Given: A sphere with diameter d

To Find: Altitude of the cone of greatest volume which can be inscribed in the sphere to the diameter of the sphere

Solution:

Let's assume a cone of height h and radius r can be inscribed in the sphere such that it touches the sphere at the base of the cone.

As per the geometry, the slant height of the cone can be given by:

l² = h² + r²

As the cone is inscribed in the sphere, the diameter of the sphere is equal to the diagonal of the cube which can be inscribed in the sphere. Hence, the diameter of the sphere will be equal to the diagonal of the cube, which can be given by:

d = √3 x a

where 'a' is the edge of the cube.

As per the geometry, the height of the cone can be given by:

h = a - r

Therefore, the slant height of the cone can be re-written in terms of 'a' and 'r' as:

l² = (a - r)² + r²

On simplifying, we get:

l² = a² - 2ar + 2r²

Now, let's find the volume of the cone as:

V = 1/3 x π x r² x h

On substituting the value of 'h' from the above equation, we get:

V = 1/3 x π x r² x (a - r)

V = 1/3 x π x r² x a - 1/3 x π x r³

Now, let's find the maximum volume of the cone by differentiating the above equation with respect to 'r' and equating it to zero.

dV/dr = 1/3 x π x a² - 2π x r² = 0

On solving, we get:

r = a/√3

Therefore, the height of the cone can be given by:

h = a - r = a - a/√3 = a/√3

As the diameter of the sphere is equal to √3 x a, the ratio of the altitude of the cone to the diameter of the sphere can be given by:

h/(√3 x a) = (a/√3)/(√3 x a) = 1/3

Hence, the correct option is (a) 2/3.