If the volume of the material used to make the container is minimum when the inner radius of the container is $10 \ mm$, then the value of $\frac{V}{250 \pi}$ is
If the volume of the material used to make the container is minimum when the inner radius of the container is $10 \ mm$, then the value of $\frac{V}{250 \pi}$ is
$\pi r ^2 \ell= V$
Volume of material be $M$
$M =\pi( r +2)^2(\ell+2)-\pi r ^2 \ell $
$\frac{ dM }{ dr }=-\frac{4 V }{ r ^2}-\frac{8 V }{ r ^3}+8 \pi+0+4 \pi r $
$\frac{ dM }{ dr }=0 \text { when } r =10 $
$\Rightarrow V =1000 \pi \Rightarrow \frac{ V }{250 \pi}=4$
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$(1)$ eccentricity of $E$ be reciprocal of the eccentricity of $H$, and
$(2)$ the line $y=\sqrt{\frac{5}{2}} x+K$ be a common tangent of $E$ and $H$ Then $4\left(a^{2}+b^{2}\right)$ is equal to
$2 x+y-z=5$
$2 x-5 y+\lambda z=\mu$
$x+2 y-5 z=7$
has infinitely many solutions, then $(\lambda+\mu)^2+(\lambda-\mu)^2$ is equal to