Class 9 NCERT Solutions Maths Chapter 11 - Surface Areas and Volumes
Ex. 11.1
Ex. 11.2
Ex. 11.3
Ex. 11.4
Surface Areas and Volumes Exercise Ex. 11.1
Solution 1
Radius of base of cone =
cm = 5.25 cm
Slant height of cone = 10 cm
CSA of cone =
= ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1161_3e154435b6c3633940f47ca0d4ba2817.png)
Thus, the curved surface area of cone is 165
.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1161_62fc857ad9fe8b9af39b85ac6c7e2364.png)
Slant height of cone = 10 cm
CSA of cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1161_6a426bd6d4839843b6a811138e3d6b88.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1161_3e154435b6c3633940f47ca0d4ba2817.png)
Thus, the curved surface area of cone is 165
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1161_4d362ce23c4fe947e211016963b3d188.png)
Solution 2
Radius of base of cone =
m = 12 cm
Slant height of cone = 21 m
Total surface area of cone =
(r + l)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1163_c2628ccf41a71a937f3a12fc50f168ae.png)
Slant height of cone = 21 m
Total surface area of cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1163_2ceecdf317f54391e92091088b7215b4.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1163_2fd03e00b7771d3aa7fd40f09592d91c.png)
Solution 3
(i) Slant height of cone = 14 cm
Let radius of circular end of cone be r.
CSA of cone =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1170_6a426bd6d4839843b6a811138e3d6b88.png)
Let radius of circular end of cone be r.
CSA of cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1170_6a426bd6d4839843b6a811138e3d6b88.png)
Thus, the radius of circular end of the cone is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base
=
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1170_8ae85efde0fd761d59ae15c9b49972c1.png)
Thus, the total surface area of the cone is 462
.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1170_4d362ce23c4fe947e211016963b3d188.png)
Solution 4
(i) Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let slant height of conical tent be l.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_ae5fb395aa20be976748f18659d994fb.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_00137473969b0b5e694e03c59f34741b.png)
Radius (r) of conical tent = 24 m
Let slant height of conical tent be l.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_ae5fb395aa20be976748f18659d994fb.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_10e34c9498fa699d2da4a908d7fc1f2a.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_d9cbcc31dd1c5737e2402e9343e90f8e.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_167ff63a3f239d4b4ec1dc52a4024700.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_00137473969b0b5e694e03c59f34741b.png)
.
Thus, the slant height of the conical tent is 26 m.
(ii) CSA of tent =
= ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_3c06420d4c571da8696346a3685b5255.png)
Cost of 1
canvas = Rs 70
Cost of
canvas =
= Rs 137280
Thus, the cost of canvas required to make the tent is Rs 137280.
(ii) CSA of tent =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_6a426bd6d4839843b6a811138e3d6b88.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_3c06420d4c571da8696346a3685b5255.png)
Cost of 1
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Cost of
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_cebd1ef1d4dbb13bb61c8704ea255271.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1171_93d7bf36a030af0708cb92b6fd9b95ab.png)
Thus, the cost of canvas required to make the tent is Rs 137280.
Solution 5
Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent =
= (3.14
6
10)
= 188.4 ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(L - 0.2 m)
3] m = 188.4 ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_20eb865cc04235a6c2d39ab1fefc1bd5.png)
L - 0.2 m = 62.8 m
L = 63 m
Thus, the length of the tarpaulin sheet will be 63 m.
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_b07db2a138b9cd3d76896950b166e7cb.png)
CSA of conical tent =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_6a426bd6d4839843b6a811138e3d6b88.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_20eb865cc04235a6c2d39ab1fefc1bd5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(L - 0.2 m)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1172_20eb865cc04235a6c2d39ab1fefc1bd5.png)
L - 0.2 m = 62.8 m
L = 63 m
Thus, the length of the tarpaulin sheet will be 63 m.
Solution 6
Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb =
= 7 m
CSA of conical tomb =
=
Cost of white-washing 100
area = Rs 210
Cost of white-washing 550
area =Rs
= Rs 1155
Thus, the cost of white washing the conical tomb is Rs 1155.
Base radius (r) of tomb =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1175_27184b040fdbd198b3d03ea2e06acef7.png)
CSA of conical tomb =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1175_6a426bd6d4839843b6a811138e3d6b88.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1175_f4faf625d2465dd52bf6063fc81bb032.png)
Cost of white-washing 100
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1175_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Cost of white-washing 550
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1175_20eb865cc04235a6c2d39ab1fefc1bd5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1175_a1ccf19f2d2f693e16a72a6f06991338.png)
Thus, the cost of white washing the conical tomb is Rs 1155.
