Class 9: Surface Area and Volume of a Cuboid and a Cube

$\displaystyle \textbf{Exercise - 21(A)}$

$\displaystyle \textbf{Question 1: }\text{Find the volume, total surface area and lateral surface area of a}$
$\displaystyle \text{rectangular solid having:}$
$\displaystyle \text{(i) length }=8.5\text{ m, breadth }=6.4\text{ m and height }=50\text{ cm.}$
$\displaystyle \text{(ii) length }=5.6\text{ dm, breadth }=22.5\text{ dm and height }=1\text{ m.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{(i) Length }(l)=8.5\text{ m, Breadth }(b)=6.4\text{ m}$
$\displaystyle \text{Height }(h)=50\text{ cm}=0.5\text{ m}$
$\displaystyle \text{(a) Volume}=lbh$
$\displaystyle =8.5\times6.4\times0.5$
$\displaystyle =27.2\text{ m}^3$
$\displaystyle \text{(b) Total surface area}=2(lb+bh+hl)$
$\displaystyle =2[(8.5\times6.4)+(6.4\times0.5)+(0.5\times8.5)]$
$\displaystyle =2[54.4+3.2+4.25]$
$\displaystyle =2\times61.85=123.7\text{ m}^2$
$\displaystyle \text{(c) Lateral surface area}=2(l+b)h$
$\displaystyle =2(8.5+6.4)\times0.5$
$\displaystyle =2\times14.9\times0.5=14.9\text{ m}^2$

