InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Find the volume and surface area of a sphere whose radius is: (i) 3.5 cm (ii) 4.2 cm (iii) 5 cm |
|
Answer» Correct Answer - (i) `179.67 cm^(3), 154 cm^(2)` (ii) `310.464 cm^(3), 221.76 cm^(2)` (iii) `523.81 m^(3), 314.28 m^(2)` |
|
| 52. |
The volume of a sphere is `38808 cm^(3)`. Find its radius and hence its surface area |
| Answer» Correct Answer - `21 cm, 5544 cm^(2)` | |
| 53. |
Find the surface area of a sphere whose volume is `60.375 m^(3)` |
| Answer» Correct Answer - `346.5 m^(2)` | |
| 54. |
The volume of a right circular cone of height 12 cm and base radius 6 cm, isA. `(12pi) cm^(3)`B. `(36pi) cm^(3)`C. `(72 pi) cm^(3)`D. `(144 pi)cm^(3)` |
| Answer» Correct Answer - D | |
| 55. |
The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find themass of the pipe, if 1 cm3 of wood has a mass of 0.6 g. |
|
Answer» Correct Answer - 3.432 kg Volume of the pipe `= {pi (R^(2) - r^(2)) xx h}` `= [(22)/(7) xx {(14)^(2) - (12)^(2)} xx 35] cm^(3) = 5720 cm^(3)` Mass of the pipe `= (5720 xx (6)/(10) xx (1)/(1000)) kg = 3.432 kg` |
|
| 56. |
The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter. |
| Answer» Correct Answer - 0.6 cm | |
| 57. |
How much cardboard is required to make 35 penholders in the shape of cylinders, each of radius 3 cm and height 10.5 cm ? |
|
Answer» Cardboard required to make 1 penholder = (curved surface area + area of the base) `= (2pi rh + pi r^(2)) cm^(2), " where " r = 3 cm and h = (21)/(2) cm` `= pi r (2h + r) cm^(2) = {(22)/(7) xx 3 xx (2 xx (21)/(2) + 3)} cm^(2)` `= ((22)/(7) xx 3 xx 24) cm^(2)` Cardbaord required to make 35 penholders `= ((22)/(7) xx 3 xx 24 xx 35) cm^(2) = 7920 cm^(2)` |
|
| 58. |
The volume of a metallic cylindrical pipe is `784 cm^(3)`. Its length is 14 cm and its external radius is 9 cm. Find its thickness |
|
Answer» Volume of the pipe, `V = 748 cm^(3)`. Length of the pipe, `h = 14 cm` External radius of the pipe, R = 9cm Let the internal radius of the pipe be r cm. Then, volume of the pipe `= (pi R^(2) h = pi r^(2)h)` `= {pi (R^(2) - r^(2)) h} cm^(3)` `= {(22)/(7) xx (81 - r^(2)) xx 14} cm^(3)` But, volume of the pipe = `748 cm^(3)` `:. (22)/(7) xx (81 - r^(2)) xx 14 = 748` `rArr 44 xx (81 - r^(2)) = 748` `rArr (81 - r^(2)) = (748)/(44) = 17 rArr r^(2) = (81 - 17) = 64 = 8^(2)` `rArr r = 8cm` `:.` thickness `= (R -r) cm = (9 - 8) cm = 1 cm` |
|
| 59. |
The radiiof two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3.Calculate the ratio of their volumes and the ratio of thericurved surfaces. |
|
Answer» Let their radii be 2R, 3R and their heights be 5H, 3H Then, `(V_(1))/(V_(2)) = (pi (2R)^(2) xx 5H)/(pi xx (3R)^(2) xx 3H) = (20)/(27) rArr V_(1) : V_(2) = 20 : 27` and `(S_(1))/(S_(2)) = (2pi (2R)(5H))/(2pi (3R) (3H)) = (10)/(9) rArr S_(1) : S_(2) = 10 : 9` Hence, their volumes are in the ratio `20: 27` adn their surface areas are in the ratio `1 : 9` |
