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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?

  1. Baij is bound to succeed if the area of the red part is $10 sq. metres$;
  2. Baij is bound to fail if the area of the red part is $10 sq. metres$;
  3. Baij is bound to fail if the area of the red part is $11 sq. metres$;
  4. Baij is bound to succeed if the area of the red part is $11 sq. metres$;
  5. None of the above.
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A similar problem which is used to solve the given question is given in the following link :

Question No. 9 here.

The solution is explained for the given problem by TA as :

Black colored surface : $x\%$

White colored surface : $(100-x)\% = y\% = y/100$

So, $x+y=100$

Now, probability that a particular vertex has black color  is = $\frac{x}{100}$

and probability that a particular vertex has white color  is = $\frac{y}{100}$

Now, define a random variable $X_i$ as :

$X_i$ = $\left\{\begin{matrix} 1 & if \;i^{th} \; vertex\; is \;colored\; white \\ 0& otherwise \end{matrix}\right.$

Now, $P(X_i=1) = \frac{y}{100}$ where $1\leqslant i \leqslant 8$

Now, consider a new random variable $X=$ $\sum_{i=1}^{8}X_i$.

So, Expected no. of vertices of color white = $E(X)= E(\sum_{i-1}^{8}X_i) = \frac{y}{100}*8$ (Using linearity of expectation)

Since, all its '8' vertices are white which touches the sphere, so expected no. of vertices of color white should be strictly greater than 7

So, $\frac{8y}{100} > 7$

$ \Rightarrow y>700/8$

$ \Rightarrow 100-x>700/8$

$ \Rightarrow x>100/8$

$ \Rightarrow x<12.5 \%$

similar link.

Now, the same concept can be applicable for the given question.

Since, here the eight vertices of the cube touching only red colored parts of the surface of the sphere.

So, here also $y > \frac{700}{8}\%$

$\Rightarrow y > \frac{7}{8}$

So, It is saying that we need 7/8 of the total surface area of sphere to be red, so that all the $8$ vertices of the cube touch the sphere.

Since, radius of the sphere is $1$ metre.

So, $\frac{7}{8}$ of the total surface area of sphere = $\frac{7}{8}*4\pi(1)^2=10.99$ $metre^2$

So, area of the red part should be greater than $10.99$ $metre^2$

So, I think answer should be (D).

same question is asked here

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When the cube is fitted inside, its side will be $a$ such that the diagonal $\sqrt{3}a = 2r$ where $r$ is the radius of the sphere.

Given that $r = 1.$

So, $\sqrt{3}a = 2 \implies a = \frac{2}{\sqrt{3}}.$

To ensure the corners of the cube can always find a red colored region we have to ensure that all regions except a face of the cube are colored in red. If this is not the case, we can always find a coloring such that at least one corner is not able to get a red colored region. 

Now, we may think that we can color the spherical region formed by a face of cube with blue color. This region is called a spherical cap and we get its area as,

Surface area $ = 2 \pi r h$

$\quad= 2 \pi  \left(r - \frac{a}{2}\right)$ $(\because$ the distance from the centre of sphere (also center of cube) to the center of a face of cube is $a/2)$ 
$\quad= 2 \pi \left(1 - \frac{1}{\sqrt{3}}\right) = 2.65\;m^2.$

Total surface area of sphere $ =4\pi r^2 = 4\pi  $

$4\pi  - 2.65 = 9.91$ which means if we color at least $9.91$ square meters of the surface area of sphere with red color we should be able to fit the cube inside it.

So, options A and D should be correct. (Official answer was only D)

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Here given , radius of sphere  $r=1m.$

So, $d=2m.$

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

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

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 chord.

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

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.]

And to check more clearly , that how much area covered by RED color just cut it as it look like 2D [i.e. 1(ii), 2,3,4 and middle curved square already coverd  by inner cube area ]

So, $\frac{5}{6}$th area of cube (upper region will not be taken, it is in blue part)$=6\times \left ( \frac{2}{\sqrt{3}} \right )^{2}\times \frac{5}{6}=6.667$

Now $\frac{5}{6}$ of hollow region of sphere $\left ( 4\pi \left ( 1 \right )^{2}-6\left ( \frac{2}{\sqrt{3}} \right )^{2} \right )\times \frac{10}{12}=3.809$

So, total area will be $=6.667+3.809=10.476m.$

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

 

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