VOLUME AND SURFACE AREAS OF SOLIDS


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1 VOLUME AND SURFACE AREAS OF SOLIDS Q.1. Find the total surface area and volume of a rectangular solid (cuboid) measuring 1 m by 50 cm by 0.5 m Ans. Length of cuboid l = 1 m, Breadth of cuboid, b = 50 cm = = m Height of cuboid, h = 0.5 m = m= m Volume of cuboid = l b h = 1 = m = 0.5 m Total surface area of cuboid = ( lb + bh + hl) = = + + = = = =.50 m 4 Hence, volume of cuboid is 0.5 m and total surface area of cuboid is.50 m Q.. The length, breadth and height of a rectangular solid are in the ratio 5 : 4 :. If the total surface area is 116 cm, find the length, breadth and height of the solid. Ans. Let length, l = 5 x, breadth, b = 4x and height, h = x. Total surface area = 116 cm ( lb + bh + hl) = 116 (5x 4x + 4x x + x 5 x) = 116 (0x + 8x + 10 x ) = 116 (8 x ) = 116 6x = x = = 16 x = 16 x = 4 l = 5x = 5 4 = 0 cm 6 b = 4x = 4 4 = 16 cm, and h = x = 4 = 8 cm Hence, length, l = 0 cm, breadth, b = 16 cm, and height h = 8 cm. Math Class IX 1 Question Bank
2 Q.. The volume of a rectangular wall is 8 m, find the width of the wall. m. If its length is 16.5 m and height Ans. Volume of rectangular wall = m Length of wall ( l ) = 16.5 m, Height of wall ( h ) = 8 m Let width of wall = b m, then volume of rectangular wall = l b h b = 1 b = = m=0.5 m Hence, width of wall is 0.5 m. Q.4. A class room is 1.5 m long, 6.4 m broad and 5 m high. How many students can accommodate if each student needs 1.6 m of floor area? How many cubic metres of air would each student get? Ans. Length of room ( l ) = 1.5 m, width of room ( b ) = 6.4 m and height of room ( h ) = 5 m Volume of air inside the room = l b h = m = 400 m Area of floor of the room = l b = m = 80 m For each student area required = 1.6 m Number of students = = = Volume of air 400 The required air received by each student = = = 8 m Number of students 50 Q.5. Find the length of the longest rod that can be placed in a room measuring 1 m 9 m 8 m. Ans. Length of room ( l ) = 1 m, breadth of room ( b ) = 9 m, and height of room ( h ) = 8 m Hence, the longest rod required to place in the room = l + b + h = (1) + (9) + (8) = m = 89 = 1 m Q.6. The volume of a cuboid is cm and its height is 15 cm. The crosssection of the cuboid is a rectangle having its sides in the ratio 5 :. Find the perimeter of the crosssection. Ans. Volume of cuboid = cm, and Height of cuboid ( h ) = 15 cm volume Length Breadth cm = = = 960 cm h 15 Ratio of remaining sides = 5 : Thus, length = 5x and breadth = x Math Class IX Question Bank
3 5x x = x = 960 x = 64 = (8) x = 8 Length = 5x = 8 5 = 40 cm and breadth = x = 8 = 4 cm Hence, perimeter of rectangular crosssection = ( l + b) = (40 + 4) cm = 64 = 18 cm. Q.. The area of path is 6500 m. Find the cost of covering it with gravel 14 cm deep at the rate of Rs 5.60 per cubic metre. 14 Ans. Area of path = 6500 m and Depth of gravel = 14 cm = m Volume of gravel = Area Depth = 6500 = 910 m 100 Rate of covering the gravel = Rs 5.60 per m Total cost = Rs = Rs = Rs Q.8. The cost of papering the four walls of a room 1 m long at Rs 6.50 per square metre is Rs 168 and the cost of matting the floor at Rs.50 per square metre is Rs 8. Find the height of the room. Ans. Rate of papering the walls = Rs 6.50 per m and Total cost = Rs Area of four walls = = m = 5 m Rate of matting the floor = Rs.50 per m and Total cost = Rs Area of floor = = m = 108 m Area of floor 108 Length of room = 1 m Breadth of room = = = 9 m Length 1 Area of four walls = ( l + b) h = 5 (1 + 9) h = 5 1h = 5 5 h = h = 6 1 Hence, height of the room is 6 m. Math Class IX Question Bank
