Question

If the radius of a sphere is 2r, then its volume will be :

• A. \(\frac{4}{3}\pi r^3\)
• B. \(\frac{32\pi r^3}{3}\)
• C. \(\frac{2\pi r^3}{3}\)
• D. \(4\pi r^3\)

Hint:

When the radius of a 3D object is multiplied by a factor of \(k\), its volume is multiplied by \(k^3\).
Here, the radius is doubled (\(k = 2\)), so the volume increases by a factor of \(2^3 = 8\).
Multiplying the standard volume \(\frac{4}{3}\pi r^3\) by 8 gives \(\frac{32}{3}\pi r^3\) instantly.

Answer 1

The Correct Option is B

Step 1: Think in terms of scaling rather than plugging straight in.
A sphere of radius $r$ has volume $\frac{4}{3}\pi r^3$. If we double the radius to $2r$, the volume does not just double, it scales by the cube of the scale factor.
Step 2: Apply the scale factor. \[ \left(\frac{2r}{r}\right)^3 = 2^3 = 8 \] So the new volume is 8 times the volume of the original radius-$r$ sphere.
Step 3: Multiply and conclude. \[ V = 8 \times \frac{4}{3}\pi r^3 = \frac{32}{3}\pi r^3 \] \[ \boxed{\dfrac{32\pi r^3}{3}} \]