions it will cover area
= 3.1680 x 500
= 1584 m2
CSA of the cylindrical pillar
= 2Ï€rh
= 2 x (22/7) x 0.25 x 3.5
= 2 x 22 x 0.25 x 0.5
= 5.50 m2
Cost of painting 1 m2= Rs 12.50
Cost of painting 550 m2 = 12.50 x 5.50
= Rs 68.75
CSA of the cylinder = 4.4 m2
2Ï€rh = 4.4 m2
2 (22/7) x 0.7 x h = 4.4
h= 1 m
Radius of the well = 3.5/2 = 1.75 m
Height of the circular well = 10 m
Inner CSA of the well
= 2Ï€rh
= 2 x (22/7) x 1.75 x 10
= 2 x 22 x 0.25 x 10
= 44 x 2.5
= 110 m2
Cost of plastering the 1 m2 of the inner CSA of the well = Rs 40
Cost of plastering the 110 m2 of the inner CSA of the well
= Rs 40 x 110
= Rs 4400
Height of the cylindrical pipe
= 28 m = 2800 cm
Radius of the pipe
= 5/2 = 2.25 cm
Total radiating surface = CSA = 2Ï€rh
= 2 x (22/7) x 2.25 x 2800
= 2 x 22 x 2.25 x 400
= 44 x 225 x 4
= 44 x 900
= 39600 cm2
(i) CSA of cylindrical petrol tank
= 2Ï€rh
= 2 x (22/7) x 2.1 x 4.5
= 44 x 0.3 x 4.5
= 59.4 m2
(ii) Let the originally used quantity of steel = n
Original − wasted = 59.4
n − (n/12) = 59.4
(11/12) x n = 59.4
n= 59.4 (12/11)
n= 5.4 x 12
n= 64.8 m2
The radius of the base of the cylinder = 10 cm
Height = 30 cm + 2.5 + 2.5 = 35 (margin of 2.5 is to top and bottom)
CSA of the cylinder
= 2Ï€rh
= 2 x (22/7) x 10 x 35
= 44 x 10 x 5
= 2200 cm2
In the case of the penholder, the area of the circle at the top is not considered
Area of cardboard required to make 1 penholder = CSA + Area of the base
= 2πrh + πr2
= (22/7) x { (2 x 3 x 10.5) + (3 x 3)}
= (22/7) x {63+9}
= (22/7) x 72
Area of card board needed for such 35 penholders
= (22/7) x 72 x 35
= 22 x 72 x 5
= 110 x 72
=7920 cm2
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