4 Q.9. The dimensions of a field are 15 m 1 m. A pit.5 m 6 m 1.5 m is dug in one corner of the field and the earth removed from it, is evenly spread over the remaining area of the field, calculate, by how much the level of the field is raised? Ans. Length of field ( l ) = 15 m and breadth of field ( b ) = 1 m Area of total field = l b = 15 1 = 180 m Length of pit =.5 m breadth of pit = 6 m and depth of pit = 1.5 m Area of pit = l b =.5 6 = 45 m and volume of earth removed = l b h = m = 6.5 m Area of field leaving pit = (180 45) m = 15 m and volume of earth removed = l b h = m = 6.5 m Area of field leaving pit = ) m = 15 m Volume of earth Level (height) of the earth in the field = Area of remaining part = = m = 0.5 m = 50 cm Q.10. The sum of length, breadth and depth of a cuboid is 19 cm, and the length of its diagonal is 11 cm. Find the surface area of the cuboid. Ans. Let l, b and h be the length, breadth and depth of the cuboid, then l + b + h = 19 cm and l + b + h = 11 cm l + b + h = (11) = 11 The surface area of the cuboid = ( lb + bh + hl) We know that ( l + b + h) = l + b + h + ( lb + bh + hl) (19) = 11+ ( lb + bh + hl) ( lb + bh + hl) = (19) 11 = = 40 cm Hence, surface area of the cuboid = 40 cm Q.11. Find the number of bricks, each measuring 5 cm 1.5 cm.5 cm, required to construct a wall 6 m long, 5 m high and 50 cm thick, while the cement and the sand mixture occupies 1 th of the volume of the wall. 0 Ans. Volume of one brick = 5 cm 1.5 cm.5 cm m 1 1 = m = m = Math Class IX 4 Question Bank
5 Length of wall ( l ) = 6 m and Height of wall ( h ) = 5 m 50 1 and thickness of wall ( b ) = m = m Volume of wall = l b h = 6 5 = 15 m 1 Volume of cement and sand = of 15 m = m 0 4 Volume of bricks = Volume of wall Volume of cement and sand 60 5 = m = m 4 4 Volume of total bricks Number of bricks = Volume of one brick = = = 19 0 = Q.1. The total surface area of a cube is 6 cm. Find its volume. Ans. Total surface area of cube = 6 cm Let edge of cube = a cm Total surface area of cube = 6a = 6 6 a = = 11 a = 11 = 11 cm 6 Edge of cube = 11 cm Volume of cube Q.1. The volume of a cube is 9 Ans. Volume of cube = 9 cm Let edge of cube = a = (side) = (11 cm) = cm = 11 cm cm. Find its total surface area. a = 9 = a = 9 cm Total surface area of cube = 6a = 6 9 cm 9 cm = 486 cm. Q.14. The edges of three cubes of metal are cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube. Ans. The edges of three cubes are cm, 4 cm and 5 cm. Volume of three cubes are (), (4) and (5) i.e. cm, 64 cm and 15 cm. Volume of new cube = = 16 cm Let edge of new cube be a Math Class IX 5 Question Bank
6 a = 16 = 6 6 6, a = 6 cm Hence, edge of new cube = 6 cm Q.15. Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 1 cm. Find x. Ans. Edges of three cubes are x cm, 8 cm and 10 cm respectively Volume of these cubes are x, (8) and (10) i.e. x cm, 51 cm and 1000 cm Edge of new cube formed = 1 cm Volume of new cube = (side) = (1) = = 18 cm According to the given problem x = 18 x = 18 x = = 16 = x = 6 cm Hence, edge of cube = x = 6 cm. Q.16. Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes. Ans. Let edge of each of three cubes be x when three cubes are placed end to end, a cuboid is formed whose dimensions are : l = x + x + x = x, b = x, h = x Total surface area of a resulting cuboid = ( lb + bh + hl) = (x x + x x + x x) Sum of total surface areas of three cubes = (x + x + x ) = x = 14x = 6x = 18x 14x 14 Ratio of suface areas of cubes = = = 18x 18 9 Hence, the required ratio is : 9. Q.1. A metal cube of edge 1 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, find the edge of third smaller cube. Ans. Edge of metal cube = 1 cm Volume of metal cube = (1) = 18 cm Edge of first smaller cube = 6 cm Volume of smaller cube = (6) = 16 cm Edge of second smaller cube = 8 cm Math Class IX 6 Question Bank
