Surface Areas and Volumes: 9 Math
NCERT Exemplar Exercise 13.3: 9th math
Solution of NCERT Exemplar Exercise 13.3 Surface Areas And Volumes Class 9 Math Question (1) Metal spheres, each of radius 2cm, are packed into a rectangular box of internal dimension 16 cm × 8 cm × 8 cm. When 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer. [use ℼ = 3.14]
Solution
Given, Radius of spheres = 2 cm
And, number of the spheres = 16
And, dimension of rectangular box = 16 cm × 8 cm × 8 cm
Thus, after packing of given 16 spheres, the volume of preservative liquid filled in the box = ?
Now, we know that, volume of a sphere = 4/3 ℼ r3
Thus, volume of given sphere
= 4/3 × 3.14 × (2 cm)3
= 4/3 × 3.14 × 8 cm3
= 33.493 cm3
Therefore, volume of 16 spheres = 33.493 cm3 × 16
= 535.893 cm3
Now, we know that volume of rectangular box = `l`bh
= 16 cm × 8 cm × 8 cm
= 1024 cm3
Now, volume of the preservative liquid filled in the box = the remaning volume of the box =
= volume of box – volume of 16 spheres
= 1024 cm 2 – 535.893 cm2
= 488.12 cm3
= 488.107 cm3
≃ 488 cm3
Thus, volume of preservative liquid ≃ 488 cm3 Answer
Solution of NCERT Exemplar Exercise 13.3 Surface Areas And Volumes Class 9 Math Question (2) A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. If the present depth of water is 1.3 m, find the volume of water already used from the tank.
Solution
Given, volume of water in the cube shaped storage tank = 15.625 m3
The present height of water in the tank = 1.3 m
Thus, volume of water used = ?
Let side of cubic tank = a
We know that, volume of a cube = side3
Thus, volume of the given cubical tank = (a)3
⇒ 15.625 m3 = a3
`=>a = root3(15.625 m^3)`
`=>a = root3(2.5mxx2.5mxx2.5m)`
⇒ a = 2.5 m
Thus, edge of the cubical tank = 2.5 m
Now, as per question, height after using water = 1.3 m
And, we know that Volume of a cuboid = `l\ b\ h`
Thus, volume of water in the given tank
= 2.5 m × 2.5 m × 1.3 m
= 8.125 m3
Now, water used = previous volume of water in the tank – present volume of water in the tank
= 15.625 m3 – 8.125 m3
= 7.5 m3
Thus, water used from the tank = 7.5 m3
Now, we know that, 1 m3 = 1000 Litre
Therefore, 7.5 m3 of water = 1000 × 7.5
= 7500 Litre
Thus, volume of water already used from the tank
= 7.5 m3 or 7500 Litre Answer
Solution of NCERT Exemplar Exercise 13.3 Surface Areas And Volumes Class 9 Math Question (3) Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.
Solution
Here, volume of water displaced by the given solid spherical ball = volume of the given spherical ball
Given, diameter of the spherical ball = 4.2 cm
Therefore, radius of the given spherical ball = 2.1 cm
Now, we know that, volume of a spherical ball = 4/3 ℼ r3
Thus, volume of the given spherical ball
= `4/3 xx 22/7` × 2.1 cm × 2.1 cm × 2.1 cm
= 4 × 22 × 0.7 cm × 0.3 cm × 2.1 cm
= 88 × 0.441 cm3
= 38.808 cm3
Thus, volume water displaced after immersing of given spherical ball
= 38.808 cm3 Answer
Solution of NCERT Exemplar Exercise 13.3 Surface Areas And Volumes Class 9 Math Question (4) How many square metres of canvas is required for a conical tent whose height is 3.5 m and radius of the base is 12 m?
Solution
Given, Height of a conical tent = 3.5 m
And radius of the base of conical tent = 12 m
Thus, square meter of canvas required to make the tent = ?
Here, the square meter of canvas required to make the given tent = Curved Surface Area of given tent
We know that, Curved Surface Area of a Cone `=pi\ r\ l`
From the formula of curved surface area of a cone it is clear that to find the curved surface area of a cone, radius and slant height are required.
And, we know that, the slant height, radius and height of a cone together make a right angle triangle.
And, we know that, [slant height of a cone (`l`)]2 = (r)2 + (h) 2
Thus, (slant height of the given conical tent)2
= (12 m)2 + (3.5 m)2
= 144 m2 + 12.25 m2
= 156.25 m2
Thus, `l=sqrt(156.25 m^2)`
⇒ slant height = 12.5 m
Now, we know that, curved surface area of a cone = ℼ r `l`
Thus, curved surface area of the given conical tent
= `22/7` × 12 m × 12.5 m
= `22/7` × 150 m2
`=3300/7\ m^2`
= 471.43 m2
Thus, curved surface area of the given conical tent = 471.43 m2 = Area of canvas required for the given tent.
Thus, canvas required for the given conical tent = 471.43 m2 Answer
Solution of NCERT Exemplar Exercise 13.3 Surface Areas And Volumes Class 9 Math Question (5) Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Solution
Given, weight of the larger sphere = 5920 g
And, weight of the larger sphere = 740 g
And, diameter of the smaller sphere = 5 cm
Therefore, radius of the smaller sphere = 5/2 = 2.5 cm
Thus, radius of the larger sphere = ?
Here, if the weight of the sphere is 740 g, then radius = 2.5 cm
Therefore, if the weight of the sphere is 1 g, then radius `=2.5/740` cm
Therefore, if the weight of the sphere is 5920 g, then radius `=2.5/740xx5920` cm
= 2.5 × 8 cm = 20 cm
Thus, radius of the larger sphere = 20 cm
Alternate Method (Shortcut Method)
Given, weight of the larger sphere = 5920 g
And, weight of the larger sphere = 740 g
And, diameter of the smaller sphere = 5 cm
Therefore, radius of the smaller sphere (r) = 5/2 = 2.5 cm
Thus, radius of the larger sphere = ?
Let, radius of larger sphere = R
The, ratio of radius of larger sphere : smaller sphere = weight of the larger sphere : weight of smaller sphere
⇒ R : 2.5 cm = 5920 g : 740 g
`=>R/(2.5 cm) = 5920/740`
`=> R = (5920 xx 2.5)/740` cm
⇒ R = 8 × 2.5 cm
⇒ R = 20 cm
Thus, radius of the larger sphere = 20 cm Answer
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