$\displaystyle \text{(ii) Length }(l)=5.6\text{ dm, Breadth }(b)=22.5\text{ dm}$
$\displaystyle \text{Height }(h)=1\text{ m}=10\text{ dm}$
$\displaystyle \text{(a) Volume}=lbh$
$\displaystyle =5.6\times22.5\times10$
$\displaystyle =1260\text{ dm}^3$
$\displaystyle \text{(b) Total surface area}=2(lb+bh+hl)$
$\displaystyle =2[(5.6\times22.5)+(22.5\times10)+(10\times5.6)]$
$\displaystyle =2[126+225+56]$
$\displaystyle =2\times407=814\text{ dm}^2$
$\displaystyle \text{(c) Lateral surface area}=2(l+b)h$
$\displaystyle =2(5.6+22.5)\times10$
$\displaystyle =2\times28.1\times10=562\text{ dm}^2$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The volume of a rectangular wall is }33\text{ m}^3.\text{ If its length is }16.5\text{ m}$
$\displaystyle \text{and height is }8\text{ m, find the width of the wall.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of rectangular wall}=33\text{ m}^3$
$\displaystyle \text{Length of wall }(l)=16.5\text{ m}$
$\displaystyle \text{Height of wall }(h)=8\text{ m}$
$\displaystyle \text{Let the width of the wall}=b\text{ m.}$
$\displaystyle \text{Volume}=l\times b\times h$
$\displaystyle 16.5\times b\times8=33$
$\displaystyle b=\frac{33}{16.5\times8}=\frac{33}{132}=0.25\text{ m}$
$\displaystyle \therefore \text{The width of the wall is }0.25\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the number of bricks, each measuring }$
$\displaystyle 25\text{ cm}\times12.5\text{ cm}\times7.5\text{ cm,} \ \text{required to construct a wall }6\text{ m long, }5\text{ m high and}$
$\displaystyle \text{50 cm thick, while the cement and sand mixture occupies }\frac{1}{20}\text{ of the volume of the wall.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of one brick}=25\times12.5\times7.5\text{ cm}^3$
$\displaystyle =\frac{25}{100}\times\frac{12.5}{100}\times\frac{7.5}{100}\text{ m}^3$
$\displaystyle =\frac{1}{4}\times\frac{1}{8}\times\frac{3}{40}=\frac{3}{1280}\text{ m}^3$
$\displaystyle \text{Length of wall }(l)=6\text{ m}$
$\displaystyle \text{Height of wall }(h)=5\text{ m}$
$\displaystyle \text{Thickness of wall }(b)=50\text{ cm}=0.5\text{ m}$
$\displaystyle \therefore \text{Volume of wall}=lbh=6\times5\times0.5=15\text{ m}^3$
$\displaystyle \text{Volume occupied by cement and sand}=\frac{1}{20}\times15=\frac{3}{4}\text{ m}^3$
$\displaystyle \therefore \text{Volume occupied by bricks}=15-\frac{3}{4}=\frac{57}{4}\text{ m}^3$
$\displaystyle \text{Number of bricks}=\frac{\text{Volume occupied by bricks}}{\text{Volume of one brick}}$
$\displaystyle =\frac{57}{4}\div\frac{3}{1280}$
$\displaystyle =\frac{57}{4}\times\frac{1280}{3}=19\times320=6080$
$\displaystyle \therefore \text{The number of bricks required is }6080.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{A classroom is }12.5\text{ m long, }6.4\text{ m broad and }5\text{ m high. How many students}$
$\displaystyle \text{can be accommodated if each student needs }1.6\text{ m}^2\text{ of floor area? How many cubic}$
$\displaystyle \text{metres of air would each student get?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of room }(l)=12.5\text{ m}$
$\displaystyle \text{Breadth of room }(b)=6.4\text{ m}$
$\displaystyle \text{Height of room }(h)=5\text{ m}$
$\displaystyle \text{Volume of air inside the room}=lbh$
$\displaystyle =12.5\times6.4\times5=400\text{ m}^3$
$\displaystyle \text{Area of floor}=l\times b=12.5\times6.4=80\text{ m}^2$
$\displaystyle \text{Floor area required by each student}=1.6\text{ m}^2$
$\displaystyle \therefore \text{Number of students}=\frac{80}{1.6}=50$
$\displaystyle \text{Air available to each student}=\frac{\text{Volume of air}}{\text{Number of students}}$
$\displaystyle =\frac{400}{50}=8\text{ m}^3$
$\displaystyle \therefore \text{The room can accommodate }50\text{ students and each student gets }8\text{ m}^3\text{ of air.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the length of the longest rod that can be placed in a room measuring}$
$\displaystyle 12\text{ m}\times9\text{ m}\times8\text{ m.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of room }(l)=12\text{ m, Breadth }(b)=9\text{ m, Height }(h)=8\text{ m}$
$\displaystyle \text{The longest rod that can be placed in the room is its space diagonal.}$
$\displaystyle \text{Length of rod}=\sqrt{l^2+b^2+h^2}$
$\displaystyle =\sqrt{12^2+9^2+8^2}$
$\displaystyle =\sqrt{144+81+64}$
$\displaystyle =\sqrt{289}=17\text{ m}$
$\displaystyle \therefore \text{The length of the longest rod is }17\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{The volume of a cuboid is }14400\text{ cm}^3\text{ and its height is }15\text{ cm. The cross-section}$
$\displaystyle \text{of the cuboid is a rectangle having its sides in the ratio }5:3.\text{ Find the perimeter}$
$\displaystyle \text{of the cross-section.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of cuboid}=14400\text{ cm}^3$
$\displaystyle \text{Height }(h)=15\text{ cm}$
$\displaystyle \text{Area of cross-section}=\frac{\text{Volume}}{\text{Height}}$
$\displaystyle =\frac{14400}{15}=960\text{ cm}^2$
$\displaystyle \text{Ratio of the sides of the cross-section}=5:3$
$\displaystyle \text{Let the length}=5x\text{ cm and breadth}=3x\text{ cm.}$
$\displaystyle 5x\times3x=960$
$\displaystyle 15x^2=960$
$\displaystyle x^2=64$
$\displaystyle x=8$
$\displaystyle \therefore \text{Length}=5\times8=40\text{ cm}$
$\displaystyle \text{Breadth}=3\times8=24\text{ cm}$
$\displaystyle \text{Perimeter of cross-section}=2(l+b)$
$\displaystyle =2(40+24)=128\text{ cm}$
$\displaystyle \therefore \text{The perimeter of the cross-section is }128\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The area of a path is }6500\text{ m}^2.\text{ Find the cost of covering it with gravel }14\text{ cm}$
$\displaystyle \text{deep at the rate of Rs. }5.60\text{ per cubic metre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of path}=6500\text{ m}^2$
$\displaystyle \text{Depth of gravel}=14\text{ cm}=\frac{14}{100}\text{ m}$
$\displaystyle \text{Volume of gravel}=\text{Area}\times\text{Depth}$
$\displaystyle =6500\times\frac{14}{100}=910\text{ m}^3$
$\displaystyle \text{Rate of covering with gravel}=\text{Rs. }5.60\text{ per m}^3$
$\displaystyle \text{Total cost}=910\times5.60=\text{Rs. }5096$
$\displaystyle \therefore \text{The cost of covering the path with gravel is Rs. }5096.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{The cost of papering the four walls of a room }12\text{ m long at Rs. }6.50\text{ per}$
$\displaystyle \text{square metre is Rs. }1638\text{ and the cost of matting the floor at Rs. }3.50\text{ per square}$
$\displaystyle \text{metre is Rs. }378.\text{ Find the height of the room.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Rate of papering the walls}=\text{Rs. }6.50\text{ per m}^2$
$\displaystyle \text{Cost of papering the four walls}=\text{Rs. }1638$
$\displaystyle \therefore \text{Area of four walls}=\frac{1638}{6.50}=252\text{ m}^2$
$\displaystyle \text{Rate of matting the floor}=\text{Rs. }3.50\text{ per m}^2$
$\displaystyle \text{Cost of matting the floor}=\text{Rs. }378$
$\displaystyle \therefore \text{Area of floor}=\frac{378}{3.50}=108\text{ m}^2$
$\displaystyle \text{Length of room}=12\text{ m}$
$\displaystyle \therefore \text{Breadth of room}=\frac{\text{Area of floor}}{\text{Length}}=\frac{108}{12}=9\text{ m}$
$\displaystyle \text{Area of four walls}=2(l+b)h$
$\displaystyle 2(12+9)h=252$
$\displaystyle 42h=252$
$\displaystyle h=6\text{ m}$
$\displaystyle \therefore \text{The height of the room is }6\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The dimensions of a field are }15\text{ m}\times12\text{ m. A pit }7.5\text{ m}\times6\text{ m}$
$\displaystyle \times1.5\text{ m} \text{ is dug in one corner of the field and the earth removed from it is evenly spread}$
$\displaystyle \text{over the remaining area of the field. Calculate by how much the level of the field is raised.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Length of field}=15\text{ m, Breadth of field}=12\text{ m}$
$\displaystyle \text{Area of field}=15\times12=180\text{ m}^2$
$\displaystyle \text{Length of pit}=7.5\text{ m, Breadth of pit}=6\text{ m}$
$\displaystyle \text{Depth of pit}=1.5\text{ m}$
$\displaystyle \text{Area of pit}=7.5\times6=45\text{ m}^2$
$\displaystyle \text{Volume of earth removed}=7.5\times6\times1.5=67.5\text{ m}^3$
$\displaystyle \text{Area over which the earth is spread}=180-45=135\text{ m}^2$
$\displaystyle \text{Let the rise in the level of the field}=h\text{ m.}$
$\displaystyle 135h=67.5$
$\displaystyle h=\frac{67.5}{135}=0.5\text{ m}$
$\displaystyle \therefore \text{The level of the field is raised by }0.5\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The sum of length, breadth and depth of a cuboid is }19\text{ cm, and the length}$
$\displaystyle \text{of its diagonal is }11\text{ cm. Find the surface area of the cuboid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }l,b\text{ and }h\text{ be the length, breadth and depth of the cuboid.}$
$\displaystyle l+b+h=19\text{ cm}$
$\displaystyle \sqrt{l^2+b^2+h^2}=11\text{ cm}$
$\displaystyle \therefore l^2+b^2+h^2=121$
$\displaystyle (l+b+h)^2=l^2+b^2+h^2+2(lb+bh+hl)$
$\displaystyle 19^2=121+2(lb+bh+hl)$
$\displaystyle 361=121+2(lb+bh+hl)$
$\displaystyle 2(lb+bh+hl)=240$
$\displaystyle \therefore \text{Surface area of the cuboid}=240\text{ cm}^2$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Three cubes, each of side }6\text{ cm, are joined end to end. Find the surface area}$
$\displaystyle \text{of the resulting cuboid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Side of each cube}=6\text{ cm}$
$\displaystyle \text{Length of the resulting cuboid}=6\times3=18\text{ cm}$
$\displaystyle \text{Breadth}=6\text{ cm, Height}=6\text{ cm}$
$\displaystyle \text{Surface area of cuboid}=2(lb+bh+hl)$
$\displaystyle =2[(18\times6)+(6\times6)+(6\times18)]$
$\displaystyle =2(108+36+108)$
$\displaystyle =2\times252=504\text{ cm}^2$
$\displaystyle \therefore \text{The surface area of the resulting cuboid is }504\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Find (i) the volume, (ii) the total surface area, (iii) the lateral surface area and}$
$\displaystyle \text{(iv) the length of the diagonal of a cube of side }10\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Side of cube }(a)=10\text{ cm}$
$\displaystyle \text{(i) Volume}=a^3=10^3=1000\text{ cm}^3$

