Question

The solid sphere is formed by metting $3$ lead spheres housing radii $7.5cm,10cm$ and $12.5cm$. The hemisphere are formed from the sphere. Find $ISA$ each part of sphere.

Answer 1

Given that,

$r_1=7.5$cm

$r_2=19$cm

$r_3=12.5$cm

Then, volume of sphere Ist

$v_1=\frac{4}{3}\pi r_1^3$

$=\frac{4}{3}\times \frac{22}{7}\times (7.5{)}^3$

$=\frac{4}{3}\pi (7.5{)}^3$

Volume of sphere IInd

$v_2=\frac{4}{3}\pi r_2^2$

$v_2=\frac{4}{3}\pi (19{)}^3$

volume of sphere IIIrd

$v_3=\frac{4}{3}\pi r_3^3$

$=\frac{4}{3}\pi (12.5{)}^3$

According to given question,

melting $3$ lead spheres $=$ one hemisphere

$v_1+v_2+v_3=v$

$\frac{4}{3}\pi (7.5{)}^3+\frac{4}{3}\pi (19{)}^3+\frac{4}{3}\pi (12.5{)}^3=\frac{2}{3}\pi R^3$

$=\frac{4}{3}\pi \left[\right(7.5{)}^3+(19{)}^3+(12.5{)}^3]=\frac{2}{3}\pi R^3$

$2\times 9234=R^3$

$R^3=18468$

$R^3=(26.43{)}^3$

$R=26.43$cm

Hence, the radius $R=26.43$cm.