7 Volume of second smaller cube = (8) = 51 cm Volume of third smaller cube = 18 ( ) = 18 8 = 1000 cm 1 1 Hence, edge of third cube = (1000) cm = [(10) ] cm = 10 cm Q.18. The dimensions of a metallic cuboid are 100 cm 80 cm 64 cm. It is melted and recast into a cube. Find (i) the edge of the cube (ii) the surface area of the cube. Ans. Length of the metallic cuboid ( l ) = 100 cm breadth of the metallic cuboid ( b ) = 80 cm and height of metallic cuboid ( h ) = 64 cm Volume of metallic cuboid = l b h = = cm Volume of cube so formed = cm (i) Edge of the cube (ii) Surface area of cube ( a ) = (51000) 1/ = 6a = 6 (80) 1/ = [(80) ] = 80 cm = cm Q.19. The square on the diagonal of cube has an area of 185 the side of the cube (ii) the total surface area of the cube. Ans. (Diagonal) of cube = 185 cm (i) Let side of cube = a ( a ) = 185 a = 65 = (5) a = 5 Hence, side of the cube is 5 cm. (ii) Total surface area of the cube a = a = = 8400 cm cm, Calculate (i) = 6 side = 6a = 6 (5) = 6 65 cm = 50 cm Q.0. Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid is 648 cm. Find the length of edge of each cube. Also find the ratio between the surface area of resulting cuboid and the surface area of a cube. Ans. Surface area of cuboid = 648 cm, let edge of each cube = x cm Length of cuboid = 4 x cm, width of cuboid = x cm and height of cuboid = x cm Surface area of cuboid = ( lb + bh + hl) 648 = (4x x + x x + x 4 x) = (4x + x + 4 x ) 648 = 18x 648 x = = 6 = (6) x = 6 18 Math Class IX Question Bank
8 Thus, length of edge of each cube = 6 cm Surface area of cube = 6 (side) = 6x = 6 (6) = 6 6 = 16 cm Hence, ratio between the surface area of cuboid and that of cube = 648 : 16 = :1 Q.1. A rectangular container has base of length 1 cm and width 9 cm. A cube of edge 6 cm is placed in the container and then sufficient water is filled into it, so that the cube is just submerged. Find the fall in level of water in the container, when the cube is removed. Ans. Length of the base of container = 1 cm and width of container = 9 cm Let fall in water level be x cm. Side of cube = 6 cm Volume of cube = = 16 cm Volume of water = 16 cm According to the given problem, we get x = 16 x = = 1 9 Hence, fall in water level is cm. Q.. A rectangular container whose base is a square of side 1 cm, contains sufficient water to submerge a rectangular solid 8 cm 6 cm cm. Find the rise in level of water in the container when the solid is in it. Ans. Let the rise in water level = x cm Length of solid cuboid = 8 cm, width of solid cuboid = 6 cm Height of solid cuboid = cm Volume of cuboid = l b h = 8 6 = 144 cm Volume of water = 144 cm Side of the square base = 1 cm Area of base = 1 1 = 144 cm According to given problem, we get 144 x = 144 x = 1 Hence, rise in water level is 1 cm. Q.. A field is 10 m long and 50 m broad. A tank 4 m long, 10 m broad and 6 m deep is dug any where in the field and the earth taken out of the tank is evenly spread over the remaining part of the field. Find the rise in level of the field. Ans. Length of rectangular field = 10 m, Breadth of rectangular field = 50 m Area of of the total field = m = 6000 m Length of tank = 4 m, breadth of tank = 10 m and depth of tank = 6 m Volume of earth dug out = m = 1440 m Math Class IX 8 Question Bank
9 Area of surface of tank = l b = 4 10 = 40 m Area of remaining part of the field = Area of total field Surface area of tank = = 560 m Let height of rise of field = x m 560 x = x = = = 0.5. Hence, rise in level of the field = 0.5 m or 5 cm Q.4. The following figure shows a solid of uniform crosssection. Find the volume of the solid. All measurements are in centimetres. Assume that all angles in the figure are rightangles. Ans. The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, cm and 9 cm respectively. Hence, volume of solid = = = 180 cm Q.5. A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and m deep respectively if the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool. Ans. The crosssection of the swimming pool is a trapezium with parallel sides 1.5 m and m (i.e. depths of both ends) and 40 m (i.e. length of pool) as the between sides. The width of pool i.e. 15 m can be taken as the length of crosssection. Length of swimming pool ( l ) = 40 m Breadth of swimming pool ( b ) = 15 m Depth of its ends = 1.5 m and.5 m 1 Volume of water [(1.5.0) 40] 15 m = m = = 150 m Capacity of water in litres = litres = litres (1 m = 1000 litres) r distance Math Class IX 9 Question Bank
