Question

Water level in a cylinder raises by 16 cm when three spheres are dropped in it. If the ratio of the radii of the spheres is 1:2:3 and the radius of the cylinder is 6 cm, what is the volume (in cm${}^3$) of the smallest sphere?

• A. $16\pi$
• B. $24\pi$
• C. $12\pi$
• D. $8\pi$

Answer 1

The correct option is A $16\pi$

Given the ratio of the radii of the spheres is 1:2:3 and the radius of the cylinder is 6 cm.

Let the common ratio be $x$.
So, the radius of the spheres are $x$, $2x$ and $3x$.

Volume of the spheres $=\frac{4}{3}\pi x^3+\frac{4}{3}\pi (2x{)}^3+\frac{4}{3}\pi (3x{)}^3=\frac{4}{3}\pi (x^3+8x^2+27x^3)=48\pi x^{3}c{m}^3$

Volume of water raise in the cylinder after putting sphere balls
$=\pi \times 6^2\times 16=576\pi c{m}^2$

Now, volume of sphere balls = volume of water raises in the cylinder
$\Rightarrow 48\pi x^3=576\pi \Rightarrow x^3=\frac{576}{48}\Rightarrow x^3=12$

Volume of the smallest sphere $=\frac{4}{3}\pi \times x^3=\frac{4}{3}\pi \times 12=16\pi c{m}^3$