$\displaystyle \text{(ii) Total surface area}=6a^2$
$\displaystyle =6\times10^2=600\text{ cm}^2$

$\displaystyle \text{(iii) Lateral surface area}=4a^2$
$\displaystyle =4\times10^2=400\text{ cm}^2$

$\displaystyle \text{(iv) Length of diagonal}=\sqrt{3}a$
$\displaystyle =10\sqrt{3}\text{ cm}\approx17.32\text{ cm}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{The surface area of a cube is }1536\text{ cm}^2.\text{ Find:}$
$\displaystyle \text{(i) the length of its edge, (ii) its volume, (iii) the volume of its material whose thickness}$
$\displaystyle \text{is }5\text{ mm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Surface area of cube}=1536\text{ cm}^2$

$\displaystyle \text{(i) Let the edge of the cube}=a\text{ cm.}$
$\displaystyle 6a^2=1536$
$\displaystyle a^2=\frac{1536}{6}=256$
$\displaystyle a=16\text{ cm}$
$\displaystyle \therefore \text{The length of its edge is }16\text{ cm.}$

$\displaystyle \text{(ii) Volume of cube}=a^3=16^3=4096\text{ cm}^3$

$\displaystyle \text{(iii) Thickness of material}=5\text{ mm}=0.5\text{ cm}$
$\displaystyle \text{Inner side}=16-2(0.5)=15\text{ cm}$
$\displaystyle \text{Volume of material}=16^3-15^3$
$\displaystyle =4096-3375=721\text{ cm}^3$
$\displaystyle \therefore \text{The volume of the material is }721\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Two cubes, each of volume }512\text{ cm}^3,\text{ are joined end to end. Find the surface}$
$\displaystyle \text{area of the resulting cuboid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of each cube}=512\text{ cm}^3$
$\displaystyle \text{Let the side of each cube}=a\text{ cm.}$
$\displaystyle a^3=512=8^3$
$\displaystyle \therefore a=8\text{ cm}$
$\displaystyle \text{Length of the resulting cuboid}=8+8=16\text{ cm}$
$\displaystyle \text{Breadth}=8\text{ cm, Height}=8\text{ cm}$
$\displaystyle \text{Surface area of cuboid}=2(lb+bh+hl)$
$\displaystyle =2[(16\times8)+(8\times8)+(8\times16)]$
$\displaystyle =2(128+64+128)$
$\displaystyle =2\times320=640\text{ cm}^2$
$\displaystyle \therefore \text{The surface area of the resulting cuboid is }640\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{(i) How many cubic centimetres of iron are there in an open box whose external}$
$\displaystyle \text{dimensions are }36\text{ cm, }25\text{ cm and }16.5\text{ cm, the iron being }1.5\text{ cm thick throughout?}$
$\displaystyle \text{(ii) If }1\text{ cm}^3\text{ of iron weighs }15\text{ g, find the weight of the empty box in kg.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{External length}=36\text{ cm, External breadth}=25\text{ cm}$
$\displaystyle \text{External height}=16.5\text{ cm, Thickness}=1.5\text{ cm}$

$\displaystyle \text{(i) Inner length}=36-2(1.5)=33\text{ cm}$
$\displaystyle \text{Inner breadth}=25-2(1.5)=22\text{ cm}$
$\displaystyle \text{Inner height}=16.5-1.5=15\text{ cm}$
$\displaystyle \text{Volume of iron}=\text{Volume of outer box}-\text{Volume of inner box}$
$\displaystyle =(36\times25\times16.5)-(33\times22\times15)$
$\displaystyle =14850-10890=3960\text{ cm}^3$
$\displaystyle \therefore \text{The volume of iron is }3960\text{ cm}^3.$

$\displaystyle \text{(ii) Weight of }1\text{ cm}^3\text{ of iron}=15\text{ g}$
$\displaystyle \text{Weight of the box}=3960\times15=59400\text{ g}$
$\displaystyle =\frac{59400}{1000}=59.4\text{ kg}$
$\displaystyle \therefore \text{The weight of the empty box is }59.4\text{ kg.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{A metal cube of edge }12\text{ cm is melted and formed into three smaller cubes. If}$
$\displaystyle \text{the edges of two smaller cubes are }6\text{ cm and }8\text{ cm, find the edge of the third smaller cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Edge of the original cube}=12\text{ cm}$
$\displaystyle \text{Volume of the original cube}=12^3=1728\text{ cm}^3$
$\displaystyle \text{Volume of the first smaller cube}=6^3=216\text{ cm}^3$
$\displaystyle \text{Volume of the second smaller cube}=8^3=512\text{ cm}^3$
$\displaystyle \text{Volume of the third smaller cube}=1728-(216+512)$
$\displaystyle =1728-728=1000\text{ cm}^3$
$\displaystyle \text{Edge of the third smaller cube}=\sqrt[3]{1000}=10\text{ cm}$
$\displaystyle \therefore \text{The edge of the third smaller cube is }10\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{The dimensions of a metallic cuboid are }100\text{ cm}\times80\text{ cm}\times64\text{ cm. It is melted}$
$\displaystyle \text{and recast into a cube. Find (i) the edge of the cube, (ii) the surface area of the cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the metallic cuboid}=100\times80\times64$
$\displaystyle =512000\text{ cm}^3$
$\displaystyle \text{Since the metal is recast, volume of the cube}=512000\text{ cm}^3$

$\displaystyle \text{(i) Let the edge of the cube}=a\text{ cm.}$
$\displaystyle a^3=512000=80^3$
$\displaystyle \therefore a=80\text{ cm}$
$\displaystyle \therefore \text{The edge of the cube is }80\text{ cm.}$

$\displaystyle \text{(ii) Surface area of the cube}=6a^2$
$\displaystyle =6\times80^2=6\times6400=38400\text{ cm}^2$
$\displaystyle \therefore \text{The surface area of the cube is }38400\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{The square on the diagonal of a cube has an area of }1875\text{ cm}^2.\text{ Calculate}$
$\displaystyle \text{(i) the side of the cube, (ii) the total surface area of the cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the square on the diagonal}=1875\text{ cm}^2$

