The document RD Sharma Solutions - Chapter 22 - Mensuration - III (Part - 4), Class 8, Maths Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

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**PAGE NO 22.27:**

**Question 37:**

**Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square meter.**

**ANSWER:**

Cost of sinking a tube well = Volume of the tube well Ã— Cost of sinking a tube well per cubic meter

= 22/7 x (1.5^{2}) x (280) mÂ³^{ }x Rs 3.6/mÂ³ = Rs 7128.

Cost of cementing = Inner surface area of the tube well Ã— Cost of cementing per square meter

= ((2 x 22/7 x 1.5x 280) m^{2}) x Rs 2.5/m^{2} = Rs 6600

**Question 38:**

**Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.**

**ANSWER:**

Since we know the weight and the volume of copper, we can calculate its density.

density of copper =

If the weight of copper wire is 13.2 kg and the density of copper is 8.4 g/cmÂ³, then:

Volume = Weight / Density = 13.2 kg x 1000 gram/kg / 8.4 gram/cmÂ³ = 1571.43 cmÂ³

The radius of copper wire is 2 mm or 0.2 cm.

So, the length of the wire can be determined in the following way:

= 125050.01 cm = 125 m

Thus, the length of 13.2 kg of copper is 125 m.

**Question 39:**

**2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.**

**ANSWER:**

Let r cm be the radius of the wire and h cm be the length of the wire.

Volume of brass = Volume of the wire

We know that the volume of brass = 2.2 dmÂ³ = 2200 cmÂ³^{ }

Volume of the wire = Ï€rÂ²h = (0.125 cm)^{2} (h)

= 44800 cm = 448 mThus, length of the wire is 448 m.

**Question 40:**

**A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.**

**ANSWER:**

Let r m be the radius and d m be the depth of the well that is dug.

Volume of the well = Ï€r^{2}d = Ï€(5 m)^{2}(8.4 m) = 660 mÂ³

An embankment has the shape of hollow cylinder with thickness. Its inner radii is equal to the well's radii,

i.e. r = 5 m, and its outer radii is R = (5 + 7.5 ) = 12.5 cm.

Then, the volume of the embankment = Ï€h(R âˆ’ r^{2})

Volume of the well = Volume of the embankment

659.73 mÂ³ = Ï€h((12.5 m)^{2} âˆ’ (5 m)^{2})

= 1.6 m

Hence, the height of the embankment is 1.6 m.

**Question 41:**

**A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.**

**ANSWER:**

Here, R = Outer radius

r = Inner radius

t = Thickness = 4 cm

w = Width = 63 cm

Girth = 440 cm = 2Ï€R

r = R âˆ’ t = 70 cm âˆ’ 4 cm = 66 cm

Volume of the iron = Ï€ (R^{2} âˆ’ r^{2}) w = 22/7 âˆ’ (70^{2} âˆ’ 66^{2}) âˆ’ (63) = 107712 cmÂ³

Hence, volume of the iron is 107712 cmÂ³.

**Question 42:**

**What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm?**

**ANSWER:**

Here, r = Internal radius

R = External radius = 10 cm

h = Length of the cylinder

t = Thickness = 0.25 cm

Volume of the hollow cylinder = Ï€h(R^{2} - r^{2})

= Ï€ (16) (10^{2} - (10-0.25)^{2}) = 79 Ï€ cmÂ³

Volume of the solid cylinder = Volume of the hollow cylinder

We know that the radius of the solid cylinder is 1 cm.

âˆ´ Ï€(1^{2})h = 79 Ï€

h = 79 cm

Hence, length of the solid cylinder that gives the same volume as the hollow cylinder is 79 cm.

**Question 43:**

**In the middle of a rectangular field measuring 30m Ã— 20m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.**

**ANSWER:**

Let r m be the radius and h m be the depth of the well that is dug.

Volume of the well = Ï€Ï€r^{2}h = 22/7 Ã— (3.5 m)^{2} Ã— (10 m) = 385 mÂ³

Volume of the well = Volume of the rectangular field

Volume of the rectangular field = 385 mÂ³ = 30 m Ã— 20 m Ã— height

Height = = 0.641 m = 64.1 cmHence, the height through which the level of the field is raised is 64.1 cm.