Question

. In the middle of a rectangular field measuring 30m x 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 1

☯ Given dimensions of a Rectanglular field is 30 m × 20 m.

• Diameter of cylindrical well = 10 m
• Therefore, Radius of well = 10/2 = 3.5 m
• Height of cylindrical well = 10 m

⠀⠀⠀⠀⠀⠀⠀

☯ We have to find, the height through which

the level of the field is raised.

━━━━━━━━━━━━━━━━━━━━━━━━━━

In the middle of a Rectanglular field a well is dug. The earth so removed is evenly spread over the remaining part of the field.

⠀⠀⠀⠀⠀⠀⠀

$\boxed{\bf{\mid{\overline{\underline{\bigstar\: According\: to \: the \: Question :}}}}\mid}$

⠀⠀⠀⠀⠀⠀⠀

★ Volume of earth dugout = Volume of well

⠀⠀⠀⠀⠀⠀⠀

We know that,

Well is in cylindrical shape.

⠀⠀⠀⠀⠀⠀⠀

Therefore, Volume of well is,

⠀⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\pink{Volume_{\;(cylinder)} = \pi r^2 h}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\dag\;{\underline{\frak{Putting\;values\;:}}}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf Volume_{\;(well)} = \dfrac{22}{7} \times 3.5 \times 3.5 \times 10$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf Volume_{\;(well)} = \dfrac{22}{ \cancel{7}} \times \cancel{3.5} \times 3.5 \times 10$

⠀⠀⠀⠀⠀⠀⠀

$:\implies{\underline{\boxed{\sf{\pink{Volume_{\;(well)} = 385\;m^3}}}}}\;\bigstar$

⠀⠀⠀⠀⠀⠀⠀

Similarly,

Volume of earth dugout = 385 m³

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now, Finding area of field -

⠀⠀⠀⠀⠀⠀⠀

Given that,

Dimensions of a Rectanglular field is 30 m × 20 m

⠀⠀⠀⠀⠀⠀⠀

Also we know that,

$\star\;{\boxed{\sf{\purple{Area_{\;(Rectangle)} = l \times b}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\dag\;{\underline{\frak{Putting\;values\;:}}}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf Area_{\;(field)} = 30 \times 20$

⠀⠀⠀⠀⠀⠀⠀

$:\implies{\underline{\boxed{\sf{\purple{Area_{\;(field)} = 600\;m^2}}}}}\;\bigstar$

⠀⠀⠀⠀⠀⠀⠀

★ Area of remaining field = Area of field - Area of base of well

⠀⠀⠀⠀⠀⠀⠀

Therefore,

$:\implies\sf 60 - \dfrac{22}{7} \times 3.5 \times 3.5$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 60 - 38.5$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\bf \red{561.5\;m^2}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that,

⠀⠀⠀⠀

★ Increase in level of remaining field × Area of remaining field = Volume of earth dugout

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf Increase\;in\;level = \dfrac{Volume\;of\;earth\;dugout}{Area\;of\; Remaining\;field}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf Increase\;in\;level = \dfrac{385}{561.5}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\bf \red{0.686\;m\;or\;68.6\;cm}$

⠀⠀⠀⠀⠀⠀⠀

$\therefore$ Hence, The height through which the level of the field is raised is 68.6 cm.