Question

Mensuration 261 TRY THIS . If the diameter of the cross-section of a wire is decreased by 5%, by what percentage should the length be increased so that the volume remains the same?
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Answer 1

10.8 %

Step-by-step explanation:

Given -

• The diameter of a cross-section of the wire is decreased by 5%.

To find -

• If the volume remains the same then we have to find the percentage increase in the length of the wire .

Solution -

The wires are generally cylindrical in shape , we will assume the same over here .

Let 'r' be the radius of the cross-section and 'h' be the length of the wire .

Now , the volume of the wire is ,

⇒ V = πr²h (1)

5% of diameter of cross-section is ,

$\large{ \frac{ \cancel5}{ \cancel{100}} } \normalsize\times \cancel{2}r$

$\large \frac{r}{10}$

$New \: \: diameter = 2r - \large{\frac{r}{10}}$

$New \: \: diameter = \large \frac{19r}{20}$

$New \: \: radius = \large{ \frac{1}{2} \times \frac{19r}{10} }$

$New \: \: radius = \large{ \frac{19r}{20} }$

Let the new length be 'h1' .

$Volume =π( \frac{19r}{20})^{2} h1 \: \: \: \: \: \: \: \: (2)$

On equating (1) and (2) , we get ,

$\cancelπ{r}^{2} h= \cancel\pi( \large{\frac{19r}{20}} ) ^{2} \normalsize \: h1$

$⇒h= \large \frac{361}{400} \normalsize \: h1$

$⇒h1= \large \frac{400}{361} \normalsize \: h$

$Increase \: \: in \: \: length = \: h1 \: − \: h$

$Increase \: \: in \: \: length = \: \large\frac{400}{361} - \normalsize h$

$Increase \: \: in \: \: length = \: \large\frac{39h}{361}$

Percentage increase in length is

$\large \frac{h1 \: - \: h}{h} \: \normalsize\times 100$

$\large \frac{39 \cancel h }{ 361} \times \frac{1}{ \cancel h } \normalsize\times 100$

$\large \frac{3900}{361}$

$10.8 \: \%$

Therefore , the length should be increased by 10.8 % .