Questions & Answers

Question

Answers

Answer
Verified

Hint: To solve this problem we will make a figure to understand the problem and then we will apply the concept of volumes to solve the problem.

Complete step-by-step answer:

We will first make the figure according to the given problem.

Now, according to the question, the well is dug in the form of a cylinder and the mud taken out is used to form an embankment around the well which is also in the form of a cylinder. Now, we have to find the width (r). To solve the problem, let ${R_2}$ be the radius of embankment, H be the height of the embankment, ${R_1}$ be the radius of well. Now first we have to find the volume of mud drawn from the well. So,

Volume of mud drawn (V) = volume of well = $\pi {r^2}h$, where r is the radius of well and h is the height of well.

Volume of well = $\pi {(1)^2}14$cubic metre.

Now, the embankment is formed from the mud drawn. So,

Volume of Embankment (Vâ€™) = $\pi \left( {{{({R_2})}^2} - {{({R_1})}^2}} \right)H$

Volume of Embankment = $\pi \left( {\dfrac{{40}}{{100}}} \right)\left( {R_2^2 - 1} \right)$

Also, the embankment is made from the mud drawn from the well. So, the volume of mud = volume of embankment. So, according to question,

V = Vâ€™

$\pi {(1)^2}14$ = $\pi \left( {\dfrac{{40}}{{100}}} \right)\left( {R_2^2 - 1} \right)$

On solving the above equation, we get

$ \Rightarrow $ $R_2^2 = 36$$6 - 1$

$ \Rightarrow $ ${R_2} = 6$

Now the radius of embankment = 6 m. To find the width (r) we have to subtract the radius of well from the radius of embankment. So,

Width (r) = ${R_2} - {R_1}$ = $6 - 1$ = 5 m.

So, the width of the embankment is 5 m.

Note: While solving such problems in which there are dimensions in different units it is important that all the dimensions should be converted in one unit, if you donâ€™t change the units of dimension you will get an incorrect answer. Draw the figure to understand and to properly solve the question without any mistake.

Complete step-by-step answer:

We will first make the figure according to the given problem.

Now, according to the question, the well is dug in the form of a cylinder and the mud taken out is used to form an embankment around the well which is also in the form of a cylinder. Now, we have to find the width (r). To solve the problem, let ${R_2}$ be the radius of embankment, H be the height of the embankment, ${R_1}$ be the radius of well. Now first we have to find the volume of mud drawn from the well. So,

Volume of mud drawn (V) = volume of well = $\pi {r^2}h$, where r is the radius of well and h is the height of well.

Volume of well = $\pi {(1)^2}14$cubic metre.

Now, the embankment is formed from the mud drawn. So,

Volume of Embankment (Vâ€™) = $\pi \left( {{{({R_2})}^2} - {{({R_1})}^2}} \right)H$

Volume of Embankment = $\pi \left( {\dfrac{{40}}{{100}}} \right)\left( {R_2^2 - 1} \right)$

Also, the embankment is made from the mud drawn from the well. So, the volume of mud = volume of embankment. So, according to question,

V = Vâ€™

$\pi {(1)^2}14$ = $\pi \left( {\dfrac{{40}}{{100}}} \right)\left( {R_2^2 - 1} \right)$

On solving the above equation, we get

$ \Rightarrow $ $R_2^2 = 36$$6 - 1$

$ \Rightarrow $ ${R_2} = 6$

Now the radius of embankment = 6 m. To find the width (r) we have to subtract the radius of well from the radius of embankment. So,

Width (r) = ${R_2} - {R_1}$ = $6 - 1$ = 5 m.

So, the width of the embankment is 5 m.

Note: While solving such problems in which there are dimensions in different units it is important that all the dimensions should be converted in one unit, if you donâ€™t change the units of dimension you will get an incorrect answer. Draw the figure to understand and to properly solve the question without any mistake.

×

Sorry!, This page is not available for now to bookmark.