Answer: B
Let the radius before melting be \(r_1\), and after melting, let the radius be \(r_2\) units.
According to the condition: \( \cfrac{4πr_1^2 - 4πr_2^2}{4πr_1^2} \times 100 = 75 \)
or, \(4(r_1^2 - r_2^2) = 3r_1^2\)
or, \(4r_1^2 - 3r_1^2 = 4r_2^2\)
or, \(r_1^2 = 4r_2^2\)
or, \(r_1 = 2r_2\).
∴ The volume reduced (or ice melted): \( \cfrac{\cfrac{4}{3}πr_1^3 - \cfrac{4}{3}πr_2^3}{\cfrac{4}{3}πr_1^3} \times 100\% \)
\(= \cfrac{r_1^3 - r_2^3}{r_1^3} \times 100\% \)
\( = \cfrac{(2r_2)^3 - r_2^3}{(2r_2)^3} \times 100\% \)
\( = \cfrac{7}{8} \times 100\% = 87.5\%. \)
Thus, 87.5% of the ice has melted.
Let the radius before melting be \(r_1\), and after melting, let the radius be \(r_2\) units.
According to the condition: \( \cfrac{4πr_1^2 - 4πr_2^2}{4πr_1^2} \times 100 = 75 \)
or, \(4(r_1^2 - r_2^2) = 3r_1^2\)
or, \(4r_1^2 - 3r_1^2 = 4r_2^2\)
or, \(r_1^2 = 4r_2^2\)
or, \(r_1 = 2r_2\).
∴ The volume reduced (or ice melted): \( \cfrac{\cfrac{4}{3}πr_1^3 - \cfrac{4}{3}πr_2^3}{\cfrac{4}{3}πr_1^3} \times 100\% \)
\(= \cfrac{r_1^3 - r_2^3}{r_1^3} \times 100\% \)
\( = \cfrac{(2r_2)^3 - r_2^3}{(2r_2)^3} \times 100\% \)
\( = \cfrac{7}{8} \times 100\% = 87.5\%. \)
Thus, 87.5% of the ice has melted.