|
| 60. |
The radius of a wire is decreased to one third. If volume remains the same, the length will becomeA. 2 timesB. 3 timesC. 6 timesD. 9 times |
|
Answer» Correct Answer - D Let the radii be R and `((R)/(3))` and the heights be h and H respectively. Then, `pi R^(2) h = pi xx ((R)/(3))^(2) xx H rArr H = 9h` |
|
| 61. |
Two circular cylinders of equal volume have their heights in the ratio `1 : 2`. The ratio of their radii isA. `1 : sqrt2`B. `sqrt2 : 1`C. `1 : 2`D. `1 : 4` |
|
Answer» Correct Answer - B Let their heights be x cm and 2x cm respectively and let their radii be `R_(1) and R_(2)` respectively. Then, `pi xx R_(1)^(2) xx x = pi xx R_(2)^(2) xx 2x` `rArr ((R_(1))/(R_(2)))^(2) = 2 rArr (R_(1))/(R_(2)) = sqrt2 = (sqrt2)/(1) = sqrt2 : 1` |
|
| 62. |
A classroom is 10 m long, 6.4 m wide and 5m high. If each student be given `1.6 m^(2)` of the floor area, how many students can be accommodated in the room ? How many cubic metres of air would each student get ? |
|
Answer» Correct Answer - `40, 8m^(3)` Required number of stundent `= (10 xx 6.4)/(1.6) = 40` Air needed for each student `= ((10 xx 6.4 xx 5)/(40)) m^(3) = 8 m^(3)` |
|
| 63. |
How many person can be accommodated in a dining hall of dimensions `(20 m xx 16 m xx 4.5 m)`, assuming that each person requires 5 cubic metres of air ? |
|
Answer» Correct Answer - 288 Required number of persons `= ((20 xx 16 xx 4.5)/(5)) = 288` |
|
| 64. |
A solidcube of side `12 c m`is cut into8 cubes of equal volume. What will be the side of the new cube? Also, findthe ratio between their surface areas. |
|
Answer» Volume of the big cube `= (12 xx 12 xx 12) cm^(3) = 1728 cm^(3)` Volume of a small (new) cube `= (1728)/(8) cm^(3) = 216 cm^(3)` Let each side of the new cube b a cm Then, its volume `= a^(3) cm^(3)` `:. a^(3) = 216 = (6)^(3) rArr a = 6` Thus, the side of the new cube is 6 cm. Surface area of the big cube `= {16 xx (12)^(2)} cm^(2) = (6 xx 144) cm^(2) = 864 cm^(2)` Surface area of the smaller cube `= {6 xx (6)^(2)} cm^(2) = (6 xx 36) cm^(2) = 216 cm^(2)` Ratio between the surface areas of big cube and small cube `= 864 : 216 = (864)/(216) = (4)/(1) = 4 : 1` |
|
| 65. |
The radius of hemispherical balloon increases from 6cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloons in two cases isA. `1 : 4`B. `1 : 3`C. `2 : 3`D. `1 : 2` |
| Answer» Correct Answer - A | |
| 66. |
A shopkeer has one spherical laddoo of radius 5 cm. With the same amount of material. How many laddoos of radius 2.5 cm cam be made ? |
|
Answer» Volume of one laddoo of radius 5 cm `= {(4)/(3) pi xx (5)^(3)} cm^(3) = ((500pi)/(3)) cm^(3)` From this laddoo, let n laddoos of radius 2.5 cm can be made. Volume of one laddoo of radius 2.5 cm `= {(4)/(3)pi xx ((5)/(2))^(3)} cm^(3) = ((125pi)/(6)) cm^(3)` `:. n = ("volume of one laddoo of radius 5 cm")/("volume of one laddoo of radius 2.5 cm")` `= ((500po)/(3) xx (6)/(125pi)) = 8` Hence, the required number of laddoos is 8 |
