Volume of a sphere varies directly with the cube of its radius. Three solid spheres having length of `1""1/2,2and 2""1/2 ` metres diameter are melted and a new sphere is formed . Find the length of diameter of the new sphere. [Consider that the volume of sphere remains same before and after melting ]

Explanation

Let the volume of the sphere =V and radius =R .
As per question , V `prop R^(3) ` or , V =k. `R^(3) ` ……………(1) [when k=non -zero variation constant ]
Now , putting , R =`(1""1/2)/(2) =3/4 , R=2/2=1 and R =(2""1/2)/(2) =5/4` respectively in (1)
we get , ` V_1 =k , (3/4)^3 or , 64 V_1 = 27 k `..............(2)
` V_2 =k .(1)^(3) or,64 V_(2) =k ...........(3)`
and `V_(3)=k.(5/4)^(3) or , 64 V_(3) =125 k ...........(4) `
when volumes of the first , seconds and third spheres are `V_(1) ,V_(2) and V_(3) ` respectively
`therefore ` Total volume , V =`V_(1)+V_(2) +V_(3) `
`=(27k)/64+k+(125k)/(64)=(27k+64k+125k)/(64)`
`=(216k)/(64)=(54k)/(16) =(27k)/(8)`
Now putting V =`(27k)/(8) ` in (1) we get , `(27k)/(8)=k.R^(3)`
or , `R^(3) =(27)/(8) [because kne0]` or , `R^(3)=(3/2)^(3) or , R =3/2`
`therefore` The radius of the new sphere =`3/2 ` metres
`therefore` Diameter of the new sphere=`2xx3/2 `metres =3 metres.
Hence the required diameter of the new sphere =3 metres .