The figure below shows an annular ring with outer and inner as b and a,…
2020
The figure below shows an annular ring with outer and inner as b and a, respectively. The annular space has been painted in the form of blue colour circles touching the outer and inner periphery of annular space. If maximum n number of circles can be painted, then the unpainted area available in annular space is _________ .
- A.
π × [ (b² − a²) − (n/4)(b − a)² ]
- B.
π × [ (b² − a²) − (n)(b − a)3 ]
- C.
π × [ (b² − a²) − (n)(a − b)² ]
- D.
π × [ (b² − a²) − (n)(b − a)1/2 ]
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Correct answer: A
Key idea: a circle tangent to both the inner and outer concentric circles has radius equal to half the radial gap.
Radius of each painted circle: r = (b - a)/2.
Area of one painted circle: π r^2 = π((b - a)^2)/4.
Area of the annulus (entire painted region before placing small circles): π(b^2 - a^2).
Total painted area by n circles: n × π((b - a)^2)/4.
Therefore the unpainted area = π(b^2 - a^2) - n × π((b - a)^2)/4 = π[(b^2 - a^2) - (n/4)(b - a)^2].
Note:
The centers of the small circles lie on the mid-circle of radius R = (a + b)/2. For adjacent painted circles to be tangent, their center-to-center chord length must equal b - a, which leads to sin(π/n) = (b - a)/(a + b).
Thus the theoretical value n = π / arcsin((b - a)/(a + b)). The maximum integer number of circles that can be painted is the floor of this value.
This derivation yields the unpainted area expression π[(b^2 - a^2) - (n/4)(b - a)^2], matching the correct answer.
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