Question

if two solid hemisphere soft same base radius or units or joined together around the basis then curved surface area of this new solid is​

Answer 1

Answer:

Explanation: Because curved surface area of a hemisphere is 2 w2 and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere. Hence, the curved surface area of new solid = 2 πr2 + 2 πr2 = 4πr2.

Answer 2

Answer:

Step 1:

Base radius of the right circular cone = r

Let the height of the cone be “h”.

∴ The volume of right circular cone = 1/3 * πr²h

And,  

Mass of the cone, m₁ = density * volume = ρ * 1/3 * πr²h …… (i)

Step 2:

The radius of uniform solid hemisphere = r

The density of cone = density of hemisphere = ρ … [given]

∴ The volume of uniform solid hemisphere = ½ * 4/3 * πr³ = 2/3 * πr³

And,  

Mass of the cone, m₂ = density * volume = ρ * 2/3 * πr³ …… (ii)

Step 3:

It is given that the centre of mass of the composite solid lies on the common face, therefore, we can say Ycm = 0.

The formula for the centre of mass of the combined system is given as,

Ycm = [m1y1 + m2y2] / [m1 + m2]

Substituting the values from eq. (i) & (ii), we get

⇒ 0 = [{ρ * 1/3 * πr²h * (h/4)}+{ ρ * 2/3 * πr³ * (-3r/8)}] / [{ρ * 1/3 * πr²h}+{ ρ * 2/3 * πr³}]

⇒ ρ * 1/3 * πr² [(h²/4) – (2r * 3r/8)] = 0

⇒ h²/4 – 3r²/4 = 0

⇒ h²/4 = 3r²/4

⇒ h = r*√3

Thus, the height of the cone is r√3.