Question

A solid sphere of radius r is melted to form a cone of the same radius.what is the height of the cone?​

Answer 1

Answer:

I hope it helps you

Step-by-step explanation:

Let the radius of the cone = R

and Height of a cone = r [Given]

Volume of cone =

1 \div 3\pi {}^{2} h1÷3π

2

h

Now,

Volume of sphere = Volume of cone

\begin{gathered}= )\frac{4}{3} \pi ^{3} = \frac{1}{3} ^{2} h \\ = ) \: 4r^{2} = r^{2} (r) \\ = )r = 2\end{gathered}

=)

3

4

π

3

=

3

1

2

h

=)4r

2

=r

2

(r)

=)r=2

thus,the radius of the base of cone =2r

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.