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Physics Question on Current electricity

A 2V2V battery, a 990 Ω990 \, \Omega resistor and a potentiometer of 2m2m length, all are connected in series of the resistance of potentiometer wire is 10 Ω10 \, \Omega , then the potential gradient of the potentiometer wire is

A

0.05 Vm−10.05\,Vm^{-1}

B

0.5 Vm−10.5\,Vm^{-1}

C

0.01 Vm−10.01\,Vm^{-1}

D

0.1 Vm−10.1\,Vm^{-1}

Answer

0.01 Vm−10.01\,Vm^{-1}

Explanation

Solution

Potential gradient
x=e(R+Rh+r)⋅RLx=\frac{e}{\left(R+R_{h}+r\right)} \cdot \frac{R}{L}
⇒x=2(990+10)×102\Rightarrow x=\frac{2}{(990+10)} \times \frac{10}{2}
=0.01Vm−1=0.01 Vm ^{-1}