$\displaystyle \text{(i) Let the side of the cube}=a\text{ cm.}$
$\displaystyle \text{Diagonal of the cube}=a\sqrt{3}\text{ cm}$
$\displaystyle (a\sqrt{3})^2=1875$
$\displaystyle 3a^2=1875$
$\displaystyle a^2=\frac{1875}{3}=625$
$\displaystyle a=25\text{ cm}$
$\displaystyle \therefore \text{The side of the cube is }25\text{ cm.}$

$\displaystyle \text{(ii) Total surface area of the cube}=6a^2$
$\displaystyle =6\times25^2=6\times625=3750\text{ cm}^2$
$\displaystyle \therefore \text{The total surface area of the cube is }3750\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{The areas of three adjacent faces of a cuboid are }x,y\text{ and }z\text{ units. If the volume}$
$\displaystyle \text{is }V\text{ cubic units, prove that }V=\sqrt{xyz}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length, breadth and height of the cuboid be }l,b\text{ and }h\text{ respectively.}$
$\displaystyle x=lb,\qquad y=bh,\qquad z=hl$
$\displaystyle \text{Also, }V=lbh$
$\displaystyle xyz=(lb)(bh)(hl)$
$\displaystyle =l^2b^2h^2$
$\displaystyle =(lbh)^2$
$\displaystyle =V^2$
$\displaystyle \therefore V=\sqrt{xyz}.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{The diagonal of a cube is }16\sqrt{3}\text{ cm. Find its surface area and volume.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the edge of the cube}=a\text{ cm.}$
$\displaystyle \text{Diagonal of a cube}=a\sqrt{3}$
$\displaystyle a\sqrt{3}=16\sqrt{3}$
$\displaystyle \therefore a=16\text{ cm}$

$\displaystyle \text{(i) Surface area of cube}=6a^2$
$\displaystyle =6\times16^2$
$\displaystyle =6\times256=1536\text{ cm}^2$

$\displaystyle \text{(ii) Volume of cube}=a^3$
$\displaystyle =16^3=4096\text{ cm}^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Water flows into a tank }150\text{ m long and }100\text{ m broad through a rectangular}$
$\displaystyle \text{pipe whose cross-section is }5\text{ dm}\times3.5\text{ dm, at the speed of }15\text{ km/hr. In what}$
$\displaystyle \text{time will the water be }7\text{ m deep?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of tank}=150\text{ m, Breadth}=100\text{ m}$
$\displaystyle \text{Depth of water}=7\text{ m}$
$\displaystyle \text{Volume of water required}=150\times100\times7$
$\displaystyle =105000\text{ m}^3$
$\displaystyle \text{Cross-section of pipe}=5\text{ dm}\times3.5\text{ dm}$
$\displaystyle =0.5\text{ m}\times0.35\text{ m}=0.175\text{ m}^2$
$\displaystyle \text{Length of water column}=\frac{\text{Volume of water}}{\text{Area of cross-section}}$
$\displaystyle =\frac{105000}{0.175}=600000\text{ m}$
$\displaystyle =600\text{ km}$
$\displaystyle \text{Speed of water flow}=15\text{ km/hr}$
$\displaystyle \text{Time taken}=\frac{\text{Distance}}{\text{Speed}}=\frac{600}{15}=40\text{ hr}$
$\displaystyle \therefore \text{The required time is }40\text{ hours.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{A rectangular tank is }25\text{ m long and }7.5\text{ m deep. If }540\text{ m}^3\text{ of water is drawn}$
$\displaystyle \text{off the tank, the level of water in the tank goes down by }1.8\text{ m. Calculate:}$
$\displaystyle \text{(i) the width of the tank, (ii) the capacity of the tank.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of tank}=25\text{ m, Depth}=7.5\text{ m}$
$\displaystyle \text{Volume of water drawn off}=540\text{ m}^3$
$\displaystyle \text{Fall in water level}=1.8\text{ m}$

$\displaystyle \text{(i) Let the width of the tank}=b\text{ m.}$
$\displaystyle 25\times b\times1.8=540$
$\displaystyle b=\frac{540}{25\times1.8}=12\text{ m}$
$\displaystyle \therefore \text{The width of the tank is }12\text{ m.}$

$\displaystyle \text{(ii) Capacity of the tank}=lbh$
$\displaystyle =25\times12\times7.5$
$\displaystyle =2250\text{ m}^3$
$\displaystyle \therefore \text{The capacity of the tank is }2250\text{ m}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{A swimming pool is }30\text{ m long and }12\text{ m broad. Its shallow and deep ends are}$
$\displaystyle 1.5\text{ m and }3.5\text{ m deep respectively. If the bottom of the pool slopes uniformly, how}$
$\displaystyle \text{many litres of water will fill the pool?}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Length of swimming pool}=30\text{ m}$
$\displaystyle \text{Breadth of swimming pool}=12\text{ m}$
$\displaystyle \text{Depths at the two ends}=1.5\text{ m and }3.5\text{ m}$
$\displaystyle \text{Area of longitudinal cross-section}=\frac{1}{2}(1.5+3.5)\times30$
$\displaystyle =\frac{1}{2}\times5\times30=75\text{ m}^2$
$\displaystyle \text{Volume of water}=75\times12=900\text{ m}^3$
$\displaystyle 1\text{ m}^3=1000\text{ litres}$
$\displaystyle \therefore 900\text{ m}^3=900\times1000=900000\text{ litres}$
$\displaystyle \therefore \text{The swimming pool will hold }900000\text{ litres of water.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{The cross-section of a piece of metal }2\text{ m in length is shown in the adjoining}$
$\displaystyle \text{figure. Calculate:}$
$\displaystyle \text{(i) the area of its cross-section; (ii) the volume of the piece of metal;}$
$\displaystyle \text{(iii) the weight of the piece of metal to the nearest kg, if }1\text{ cm}^3\text{ of the metal weighs }6.5\text{ g.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{From }C,\text{ draw }CF\parallel AB.$
$\displaystyle CF=AB=13\text{ cm}$
$\displaystyle AF=CB=8\text{ cm}$
$\displaystyle EF=EA-FA=12-8=4\text{ cm}$