10 Q.6. The crosssection of a piece of metal m in length is shown in the adjoining figure. Calculate : (i) The area of its crosssection. (ii) The volume of piece of metal. (iii) The weight of piece of metal to the nearest kg, if 1 cm of the metal weighs 6.5 g. Ans. From C, draw CF AB then CF = AB = 1 cm AF = CB = 8 cm EF = EA FA = 1 8 = 4 cm (i) Now area of figure ABCDE (crosssection) = ar. (Rectangle ABCF) + ar. (Trapezium FCDE) 1 = 1 cm 8 cm + (1 + 16) 4 cm = = = 16 cm (ii) Length of the piece = m = 00 cm Volume of the metal piece = Area Length = cm = 400 cm 400 = =.4 m (iii) Weight of 1 cm metal = 6.5 g Total weight of the metal piece = g = g = 11 kg (approx) Q.. A square brass plate of side x cm is 1 mm thick and weighs 5.44 kg. If 1 cm of brass weighs 8.5 g, find the value of x. Ans. Side of square plate = x cm Thickness of square plate = 1 mm Total weight of square plate = 5.44 kg Weight of 1 cm = 8.5 g Total volume of the plate cm = 8.5 = 640 cm Math Class IX 10 Question Bank
11 According to the given problem x = 6400 x = 6400 = 80 cm 1 = x x = Q.8. The area of crosssection of a rectangular pipe is 5.4 cm and water is pumped out of it at the rate of kmph. Find, in litres, the volume of water which flows out of the pipe in 1 minute. Ans. Area of crosssection of rectangular pipe = 5.4 cm Speed of water pumped out = kmph Time = 1 minute Length of water flow = 1000 m = 450 m 60 Volume of water = Area Length 5.4 = 450 m = m m = m = Volume of water in litres 1000 = (1 m = 1000 litres) 1000 = 4 litres Q.9. The crosssection of a tunnel, perpendicular to its length is a trapezium ABCD in which AB = 8 m, DC = 6 m and AL = BM. The height of the tunnel is.4 m and its length is 40 m. Find : (i) The cost of paving the floor of the tunnel at Rs 16 per m. (ii) The cost of painting the internal surface of the tunnel, excluding the floor at the rate of Rs 5 per m. Ans. The crosssection of a tunnel is of the trapezium shaped ABCD in which AB = 8 m, DC = 6 m and AL = BM. The height is.4 m and its length is 40 m. (i) Area of floor of tunnel = l b = 40 8 = 0 m Math Class IX 11 Question Bank
12 Rate of cost of paving = Rs 16 per m Total cost = 0 16 = Rs (ii) AL = BM = = = 1 m In ADL, AD = AL + DL (Using Pythagoras Theorem) = (1) + (.4) = = 6.6 = (.6) AD =.6 m Perimeter of the crosssection of the tunnel = ( ) m = 18.8 m Length = 40 m Internal surface area of the tunnel (except floor) = ( ) m = (5 0) m = 4 m Rate of painting = Rs 5 per m Hence, total cost of painting = Rs 5 4 = Rs 160 Q.0. A stream, which flows at a uniform rate of 4 km/hr, is 10 metres wide and 1. m deep at a certain point. If its crosssection is rectangular is shape find, in litres, the volume of water that flows in a minute Ans. Speed of stream = 4 km/hr = 4 m/s = m/s 18 9 Area of crosssection of stream = = 1 m Volume of water discharged in 1 second = 1 = m 9 But 1 m = 1000 litres m = 1000 = litres Volume of water discharged in 1 minutes i.e. 60 sec. = = litres. Math Class IX 1 Question Bank