|
| 67. |
In Fig. 13.12, you see the frame of a lampshade. It isto be covered with a decorative cloth. The frame has a base diameter of 20 cmand height of 30 cm. A margin of 2.5 cmis to be given for folding it over the top and bottomof the frame. Find |
|
Answer» Clearly, the decorative cloth forms a cylinder of base diameter 20 cm and height `= (30 + 2.5 + 2.5) cm = 35 cm` Thus, r = 10 cm and h = 35 cm Area of the decorative cloth = lateral surface area of a cylinder having r = 10 cm and h = 35 cm `= (2pi rh) cm^(2)` `= (2 xx (22)/(7) xx 10 xx 35) cm^(2) = 2200 cm^(2)` |
|
| 68. |
A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water. Find the increase in the level of water. |
|
Answer» Correct Answer - 4.32 cm Let the increase in level be h cm. Increase in volume of water = volume of the ball `:. pi xx 15 xx 15 xx h = (4)/(3) pi xx 9 xx 9 xx 9`. Find h. |
|
| 69. |
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere. ? |
|
Answer» Correct Answer - 9 units `(2)/(3) pi r^(3) = 3pi r^(2) rArr r = (9)/(2) rArr 2r = 9` units |
|
| 70. |
A metallic sphere of radius 21 cm is dropped into a cylindrical vessel, which is partly, filled with water. The diameter of the vessel is 1.68 m. If the sphere is completely submerged, find by how much the surface of water will rise. |
|
Answer» Radius of the sphere, r = 21 cm Volume of the sphere `= ((4)/(3)pir^(3)) cm^(3)` `= ((4)/(3) pi xx 21 xx 21 xx 21) cm^(3) = (12348pi) cm^(3)` Volume of water displaced by the sphere = `(12348pi) cm^(3)` Suppose that the water rises by h cm. Then, volume of water displacd = volume of cylinder of radius 84 cm and height h cm `:. pi xx 84 xx 84 xx h = 12348pi rArr h = ((12348)/(84 xx 84)) = (7)/(4) cm` `= 1.75 cm` Hence, the surface of water will rise by 1.75 cm |
|
| 71. |
Find the volime, curved surface area and the total surface area of a hemisphere of diameter 7 cm |
|
Answer» Radius of the hemisphere, r = 3.5 cm Volume of the hemisphere `= ((2)/(3) pi r^(3)) cm^(3)` `= ((2)/(3) xx (22)/(7) xx (7)/(2) xx (7)/(2) xx (7)/(2)) cm^(3)` `= (539)/(6)cm^(3) = 89.83 cm^(3)` Curved surface area of the hemisphere `= (2pi r^(2)) cm^(2)` `= (2 xx (22)/(7) xx (7)/(2) xx (7)/(2)) cm^(2) = 77 cm^(2)` Total surface area of the hemisphere `= (3pir^(2)) cm^(2)` `= (3 xx (22)/(7) xx (7)/(2) xx (7)/(2)) cm^(2) = (231)/(2) cm^(2) = 115.5 cm^(2)` |
|
| 72. |
A hollowsphere of internal and external diameters 4 cm and 8 cm respectively ismelted into a cone of base diameter 8 cm. Calculate the height of the cone. |
|
