Question

Chapter-13 assiment Q 10 The outer and inner diameters of a hemispherical bowl are 17 cm and 15 cm find the cost of polishing it over at 25 paise percm²

Answer 1

Step-by-step explanation:

Radius of inner bowl = 17/2

= 8.5cm

Radius of outer bowl = 15/2

= 7.5cm

C.S.A of outer hemisphere

$= 2 \times 3.14 \times {8.5}^{2} \\ = 6.28 \times 72.25 \\ = 453.73 {cm}^{2}$

C.S.A of inner hemisphere

$= 2 \times 3.14 \times {7.5}^{2} \\ = 6.28 \times 56.25 \\ = 353.25 {cm}^{2}$

C.S.A of ring

=

$2 \times \pi \times ( {R - r})^{2} \\ = 2 \times 3.14 \times ({8.5- 7.5}^){2}$

$6.28 \times ( {8.5}^{2} + {7.5}^{2} - 2 \times 8.5 \times 7.5) \\ = 6.28(72.25 + 56.25 - 127.5)$

$6.28(128.5 - 127.5) \\ = 6.28 {cm}^{2}$

Total area = (area of outer hemisphere) + (area of inner hemisphere) + ( area of ring)

Total area = (453.73 + 353.25 + 6.28) cm^2

= 813.26 cm^2

Cost of polishing = 25 paise/cm^2

= 0.4 rs/cm^2

Total cost = 813.26 × 0.4

= 325.304 Rs

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