13 Q.1. If cubic metres of sand be thrown into a rectangular tank 14 m long and 5 m wide; find the rise in level of the water. Ans. Length of rectangular tank = 14 m Width of rectangular tank = 5 m Let rise in level = x m Volume = 14 5 x = 0 x m But volume = m (Given) 0x = x = x = x = x = 1.44 m 100 Q.. A rectangular cardboard sheet has length cm and breadth 6 cm. Squares of each side cm are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed. Ans. Length of sheet = cm Breadth of sheet = 6 cm Side of each square = cm Inner length = = 6 = 6 cm Inner breadth = 6 = 6 6 = 0 cm By folding the sheet, the length of the container = 6 cm Breadth of the container = 0 cm and height of container = cm Volume of the container = l b h = 6 cm 0 cm cm = 1560 cm Q.. A swimming pool is 18 m long and 8 m wide its deep and shallow ends are m and 1. m respectively find the capacity of the pool, assuming that the bottom of the pool slopes uniformly. Ans. Length of pool = 18 m Breadth of pool = 8 m Height of one side = m Height on second side = 1. m ( + 1.) Volume of pool = 18 8 m = = 0.4 m Math Class IX 1 Question Bank
14 Q.4. A plot is 8 m long and 5 m broad. A tank 15 m long, m broad and 4 m deep is dug anywhere in the plot and the earth so removed is evenly spread over the remaining part of the plot. Find the rise in level of the plot. Ans. Length of plot = 8 m and width of plot = 5 m Area of plot = 8 5 m = 905 m Area of tank = l b = 15 m = 105 m Area of remaining plot = ( ) m = 800 m Volume of earth dug out = l b d = 15 4 m = 40 m Volume of earth Height of level of earth spread over the remaining part = Area of plot 40 = m = m = 100 = 15 cm Q.5. Find the volume of the cylinder in which : (i) Height = 1 cm and Base radius = 5 cm (ii) Diameter = 8 cm and Height = 40 cm. Ans. (i) Height of cylinder ( h ) = 1 cm and radius of the base ( r ) = 5 cm Volume of cylinder = π r h cm = = 1650 cm 8 (ii) Radius of the base of cylinder, r = = 14 cm Height of cylinder ( h ) = 40 cm Volume of cylinder = π r h cm = = 8 40 = 4640 cm Q.6. Find the weight of the solid cylinder of radius 10.5 cm and height 60 cm, if the material of the cylinder weighs, 5 grams per cu. cm. Ans. Radius of solid cylinder ( r ) = 10.5 cm Height of cylinder ( h ) = 60 cm Volume of cylinder = π r h cm = = = 090 cm Weight of 1 cubic cm (i.e. 1 cm ) = 5 g Math Class IX 14 Question Bank
15 10950 Total weight of the cylinder = g = g = kg = kg 1000 Q.. A cylindrical tank has a capacity of 6160 m. Find its depth if its radius is 14 m. Also find the cost of painting its curved surface at Rs per m. Ans. Volume of cylinder = 6160 m and radius of cylinder ( r ) = 14 m Let depth of cylinder be h m. Volume of cylinder = π r h = h = 6160 h = = 10 m Curved surface area of cylinder = π rh = m = 880 m Rate of painting its curved surface = Rs per m Total cost = 880 = Rs 640. Q.8. The curved surface area of a cylinder is 4400 cm and the circumference of its base is 110 cm. Find the height and the volume of the cylinder. Ans. Curved surface area = 4400 cm and circumference of its base = 110 cm Let radius be r and height be h Circumference of base = π r = r = 110 r = = 5 cm Curved surface area of cylinder = π rh = h = h = = 40 cm Volume of the cylinder = π r h 40 cm = = 8500 cm Q.9. The total surface area of a solid cylinder is 46 cm and its curved surface area is onethird of its total surface area. Find the volume of the cylinder. Ans. Total surface area of cylinder = 46 cm 1 1 and curved surface area = of total surface area cm = = Let r be the radius and h be its height, then π rh = rh = 49 rh = 154 =...(i) 44 and Area of two bases π r = = 08 r = r = = 49 = () r = cm Math Class IX 15 Question Bank
16 Now putting the value of r in (i), we get h = h = = cm Hence, volume of cylinder = π r h = cm = 59 cm Q.40. The sum of the radius of the base and the height of a solid cylinder is m. If the total surface area of the cylinder be 168 m, find its volume. Ans. Let r be the radius and h be the height of the cylinder. Then total surface area of cylinder = 168 m π r ( r + h) = r = 168 r = = h = = 0 m. But r + h = + h = Volume of the cylinder = π r h 0 m = = 460 m Q.41. A road roller is cylindrical in shape. Its circular end has a diameter 0. m and its width is m. Find the least number of complete revolutions that the roller must make in order to level a playground measuring 80 m by 44 m. 0. Ans. Diameter of cylinder = 0. m Radius of cylinder, r = = 0.5 m Width of roller, h = m Curved surface area of roller = π rh = 0.5 = 4.40 m In 1 revolution, area covered by roller = 4.40 m Area of playground = = 50 m Number of revolutions = = 100 = = Q.4. How many cubic metres of earth must be dug out to make a well 8 m deep and.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq. metre. Ans. Depth of well, h = 8 m and diameter of well =.8 m.8 Radius of well, r = m = 1.4 m Volume of earth dug out = π r h m = = Math Class IX 16 Question Bank
