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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2015 · Paper 1 · Mathematics · Application of Derivatives

A cylindrical container is to be made from certain solid material...

A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of Vmm3Vm{m}^{3}, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2mm and is of radius equal to the outer radius of the container.
If the volume of the material used to make the container is minimum when the inner radius of the container is 10mm, then the value of V250π\dfrac{V}{250\pi} is

Correct answer

0

Step-by-step solution

Volume of material used
V1=π(r+2)2h−πr2h+π(r+2)2.2{V}_{1} = \pi{\left(r + 2\right)}^{2}h - \pi{r}^{2}h + \pi{\left(r + 2\right)}^{2} . 2
=4\pih(r+1)+2π(r+2)2= 4\pih\left(r + 1\right) + 2\pi{\left(r + 2\right)}^{2}
=4Vr2(r+1)+2π(r+2)2= \dfrac{4V}{{r}^{2}}\left(r + 1\right) + 2\pi{\left(r + 2\right)}^{2}
\left(∵ h = \dfrac{V}{\pi{r}^{2}}\right)<br><br >\dfrac{d{V}_{1}}{dr} = 4V\left(- \dfrac{1}{{r}^{2}} - \dfrac{2}{{r}^{3}}\right) + 4\pi\left(r + 2\right)<br>Where<br >Wherer = 10, \dfrac{d{V}_{1}}{dr} = 0<br><br >⇒ 4V\left(- \dfrac{1}{100} - \dfrac{- 2}{1000}\right) + 48\pi = 0<br><br >⇒ 4V\left(\dfrac{12}{1000}\right) = 48\pi<br><br >∴ \dfrac{V}{250\pi} = 4
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