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प्रश्न
A hollow hemisphere bowl of thickness 1 cm has an inner radius of 6 cm. Find the volume of metal required to make the bowl.
Inner Radius r = 6 cm
Thickness, t = 1 cm
∴ Outer Radius (R) = 6 + 1 = 7 cm
∴ Volume of steel required = `2/3 πr^3 - 2/3 πr^3`
= `2/3 xx 22/7 xx square^3 - 2/3 xx 22/7 xx square^3`
= `44/21 (square - 6^3)`
= `44/21 xx (square - square)`
= `44/21 xx square`
= `5588/21` cm2
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उत्तर
Inner Radius r = 6 cm
Thickness, t = 1 cm
∴ Outer Radius (R) = 6 + 1 = 7 cm
∴ Volume of steel required = `2/3 πr^3 - 2/3 πr^3`
= \[\frac{2}{3} \times \frac{22}{7} \times \boxed{7}^3 - \frac{2}{3} \times \frac{22}{7} \times \boxed{6}^3\]
= \[\frac{44}{21} (\boxed{7}^3 - 6^3)\]
= \[\frac{44}{21} \times (\boxed{343} - \boxed{216})\]
= \[\frac{44}{21} \times \boxed{127}\]
= `5588/21` cm2
