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In a Corner of a Rectangular Field with Dimensions 35m × 22 M, a Well with 14 M Inside Diameter is Dug 8 M Deep. the Earth Dug Out is Spread Evenly Over the Remaining Part - Mathematics

Sum

In a corner of a rectangular field with dimensions 35m × 22 m, a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.

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Solution

We have,

Length of the field, l = 35 m,

Width of the field, b = 22 m,

Depth of the well, H = 8 m and 

Radius of the well, `"R" = 14/7 = 7 "m"`

Let the rise in the level of the field be h.

Now,

Volume of the earth on remaining part of the field = Volume of earth dug out

⇒ Area of the earth on remaining part of the field = Volume of earth

⇒ (Area of the field - Area of base of the well ) × h = πR2H

⇒ (lb - πR2) × h = πR2

`rArr (35xx22-22/7xx7xx7)xx"h" =22/7xx7xx7xx8 `

⇒ (770 - 154) × h = 1232

⇒ 616 × h = 1232

`⇒ "h" = 1232/616`

∴ h = 2 m

So, the rise in the level of the field is 2 m.

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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 19 Volume and Surface Area of Solids
Exercise 19B | Q 32 | Page 900
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