$\displaystyle \text{(i) Area of cross-section}=\text{Area of rectangle }ABCF+\text{Area of trapezium }FCDE$
$\displaystyle =(13\times8)+\frac{1}{2}(13+16)\times4$
$\displaystyle =104+\frac{1}{2}\times29\times4$
$\displaystyle =104+58=162\text{ cm}^2$
$\displaystyle \therefore \text{The area of the cross-section is }162\text{ cm}^2.$

$\displaystyle \text{(ii) Length of the metal piece}=2\text{ m}=200\text{ cm}$
$\displaystyle \text{Volume of the metal piece}=\text{Area of cross-section}\times\text{Length}$
$\displaystyle =162\times200=32400\text{ cm}^3$
$\displaystyle =\frac{32400}{1000000}\text{ m}^3=0.0324\text{ m}^3$
$\displaystyle \therefore \text{The volume of the metal piece is }32400\text{ cm}^3\text{ or }0.0324\text{ m}^3.$

$\displaystyle \text{(iii) Weight of }1\text{ cm}^3\text{ of metal}=6.5\text{ g}$
$\displaystyle \text{Weight of the metal piece}=32400\times6.5$
$\displaystyle =210600\text{ g}=210.6\text{ kg}$
$\displaystyle \therefore \text{The weight of the metal piece to the nearest kg is }211\text{ kg.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{The adjoining figure shows the cross-section of a concrete wall to be}$
$\displaystyle \text{constructed. It is }1.9\text{ m wide at the bottom, }90\text{ cm wide at the top and }5\text{ m high. If its}$
$\displaystyle \text{length is }20\text{ m, find: (i) the cross-sectional area; (ii) the volume of concrete in the wall.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Width at the bottom}=1.9\text{ m}$
$\displaystyle \text{Width at the top}=90\text{ cm}=0.9\text{ m}$
$\displaystyle \text{Height}=5\text{ m}$

$\displaystyle \text{(i) The cross-section is in the form of a trapezium.}$
$\displaystyle \text{Area of cross-section}=\frac{1}{2}(1.9+0.9)\times5$
$\displaystyle =\frac{1}{2}\times2.8\times5=7\text{ m}^2$
$\displaystyle \therefore \text{The cross-sectional area is }7\text{ m}^2.$

$\displaystyle \text{(ii) Length of the wall}=20\text{ m}$
$\displaystyle \text{Volume of concrete}=\text{Area of cross-section}\times\text{Length}$
$\displaystyle =7\times20=140\text{ m}^3$
$\displaystyle \therefore \text{The volume of concrete in the wall is }140\text{ m}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{A square brass plate of side }x\text{ cm is }1\text{ mm thick and weighs }5.44\text{ kg.}$
$\displaystyle \text{If }1\text{ cm}^3\text{ of brass weighs }8.5\text{ g, find the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Side of square plate}=x\text{ cm}$
$\displaystyle \text{Thickness}=1\text{ mm}=0.1\text{ cm}$
$\displaystyle \text{Weight of the plate}=5.44\text{ kg}=5440\text{ g}$
$\displaystyle \text{Weight of }1\text{ cm}^3\text{ of brass}=8.5\text{ g}$
$\displaystyle \therefore \text{Volume of the plate}=\frac{5440}{8.5}=640\text{ cm}^3$
$\displaystyle x\times x\times0.1=640$
$\displaystyle \frac{x^2}{10}=640$
$\displaystyle x^2=6400$
$\displaystyle x=\sqrt{6400}=80\text{ cm}$
$\displaystyle \therefore \text{The value of }x\text{ is }80\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{The area of cross-section of a rectangular pipe is }5.4\text{ cm}^2$
$\displaystyle \text{ and water is pumped} \ \text{out of it at the rate of }27\text{ kmph. Find, in litres, the}$
$\displaystyle \text{volume of water which flows out of the pipe in }1\text{ minute.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of cross-section of pipe}=5.4\text{ cm}^2$
$\displaystyle \text{Speed of water}=27\text{ km/hr}$
$\displaystyle \text{Time}=1\text{ minute}=\frac{1}{60}\text{ hr}$
$\displaystyle \text{Length of water column}=27\times\frac{1}{60}\text{ km}$
$\displaystyle =0.45\text{ km}=450\text{ m}$
$\displaystyle 5.4\text{ cm}^2=\frac{5.4}{10000}\text{ m}^2$
$\displaystyle \text{Volume of water}=\text{Area of cross-section}\times\text{Length}$
$\displaystyle =\frac{5.4}{10000}\times450=0.243\text{ m}^3$
$\displaystyle 1\text{ m}^3=1000\text{ litres}$
$\displaystyle \therefore 0.243\text{ m}^3=0.243\times1000=243\text{ litres}$
$\displaystyle \therefore \text{The volume of water flowing out in }1\text{ minute is }243\text{ litres.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{The adjoining figure shows a solid of uniform cross-section.}$
$\displaystyle \text{Find the volume of the solid. It is given that all the measurements are in cm}$
$\displaystyle \text{and each angle in the figure is a right angle.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{The solid can be divided into two cuboids of dimensions}$
$\displaystyle 4\text{ cm}\times6\text{ cm}\times3\text{ cm and }4\text{ cm}\times3\text{ cm}\times9\text{ cm.}$
$\displaystyle \text{Volume of the solid}=(4\times6\times3)+(4\times3\times9)$
$\displaystyle =72+108=180\text{ cm}^3$
$\displaystyle \therefore \text{The volume of the solid is }180\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{The cross-section of a tunnel, perpendicular to its length, is a trapezium }ABCD$
$\displaystyle \text{in which }AB=8\text{ m, }DC=6\text{ m and }AL=BM.\text{ The height of the tunnel is }2.4\text{ m}$
$\displaystyle \text{and its length is }40\text{ m. Find:}$
$\displaystyle \text{(i) the cost of paving the floor of the tunnel at Rs. }16\text{ per m}^2;$
$\displaystyle \text{(ii) the cost of painting the internal surface of the tunnel, excluding the floor, at Rs. }5\text{ per m}^2.$ $\displaystyle \text{Answer:}$
$\displaystyle AB=8\text{ m, }DC=6\text{ m, Height}=2.4\text{ m, Length}=40\text{ m}$

$\displaystyle \text{(i) Area of floor}=8\times40=320\text{ m}^2$
$\displaystyle \text{Cost of paving}=\text{Rs. }16\times320=\text{Rs. }5120$
$\displaystyle \therefore \text{The cost of paving the floor is Rs. }5120.$

