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 =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85881_af6215bbad1f2428bd8e2aa8b877d5eb.png)
CSA of cylindrical pipe =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85881_fc3f6bef2a68e31aeedb1f9034513f6f.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85881_72f1af771aca82a5db589180340b0cdd.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85881_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Thus, the area of radiating surface of the system is 4.4 m2 or 44000 cm2.
Solution 3
Radius of the circular end of the pillar =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85891_6680664022c132e085b31a8ff53a1678.png)
CSA of pillar =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85891_fc3f6bef2a68e31aeedb1f9034513f6f.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85891_39ee3741859b5ad04fd71ac1d20c1412.png)
Cost of painting 1
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85891_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Cost of painting 5.5
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85891_20eb865cc04235a6c2d39ab1fefc1bd5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85891_b9ccb383b0c73c26cfb3ff194a86e235.png)
Thus, the cost of painting the CSA of pillar is Rs 68.75.
Solution 4
Base radius (r) of cylindrical tank =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85901_2f13cc6f4f13166b3713e322c5f41164.png)
Area of sheet required = total surface area of tank =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85901_8c956b5fe65cc8e39c09fa758148f942.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85901_46871a71e330b1206153fc06e2e0178f.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85901_554f6475bc40bb441e247f45116c79da.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85901_9ac408babc96343da36d19e69fc01a93.png)
So, it will require 7.48
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85901_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Solution 5
Solution 6
Solution 7
Solution 8
Depth (h) of circular well = 10 m
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85911_fc3f6bef2a68e31aeedb1f9034513f6f.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85911_0cdb1c0a34373f7d6d17a85dbc5d5178.png)
= (44 x 0.25 x 10)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85911_20eb865cc04235a6c2d39ab1fefc1bd5.png)
= 110 m2
Cost of plastering 110 m2 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
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85931_fc3f6bef2a68e31aeedb1f9034513f6f.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85931_54d799619cb208abac9c32ad9ad9ad79.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85931_84dd810415f636effbc5d92aa6de3874.png)
Area of cardboard sheet used by 1 competitor =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85931_427cefd18c9ee168733c514422d51fab.png)
Area of cardboard sheet used by 35 competitors
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85931_cbecf4a9cc98afee27465b26e7709de4.png)
Thus, 7920 cm2 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 =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85921_963bc27d98a72229aee7c3948a088b15.png)
(i) Lateral or curved surface area of tank =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85921_fc3f6bef2a68e31aeedb1f9034513f6f.png)
=
= 59.4 m2
(ii) Total surface area of tank = 2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85921_2ceecdf317f54391e92091088b7215b4.png)
=
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85921_53d10ecd20d7d1088a7d7e10363e1255.png)
= 87.12 m2
Let A m2 steel sheet be actually used in making the tank.
Thus, 95.04
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85921_20eb865cc04235a6c2d39ab1fefc1bd5.png)
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
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_4a463b6ff5e6c2b4db604fe74427d772.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85941_ad433fab832ca5f247afc6b4b1aa1a08.png)
Difference in capacity = (385 - 300) cm3 = 85 cm3
Solution 2
Solution 3
Solution 4
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm2
2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b30c14c35f7e20752f646111109b139b.png)
(2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b9ccb383b0c73c26cfb3ff194a86e235.png)
r = 3 cm
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85961_b9ccb383b0c73c26cfb3ff194a86e235.png)
Solution 5
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85971_1da24ff3bf1acd9caf94306bc7c5376c.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85971_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85971_8a21452002d7923285f45bca13aad9e7.png)
Solution 6
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85981_1bbff04738f69ed2751f6ca18af28e0d.png)
Height (h) up to which the bowl is filled with soup = 4 cm
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85981_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85981_41f56a7ba4cd75d1e9d1cbbac41fd411.png)
Volume of soup in 250 bowls = (250
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/85981_b9ccb383b0c73c26cfb3ff194a86e235.png)
Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
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
= (∏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 9
Solution 14
Solution 15
Solution 16
Solution 10
Solution 11
Solution 12
Solution 13
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).