The height of the cone of maximum volume inscribed in a sphere of radius R is

\frac{R}{3}

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\frac{2 R}{3}

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\frac{4 R}{3}

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\frac{4 R}{\sqrt{3}}

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Solution

The correct option is D $\frac{4 R}{3}$
$t a n θ = \frac{r}{h}$
$c o t θ = \frac{r}{2 R - h}$

Now, tanx .cotx=1

$r^{2} = h (2 R - h)$
Volume of cone $= \frac{1}{3} π r^{2} h$
$V = \frac{1}{3} π h^{2} (2 R - h)$
For maximum $\frac{d v}{d h}$ =0
$\frac{d v}{d h} = \frac{1}{3} π (4 R h - 3 h^{2})$
$0 = h (4 R - 3 h)$
$h = \frac{4 R}{3}$

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The height of the cone of maximum volume inscribed in a sphere of radius R is

The correct option is D 4R3tanθ=rhcotθ=r2R−hNow, tanx .cotx=1r2=h(2R−h)Volume of cone =13πr2hV=13πh2(2R−h)For maximum dvdh=0dvdh=13π(4Rh−3h2)0=h(4R−3h)h=4R3