A) 1 cm
B) 5.2 cm
C) 2.3 cm
D) 3.7 cm
Correct Answer: A
Solution :
Let the thickness of the pipe = x cm |
\[\therefore \] If the external radius \[=9\,\,\text{cm}\] |
then, in radius \[=(9-x)\,\,\text{cm}\] |
According to the question, |
\[\pi \times {{9}^{2}}\times 14-\pi \times 14\times {{(9-x)}^{2}}=748\] |
\[\Rightarrow \] \[\pi \times 14\,\,(81-(81+{{x}^{2}}-18x))=748\] |
\[\Rightarrow \] \[\pi \times 14\,\,(-{{x}^{2}}+18x)=748\] |
\[\Rightarrow \] \[-{{x}^{2}}+18x=\frac{748}{\pi \times 14}=\frac{748\times 7}{22\times 14}\] |
\[\Rightarrow \] \[-{{x}^{2}}+18x=17\] |
\[\Rightarrow \] \[{{x}^{2}}-18x+17=0\] |
\[\Rightarrow \] \[{{x}^{2}}-17x-x+17=0\] |
\[\Rightarrow \] \[x\,\,(x-17)-1\,\,(x-17)=0\] |
\[(x-1)(x-17)=0\] |
\[\therefore \] \[x=1\] |
or 17 but x = 17 is inadmissible |
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