17 Curved surface of well = π rh = = 46.4 m Cost of platering at the rate of Rs 4.50 per sq. metre = = Rs Q.4. Find the length of 11 kg copper wire of diameter 0.4 cm. Given one cubic cm of copper weighs 8.4 gram. Ans. Let length of copper wire be h cm. Diameter of copper wire = 0.4 cm then radius of copper wire ( r ) = 0. cm Volume of copper wire (cylinder) = π r h = h 88 = h = h cm Weigth of 1 cm of wire = 8.4 grams 88 Total weight of wire = h = h grams But, weight of wire = 11 kg = = gm h = h = h = = h = cm = m = m h = m Q.44. A cylinder has a diameter of 0 cm. The area of curved surface is 100 (sq. cm). Find : (i) the height of the cylinder correct to one decimal place. (ii) the volume of the cylinder correct to one decimal place. 0 Ans. (i) Diameter of cylinder = 0 cm then radius r = = 10 cm Let height of cylinder be h cm. Curved surface area of cylinder = 100 cm π rh = h = h = = cm = cm Math Class IX 1 Question Bank
18 (ii) If h is taken as 1.6 cm Then volume of cylinder = π r h cm = = Hence, (i) height of cylinder = 1.6 cm (ii) volume of cylinder 50.9 cm. Q.45. A metal pipe has a bore (inner diameter) of 5 cm. The pipe is mm thick all round. Find the weight in kilogram, of metres of the pipe, if 1 cm of the metal weighs. g. 5 Ans. Inner diameter = 5 cm Inner radius r 1 = =.5 cm Thickness = mm = 0. cm Outer radius, r =.5 cm + 0. cm =. cm Length of pipe = metres = 00 cm π 1 Volume of metal in the pipe = π r h r h = π h [ r r1 ] = 00 [(.) (.5) ] 104 = = 00 = 100 = 00 [.9 6.5] cm 1 cm of metal weighs =. gm of metal weighs =. = = 50.6 gm 50.6 = kg = 5.06 kg = 5.04 kg 1000 Q.46. Find the total surface area of a hollow cylinder open at both ends, if its length is 1 cm, external diameter is 8 cm and the thickness is cm. Ans. Length of hollow cylinder ( h ) = 1 cm External diameter of cylinder = 8 cm 8 External radius of cylinder ( r ) = cm = 4 cm Thickness of the cylinder = cm Inner radius of cylinder = 4 = cm Thus, total surface area of cylinder = [π Rh + π rh + ( π R π r )] cm = cm Math Class IX 18 Question Bank
19 = + + cm = + cm = + = = 58 cm Q.4. Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is cm, into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in 0 minutes. Ans. Diameter of pipe = cm, Radius of pipe = 1 cm Rate of water flow = 8 m/sec Time taken = 0 minutes Length of water flow into the pipe ( h ) = = m Volume of water = π r h = m m = Radius of the base of the cylindrical tank = 40 cm = m = m Let h be the water increased in the tank π r h = h = h = = 9 m 00 Hence, height of water is 9 m. Q.48. The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm, and the volume of material in it is 1056 cm. Find the internal and external radii. Given that the height of the cylinder is 1 cm. Ans. Let r and R be the inner and outer radii and h be their height π Rh + π rh = 1056 π h ( R + r) = ( R + r) = ( R + r) = 1056 R + r = = 8...(i) 1 Again π R h π r h = 1056 π h ( R r ) = ( ) = 1056 R 66 ( R r ) = 1056 r Math Class IX 19 Question Bank
20 1056 R r = = ( R + r) ( R r) = 16 8 ( R r) = R r = = 8 R + r = 8...(ii) R r =...(iii) Adding (ii) and (iii), we get R = 10 R = 5 Subtracting eqn. (iii) from (ii), we get r = 6 r = Hence, outer radius of cylinder = 5 cm and inner radius of cylinder = cm. Math Class IX 0 Question Bank
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