A) 3.5 cm, 3 cm
B) 2.5 cm, 2 cm
C) 4.5 cm, 4 cm
D) 4.75 cm, 4.25 cm
Correct Answer: B
Solution :
Let \[{{R}_{1}}\] is inner radius and \[{{R}_{2}}\] is outer radius of cylinder, then \[2\pi \,\,{{R}_{2}}h-2\pi \,\,{{R}_{1}}h=44\] (Given) \[2\times \frac{22}{7}\times {{R}_{2}}\times 14-2\times \frac{22}{7}\times {{R}_{1}}\times 14=44\] \[\Rightarrow \]\[2\times \frac{22}{7}\times 14({{R}_{2}}-{{R}_{1}})=44\] \[\Rightarrow \]\[{{R}_{2}}-{{R}_{1}}=\frac{1}{2}\] ?(i) Volume of pipe \[=99\,\,\text{c}{{\text{m}}^{3}}\] \[\pi R_{2}^{2}h-\pi R_{1}^{2}h=99\] \[\Rightarrow \]\[R_{2}^{2}-R_{1}^{2}=99\times \frac{7}{22}\times \frac{1}{14}\]\[\Rightarrow \]\[({{R}_{2}}+{{R}_{1}})({{R}_{2}}-{{R}_{1}})=\frac{9}{4}\] \[({{R}_{2}}+{{R}_{1}})=\frac{1}{2}=\frac{9}{4}\] [from Eq. (i)] \[{{R}_{2}}+{{R}_{1}}=\frac{9}{2}\] ?(ii) On adding Eqs. (i) and (ii), we get \[{{R}_{2}}=\frac{5}{2}=2.5\,\,\text{cm}\] \[\therefore \frac{5}{2}+{{R}_{1}}=\frac{9}{2}\] \[\Rightarrow \]\[{{R}_{1}}=2\] \[\therefore \]\[{{R}_{2}}=2.5\,\,\text{cm,}\]\[{{R}_{1}}=2\,\,\text{cm}\]
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