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प्रश्न
A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.
उत्तर
Radius of hemisphere = 9 cm
Volume of hemisphere = `2/3pir^3= 2/3pixx9xx9xx9 "cm"^3`
Height of cone = 72 cm
Let the radius of the cone be r cm.
Volume of the cone `= 1/3pir^2h=1/3pir^2xx 72 "cm"^3`
The Volume of the hemisphere and cone are equal.
Therefore,
`1/3pir^2xx72xx =2/3xx9xx9xx9xx9`
`r^2 =(2xx9xx9xx9)/72 = 81/4`
`r = sqrt(81/4) = 9/2 = 4.5 "cm"`
The radius of the base of the cone is 4.5 cm.
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