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# Class 9 RD SHARMA Solutions Maths Chapter 19 - Surface Areas and Volume of a Circular Cylinder

## Surface Areas and Volume of a Circular Cylinder Exercise Ex. 19.1

### Solution 2

Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe =  cm = 2.5 cm = 0.025 m
CSA of cylindrical pipe =   = 4.4
Thus, the area of radiating surface of the system is 4.4 m2 or 44000 cm2.

### Solution 3

Height of the pillar = 3.5 m
Radius of the circular end of the pillar = cm = 25 cm  = 0.25 m
CSA of pillar =  =
Cost of painting 1  area = Rs 12.50
Cost of painting 5.5  area = Rs (5.5 12.50) = Rs 68.75

Thus, the cost of painting the CSA of pillar is Rs 68.75.

### Solution 4

Height (h) of cylindrical tank = 1 m.
Base radius (r) of cylindrical tank =  = 70 cm = 0.7 m

Area of sheet required = total surface area of tank =

So, it will require 7.48 of metal sheet.

### Solution 8

Inner radius (r) of circular well = 1.75 m
Depth (h) of circular well = 10 m
(i) Inner curved surface area =

= (44 x 0.25 x 10)
= 110 m2
(ii) Cost of plastering 1 m2 area = Rs 40
Cost of plastering 110 m2 area = Rs (110 x 40) = Rs 4400

### Solution 9

Radius of circular end of cylindrical penholder = 3 cm
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of
SA of 1 penholder =  +

Area of cardboard sheet used by 1 competitor =
Area of cardboard sheet used by 35 competitors

= 7920 cm2
Thus, 7920 cm2 of cardboard sheet will be required for the competition.

### Solution 14

Height (h) cylindrical tank = 4.5 m
Radius (r) of circular end of cylindrical tank =m = 2.1m
(i)    Lateral or curved surface area of tank =

=
= 59.4 m2

(ii)    Total surface area of tank = 2 (r + h)

=
= 87.12 m2

Let A m2 steel sheet be actually used in making the tank.

Thus, 95.04  steel was used in actual while making the tank.

## Surface Areas and Volume of a Circular Cylinder Exercise Ex. 19.2

### Solution 1

The tin can will be cuboidal in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l  h = (5  15) cm3 = 300 cm3
Radius (R) of circular end of plastic cylinder =
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = R2H  ==385 cm3
Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 - 300) cm3 = 85 cm3

### Solution 4

(i)    Height (h) of cylinder = 5 cm
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm2
2rh = 94.2 cm2
(2  3.14  5) cm = 94.2 cm2
r = 3 cm

(ii)    Volume of cylinder = r2h = (3.14  (3)2  5) cm3 = 141.3 cm3

### Solution 5

Let radius of the circular ends of the cylinder be r.
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
Total  Surface area of vessel = 2 r(r+h)

Thus, 0.4708 m2 of metal sheet would be needed to make the cylindrical vessel.

### Solution 6

Radius (r) of cylindrical bowl = cm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = r2h=
Volume of soup in 250 bowls = (250  154) cm3 = 38500 cm3 = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

## Surface Areas and Volume of a Circular Cylinder Exercise 19.28

### Solution 3

Number of Surfaces In a Right cylinder are 3.

Top surface, bottom surface and curved surface.

Hence, correct option is (c).

### Solution 4

Vertical cross-section of cylinder will always be a Rectangle of sides 'h', and 'r',

where h is the height of a cylinder and r is the radius of a cylinder.

Hence, correct option is (b).

### Solution 5

Volume of cylinder

= Area of Base × Height

= (∏r2) × h

V = ∏r2h

Hence, correct option is (b).

### Solution 6

A Hollow cylinder has only 2 surfaces.

One is outer-curved surface and another is inner-curved surface.

Hence, correct option is (b).

## Surface Areas and Volume of a Circular Cylinder Exercise 19.29

### Solution 7

Volume  of a cylinder = V = ∏r2h

If r' = 2r and h' = h, then

V' = ∏(2r)2h   = 4∏r2h

V' = 4V

Hence, correct option is (d).

### Solution 8

Volume of cylinder V = ∏r2h

If h' = 2h and r' = r, then

V' = ∏(r)2(2h) = 2∏r2h = 2V

Hence, correct option is (a).

### Solution 17

Total surface Area = Area of Top + Area of bottom + Curved Surface Area

T.S.A. = ∏r2 + ∏r2 + 2∏rh = 2∏r2 + 2∏rh = 2∏r (r + h)

Hence, correct option is (a).