The radii of two solid metallic spheres are $r_{1}$ and $r_{2}$ . The spheres are melted together and recast into a solid cone of height $(r_{1} + r_{2})$ . Show that the radius of the cone is $2 \sqrt{r_{1}^{2} + r_{2}^{2} - r_{1} r_{2}}$ .

Open in App

Solution

Volume of (sphere $1$ + sphere $2$ )= Volume of cone
$\frac{4}{3} π r_{1}^{3} + \frac{4}{3} π r_{2}^{3} = \frac{1}{3} π (r_{c})^{2} \times (r_{1} + r_{2})$
$\frac{4}{3} π (r_{1}^{3} + r_{2}^{3}) = \frac{1}{3} π r_{c}^{2} (r_{1} + r_{2})$
$r_{c} = \frac{\frac{4}{3} π (r_{1}^{3} + r_{2}^{3})}{\frac{1}{3} π (r_{1} + r_{2})} = \frac{4 (r_{1} + r_{2}) (r_{1}^{2} - r_{1} r_{2} + r_{2}^{2})}{(r_{1} + r_{2})}$
$r_{c}^{2} = 4 (r_{1}^{2} - r_{1} r_{2} + r_{2}^{2})$
$r_{c} = \sqrt{4 (r_{1}^{2} - r_{1} r_{2} + r_{2}^{2})} = 2 \sqrt{r_{1}^{2} - r_{1} r_{2} + r_{2}^{2}}$
$∴$ Radius of the base of cone $= 2 \sqrt{r_{1}^{2} - r_{1} r_{2} + r_{2}^{2}}$

Suggest Corrections

Similar questions

Q. The radii of the bases of two right circular solid cones of same height are

r_{1}

and

r_{2}

respectively. The cones are melted and recast into a solid sphere of radius R. Show that the height of each cone is given by

h = \frac{4 R^{3}}{r_{1}^{2} + r_{2}^{2}}

The radii of the bases of two solid right circular cones of same height are $r_{1}$ and $r_{2}$ respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms of $r_{1}, r_{2}$ and R.

Q. Two solid metallic right circular cones have same height

h

. The radii of their bases are

r_{1}

and

r_{2}

. The two cones are melted together and recast into a right circular cylinder of height

h

. Show that radius of the base of the cylinder is

\sqrt{\frac{1}{3} (r_{1}^{2} + r_{2}^{2})}

Q. If r₁ and r₂ be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is

(r_{1}^{3} + r_{2}^{3}) .^{\frac{1}{3}}

Q. The radius of a solid metallic sphere is

r

. A solid metallic cone of height

h

has base radius

r

. The two are melted together and recast into a solid right circular cone with base radius

r

. Prove that the height of the resulting cone is

4 r + h .

Join BYJU'S Learning Program

The radii of two solid metallic spheres are r1 and r2. The spheres are melted together and recast into a solid cone of height (r1+r2). Show that the radius of the cone is 2√r21+r22−r1r2.

Volume of (sphere 1 + sphere 2)= Volume of cone43πr31+43πr32=13π(rc)2×(r1+r2)43π(r31+r32)=13πr2c(r1+r2)rc=43π(r31+r32)13π(r1+r2)=4(r1+r2)(r21−r1r2+r22)(r1+r2)r2c=4(r21−r1r2+r22)rc=√4(r21−r1r2+r22)=2√r21−r1r2+r22∴ Radius of the base of cone =2√r21−r1r2+r22

The radii of two solid metallic spheres are $r_{1}$ and $r_{2}$ . The spheres are melted together and recast into a solid cone of height $(r_{1} + r_{2})$ . Show that the radius of the cone is $2 \sqrt{r_{1}^{2} + r_{2}^{2} - r_{1} r_{2}}$ .