3 solid spheres of diameters 6 cm, 8 cm and 10 cm are melted together to form a single solid sphere. Find its radius.

The correct option is 6 cm.

To find the radius of the newly formed solid sphere, we use the principle of conservation of volume. The total volume of the three smaller solid spheres combined must be equal to the volume of the single large solid sphere formed after melting them.

Step 1: Find the radii of the three smaller spheres.
Recall that radius is half of the diameter ($r=\frac{d}{2}$).

For the first sphere:
Diameter $d_1=6\text{cm}$
Radius $r_1=\frac{6}{2}=3\text{cm}$

For the second sphere:
Diameter $d_2=8\text{cm}$
Radius $r_2=\frac{8}{2}=4\text{cm}$

For the third sphere:
Diameter $d_3=10\text{cm}$
Radius $r_3=\frac{10}{2}=5\text{cm}$

Step 2: Apply the volume conservation formula.
The formula for the volume of a sphere of radius $r$ is given by:

$V=\frac{4}{3}\pi r^3$

Let $R$ be the radius of the new solid sphere. The sum of the volumes of the three smaller spheres equals the volume of the new sphere:

$\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^3+\frac{4}{3}\pi r_3^3=\frac{4}{3}\pi R^3$

Factoring out the common term $\frac{4}{3}\pi$ from both sides:

$\frac{4}{3}\pi (r_1^3+r_2^3+r_3^3)=\frac{4}{3}\pi R^3$

Dividing both sides by $\frac{4}{3}\pi$:

$r_1^3+r_2^3+r_3^3=R^3$

Step 3: Calculate the value of $R$.
Substitute the values of $r_1,r_2,r_3$ into the equation:

$3^3+4^3+5^3=R^3$

$27+64+125=R^3$

$216=R^3$

Taking the cube root of both sides:

$R=\sqrt[3]{216}=6\text{cm}$

Thus, the radius of the newly formed solid sphere is 6 cm.