| The sum of radii of two spheres is 10 cm and the sum of their volumes is \[880\,\,c{{m}^{3}}.\] What will be the product of their radii? [SSC (10+2) 2005] |
A) \[21\]
B) \[26\frac{1}{3}\]
C) \[33\frac{1}{3}\]
D) \[70\]
Correct Answer: B
Solution :
| Let \[{{r}_{1}}\] and \[{{r}_{2}}\]be the radii of sphere. |
| Given, \[{{r}_{1}}+{{r}_{2}}=10\] (i) |
| and volume = 880 |
| \[\Rightarrow \]\[\frac{4}{3}\pi (r_{1}^{3}+r_{2}^{3})=880\] |
| \[\Rightarrow \]\[r_{1}^{3}+r_{2}^{3}=\frac{880\times 3\times 7}{22\times 4}=210\] (ii) |
| On cubing both sides of Eq. (i), we get |
| \[{{({{r}_{1}}+{{r}_{2}})}^{3}}=1000\] |
| \[\Rightarrow \]\[r_{1}^{3}+r_{2}^{3}+3{{r}_{1}}{{r}_{2}}({{r}_{1}}+{{r}_{2}})=1000\] |
| \[\Rightarrow \]\[210+3{{r}_{1}}{{r}_{2}}(10)=1000\] |
| \[\Rightarrow \] \[30{{r}_{1}}{{r}_{2}}=1000-210=790\] |
| \[\Rightarrow \] \[{{r}_{1}}{{r}_{2}}=\frac{790}{30}=\frac{79}{3}=26\frac{1}{3}\] |
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