$\displaystyle \text{(ii) }AL=BM=\frac{AB-DC}{2}=\frac{8-6}{2}=1\text{ m}$
$\displaystyle \text{In right-angled }\triangle ADL,$
$\displaystyle AD^2=AL^2+DL^2$
$\displaystyle =1^2+2.4^2=1+5.76=6.76$
$\displaystyle \therefore AD=\sqrt{6.76}=2.6\text{ m}$
$\displaystyle \text{Similarly, }BC=2.6\text{ m}$
$\displaystyle \text{Internal surface area excluding floor}=(AD+DC+BC)\times40$
$\displaystyle =(2.6+6+2.6)\times40$
$\displaystyle =11.2\times40=448\text{ m}^2$
$\displaystyle \text{Cost of painting}=\text{Rs. }5\times448=\text{Rs. }2240$
$\displaystyle \therefore \text{The cost of painting the internal surface is Rs. }2240.$
$\displaystyle \\$

$\displaystyle \textbf{Exercise - 21(B)}$

$\displaystyle \textbf{Question 1: }\text{Find the curved surface area and the total surface area of the cylinder for which:}$
$\displaystyle \text{(i) }h=16\text{ cm},\ r=10.5\text{ cm}$
$\displaystyle \text{(ii) }h=5\text{ cm},\ r=21\text{ cm}$
$\displaystyle \text{(iii) }h=20\text{ cm},\ r=14\text{ cm}$
$\displaystyle \text{(iv) }h=1\text{ m},\ r=1.4\text{ cm}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Height }(h)=16\text{ cm, Radius }(r)=10.5\text{ cm}$
$\displaystyle \text{Curved surface area}=2\pi rh$
$\displaystyle =2\times\frac{22}{7}\times10.5\times16$
$\displaystyle =1056\text{ cm}^2$
$\displaystyle \text{Total surface area}=2\pi r(h+r)$
$\displaystyle =2\times\frac{22}{7}\times10.5\times(16+10.5)$
$\displaystyle =44\times1.5\times26.5=1749\text{ cm}^2$

$\displaystyle \text{(ii) Height }(h)=5\text{ cm, Radius }(r)=21\text{ cm}$
$\displaystyle \text{Curved surface area}=2\pi rh$
$\displaystyle =2\times\frac{22}{7}\times21\times5$
$\displaystyle =660\text{ cm}^2$
$\displaystyle \text{Total surface area}=2\pi r(h+r)$
$\displaystyle =2\times\frac{22}{7}\times21\times(5+21)$
$\displaystyle =44\times3\times26=3432\text{ cm}^2$

$\displaystyle \text{(iii) Height }(h)=20\text{ cm, Radius }(r)=14\text{ cm}$
$\displaystyle \text{Curved surface area}=2\pi rh$
$\displaystyle =2\times\frac{22}{7}\times14\times20$
$\displaystyle =1760\text{ cm}^2$
$\displaystyle \text{Total surface area}=2\pi r(h+r)$
$\displaystyle =2\times\frac{22}{7}\times14\times(20+14)$
$\displaystyle =44\times2\times34=2992\text{ cm}^2$

$\displaystyle \text{(iv) Height }(h)=1\text{ m}=100\text{ cm, Radius }(r)=1.4\text{ cm}$
$\displaystyle \text{Curved surface area}=2\pi rh$
$\displaystyle =2\times\frac{22}{7}\times1.4\times100$
$\displaystyle =44\times0.2\times100=880\text{ cm}^2$
$\displaystyle \text{Total surface area}=2\pi r(h+r)$
$\displaystyle =2\times\frac{22}{7}\times1.4\times(100+1.4)$
$\displaystyle =44\times0.2\times101.4=892.32\text{ cm}^2$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the volume of the cylinder in which:}$
$\displaystyle \text{(i) Height }=21\text{ cm and base radius }=5\text{ cm}$
$\displaystyle \text{(ii) Diameter }=28\text{ cm and height }=40\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Height }(h)=21\text{ cm, Radius }(r)=5\text{ cm}$
$\displaystyle \text{Volume}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times5\times5\times21$
$\displaystyle =1650\text{ cm}^3$

