A) 8 R
B) 4 R
C) 2 R
D) R
Correct Answer: B
Solution :
As, \[{{\operatorname{T}}^{2}}\,\,\propto \,\,{{R}^{3}}\] \[\therefore \,\,\,\,\frac{T_{A}^{2}}{T_{B}^{2}}=\frac{{{R}_{{{A}^{3}}}}}{{{R}_{{{B}^{3}}}}}\] \[{{\left( \frac{T_{A}^{{}}}{8T_{B}^{{}}} \right)}^{2}}\,\,=\,\,{{\left( \frac{{{R}_{{{A}^{{}}}}}}{{{R}_{{{B}^{{}}}}}} \right)}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,[\because \,\,{{T}_{B}}=8{{T}_{A}}]\] \[{{\left( \frac{1}{8} \right)}^{2}}={{\left( \frac{{{R}_{A}}}{{{R}_{B}}} \right)}^{3}}\] \[\Rightarrow \,\,\,\frac{1}{64}={{\left( \frac{{{R}_{A}}}{{{R}_{B}}} \right)}^{3}}\] \[\Rightarrow \,\,\,{{\left( \frac{1}{4} \right)}^{3}}\,\,=\,\,{{\left( \frac{{{R}_{A}}}{{{R}_{B}}} \right)}^{3}}\] \[\Rightarrow \,\,\,{{\left( \frac{1}{4} \right)}^{3}}={{\left( \frac{{{R}_{A}}}{{{R}_{B}}} \right)}^{3}}\] \[\Rightarrow \,\,\,\frac{1}{4}=\,\,\frac{{{R}_{A}}}{{{R}_{B}}}\] \[\Rightarrow \,\,\,{{\operatorname{R}}_{B}}=4{{R}_{A}}=4R\]You need to login to perform this action.
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