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Question
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area.
[Use π=22/7]
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Solution
No. of cones = 504
Diameter of a cone = 3.5 cm
Radius of the cone, r = 1.75 cm
Height of the cone, h = 3 cm
Volume of a cone
`=1/3pir^2h`
`=1/3xxpixx(3.5/2)^2xx3`
`=1/3xxpixx3.5/2xx3.5/2xx3 " cm"^2`
Volume of 504 cones
`=504xx1/3xxpixx3.5/2xx3.5/2xx3 " cm"^2`
Let the radius of the new sphere be ‘R’.
Volume of the sphere `=4/3 pir^3 `
Volume of 504 cones = Volume of the sphere
`504xx1/3xxpixx3.5/2xx3.5/2xx3=4/3 piR^3`
`(504xx1xxpixx3.5xx3.5xx3xx3)/(3xx2xx2xx4xxpi)=R^3`
`R^3=(504xx3xx49)/64`
`R^3=(8xx27xx7^3)/64`
`R=(2xx3xx7)/4`
`R=21/2=10.5 cm`
Radius of the new sphere = 10.5 cm
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Statement A (Assertion): Total Surface area of the top is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
Statement R( Reason): Top is obtained by joining the plane surfaces of the hemisphere and cone together.