$\displaystyle \text{(ii) Diameter}=28\text{ cm}$
$\displaystyle \therefore \text{Radius }(r)=\frac{28}{2}=14\text{ cm}$
$\displaystyle \text{Height }(h)=40\text{ cm}$
$\displaystyle \text{Volume}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times14\times14\times40$
$\displaystyle =24640\text{ cm}^3$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the weight of the solid cylinder of radius }10.5\text{ cm and height }60\text{ cm,}$
$\displaystyle \text{if the material of the cylinder weighs }5\text{ grams per cubic centimetre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of cylinder }(r)=10.5\text{ cm}$
$\displaystyle \text{Height of cylinder }(h)=60\text{ cm}$
$\displaystyle \text{Volume}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times10.5\times10.5\times60$
$\displaystyle =20790\text{ cm}^3$
$\displaystyle \text{Weight of }1\text{ cm}^3\text{ of the material}=5\text{ g}$
$\displaystyle \text{Weight of the cylinder}=20790\times5=103950\text{ g}$
$\displaystyle =\frac{103950}{1000}=103.95\text{ kg}$
$\displaystyle \therefore \text{The weight of the cylinder is }103.95\text{ kg.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{A cylindrical tank has a capacity of }6160\text{ m}^3.\text{ Find its depth if its radius is}$
$\displaystyle 14\text{ m.} \ \text{Also find the cost of painting its curved surface at Rs. }3\text{ per m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of cylinder}=6160\text{ m}^3$
$\displaystyle \text{Radius }(r)=14\text{ m}$
$\displaystyle \text{Let the depth}=h\text{ m.}$
$\displaystyle \pi r^2h=6160$
$\displaystyle \frac{22}{7}\times14\times14\times h=6160$
$\displaystyle h=\frac{6160\times7}{22\times14\times14}=10\text{ m}$
$\displaystyle \text{Curved surface area}=2\pi rh$
$\displaystyle =2\times\frac{22}{7}\times14\times10=880\text{ m}^2$
$\displaystyle \text{Rate of painting}= \text{Rs. }3\text{ per m}^2$
$\displaystyle \text{Cost of painting}=880\times3=\text{Rs. }2640$
$\displaystyle \therefore \text{The depth of the tank is }10\text{ m and the cost of painting is Rs. }2640.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The curved surface area of a cylinder is }4400\text{ cm}^2\text{ and the circumference}$
$\displaystyle \text{of its base is }110\text{ cm. Find the height and the volume of the cylinder.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Curved surface area}=4400\text{ cm}^2$
$\displaystyle \text{Circumference of the base}=110\text{ cm}$
$\displaystyle \text{Let the radius}=r\text{ cm and height}=h\text{ cm.}$
$\displaystyle 2\pi r=110$
$\displaystyle 2\times\frac{22}{7}\times r=110$
$\displaystyle r=\frac{110\times7}{44}=\frac{35}{2}\text{ cm}$
$\displaystyle 2\pi rh=4400$
$\displaystyle 2\times\frac{22}{7}\times\frac{35}{2}\times h=4400$
$\displaystyle h=40\text{ cm}$
$\displaystyle \text{Volume}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times\frac{35}{2}\times\frac{35}{2}\times40$
$\displaystyle =38500\text{ cm}^3$
$\displaystyle \therefore \text{The height is }40\text{ cm and the volume is }38500\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{The total surface area of a solid cylinder is }462\text{ cm}^2\text{ and its curved surface}$
$\displaystyle \text{area is one-third of its total surface area. Find the volume of the cylinder.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Total surface area}=462\text{ cm}^2$
$\displaystyle \text{Curved surface area}=\frac{1}{3}\times462=154\text{ cm}^2$
$\displaystyle \text{Let the radius}=r\text{ cm and height}=h\text{ cm.}$
$\displaystyle 2\pi rh=154$
$\displaystyle 2\times\frac{22}{7}\times rh=154$
$\displaystyle rh=\frac{49}{2}\qquad\cdots(1)$
$\displaystyle 2\pi r^2=462-154=308$
$\displaystyle 2\times\frac{22}{7}\times r^2=308$
$\displaystyle r^2=49$
$\displaystyle \therefore r=7\text{ cm}$
$\displaystyle \text{From (1), }7h=\frac{49}{2}$
$\displaystyle \therefore h=\frac{7}{2}\text{ cm}$
$\displaystyle \text{Volume}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times7\times7\times\frac{7}{2}$
$\displaystyle =539\text{ cm}^3$
$\displaystyle \therefore \text{The volume of the cylinder is }539\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The sum of the radius of the base and the height of a solid cylinder is }37\text{ m.}$
$\displaystyle \text{If the total surface area of the cylinder is }1628\text{ m}^2,\text{ find its volume.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }r\text{ be the radius and }h\text{ be the height of the cylinder.}$
$\displaystyle r+h=37\text{ m}$
$\displaystyle \text{Total surface area}=1628\text{ m}^2$
$\displaystyle 2\pi r(r+h)=1628$
$\displaystyle 2\times\frac{22}{7}\times r\times37=1628$
$\displaystyle r=\frac{1628\times7}{2\times22\times37}=7\text{ m}$
$\displaystyle r+h=37$
$\displaystyle 7+h=37$
$\displaystyle \therefore h=30\text{ m}$
$\displaystyle \text{Volume of cylinder}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times7\times7\times30$
$\displaystyle =4620\text{ m}^3$
$\displaystyle \therefore \text{The volume of the cylinder is }4620\text{ m}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find the height of the solid circular cylinder having total surface area of}$
$\displaystyle 660\text{ cm}^2\text{ and radius }5\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the cylinder}=h\text{ cm.}$
$\displaystyle \text{Radius }(r)=5\text{ cm}$
$\displaystyle \text{Total surface area}=660\text{ cm}^2$
$\displaystyle 2\pi r(h+r)=660$
$\displaystyle 2\times\frac{22}{7}\times5(h+5)=660$
$\displaystyle h+5=\frac{660\times7}{2\times22\times5}=21$
$\displaystyle h=21-5=16\text{ cm}$
$\displaystyle \therefore \text{The height of the cylinder is }16\text{ cm.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the total surface area of a hollow cylinder open at both ends, if its length}$
$\displaystyle \text{is }12\text{ cm, external diameter is }8\text{ cm and the thickness is }2\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of hollow cylinder }(h)=12\text{ cm}$
$\displaystyle \text{External diameter}=8\text{ cm}$
$\displaystyle \therefore \text{External radius }(R)=\frac{8}{2}=4\text{ cm}$
$\displaystyle \text{Thickness}=2\text{ cm}$
$\displaystyle \therefore \text{Inner radius }(r)=4-2=2\text{ cm}$
$\displaystyle \text{Total surface area}=2\pi Rh+2\pi rh+2\pi(R^2-r^2)$
$\displaystyle =2\pi[Rh+rh+(R^2-r^2)]$
$\displaystyle =2\times\frac{22}{7}[4\times12+2\times12+(4^2-2^2)]$
$\displaystyle =\frac{44}{7}[48+24+12]$
$\displaystyle =\frac{44}{7}\times84=528\text{ cm}^2$
$\displaystyle \therefore \text{The total surface area of the hollow cylinder is }528\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Water is flowing at the rate of }3\text{ km/hr through a circular pipe of }20\text{ cm}$
