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The diameter of a sphere is decreased by $ 25 \% $. By what per cent does its curved surface area decrease?
Given:
The diameter of a sphere is decreased by \( 25 \% \).
To do:
We have to find the per cent by which the curved surface area decreases.
Solution:
Let $d$ be the diameter of the sphere initially.
This implies,
Radius of the sphere initially $r=\frac{d}{2}$
Surface area of the sphere initially $=4 \pi(\frac{d}{2})^{2}$
$=4\times \pi \times \frac{d^{2}}{4}$
$=\pi d^2$
On decreasing the diameter by $25 \%$, the new diameter formed $d_{1}=d-\frac{25 d}{100}$
$=\frac{75}{100} \times d$
$=\frac{3 d}{4}$
The surface area of the new sphere $=4 \pi(\frac{d_{1}}{2})^{2}$
$=4 \pi(\frac{1}{2} \times \frac{3 d}{4})^{2}$
$=4 \pi \frac{9 d^{2}}{64}$
$=\frac{\pi d^{2} 9}{16}$
Decrease in the surface area of the sphere $=\pi d^{2}(1-\frac{9}{16})$
$=\pi d^{2}(\frac{7}{16})$
Therefore,
Percentage decrease in curved surface area $=\frac{\pi d^{2}(\frac{7}{16})}{\pi d^{2}} \times 100 \%$
$=\frac{700}{16} \%$
$=43.75 \%$
The curved surface area decreases by $43.75 \%$.