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 -
JEE MAIN 2021, Diffcult
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$\frac{20- V _{0}}{5}+\frac{0- V _{0}}{5}+\frac{20- V _{0}}{2}=0$
$4+10=\frac{2 V _{0}}{5}+\frac{ V _{0}}{2}$
$14=\frac{4 V _{0}+5 V _{0}}{10}$
$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
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