Answer» External radius of the sphere = 4 cm Internal radius of the sphere = 2 cm Volume of the material of the sphere `= (4)/(3) pi (4^(3) - 2^(3)) cm^(3)` `= ((224pi)/(3)) cm^(3)` Radius of the resulting cone, r = 4cm Let the height of this cone formed be h cm Volume of the cone `= ((1)/(3) pir^(2)h) cm^(3)` `= ((1)/(3) pi xx 4 xx 4 xx h) cm^(3) = ((16pih)/(3)) cm^(3)` `:. (16pih)/(3) = (224pi)/(3) rArr h = ((223)/(3) xx (3)/(16)) = 14` Hence, the height of the cone is 14 cm |
|
| 73. |
A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone. |
|
Answer» Correct Answer - 15.6 cm `(4)/(3) pi xx 7.8 xx 7.8 xx 7.8 = (1)/(3) xx pi xx r^(2) xx 31.2 rArr r^(2) = (4 xx 7.8 xx 7.8 xx 7.8)/(31.2)` `:. r = 7.8 rArr d= 2r = 15.6 cm` |
|
| 74. |
The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other end. Given that common radius =3.5 cm the height of the cylinder =6.5 cm and the total height =12.8 cm , claculate the volume of the solid correct to the nearest cm. |
|
Answer» Radius of the hemisphere =3.5 cm=`(7)/(2)`cm Height of the cylinder = 6.5 cm =`(13)/(2)`cm Given total height -12.8 cm Height of the cone =(12.8-3.5-6.5)cm =2.8 cm=`(14)/(5)`cm Radius of cylinder =`(7)/(2)` cm=Radius of cone Volume of the hemisphere=`(2)/(3)pi((7)/(2))^(3)cm^(3)` volume of the cylinder =`pi((7)/(2))^(2)((13)/(2))cm^(3)` Volume of the given solid `=[(2)/(3)pixx(7)/(2^(3))+pixx(7)/(2^(2))xx(13)/(2)+(1)/(3)pixx(7)/(2^(2))xx(14)/(5)]cm^(3)` `=pi[(7)/(2^(2)) ((2)/(3))xx(7)/(2^(2))xx(13)/(2)+(1)pixx(7)/(2^(2))xx(14)/(5)]cm^(3)` `=pi[(7)/(2^(2))(2)/(3)xx(7)/(2^(2))xx(13)/(2)+(1)/(3)pixx(7)/(2^(2))xx(14)/(5)]cm^(3)` `=pi(7)/(2^(2))(2)/(3)xx(7)/(2)+(13)/(2)+(1)/(3)xx((14)/(5))cm^(3)` `=(77)/(2)xx(70+195+28)/(30)cm^(3)=(77xx293)/(60)cm^(3)` `=376 cm^(3)` (approx) |
|
| 75. |
A cone, a hemisphere and a cylinder stand on equalbases and have the same height. Show that their volumes are in the ratio1:2:3.A. `1 : 2 : 3`B. `2 : 1 : 3`C. `2 : 3 : 1`D. `3 : 2 : 1` |
|
Answer» Correct Answer - A Let R be the radius of each Height of hemisphere = its radius = R cm `:.` height of each = R cm Ratio of their volumes `= ((1)/(3) pi R^(2) xx R) : ((2)/(3)pi R^(3)) : (pi R^(2) xx R)` |
|
| 76. |
A rectangular sheet of paper `30 cm xx 18 cm` can be transformed into the curved surface of a right circular cylinder in two ways namely, either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volume of the two cylinders, thus formed. |
|
Answer» Correct Answer - `5 : 3` Case I When the sheet is folded along its length: In this case, it forms a cylinder having height `h_(1) = 18cm` and the circumference of its base equal to 30 cm Let the radius of its base be `r_(1)`. Then, `2pir_(1) = 30 rArr r_(1) = (15)/(pi)` `:. V_(1) = pir_(1)^(2) h_(1) = {pi xx ((15)/(pi))^(2) xx 18} cm^(3) = (4050)/(pi) cm^(3)` Case II When the sheet is folded along its breadth: In this case, we get a cylinder having height `h_(2)` = 30 cm and the circumference of its base equal to 18 cm. Let the radius of its base be `r_(2)`. Then, `2pir_(2) = 18 rArr r_(2) = (9)/(pi)` `:. V_(2) = pi r_(2)^(2) h_(2) = {pi xx ((9)/(11))^(2) xx 30} cm^(3) = (2430)/(pi) cm^(3)` `:. (V_(1))/(V_(2)) = ((4050)/(pi) xx (pi)/(2430)) = (4050)/(2430) = (5)/(3)` |
