JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 18)
If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :
$$6\sqrt 2 \pi $$
$$6\sqrt 3 \pi $$
$$8\sqrt 2 \pi $$
$$8\sqrt 3 \pi $$
Explanation
Sphere of radius r = 3 cm
Let b, h be base radius and height of cone respectively.
So, volume of cone = $${1 \over 2}$$ $$\pi $$b2h
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In right $$\Delta $$ABC by Pythagoras theorem
(h $$-$$ r)2 + b2 = r2
$$ \Rightarrow $$ b2 = r2 $$-$$ (h $$-$$ r)2 = r2 $$-$$ (h2 $$-$$ 2hr + r2) = 2hr $$-$$ h2
$$\therefore\,\,\,$$ Volume (v) = $${1 \over 3}$$ $$\pi $$h[2hr $$-$$ h2] = $${1 \over 3}$$ [ 2h2r $$-$$ h3]
$${{dv} \over {dh}}$$ = $${1 \over 3}$$ [4hr $$-$$ 3h2] = 0
$$ \Rightarrow $$ h (4r $$-$$ 3h) = 0
$${{{d^2}v} \over {d{h^2}}}$$ = $${1 \over 3}$$ [4r $$-$$ 6h]
At h = $${{4r} \over 3}$$, $${{{d^2}y} \over {d{h^2}}}$$ = $${1 \over 3}$$ $$\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0$$
$$ \Rightarrow $$ maximum volume cours at h = $${{4r} \over 3}$$ = $${4 \over 3}$$ $$ \times $$ 3 = 4 cm
As from (1),
(h $$-$$ r)2 + b2 = r2
$$ \Rightarrow $$ b2 = 2hr $$-$$ h2 = 2.$${{4r} \over 3}$$ r $$-$$ $${{16{r^2}} \over 9}$$ = $${{8{r^2}} \over 3}$$ $$-$$ $${{16{r^2}} \over 9}$$
= $${{\left( {24 - 16} \right){r^2}} \over 9}$$ = $${{8{r^2}} \over 9}$$
$$ \Rightarrow $$ b = $${{2\sqrt 2 } \over 3}$$ r = 2 $$\sqrt 2 \,\,m$$
Therefore curved surface area = $$\pi bl$$
= $$\pi $$b$$\sqrt {{h^2} + {r^2}} $$ = $$\pi $$2$$\sqrt 2 $$ $$\sqrt {{4^2} + 8} $$ = 8$$\sqrt 3 $$$$\pi $$ cm2
Let b, h be base radius and height of cone respectively.
So, volume of cone = $${1 \over 2}$$ $$\pi $$b2h
In right $$\Delta $$ABC by Pythagoras theorem
(h $$-$$ r)2 + b2 = r2
$$ \Rightarrow $$ b2 = r2 $$-$$ (h $$-$$ r)2 = r2 $$-$$ (h2 $$-$$ 2hr + r2) = 2hr $$-$$ h2
$$\therefore\,\,\,$$ Volume (v) = $${1 \over 3}$$ $$\pi $$h[2hr $$-$$ h2] = $${1 \over 3}$$ [ 2h2r $$-$$ h3]
$${{dv} \over {dh}}$$ = $${1 \over 3}$$ [4hr $$-$$ 3h2] = 0
$$ \Rightarrow $$ h (4r $$-$$ 3h) = 0
$${{{d^2}v} \over {d{h^2}}}$$ = $${1 \over 3}$$ [4r $$-$$ 6h]
At h = $${{4r} \over 3}$$, $${{{d^2}y} \over {d{h^2}}}$$ = $${1 \over 3}$$ $$\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0$$
$$ \Rightarrow $$ maximum volume cours at h = $${{4r} \over 3}$$ = $${4 \over 3}$$ $$ \times $$ 3 = 4 cm
As from (1),
(h $$-$$ r)2 + b2 = r2
$$ \Rightarrow $$ b2 = 2hr $$-$$ h2 = 2.$${{4r} \over 3}$$ r $$-$$ $${{16{r^2}} \over 9}$$ = $${{8{r^2}} \over 3}$$ $$-$$ $${{16{r^2}} \over 9}$$
= $${{\left( {24 - 16} \right){r^2}} \over 9}$$ = $${{8{r^2}} \over 9}$$
$$ \Rightarrow $$ b = $${{2\sqrt 2 } \over 3}$$ r = 2 $$\sqrt 2 \,\,m$$
Therefore curved surface area = $$\pi bl$$
= $$\pi $$b$$\sqrt {{h^2} + {r^2}} $$ = $$\pi $$2$$\sqrt 2 $$ $$\sqrt {{4^2} + 8} $$ = 8$$\sqrt 3 $$$$\pi $$ cm2
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