Question

Plz tell me Ques no. 7

Answer 1

If rr is the radius of the sphere, then we have:

Surface area of a sphere: SA=4πr2SA=4πr2

Volume of a sphere: V=43πr3V=43πr3

Setting the two equal:

4πr2=43πr34πr2=43πr3

⟹4πr2−43πr3=0⟹4πr2−43πr3=0

⟹4πr2(1−13r)=0⟹4πr2(1−13r)=0

Clearly, this gives the two solutions:

r=0r=0, or r=3r=3

Since diameter, DD, is just twice the radius, this means the two solutions requested are:

D=0D=0, or D=6D=6

Note that strictly speaking, a “sphere” of diameter 00isn’t much of a sphere, but it is the limiting case as the radius vanishes

Basically its this

Since surface area of sphere = volume of sphere4πr2 = 4/3 * πr3 3r2 = r3 r = 3 therefore, diameter = 6


If still CONFUSED

Let the diameter of the sphere be D = 2r

Then the surface of the sphere is 4π∗r24π∗r2

The volume is given as 4/3π∗r34/3π∗r3

These two expressions should be the same, i.e.

4/3π∗r3=4π∗r24/3π∗r3=4π∗r2

Divide with what’s common on each side (supposing that r≠3r≠3)

Then we get

r/3=1r/3=1

or D =2r=6=2r=6

Test: Surface: 4π∗32=36π4π∗32=36π

Volume: 4/3π∗33=36π4/3π∗33=36π

The two are equal, so we have found our answer.