|
| 77. |
A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone. |
|
Answer» Correct Answer - 4.5 cm `(2)/(3) pi xx (9)^(3) = (1)?(3) pi r^(2) xx 72 rArr r^(2) = (1458)/(72) = 20.25 rArr r= 4.5 cm` |
|
| 78. |
A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights isA. `1 : 2`B. `2 : 1`C. `4 : 1`D. `sqrt2 : 1` |
| Answer» Correct Answer - B | |
| 79. |
The ratio between the curved surface area and the total right circular cylinder is `1 : 2`. Find the volume of the cylinder if its total surface area is 616 `cm^(2)` |
| Answer» Correct Answer - `1078 cm^(3)` | |
| 80. |
The total surface area of solid cylinder is `231 cm^(2)` and its curved surface area is `(2)/(3)` of the total surface area. Find the volume of the cylinder. |
| Answer» Correct Answer - `269.5 cm^(3)` | |
| 81. |
The curved surface area of a cylinder is `4400 cm^(2)` and the circumference of its base is 110 cm. Find the height and the volume of the cylinder |
| Answer» Correct Answer - 40 cm, `38500 cm^(3)` | |
| 82. |
The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm isA. 540B. 450C. 380D. 472 |
| Answer» Correct Answer - B | |
| 83. |
A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained ? |
|
Answer» Correct Answer - 126 Volume of metallic sphere `= ((4)/(3)pi xx (21)/(2) xx (21)/(2) xx (21)/(2)) cm^(3) = ((3087pi)/(2)) cm^(3)` VOlume of 1 smaller cone `= ((1)/(3)pi xx (7)/(2) xx (7)/(2)xx 3) cm^(3) =((49pi)/(4)) cm^(3)` Number of cones `= ("volume of sphere")/("volume of 1 cone")` |
|
| 84. |
A metallic cylinder of diameter 16 cm and height 9 cm is melted and recast in to sphere of diameter 6cm .How many such spheres can be formed? |
|
Answer» Correct Answer - (i)5796 `cm^(2)` (ii) 23968 `cm^(3)` For the cylinder, Radius =`(16)/(2)=8 cm` Height =9cm Diameter of sphere=6cm Radius of sphere=`(6)/(2)=3cm` Now volume of one sphere=`(4)/(3)pi(3)^(3)=36picm^(3)` `therefore` Number of spheres formed=`("volume of cylinder")/("volume of one sphere")=(576pi)/(36pi)=16` |
|
| 85. |
The radius of a metallic sphere is 60 mm .It is melted and recast in to wire of diameter 0.8 mm .Find the length of the wire. |
|
Answer» Correct Answer - 502.07 `cm^(2)` Radius of sphere =60 mm=6cm Volume of spherre=`(4)/(3)pi(6)^(3)=288pi cm^(3)` Let, length o fwire =l cm Radius of wire r = `(0.8)/(2)mm 0.4 mm=(0.4)/(10) cm=(4)/(100)cm` Volume of wire =`pir^(2)l=pi((4)/(100))^(2).l` Now volume of wire =volume of sphere `pixx(4)/(100)xx(4)/(100)xxl=288pi` `l=(288xx100xx100)/(4xx4)=180000 cm =1800 m` |
|
| 86. |
A cone is 8.4 cm heigh and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere isA. 4.2 cmB. 2.1 cmC. 2.4 cmD. 1.6 cm |
| Answer» Correct Answer - B | |
| 87. |
A solid metallic cuboid of dimensions `9m xx 8m xx2` is melted and recast in to solid cubes of edge 2 m .find the number of cubes so formed. |
