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The late painter Maqbool Fida Husain once coloured the surface of a huge hollow steel sphere, of radius $1$ metre, using just two colours, Red and Blue. As was his style however, both the red and blue areas were a bunch of highly irregular disconnected regions. The late sculptor Ramkinkar Baij then tried to fit in a cube inside the sphere, the eight vertices of the cube touching only red coloured parts of the surface of the sphere. Assume $\pi=3.14$ for solving this problem. Which of the following is true?

A)Baij is bound to succeed if the area of the red part is $10 sq. metres$;

B)Baij is bound to fail if the area of the red part is $10 sq. metres$;

C)Baij is bound to fail if the area of the red part is $11 sq. metres$;

D)Baij is bound to succeed if the area of the red part is $11 sq. metres$;

E)None of the above.


My solution is: Here given , radius of sphere  $r=1m.$

So, $d=2m.$

Now, say side of a cube is $a$ $m.$

So, diagonal of a square $a\sqrt{2}m.$

enter image description here

Now, see using Pythagoras Theorem 

$a^{2}+\left ( a\sqrt{2} \right )^{2}=2^{2}$

$\Rightarrow a=\frac{2}{\sqrt{3}}$

Now, we have to divide the sphere such a way $a=\frac{2}{\sqrt{3}}$ is maximum cord.

Check the diagram, where red area inscribe the cube. and blue area outside the cube.

enter image description here Now see the sphere divided inside hollow region of the cube in $6$ equal parts (1,2,3,4, shown as in diagram and 5th one in front and 6th one is back. )

So, we can say , it takes the (area of cube+$\frac{5}{6}$ . area of hollow sphere) [As only upper side of cube is blue region.]

So, area of cube$=6\times \left ( \frac{2}{\sqrt{3}} \right )^{2}=8m.$

Now $\frac{5}{6}$ of hollow region of sphere $\left ( 4\pi \left ( 1 \right )^{2}-6\left ( \frac{2}{\sqrt{3}} \right )^{2} \right )\times \frac{5}{6}=3.809$

So, total area will be $=8+3.809=11.809m.$

So, nearest option will be $D)$ or exactly $E)$


Is this correct approach?

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  • $\begingroup$ gateoverflow.in/25387/tifr2013-a-5 $\endgroup$ – Will Jagy Jun 17 at 17:46
  • $\begingroup$ @WillJagy haha, that is me only :). I want to know, is my approach correct? $\endgroup$ – Srestha Jun 17 at 17:58
  • $\begingroup$ @WillJagy Is the ans correct? $\endgroup$ – Srestha Jun 18 at 2:10

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