Solution 7
Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 cm
Slant height (l) of conical cap =
CSA of 1 conical cap =
= ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_7129eae86e6459a3dcc727337303280d.png)
CSA of 10 such conical caps = (10
550)
= 5500 ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_4d362ce23c4fe947e211016963b3d188.png)
Thus, 5500
sheet will be required to make the 10 caps.
Height (h) of conical cap = 24 cm
Slant height (l) of conical cap =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_0d5a75fef36c1a0dfc09811c0222d3f0.png)
CSA of 1 conical cap =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_6a426bd6d4839843b6a811138e3d6b88.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_7129eae86e6459a3dcc727337303280d.png)
CSA of 10 such conical caps = (10
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_4d362ce23c4fe947e211016963b3d188.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_4d362ce23c4fe947e211016963b3d188.png)
Thus, 5500
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1178_4d362ce23c4fe947e211016963b3d188.png)
Solution 8
Radius (r) of cone =
= 0.2 m
Height (h) of cone = 1 m
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_ee14159fea1d24685d2b102eae0ae1b6.png)
Height (h) of cone = 1 m
Slant height (l) of cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_54edddcf7eb9936e63d6281bef275624.png)
CSA of each cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_6a426bd6d4839843b6a811138e3d6b88.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_20eb865cc04235a6c2d39ab1fefc1bd5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_20eb865cc04235a6c2d39ab1fefc1bd5.png)
CSA of 50 such cones = (50
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_b9ccb383b0c73c26cfb3ff194a86e235.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_20eb865cc04235a6c2d39ab1fefc1bd5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Cost of painting 1
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_20eb865cc04235a6c2d39ab1fefc1bd5.png)
Cost of painting 32.028
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_20eb865cc04235a6c2d39ab1fefc1bd5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1182_b9ccb383b0c73c26cfb3ff194a86e235.png)
Thus, it will cost Rs 384.34 (approximately) in painting the 50 hollow cones.
Surface Areas and Volumes Exercise Ex. 11.2
Solution 1
(i) Radius of sphere = 10.5 cm
Surface area of sphere =
(ii) Radius of sphere = 5.6 cm
Surface area of sphere = =
(iii) Radius of sphere = 14 cm
Surface area of sphere = =
Solution 2
(i) Radius of sphere
Surface area of sphere
(ii) Radius of sphere
Surface area of sphere
(iii) Radius of sphere
Surface area of sphere =
Solution 3
Radius of hemisphere = 10 cm
Total surface area of hemisphere![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1188_93acc9d7bbc2775efa08ae05f5249a62.png)
Total surface area of hemisphere
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1188_93acc9d7bbc2775efa08ae05f5249a62.png)
Solution 4
Radius
of spherical balloon = 7 cm
Radius
of spherical balloon, when air is pumped into it = 14 cm
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1190_bf2a7b4d7a1dde2b71252208831b89d3.png)
Radius
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1190_290067d73320802f9ea61e171f223566.png)
Solution 5
Inner radius (r) of hemispherical bowl = ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_282c3662e8141f566027787ee4ba0690.png)
Surface area of hemispherical bowl =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_199b0809bd4e5a058a9f82a040751b58.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_282c3662e8141f566027787ee4ba0690.png)
Surface area of hemispherical bowl =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_199b0809bd4e5a058a9f82a040751b58.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_6b728a860f34430b9d263447ccaed612.png)
Cost of tin-plating 100
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_4d362ce23c4fe947e211016963b3d188.png)
Cost of tin-plating 173.25
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_4d362ce23c4fe947e211016963b3d188.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1192_c0442ceaccee195b2d0e3fa899c88b59.png)
Thus, the cost of tin-plating the inner side of hemispherical bowl is Rs 27.72
Solution 6
Let radius of the sphere be r.
Surface area of the sphere = 154![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1195_4d362ce23c4fe947e211016963b3d188.png)
![](https://images.topperlearning.com/topper/bookquestions/1195_im19.JPG)
= 154 cm2
Surface area of the sphere = 154
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1195_4d362ce23c4fe947e211016963b3d188.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1195_25ccea34279c7b2bdb1a23bce06199ae.png)
Thus, the radius of the sphere is 3.5 cm.
Solution 7
Let diameter of earth be d. Then, diameter of moon will be
.
