This question is from CAT Mensuration in CAT Geometry. A spherical rubber ball is cut into two pieces. Some conditions are given and to achieve that condition where we could cut the ball. Mensuration Questions in the CAT exam can be solved with practical approach and thought process rather than just knowing theorems from CAT Geometry. Even if one is not completely prepared for CAT Geometry exam, he or she should be able to do well in Mensuration problems tested by the CAT Exam. Knowing Mensuration formulas is important. Make sure you master Mensuration questions for CAT exam.

Question 17: A spherical rubber ball of radius 14 cm is cut by a knife at a distance of “x” cm from its centre, into 2 different pieces. What should be the value of “x” such that the cumulative surface area of the newly formed pieces is 3/28 more than the rubber ball’s original surface area?

- 11.4 cm
- 3 cm
- 12.4 cm
- 7.5 cm

12.4 cm.

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Area of the ball = 4πr^{2}

= 4π * 14 * 14

= 784π

Cumulative area of the 2 pieces = 3/28 more than the ball's surface area.

Therefore, extra area = 3/28 * 784π = 84π

Now this extra area = areas of 2 new circles of those 2 pieces

Let area of each new circle = πr_{1}^{2}

πr_{1}^{2} = 84π / 2 = 42π

Or r_{1}^{2} = 42

Now, r_{1}, x and r form a right angled triangle.

r^{2} = x^{2} + r_{1}^{2}

Or x^{2} = 14*14 - 42

= 196 – 42 = 154

Or x = 12.4 cm

The question is **"What should be the value of “x” such that the cumulative surface area of the newly formed pieces is 3/28 more than the rubber ball’s original surface area? "**

Choice C is the correct answer.

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