|
Answer» Correct Answer - 18 Required number of cubes `= ((9 xx 8 xx 2)/(2 xx 2 xx 2)) = 18` |
|
| 88. |
The radius of a sphericla ball of iron is 1.5 cm .It is melted and recast into three sperical balls .If the radii of two such balls are 0.75 cm and 1 cm , then find the radius of 3rd bal . |
|
Answer» Radius of given ball r = 1.5 cm Radius of first ball `r_(1)=0.75 cm` Radius of second ball `r_(2)=1cm` Let the radius of third ball =`r_(3)` Now, Volume of first ball+ Volume of second ball + Volume of third ball =volume of given ball `(4)/(3)pir_(1)^(3)+(4)/(3)pir_(2)^(3)+(4)/(3)pi_(3)/(3)=(4)./(3)pir^(3)` `r_(1)^(2)+r_(2)^(3)+r_(3)^(3)=r^(3)` `r_(3)^(3)=r^(3)-r_(1)^(3)-r_(1)^(3)=(1.5)^(3)-(0.75)^(3)-(1)^(3)` `=((3)/(2))^(3)-((3)/(4))^(4)-(1)^(3)=(27)/(8)-(27)/(64)-(27)/(64)-1=(216-27-64)/(64)` `r=(216-91)/(64)=(125)/(64)=((5)/(4))^(3)=(1.25)^(3)` `r^(3)`=1.25cm Radius of third ball =1.25 cm Radius of third ball = 1.25 |
|
| 89. |
A man uses a piece of canvas having an area of `551 m^(2)`, to make a conical tent of base radius 7m. Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately `1m^(2)`, find the volume of the tent that can be made with it. |
|
Answer» Correct Answer - `1232 m^(3)` Area of canvas in making the tent `(551 - 1) m^(2) = 550 m^(2)` Curved surface area of the tent = `550 m^(2)` `:. pi rl = 550 rArr (22)/(7) xx 7 xx l = 550 rArr l = 25m` `h^(2) = (l^(2) - r^(2)) = (625- 49) = 576` |
|
| 90. |
50 circular paltes each of radius 7 cm and thickness 0.5 cm are placed one above the other to form a solid right circular cylinder .Find (i) the total surface area and (ii) the volume of the cylinder so formed. |
|
Answer» Correct Answer - 60508.8 `m^(2)` The height of the cylinder formed by placing 50 plates =`50xx0.5`=25 cm Radius of cylinder formed =Radius of plate =7cm (i) Total surface area of cylinder =`2pirh+2pir^(2)` `=[(2xx(22))/(7)xx7xx25+2xx(22)/(7)xx(7^(2))]cm^(2)` `=(1100+308)cm^(2)=1408 cm^(2)` (ii) volume of cylinder =`pir^(2)h=(22)/(7)xx(7^(2))xx25`=3850 `cm^(3)` |
|
| 91. |
The volume of a cone is `1570 cm^(3)` and its height is 15 cm. What is the radius of the cone ? (Use `pi = 3.14`)A. 10 cmB. 9 cmC. 12 cmD. 8.5 cm |
|
Answer» Correct Answer - A `(1)/(3) pi R^(2)h = 1570 rArr (1)/(3) xx 3.14 xx R^(2) xx 15 = 1570` `rArr 15.7R^(2) = 1570 rArr R^(2) = (1570 xx (10)/(157)) = 100 = (10)^(2)` |
|
| 92. |
Ravishwanted to make a temporary shelter for his car by making a box-like structurewith tarpaulin that covers all the four sides and the top of the car (withthe front face as a flap which can be rolled up). Assuming that the stitchingmargins are very small, and therefore negligible, how much tarpaulin would berequired to make the shelter of height 2.5 metre with base dimension `4m x 3m ?` |
|
