TG EAMCET 3 May 2025 Shift 2 Paper

Step 1: Total length of the wireThe total length of the wire is the length of the semicircle plus the diameter. That is:$l=\pi R+2R$Step 2: Find resistance per unit lengthSince the total resistance is $36\ \Omega$ for the entire wire, resistance per unit length is:$\sigma=\frac{36}{\pi R+2R}$Step 3: Resistance along the diameter (straight part)The length of the diameter is $2R$. So, resistance along the diameter is:$R_1=\sigma \times 2R=\frac{36}{\pi R+2R}\times 2R=\frac{72}{\pi+2}$Step 4: Resistance along the semicircleThe length of the semicircle is $\pi R$. So, resistance along the semicircle is:$R_2=\sigma \times \pi R=\frac{36}{\pi R+2R}\times \pi R=\frac{36\pi}{\pi+2}$Step 5: Connect the two paths in parallelThe diameter and semicircular parts connect at both ends, so they are in parallel. Use the formula for parallel resistances:$R_{eq}=\frac{R_1R_2}{R_1+R_2}$Step 6: Substitute the values for $R_1$ and $R_2$$R_{eq}=\frac{\frac{72}{\pi+2} \times \frac{36\pi}{\pi+2}}{\frac{72}{\pi+2}+\frac{36\pi}{\pi+2}}=\frac{72 \times 36\pi}{(\pi+2)(36\pi+72)}$Step 7: Simplify the expression further$=\frac{72 \times 36 \times 22 \times 7}{36 \times(36 \times 22+72 \times 7)}=\frac{72 \times 22 \times 7}{36(22+14)}=\frac{77}{9}\ \Omega$