
I1=4MR12,I2=2MR22,I3=52MR12 According to problem I2I1=2.5⇒2MR224MR12=25⇒R22R12=5… Now we are provided with information that $\begin{aligned}
& \frac{\mathrm{I}_3}{\mathrm{I}_2}=\mathrm{n} \
& \Rightarrow \frac{\frac{2 \mathrm{MR}_1^2}{5}}{\frac{\mathrm{MR}_2^2}{2}}=\mathrm{n} \Rightarrow \frac{4 \mathrm{R}_1^2}{5 \mathrm{R}_2^2}=\mathrm{n}
\end{aligned}FromEq′,(1)and(2)\Rightarrow n=4$