Radius of earth =
Radius of moon =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_ebacaf1044d39c45c265b8d2fd571d80.png)
Surface area of moon =
Surface area of earth =
Required ratio =
Thus, the required ratio of the surface areas is 1:16.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_a29f36500f55f3bb56aa4db34a0075f9.png)
Radius of earth =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_81c5eae8441bfe3f2d203b62871d1bae.png)
Radius of moon =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_ebacaf1044d39c45c265b8d2fd571d80.png)
Surface area of moon =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_43dc12b09a34ae14e2186dc59bef4cdd.png)
Surface area of earth =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_1918ada79aeac1f11e914f59710728e5.png)
Required ratio =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1197_79d7382e0a8e5b7b0bebef00016bfdd3.png)
Thus, the required ratio of the surface areas is 1:16.
Solution 8
Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1198_7753957243ea072248e9c2661ac60451.png)
Thickness of the bowl = 0.25 cm
Outer CSA of hemispherical bowl =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1198_7753957243ea072248e9c2661ac60451.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1198_14b0ca1b1b5fe8fc5c8f163928c3741d.png)
Thus, the outer curved surface area of the bowl is 173.25
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1198_4d362ce23c4fe947e211016963b3d188.png)
Solution 9
(i) Surface area of sphere = ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1199_25ccea34279c7b2bdb1a23bce06199ae.png)
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1199_e6394ab15f37c13f48870d208ed58410.png)
(iii) Required ratio =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1199_25ccea34279c7b2bdb1a23bce06199ae.png)
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1199_e6394ab15f37c13f48870d208ed58410.png)
(iii) Required ratio =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1199_bbc65b2d89f661c023354ca50df4e775.png)
Surface Areas and Volumes Exercise Ex. 11.3
Solution 1
(i) Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone
(ii) Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone
Height (h) of cone = 7 cm
Volume of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1261_aaeabda3b59df0273387ac3d17269267.png)
(ii) Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1261_f818229bce6a326d0f2f3e99a2043af4.png)
Solution 2
(i) Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height (h) of cone
Volume of cone![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_30025e93867316ab9782d5ea9973415b.png)
Capacity of the conical vessel =
litres= 1.232 litres
Slant height (l) of cone = 25 cm
Height (h) of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_590e5c1d16b783eafb2e77b26c66567c.png)
Volume of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_30025e93867316ab9782d5ea9973415b.png)
Capacity of the conical vessel =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_e4666c2db348d4c24e377a25264ea164.png)
(ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Radius (r) of cone![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_7451b68824dabf61267a45c7253d6fb2.png)
Volume of cone![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_e07f5c2f810a1717ab5cf213299b2dab.png)
Capacity of the conical vessel =
litres =
litres.
Slant height (l) of cone = 13 cm
Radius (r) of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_7451b68824dabf61267a45c7253d6fb2.png)
Volume of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_e07f5c2f810a1717ab5cf213299b2dab.png)
Capacity of the conical vessel =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_5704bce703eaef536336b500ae64f34d.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1262_ae942a3957165a5ae02f01c6b32f765f.png)
Solution 3
Height (h) of cone = 15 cm
Let radius of cone be r.
Volume of cone = 1570 cm3
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1263_64294e4dd3bcb53e6de87873f78e5da5.png)
Let radius of cone be r.
Volume of cone = 1570 cm3
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1263_64294e4dd3bcb53e6de87873f78e5da5.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1263_5f390ce35d95e9d15ecae77afb60a7b2.png)
r = 10 cm
Thus, the radius of the base of the cone is 10 cm.
Thus, the radius of the base of the cone is 10 cm.
Solution 4
Height (h) of cone = 9 cm
Let radius of cone be r.
Volume of cone = 48
cm3
Let radius of cone be r.
Volume of cone = 48
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1264_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1264_eb1865309b0fe902d3d05cd2c3694057.png)
Thus, the diameter of the base of the cone is 2r = 8 cm.
Solution 5
Radius (r) of pit =
Depth (h) of pit = 12 m
Volume of pit =
= 38.5 m3
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1265_2ed4302c4d909855c376f269044fe705.png)
Depth (h) of pit = 12 m
Volume of pit =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1265_e8f25b81393de8e8aaebc975ab91e64d.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1265_b9ccb383b0c73c26cfb3ff194a86e235.png)
Solution 6
(i) Radius of cone =
=14 cm
Let height of cone be h.
Volume of cone = 9856 cm3
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_ee3cc9dbfb1759a34d8e3e5e5eae8bb9.png)
Let height of cone be h.
Volume of cone = 9856 cm3
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_7eb133138a09491c1e7fb3572edb3660.png)
h = 48 cm
Thus, the height of the cone is 48 cm.
(ii) Slant height (l) of cone ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_18d37fa0789e01276792a27641a5e598.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_18d37fa0789e01276792a27641a5e598.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_b672a55d6524427d204a70933fc0696a.png)
Thus, the slant height of the cone is 50 cm.
