The given potentiometer has its wire of resistance \( 10\,\Omega \). When the sliding contact is in the middle of the potentiometer wire, the potential drop across \( 2\,\Omega \) resistor is:
\( 10 \, \text{V} \)
\( \frac{40}{11} \, \text{V} \)
\( 5 \, \text{V} \)
\( \frac{40}{9} \, \text{V} \)