The equivalent resistance between P and Q is:

The correct answer is Option (1) → \( 8\,\Omega \)

The circular wire has no resistance and is broken only between P and Q. So the arc from P is one node, joined to the top end of the \( 3\,\Omega \) resistor and to the left end of the \( 6\,\Omega \) resistor. The arc from Q is the other node, joined to the right end of the \( 10\,\Omega \) resistor. The \( 24\,\Omega \) resistor on the lower arc connects these two nodes directly.

The \( 3\,\Omega \) and \( 6\,\Omega \) resistors both run from node P to the centre, so they are in parallel:

    \( \displaystyle R_1 = \frac{3 \times 6}{3 + 6} = 2\,\Omega \)

This is in series with the \( 10\,\Omega \) resistor from the centre to Q:

    \( \displaystyle R_2 = 2 + 10 = 12\,\Omega \)

This branch is in parallel with the \( 24\,\Omega \) resistor:

    \( \displaystyle R_{PQ} = \frac{12 \times 24}{12 + 24} = 8\,\Omega \)