$\displaystyle \text{internal diameter into a circular cistern of diameter }10\text{ m and depth }2\text{ m. In how}$
$\displaystyle \text{much time will the cistern be filled?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter of cistern}=10\text{ m}$
$\displaystyle \therefore \text{Radius of cistern}=5\text{ m}$
$\displaystyle \text{Depth of cistern}=2\text{ m}$
$\displaystyle \text{Volume of cistern}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times5\times5\times2=\frac{1100}{7}\text{ m}^3$
$\displaystyle \text{Internal diameter of pipe}=20\text{ cm}$
$\displaystyle \therefore \text{Radius of pipe}=10\text{ cm}=0.1\text{ m}$
$\displaystyle \text{Let the length of water column required}=l\text{ m.}$
$\displaystyle \pi(0.1)^2l=\frac{1100}{7}$
$\displaystyle \frac{22}{7}\times\frac{1}{100}\times l=\frac{1100}{7}$
$\displaystyle l=5000\text{ m}=5\text{ km}$
$\displaystyle \text{Speed of water}=3\text{ km/hr}$
$\displaystyle \text{Time taken}=\frac{5}{3}\text{ hr}$
$\displaystyle =1\text{ hr }40\text{ minutes}$
$\displaystyle \therefore \text{The cistern will be filled in }1\text{ hour }40\text{ minutes.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{A swimming pool }70\text{ m long, }44\text{ m wide and }3\text{ m deep is filled by water}$
$\displaystyle \text{issuing from a pipe of diameter }35\text{ cm at }6\text{ m per second. How many hours does it}$
$\displaystyle \text{take to fill the pool?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of pool}=70\text{ m, Breadth}=44\text{ m, Depth}=3\text{ m}$
$\displaystyle \text{Volume of pool}=70\times44\times3=9240\text{ m}^3$
$\displaystyle \text{Diameter of pipe}=35\text{ cm}$
$\displaystyle \therefore \text{Radius of pipe}=\frac{35}{2}\text{ cm}=\frac{35}{200}\text{ m}$
$\displaystyle \text{Let the length of the water column required}=l\text{ m.}$
$\displaystyle \pi r^2l=9240$
$\displaystyle \frac{22}{7}\times\frac{35}{200}\times\frac{35}{200}\times l=9240$
$\displaystyle l=\frac{9240\times7\times200\times200}{22\times35\times35}=96000\text{ m}$
$\displaystyle \text{Speed of water}=6\text{ m/s}$
$\displaystyle \text{Time taken}=\frac{96000}{6}=16000\text{ seconds}$
$\displaystyle =\frac{16000}{60\times60}=\frac{40}{9}\text{ hours}=4\frac{4}{9}\text{ hours}$
$\displaystyle \therefore \text{The pool will be filled in }4\frac{4}{9}\text{ hours.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Water is flowing at the rate of }8\text{ m per second through a circular pipe whose}$
$\displaystyle \text{internal diameter is }2\text{ cm, into a cylindrical tank, the radius of whose base is }40\text{ cm.}$
$\displaystyle \text{Determine the increase in the water level in }30\text{ minutes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter of pipe}=2\text{ cm}$
$\displaystyle \therefore \text{Radius of pipe}=1\text{ cm}=\frac{1}{100}\text{ m}$
$\displaystyle \text{Speed of water}=8\text{ m/s}$
$\displaystyle \text{Time}=30\text{ minutes}=30\times60=1800\text{ seconds}$
$\displaystyle \text{Length of water column}=8\times1800=14400\text{ m}$
$\displaystyle \text{Volume of water}=\pi r^2l$
$\displaystyle =\frac{22}{7}\times\frac{1}{100}\times\frac{1}{100}\times14400$
$\displaystyle =\frac{144\times22}{700}\text{ m}^3$
$\displaystyle \text{Radius of cylindrical tank}=40\text{ cm}=\frac{2}{5}\text{ m}$
$\displaystyle \text{Let the increase in water level}=h\text{ m.}$
$\displaystyle \frac{22}{7}\times\frac{2}{5}\times\frac{2}{5}\times h=\frac{144\times22}{700}$
$\displaystyle h=9\text{ m}$
$\displaystyle \therefore \text{The increase in the water level is }9\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{A }20\text{ m deep well with diameter }7\text{ m is dug up and the earth from digging is}$
$\displaystyle \text{spread evenly to form a platform }22\text{ m}\times14\text{ m. Determine the height of the platform.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter of well}=7\text{ m}$
$\displaystyle \therefore \text{Radius }(r)=\frac{7}{2}\text{ m}$
$\displaystyle \text{Depth of well}=20\text{ m}$
$\displaystyle \text{Volume of earth dug out}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times20=770\text{ m}^3$
$\displaystyle \text{Let the height of the platform}=h_1\text{ m.}$
$\displaystyle 22\times14\times h_1=770$
$\displaystyle h_1=\frac{770}{22\times14}=\frac{5}{2}=2.5\text{ m}$
$\displaystyle \therefore \text{The height of the platform is }2.5\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Find the mass of a metallic hollow cylindrical pipe }24\text{ cm long with internal}$
$\displaystyle \text{diameter }10\text{ cm and made of }5\text{ mm thick metal, if }1\text{ cm}^3\text{ of the metal weighs} \ 7.5\text{ grams.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Internal diameter}=10\text{ cm}$
$\displaystyle \therefore \text{Internal radius }(r)=5\text{ cm}$
$\displaystyle \text{Thickness of metal}=5\text{ mm}=0.5\text{ cm}$
$\displaystyle \therefore \text{Outer radius }(R)=5+0.5=5.5\text{ cm}$
$\displaystyle \text{Length of pipe }(h)=24\text{ cm}$
$\displaystyle \text{Volume of metal}=\pi h(R^2-r^2)$
$\displaystyle =\frac{22}{7}\times24\times\left[(5.5)^2-5^2\right]$
$\displaystyle =\frac{22}{7}\times24\times(5.5+5)(5.5-5)$
$\displaystyle =396\text{ cm}^3$
$\displaystyle \text{Weight of }1\text{ cm}^3\text{ of metal}=7.5\text{ g}$
$\displaystyle \text{Mass of the pipe}=396\times7.5=2970\text{ g}$
$\displaystyle =2.97\text{ kg}$
$\displaystyle \therefore \text{The mass of the pipe is }2.97\text{ kg.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{A well with }10\text{ m inside diameter is dug }8.4\text{ m deep. Earth taken out of it is}$
$\displaystyle \text{spread all around it to a width of }7.5\text{ m to form an embankment. Find the height of}$
$\displaystyle \text{the embankment.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Inside diameter of well}=10\text{ m}$
$\displaystyle \therefore \text{Radius of well }(r)=5\text{ m}$
$\displaystyle \text{Depth of well}=8.4\text{ m}$
$\displaystyle \text{Volume of earth dug out}=\pi r^2h$
$\displaystyle =\frac{22}{7}\times5\times5\times8.4$
$\displaystyle =660\text{ m}^3$
$\displaystyle \text{Width of embankment}=7.5\text{ m}$
$\displaystyle \therefore \text{Outer radius }(R)=5+7.5=12.5\text{ m}$
$\displaystyle \text{Let the height of the embankment}=h_1\text{ m.}$
$\displaystyle \text{Volume of embankment}=\pi h_1(R^2-r^2)$
$\displaystyle =\frac{22}{7}\times h_1\left[(12.5)^2-5^2\right]$
$\displaystyle =\frac{22}{7}\times h_1(12.5+5)(12.5-5)$
$\displaystyle =660$
$\displaystyle h_1=\frac{660\times7}{22\times17.5\times7.5}=1.6\text{ m}$
$\displaystyle \therefore \text{The height of the embankment is }1.6\text{ m.}$
$\displaystyle \\$