Question

9. A 20 m deep pit with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m x 14 m. The height of the platform is
(a) 1.5 m
(b)2 m
(c)2.5 m
(d)3.5 m​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A 20 m deep pit with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m x 14 m.

Let assume that ABCD be the rectangular platform having dimensions as Length = 22 m and Breadth = 14 m

Now, A 20 m deep pit with diameter 7 m is dug up.

So, volume of earth dug out is equals to volume of cylinder of diameter 7 m and height 20 m.

So, Radius of cylinder, r = 7/2 m

Height of cylinder, h = 20 m

So,

$\rm \: Volume_{(earth\:dug\:out)} = Volume_{(cylinder)}$

$\rm \: Volume_{(earth\:dug\:out)} = \pi \: {r}^{2} \: h$

$\rm \: Volume_{(earth\:dug\:out)} =\dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times 20$

$\rm\implies \:Volume_{(earth\:dug\:out)} = 770 \: {m}^{2}$

Now, This earth taken out is spread uniformly over the platform of dimensions 22 m × 14 m.

Let assume that the level of the platform rises by 'h' m.

Volume of the platform where earth spread out is

$\rm \: Volume_{(platform)} = l \times b \times h$

$\rm \: Volume_{(platform)} = 22 \times 14 \times h$

$\rm\implies \:\rm \: Volume_{(platform)} = 308 \: h \: {m}^{2}$

Now,

$\rm \: Volume_{(platform)} = Volume_{(earth\:dug\:out)}$

$\rm \: 308h = 770$

$\rm\implies \:h \: = \: 2.5 \: m$

Hence, Option (c) is correct.

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Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Step-by-step explanation:

Given that,

The diameter of the cylinder is 7 m.

The radius of the cylinder is,

$\pmb{r \: = \frac{7}{2} m}$

The height of the cylinder is20 m.

Calculate the volume of the earth dug

$\pmb{V = \pi R ^ 2 h}$

$\bf \: \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 {m}^{3}$

The length of the platform is 22 m.

The breadth of the platform is 14 m.

Now,

Calculate the area of the embankment

$\pmb{A = 22 \times 14 m ^ 2}$

The volume of the embankment Volume of the earth dugout

$\bf \: \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20$

Now,

Consider the height of the embankment is hm

Calculate the height of the embankment

$\bf \: 22 \times 14 \times h =\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20$

$\bf \: h = \frac{ \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 }{22 \times 14}$

$\bf \: = 2.5$

$\boxed{ \red{ \pmb{The \: height \: of \: the \: embankment \: is \: 2.5 m.}}}$