**Solution:**

Given that in a right angled triangle, whose sides are 3 cm and 4 cm(other than hypotenuse).

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse .

Hypotenuse AC = Ö( 32 + 42) = 5 cm

Area of ΔABC = (1/2) x AB xAC

⇒ (1/2) x AC x OB = (1/2)x 4 x3

⇒ (1/2) x 5 x OB = 6

⇒ OB = 12 / 5 = 2.4 cm.

Volume of double cone = Volume of cone 1 + Volume of cone 2

= (1/3) πr^{2}h1 + (1/3)πr^{2}h2

= (1/3) πr^{2}(h1 + h2)

= (1/3) πr^{2}(OA +OC)

= (1/3) x 3.14x(2.4)^{2} x(5)

= 30.14 cm^{3}

surface area of double cone = surface area of cone ABD + surface area of cone BCD

= πrl_{1} + πrl_{2} [where l_{1} and l_{2} are the slant heights of the cone ABD and BCD respectively]

thus the volume of the double cone is 30.14 cubic cm and surface area of double cone is 52.8 sq cm