The volume $V$ of a sphere with diameter $d$ is given by the formula:

$V = \frac{\pi}{6} d^3$

If the diameter $d$ is increased by 60%, then the new diameter $d'$ will be:

$d' = d \times 1.6 = 1.6d$

Now, let’s calculate the new volume $V'$ with the new diameter $d'$:

$V' = \frac{\pi}{6} (d')^3 = \frac{\pi}{6} (1.6d)^3 = \frac{\pi}{6} \times 1.6^3 \times d^3$

Calculating $1.6^3$:

$1.6^3 = 4.096$

So,

$V' = \frac{\pi}{6} \times 4.096 \times d^3 = 4.096 \times \frac{\pi}{6} d^3$

Now, the increase in volume is:

$\text{Increase in Volume} = V' - V = 4.096V - V = (4.096 - 1)V = 3.096V$

To find the percentage increase:

$\text{Percentage Increase} = \frac{\text{Increase in Volume}}{V} \times 100 = 3.096 \times 100 = 309.6\%$

Answer:

If the diameter of the sphere is increased by 60%, the volume of the sphere increases by 309.6%.

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Related Questions:

1. What happens to the surface area of the sphere if the diameter is increased by 60%?
2. If the radius of the sphere is increased by 60%, what is the percentage increase in volume?
3. How does the formula for volume change if the shape were a cylinder instead of a sphere?
4. What percentage increase in diameter would double the volume of a sphere?
5. How would an increase in diameter affect the density of the sphere if mass remains constant?

Tip:

Remember that volume scales with the cube of the diameter (or radius) for spheres, so even small increases in diameter result in large increases in volume.