# 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 1

### Solution 2

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 m

^{2}or 44000 cm

^{2}.

### Solution 3

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

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 5

### Solution 6

### Solution 7

### Solution 8

Depth (h) of circular well = 10 m

= (44 x 0.25 x 10)

= 110 m

^{2}

^{2}area = Rs 40

Cost of plastering 110 m

^{2}area = Rs (110 x 40) = Rs 4400

### Solution 9

Height of penholder = 10.5 cm

Surface area of 1 penholder = CSA of penholder + Area of base of

Area of cardboard sheet used by 1 competitor =

Area of cardboard sheet used by 35 competitors

= 7920 cm2

Thus, 7920 cm

^{2}of cardboard sheet will be required for the competition.

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

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 m

^{2}

Let A m

^{2}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

Breadth (b) of tin can = 4 cm

Height (h) of tin can = 15 cm

Capacity of tin can = l b h = (5 4 15) cm

^{3}= 300 cm

^{3}

^{2}H ==385 cm

^{3}

Difference in capacity = (385 - 300) cm

^{3}= 85 cm

^{3}

### Solution 2

### Solution 3

### Solution 4

Let radius of cylinder be r.

CSA of cylinder = 94.2 cm

^{2}

2rh = 94.2 cm

^{2}

(2 3.14 r 5) cm = 94.2 cm

^{2}

r = 3 cm

^{2}h = (3.14 (3)

^{2}5) cm

^{3}= 141.3 cm

^{3}

### Solution 5

Height (h) of the cylindrical vessel = 1 m

Volume of cylindrical vessel = 15.4 litres = 0.0154 m

^{3}

^{}

^{2}of metal sheet would be needed to make the cylindrical vessel.

### Solution 6

Height (h) up to which the bowl is filled with soup = 4 cm

^{2}h=

Volume of soup in 250 bowls = (250 154) cm

^{3}= 38500 cm

^{3}= 38.5 litres

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

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

### Solution 1

### Solution 2

### 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

= (∏r^{2}) × h

V = ∏r^{2}h

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 = ∏r^{2}h

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

V' = ∏(2r)^{2}h = 4∏r^{2}h

V' = 4V

Hence, correct option is (d).

### Solution 8

Volume of cylinder V = ∏r^{2}h

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

V' = ∏(r)^{2}(2h) = 2∏r^{2}h = 2V

Hence, correct option is (a).

### Solution 9

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### Solution 13

### Solution 17

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

T.S.A. = ∏r^{2} + ∏r^{2} + 2∏rh = 2∏r^{2} + 2∏rh = 2∏r (r + h)

Hence, correct option is (a).

## Surface Areas and Volume of a Circular Cylinder Exercise 19.30

### Solution 18

### Solution 19

### Solution 20