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Q.
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 :
$\frac{20-V_{0}}{5}+\frac{0-V_{0}}{5}+\frac{20-V_{0}}{2}=0$
$4+10=\frac{2 V_{0}}{5}+\frac{V_{0}}{2}$
$14=\frac{4 V_{0}+5 V_{0}}{10}$
$V_{0}=\frac{140}{9}$ Volt
Potential difference across $2\, \Omega$ resistor is $20-V_{0}$
That is $\left(20-\frac{140}{9}\right)$ Volt
Hence answer is $\left(\frac{40}{9}\right)$ Volt
