10th Maths 13.3

Exercise 13.3

Take

$pi = 22/7$ unless stated otherwise.

Q1. A metallic sphere of radius

$text{4.2 cm}$ is melted and recast into the shape of a cylinder of radius

$text{6 cm}$ . Find the height of the cylinder.

$text{Sol. :}$

$text{Given :}$

$text{Radius of Sphere }R = 4.2 text{ cm}$

$text{Radius of Cylinder }r = 6 text{ cm}$

$text{To find :}$

$text{Height of the Cylinder}$

$text{Solve :}$

$text{According to question}$

$text{Volume of the Cylinder = Volume of the Sphere}$

$pir^2h = 4/3piR^3$

$h = (4piR^3)/(3pir^2)$

$h = (4R^3)/(3r^2)$

$h = (4times(4.2)^3)/(3(6)^2)$

$h = (4times4.2times4.2times4.2)/(3times6times6)$

$h = (4times4.2times0.7times0.7)/3$

$h = (8.232)/3$

$h = 2.744text{ cm}$

$text{Answer :}$

$text{The height of the cylinder 2.744 cm.$

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Q2. Metallic spheres of radii

$text{6 cm, 8 cm}$ and

$text{10 cm,}$ respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

$text{Sol. :}$

$text{Given :}$

$text{Radius of metal sphere }r_1 = 6, r_2 = 8, r_3 = 10 text{cm}$

$text{To find :}$

$text{The radius of the resulting sphere}$

$text{Solve :}$

$text{Let Radius of resulting sphere = R}$

$text{According to question}$

$text{Vol. of result Sphere = Sum of Vol. of 3 Spheres}$

$4/3piR^3 = 4/3pir_1^3 + 4/3pir_2^2 + 4/3pir_3^3$

$4/3piR^3 = 4/3pi(r_1^3 + r_2^3 + r_3^3)$

$R^3 = (r_1^3 + r_2^3 + r_3^3)$

$R^3 = ((6)^3 + (8)^3 + (10)^3)$

$R^3 = (216 + 512 + 1000)$

$R^3 = 1728$

$R^3 = (12)^3$

$R = 12 text{ cm}$

$text{Answer :}$

$text{The radius of resulting sphere 12 cm}.$

Q3. A

$text{20 m}$ deep well with diameter

$text{7 m}$ is dug and the earth from digging is evenly spread out to form a platform

$text{22 m}$ by

$text{14 m}$ . Find the height of the platform.

$text{Sol. :}$

$text{Given :}$

$text{Depth (height) of the well (cylinder) } h = 20 text{cm}$

$text{Diameter of the well (cylinder) }d = 7text{ cm}$

$text{Radius of the well (cylinder) } r = 7/2text{ cm}$

$text{To find :}$

$text{The height of the platform}$

$text{Solve :}$

$text{According to question}$

$text{Vol. of platform(cuboid) = Vol. of well(cylinder)}$

$LtimesBtimesH = pir^2h$

$22times14timesh = 22/7times(7/2)^2times20$

$h = (22times49times20)/(7times4times22times14)$

$h = 5/2$

$h = 2.5 text{ m}$

$text{Answer :}$

$text{The height of the platform is 2.5 m}.$

Q4. A well of diameter

$text{3 m}$ is dug

$text{14 m}$ deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width

$text{4 m}$ to form an embankment. Find the height of the embankment.

$text{Sol. :}$

$text{Given :}$

$text{Diameter of well (cylinder) }d = 3text{ m}$

$text{Radius of well (cylinder) }r_1 = 3/2 = 1.5 text{ m}$

$text{Depth (height) of well (cylinder) }h = 14text{ m}$

$text{Width of embankment} = 4text{m }$

$text{Radius of big circle }r_2 = 4 + 1.5 = 5.5text{ m}$

$text{To find :}$

$text{The height of the embankment.}$

$text{Solve :}$

$text{height of the embankment} = x$

$text{According to question}$

$text{Volume of embankment = Volume of well}$

$pi(r_2^2 - r_1^2)x = pir_1^2h$

$[(5.5)^2 - (1.5)^2]x = (1.5)^2times14$

$[(5.5+1.5)(5.5 - 1.5)]x = 2.25times14$

$[7times4]x = 2.25times14$

$x = (2.25times14)/(7times4)$

$x = 1.125 text{ m}$

$text{Answer :}$

$text{The height of the embankment 1.125 m.}$

Q5. A container shaped like a right circular cylinder having diameter

$text{12 cm}$ and height

$text{15 cm}$ is full of ice cream. The ice cream is to be filled into cones of height

$text{12 cm}$ and diameter

$text{6 cm},$ having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

$text{Sol. :}$

$text{Given :}$

$text{Diameter of Cylinder }d = 12text{ cm}$

$text{Radius of Cylinder }r = 6text{ cm}$

$text{Height of Cylinder }h =15text{ cm}$

$text{Hight of Cone }h' = 12text{ cm}$

$text{Diameter of Cone }d' = 6text{ cm}$

$text{Radius of Cone & hemisphere }r' = 3text{ cm}$

$text{To find :}$

$text{The of ice-cream cones.}$

$text{Solve :}$

$text{According to question}$

$text{Volume of ice-cream Cylinder}= pir^2h$

$text{V}_1 = pi(6)^(2)15$

$text{V}_1 = 540pi$

$text{Vol. of ice-cream cone = Vol. of cone}$

$text{ + Vol. of hemisphere}$

$text{V}_2 = 1/3pi(r')^2h' + 2/3pi(r')^3$

$text{ } = (pi(r')^2)/3(h' + 2r')$

$text{ } = (pi(3)^2)/3(12+2times3)$

$text{ } = 3times18pi$

$text{ } = 54pi$

$text{ Number of ice-cream cones }= V_1/V_2$

$text{ Number of ice-cream cones }= (540pi)/(54pi)$

$text{ = }10$

$text{Answer :}$

$text{The number of ice-cream cones 10.}$

Q6. How many silver coins,

$text{1.75 cm}$ in diameter and of thickness

$text{2 mm}$ , must be melted to form a cuboid of dimensions

$text{5.5 cm}times text{10 cm} times text{3.5 cm}$ ?

