Question

If the diameter of a cross-section of the wire is decreased by $5\% $.How much percent will the length be increased so that the volume remains the same?

Answer 1

Hint: Assume the wire to be in a cylindrical shape and Use the formula of the Volume of the cylinder, Volume of cylinder=$\pi {r^2}h$ where r is the radius of the cross-section of wire and h is the height or length of wire. By using the given conditions we will get the required solution.

Complete step by step Answer:

Given, the diameter of a cross-section of the wire is decreased by$5\% $. If the volume remains the same then we have to find the percentage increase in the length of the wire.

Since the wire is in cylindrical shape then let r be the radius of the cross-section and h be the length of the wire then the volume of the wire is given by-
$ \Rightarrow $ V=$\pi {r^2}h$
Now since the diameter of the cross-section is decreased by $5\% $ so let the new Radius be R then,
We know that diameter is twice the radius so we can write $5\% $ of the diameter of cross section=$\dfrac{5}{{100}} \times 2r$
Since the diameter is decreased then the new diameter will be is equal to $2r - \dfrac{{5 \times 2r}}{{100}}$
On solving we get,
The new diameter=$2r - \dfrac{{10r}}{{100}} = \dfrac{{200r - 10r}}{{100}}$
New diameter=$\dfrac{{190r}}{{100}} = \dfrac{{19r}}{{10}}$
We know the new radius R =$\dfrac{{{\text{New diameter}}}}{2}$
On putting the value we get,
$ \Rightarrow {\text{R = }}\dfrac{{19r}}{{20}}$
Now let the new length be${h_1}$. Then the New volume will be=$\pi {{\text{R}}^2}{h_1}$
But according to Question, the volume will remain the same so we can write,
$ \Rightarrow \pi {r^2}h = \pi {{\text{R}}^2}{h_1}$
On putting given values we get,
$ \Rightarrow \pi {r^2}h = \pi {\left( {\dfrac{{19r}}{{20}}} \right)^2}{h_1}$
On simplifying we get,
$ \Rightarrow \pi {r^2}h = \pi \dfrac{{361{r^2}}}{{400}}{h_1}$
On cancelling the common terms we get,
$ \Rightarrow h = \dfrac{{361}}{{400}}{h_1}$
Then the new height is${h_1} = \dfrac{{400}}{{361}}h$
Now Increase in the length=new length-old length=${h_1} - h$
Increase length=$\dfrac{{400h}}{{361}} - h$
On solving we get,
Increase length=$\dfrac{{\left( {400 - 361} \right)h}}{{361}} = \dfrac{{39h}}{{361}}$
Then the percentage increase in length=$\dfrac{{{h_1} - h}}{h} \times 100$
On putting the values we get,
The percentage increase in length=$\dfrac{{39h}}{{391h}} \times 100$
On solving it we get,
The percentage increase in length=$\dfrac{{3900}}{{361}} = 10.8\% $
Answer- The length should be increased $10.8\% $ so that the volume remains the same.

Note: Here the students may mistake the height obtained from the formula of volume to be the increase in the length but it is the new length obtained after increasing not the increase in the length. There is a small difference between both. It can be written as-
New length=old length+ increase in length
So from here the increase in length =new length-old length