The radius of a sphere is decreased by 10% by what percent will its volume decrease

Question: The radius of a sphere is decreased by 10% by what percent will its volume decrease ?

• 25%
• 27%
• 29 % 
• 35 %

Answer: 

$$\% \text{ change in volume} \approx 3 \times \% \text{ change in radius} + \frac{3}{2} \times (\% \text{ change in radius})^2$$

Given:

• The radius decreases by 10%, so A = -10.

✅Applying the AB Method:-

$$\% \text{ change in volume} \approx 3(-10) + \frac{3}{2} (-10)^2$$ $$= -30 + \frac{3}{2} \times 100$$ $$= -30 + 150$$ $$= -27$$

📌 So, the volume decreases by 27% (approximately, compared to the exact value of 27.1%).

Detail Solution:

The volume of a sphere is given by the formula:

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

If the radius $r$ is decreased by 10%, the new radius becomes:

$$r' = 0.9r$$

The new volume :

$$V' = \frac{4}{3} \pi (0.9r)^3$$ $$V' = \frac{4}{3} \pi (0.729r^3)$$ $$V' = 0.729 V$$

The percentage decrease in volume :

$$\frac{V - V'}{V} \times 100 = \frac{V - 0.729V}{V} \times 100$$ $$= (1 - 0.729) \times 100$$ $$= 27.1\%$$

'_' the volume decreases by 27.1%.

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