$text{Sol. :}$

$text{Given :}$

$text{Dimensions of cuboid are }5.5 times 10 times 3.5text{ cm}$

$text{Thickness (height) of coin }h = 2 text{ mm} = 0.2text{ cm}$

$text{Diameter of coin }d = 1.75text{ cm}$

$text{Radius of coin }r = 1.75/2text{ cm}$

$text{To find :}$

$text{Number of coins}$

$text{Solve :}$

$text{Vol. of Cuboid =}Ltimes B times H$

$text{V}_1 = 5.5times10times3.5$

$text{ } = 192.5text{cm}^3$

$text{Vol. of Coin (cylinder) } = pir^2h$

$text{V}_2 = 22/7times(1.75/2)^2times0.2$

$text{ } = 13.475/28$

$text{Number of coins }= V_1/V_2$

$text{ }= (192.5times28)/13.475$

$text{ }= 5390/13.475$

$text{ }= 5390000/13475$

$text{ }= 400$

$text{Answer :}$

$text{The number of coins 400}.$

Q7. A cylindrical bucket,

$text{32 cm}$ height and with radius of base

$text{18 cm,}$ is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is

$text{24 cm,}$ find the radius and slant height of the heap.

$text{Sol. :}$

$text{Given :}$

$text{Height of cylindrical bucket }h = 32text{ cm}$

$text{Radius of cylindrical bucket }r = 18text{cm}$

$text{Height of conical heap}h_1 = 24text{ cm}$

$text{To find :}$

$text{The radius & slant height of heap.}$

$text{Solve :}$

$text{According to question}$

$text{Vol. of conical heap = Vol. of Cylindrical bucket}$

$1/3pi(r_1)^2h_1 = pir^2h$

$1/3(r_1)^2times24 = (18)^2times32$

$(r_1)^2 = (3times(18)^2times32)/24$

$(r _1)^2 = (18)^2times4$

$r_1 = sqrt((18)^2times4)$

$r_1 = 18times2$

$r_1 = 36text{ cm}$

$text{slant height }l = sqrt((h_1)^2 + (r_1)^2)$

$text{ }l = sqrt((24)^2+(36)^2)$

$text{ }l = sqrt((12)^2(2^2 + 3^2)$

$text{ }l = 12sqrt(4 + 9)$

$text{ }l = 12sqrt13$

$text{Answer :}$

$text{The radius of conical heap 36 cm and slant height }12sqrt13text{ cm}.$

Q8. Water in a canal,

$text{6 m}$ wide and

$text{1.5 m}$ deep, is flowing with a speed of

$text{10 km/h}$ . How much area will it irrigate in

$text{30 minutes},$ if

$text{8 cm}$ of standing water is needed?

$text{Sol. :}$

$text{Given :}$

$text{Cuboid shap canal width }B = 6text{ m}$

$text{Cuboid shap canal depth }H = 1.5text{ m}$

$text{Cuboid shap canal length L = 10km/h}times30/60$

$text{ = 5 km = 5000 m}$

$text{To find :}$

$text{Area irrigate in 30 minutes.}$

$text{Solve :}$

$text{According to question}$

$text{Vol. of water in field = Vol. of water in canal}$

$l times b times h = L times B times H$

$text{Area } times h = 5000 times 6 times 1.5$

$text{Area } times 0.08 = 45000$

$text{Area =}45000/0.08$

$text{ =}(45000 times 100)/(8)$

$text{ =}22500 times 25 text{ m}^2$

$text{ =}562500 text{ m}^2$

$text{ =}56.25 text{ hectare}$

$text{Answer :}$

$text{The area irrigate in 30 minutes 56.25 hectare.}$

Q9. A farmer connects a pipe of internal diameter

$text{20 cm}$ from a canal into a cylindrical tank in her field, which is

$text{10 m}$ in diameter and

$text{2 m}$ deep. If water flows through the pipe at the rate of

$text{3 km/h}$ , in how much times will the tank be filled?

$text{Sol. :}$

$text{Given :}$

$text{Depth of Cylindrical tank } h = 2 text{ m}$

$text{Diameter of Cylinder tank } d = 10 text{ m}$

$text{Radius of Cylindrical tank }r = 5 text{ m}$

$text{Internal diameter of pipe }d_1 = 20 text{cm =0.2 m}$

$text{Internal radius of pipe }r_1 = 0.1 text{ m}$

$text{The length of water in the pipe in 1 hour }$

$text{ }h = 3text{km = 3000 m}$

$text{To find :}$

$text{The time fill the tank.}$

$text{Solve :}$

$text{Let time to fill is n minutes}$

$text{According to question}$

$text{Vol. of tank = n(Vol. of water in pipe in 1 hour)}$

$pir^2h = n(pir_1^2h_1)$

$r^2h = n(r_1^2h_1)$

$(5)^2times2 = n((0.1)^2times3000)$

$50 = n(30)$

$n = 50/30 text{hours}$

$n = 5/3times60text{ minutes}$

$n = 100text{ minutes}$

$text{Answer :}$

$text{The time fill the tank in 100 minutes}$

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10th Maths 13.3

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NCERT Class 10th solution of Exercise 13.2

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Exercise 13.3

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