A sphere of radius r cm is cut into three parts by two parallel planes such that the total surface area is increased by 2πr^{2} due to the formation of new figures. One cut is at height h_{1}, while other cut is at height h_{2} from the center. If the ratio of the heights h_{1} and h_{2} is 1 : 2, then what will be the value of h_{2} in terms of radius of the sphere?

## Answer

Correct Answer : b ) \(\frac{2}{\sqrt{5}}\: r \)

Explanation :Depicting the given information in the figure, we get

According to the question, the total surface area of the resulting figures is increased by 2πr^{2}. As increase in the area can only be because of addition of the new circular faces created because of the two cuts, therefore,

Increase in Total surface area of Sphere = 2* (Area of C_{1} + Area of C_{2})

2πr^{2} = 2πr_{1}^{2} + 2πr_{2}^{2}

r_{1}^{2 }+ r_{2}^{2} = r^{2}........... (1)

Now, applying Pythagoras theorem in Δ OFB

r_{1}^{2} = r^{2} – h_{1}^{2}.......... (2)

And, applying Pythagoras theorem in Δ OEC

r_{2}^{2} = r^{2} – h_{2}^{2}.......... (3)

Putting the values of r_{1} and r_{2} from equation 2 and 3 in equation 1, we get

r^{2} – h_{1}^{2} + r^{2} – h_{2}^{2} = r^{2}

h_{2}^{2} + h_{1}^{2} = r^{2}..... (4)

As, ratio of the heights h_{1} and h_{2} is 1 : 2 So,

\(\frac{h_{1}}{h_{2}}\: = \frac{1}{2}\: \) ........(5)

From equation 4 and 5, we get

\(h_{2}\, ^{2}+\: \frac{1}{4}{h_{2}\, ^{2}}\: = r^{2}\)

\(h_{2}\, =\frac{2}{\sqrt{5}}\: r \)

Hence, (b) is the correct answer.

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