A) 85.5
B) 92.5
C) 90.5
D) 87.5
Correct Answer: D
Solution :
| [d] Let the radius of solid spheres be \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively. According to the question, Surface area of B = 400% of surface area of A \[\Rightarrow \] \[4\pi r_{2}^{2}=16\pi r_{1}^{2}\]\[\Rightarrow \]\[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\,\text{or}\,1:2\] Given, Volume of \[B\times \frac{(100-k)}{100}\] Volume of A \[\Rightarrow \] \[\frac{4}{3}\pi r_{2}^{3}\left( \frac{100-k}{100} \right)=\frac{4}{3}\pi r_{1}^{3}\] \[\Rightarrow \] \[{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}=\frac{100-k}{100}\]\[\Rightarrow \]\[\frac{1}{8}=\frac{100-k}{100}\] \[\Rightarrow \] \[k=\frac{700}{8}=87.5\] |
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