(iii) CSA of cone =
rl=
= 2200 cm2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1266_8df161e52a75970cc58b4af4ef6c05b1.png)
Solution 7
When the right angled
ABC is revolved about its side 12 cm, a cone of height (h) 12 cm, radius (r) 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone
= 100
cm3
Thus, the volume of cone so formed by the triangle is 100
cm3.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1267_ec7fd5b2b4109644a398eba4d25ebf00.png)
Volume of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1267_e57cc5ff343d401c290822ebf879ea45.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1267_b30c14c35f7e20752f646111109b139b.png)
Thus, the volume of cone so formed by the triangle is 100
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1267_b30c14c35f7e20752f646111109b139b.png)
Solution 8
When the right angled
ABC is revolved about its side 5 cm, a cone of radius (r) 12 cm, height (h) 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1268_ec7fd5b2b4109644a398eba4d25ebf00.png)
Volume of cone =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1268_01cafca5740000b0e35677a316c54dd1.png)
Required ratio ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1268_d9fcfbcfbbd3fc88676310d6073bb53c.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1268_d9fcfbcfbbd3fc88676310d6073bb53c.png)
Solution 9
Radius (r) of heap![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1269_4d0f31c3c79b48da3ef3c7baae6b3aff.png)
Height (h) of heap = 3 m
Volume of heap=
Slant height (l) =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1269_422d6e050a2d1412be530940ab1960e8.png)
Area of canvas required = CSA of cone
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1269_4d0f31c3c79b48da3ef3c7baae6b3aff.png)
Height (h) of heap = 3 m
Volume of heap=
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1269_97282f366cd3688b4becddb1823156cb.png)
Slant height (l) =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1269_422d6e050a2d1412be530940ab1960e8.png)
Area of canvas required = CSA of cone
![](https://images.topperlearning.com/topper/bookquestions/1269_Ex_13.7_9.png)
Surface Areas and Volumes Exercise Ex. 11.4
Solution 1
(i) Radius of sphere = 7 cm
Volume of sphere =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1270_afc0de50ef6e8ffe1cab661962d1338b.png)
Volume of sphere =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1270_afc0de50ef6e8ffe1cab661962d1338b.png)
(ii) Radius of sphere = 0.63 m
Volume of sphere =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1270_d62cb45265f4c9dd2c656ef71cbc627c.png)
Volume of sphere =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1270_d62cb45265f4c9dd2c656ef71cbc627c.png)
m3(approximately)
Solution 2
(i) Radius (r) of ball = ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_0dc0f97f2e1ba036dd9bcc8734782003.png)
Volume of ball =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_f103517947bde6df152675fb0039b315.png)
Thus, the amount of water displaced is
.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_0dc0f97f2e1ba036dd9bcc8734782003.png)
Volume of ball =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_f103517947bde6df152675fb0039b315.png)
Thus, the amount of water displaced is
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_fd70eb7aa071f8e2bb95c122c874c397.png)
(ii) Radius (r) of ball =
= 0.105 m
Volume of ball =
Thus, the amount of water displaced is 0.004851 m3.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_a0d2705161eb5a6a20f4479cede2516a.png)
Volume of ball =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1271_a30f4e1feb70c688da67f085bab3bcee.png)
Thus, the amount of water displaced is 0.004851 m3.
Solution 3
Radius (r) of metallic ball = ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1272_ce19e6b3bf1db3356dc06059ee34c5a1.png)
Volume of metallic ball =
Mass = Density
Volume = (8.9 * 38.808) g = 345.3912 g
Thus, the mass of the ball is approximately 345.39 g.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1272_ce19e6b3bf1db3356dc06059ee34c5a1.png)
Volume of metallic ball =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1272_dcce32c19a8e8085ab350f0a1abe99a5.png)
Mass = Density
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1272_b9ccb383b0c73c26cfb3ff194a86e235.png)
Thus, the mass of the ball is approximately 345.39 g.
Solution 4
Let diameter of earth be d. So, radius earth will be
.
Then, diameter of moon will be
. So, radius of moon will be
.