Answer» Area of tarpaulin required = (area of four sides) + (area of the top) `= {2(l + b) xx h} + ( l + b)` `= {2 xx (4 + 3) xx 2.5 } m^(2) + (4 xx 3) m^(2) = (25 + 12) m^(2) = 47 m^(2)` |
|
| 93. |
A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone isA. `3 : 5`B. `2 : 5`C. `3 : 1`D. `1 : 3` |
|
Answer» Correct Answer - D `pi r^(2) h = (1)/(3) pi r^(2) H rArr (h)/(H) = (1)/(3) = 1 : 3` |
|
| 94. |
Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weight 5g per `cm^(3)` |
| Answer» Correct Answer - 103.95 kg | |
| 95. |
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid. (Take `pi = 3.14`) |
|
Answer» Correct Answer - `753.6 cm^(3)` Volume of the remaining solid `= (pi xx 6 xx 6 xx 10) cm^(3) - ((1)/(3) pi xx 6 xx 6 xx 10) cm^(3)` `= (2)/(3)pi xx 6 xx 6 xx 10 cm^(3) = ((2)/(3) xx 3.14 xx 360) cm^(3) = 753.6 cm^(3)` |
|
| 96. |
The ratio between the radius of the base and the height of the cylinder is 2:3. If its volume is 1617 `cm^3` , the total surface area of the cylinderical isA. `308 cm^(2)`B. `462 cm^(2)`C. `540 cm^(2)`D. `770 cm^(2)` |
|
Answer» Correct Answer - D Let the radius be 2x cm and the height be 3x cm. Then, volume `= pi xx (2x)^(2) xx 3x = (22)/(7) xx 12 xx x^(3)` `:. (22)/(7) xx 12 xx x^(3) = 1617 rArr x^(3) = (1617 xx (1)/(12) xx (7)/(22)) = (49 xx 7)/(8) = ((7)/(2))^(3)` `rArr x = (7)/(2) rArr r = 7 cm and h = (21)/(2) cm` `:.` total surface area `= 2pi rh + 2pi r^(2) = 2pi r (h + r)` `= {2 xx (22)/(7) xx 7 xx ((21)/(2) + 7)} cm^(2) = 770 cm^(2)` |
|
| 97. |
If the lateral surface of a cylinder is `94. 2 c m^2`and its height is 5 cm, then find(i) radius of its base (ii) its volume. |
| Answer» Correct Answer - (i) 3cm (ii) `141.3 cm^(3)` | |
| 98. |
The ratio between the radius of the base and the height of a cylinder si `2 : 3`. If its volume is `1617 cm^(3)` then its total surface area is |
|
Answer» Correct Answer - `770 cm^(2)` Suppose that radius `= 2x cm` and height = 3x cm Volume `= pir^(2) h = ((22)/(7) xx 4x^(2) xx 3x) = ((264x^(3))/(7)) cm^(3)` `:. (264x^(3))/(7) = 1617 rArr x^(3) = ((1617 xx 7)/(264)) = ((7)/(2))^(3) rArr x = (7)/(2)` Thus, r = 7 cm and h = 10.5 cm Total surface area `= (2pi rh + 2pi r^(2)) = 2pi r(h + r)` |
|
| 99. |
In a hot water heating system, there is a cylindricalpipe of length 28 m and diameter5 cm. Find the total radiating surface inthe system. |
|
Answer» Correct Answer - `44000 cm^(2)` Given, `r = 2.5 cm and h = 2800 cm` Total radiating surface area `= (2pi rh) cm^(2)` |
|
| 100. |
A granary is in the shape of a cuboid of size 8m `xx` 6m `xx`3x .If a bag of grain occupies a space of 0.65 `m^(3)` how many bags can be stored in the granary? |
|
Answer» Correct Answer - `7:9` The size of granary is 8m`xx`6m`xx`3m Volume of one bag of grain =0.65 `m^(3)` The number of bags which can be strored in the granary `=("volume of granary")/("volume of each bag")=(144)/(0.65)`=221.54 or 221 bags. |
|