A) 14 cm
B) 16 cm
C) 12 cm
D) 18 cm
Correct Answer: B
Solution :
\[\frac{mR_{1}^{2}+R_{2}^{2}}{2}=m{{K}^{2}}\] \[\frac{m({{10}^{2}}+{{20}^{2}})}{2}=m{{K}^{2}}\Rightarrow \frac{100+400}{2}={{K}^{2}}\] \[{{K}^{2}}=250\] \[\Rightarrow K\,\underline{\approx }\,16\,cm\]
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