Volume of moon =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_b30fd637da0ce6a5caaa0fd4c86b15cf.png)
Volume of earth =![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_5d2906a48b699716e4c0cdf7c30763ee.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_4f7783e1e3ce72b06ce3a3312aaab7f4.png)
Thus, the volume of moon is
of volume of earth.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_81c5eae8441bfe3f2d203b62871d1bae.png)
Then, diameter of moon will be
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_a29f36500f55f3bb56aa4db34a0075f9.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_ebacaf1044d39c45c265b8d2fd571d80.png)
Volume of moon =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_b30fd637da0ce6a5caaa0fd4c86b15cf.png)
Volume of earth =
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_5d2906a48b699716e4c0cdf7c30763ee.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_4f7783e1e3ce72b06ce3a3312aaab7f4.png)
Thus, the volume of moon is
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1273_7850d85a50688ff136e4371f7612ae1c.png)
Solution 5
Radius (r) of hemispherical bowl =
= 5.25 cm
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1274_f86f9d9b4b7d440c11304ca40c270ce9.png)
Volume of hemispherical bowl![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1274_a5df8801a9e147ad1832b45609b14f40.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1274_a5df8801a9e147ad1832b45609b14f40.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1274_bf82296314f902ea8c7bd6c2d513c2a8.png)
= 303.1875 cm3
Capacity of the bowl
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1274_a4b0c83907d23878dff2d5acf6d5f228.png)
= 0.3031875 litre = 0.303 litre (approximately)
Thus, the hemispherical bowl can hold 0.303 litre of milk.
Solution 6
Inner radius (r1) of hemispherical tank = 1 m
Thickness of hemispherical tank = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Thickness of hemispherical tank = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Volume of iron used to make the tank = ![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1275_6019154a0d40f72c3b77daa9fa7594b1.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1275_6019154a0d40f72c3b77daa9fa7594b1.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1275_884ca2ad0b3d886d078078f4c865b007.png)
Solution 7
Let the radius of the sphere be r.
Surface area of sphere = 154 cm2
r2 = 154 cm2
Surface area of sphere = 154 cm2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1276_429ae57cb02760a4b2aa0e3e20815697.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1276_61811ef9de27d7a128198cd1542b5be3.png)
Volume of sphere
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1276_166291d6af9e72c3aaade6183170f50a.png)
Solution 8
(i) Cost of white washing the dome from inside = Rs 498.96
Cost of white washing 1 m2 area = Rs 2
CSA of inner side of dome =
= 249.48 m2
Cost of white washing 1 m2 area = Rs 2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1277_301e6c7a1fe1cad047d40996371567ac.png)
(ii) Let inner radius of hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
2
r2 = 249.48 m2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1277_882f2fe74fa250080ac7378875b42aa0.png)
Volume of air inside the dome = Volume of the hemispherical dome
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1277_269866a6b9307717ab8fa25bea419002.png)
= 523.908 m3
Thus, the volume of air inside the dome is approximately 523.9 m3.
CSA of inner side of dome = 249.48 m2
2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1277_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1277_882f2fe74fa250080ac7378875b42aa0.png)
Volume of air inside the dome = Volume of the hemispherical dome
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1277_269866a6b9307717ab8fa25bea419002.png)
= 523.908 m3
Thus, the volume of air inside the dome is approximately 523.9 m3.
Solution 9
(i) Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere
Volume of 27 solid iron spheres![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_5c901bdce3de23d1ddcf5185f4639d90.png)
It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of
Volume of 1 solid iron sphere
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_ed678aa327bbee7525d0a418ae9e3f17.png)
Volume of 27 solid iron spheres
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_5c901bdce3de23d1ddcf5185f4639d90.png)
It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of
this iron sphere will be equal to volume of 27 solid iron spheres.
Radius of the new sphere = r'.
Volume of new sphere![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_a903ddc98a206f6140e4f1618db00c47.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_7132cca5f57bf6751364c1a8e895c120.png)
Radius of the new sphere = r'.
Volume of new sphere
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_a903ddc98a206f6140e4f1618db00c47.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_7132cca5f57bf6751364c1a8e895c120.png)
(ii) Surface area of 1 solid iron sphere of radius r = 4
r2
Surface area of iron sphere of radius r' = 4
(r')2 = 4
(3r)2 = 36
r2
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_d67e823447f822a0d1694637d7a35864.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_b30c14c35f7e20752f646111109b139b.png)
Surface area of iron sphere of radius r' = 4
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_b30c14c35f7e20752f646111109b139b.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1278_d67e823447f822a0d1694637d7a35864.png)
Solution 10
Radius (r) of capsule
Volume of spherical capsule![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1279_ed678aa327bbee7525d0a418ae9e3f17.png)
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1279_aa19ce6a39b0abb49d5e097f036f9621.png)
Volume of spherical capsule
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1279_ed678aa327bbee7525d0a418ae9e3f17.png)
![Double click to edit](https://images.topperlearning.com/topper/bookquestions/1279_4f9203f70c8b17e23b0300a9f98062b2.png)
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.