GATE 2022 EE Q.14: solution and answer

Official answer: DSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. Each solid round strand of radius $r$ has self-GMD $r^{\prime }=e^{-1/4}r\approx 0.7788r$. The four strands touch, so their centres lie at the corners of a square of side $2r$.

2. From one strand, the distances to the other three are $2r$, $2r$ and the diagonal $2\sqrt{2}r$.

3. The GMR of the group of $n=4$ strands is the $n^2$-th root of the product of all $n^2$ distances, with $D_{ii}=r^{\prime }$: $$\text{GMR}={\left(\prod _{i=1}^4\prod _{j=1}^4{D}_{ij}\right)}^{1/16}$$

4. By symmetry every strand gives the same product, so $$\text{GMR}={\left[{\left(r^{\prime }\cdot 2r\cdot 2r\cdot 2\sqrt{2}r\right)}^4\right]}^{1/16}={\left(r^{\prime }\cdot 8\sqrt{2}{r}^3\right)}^{1/4}$$

5. Substitute $r^{\prime }=0.7788r$ and $8\sqrt{2}=11.314$: $$\text{GMR}=r{\left(0.7788\times 11.314\right)}^{1/4}=r(8.811{)}^{1/4}$$

6. $(8.811{)}^{1/2}=2.968$ and $\sqrt{2.968}=1.723$, so $\text{GMR}\approx 1.723r$. Option (D).