A potentiometer wire is 100cm long and a constant potential...

A potentiometer wire is

100cm

long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at

50cm

and

10cm

from the positive end of the wire in the two cases. The ratio of EMF's is : A potentiometer wire is 100cm long and a constant potential difference is maintained across it. Two cells are connected

A potentiometer wire is 100cm long and a constant potential difference is maintained across it. Two cells are connected

Options

A5 : 1
B5 : 4
C3 : 4
D3 : 2

Correct answer

D. 3 : 2

Step-by-step solution

A potentiometer wire is 100cm long and a constant potential difference is maintained across it. Two cells are connected

as $E ∝ L$

$\dfrac{{\text{E}}_{1} + {\text{E}}_{2}}{{\text{E}}_{1} - {\text{E}}_{2}} = \dfrac{\text{\lambda}50}{\text{\lambda}10}$

${E}_{1} + {E}_{2} = 5{E}_{1} - 5{E}_{2}$

$6{E}_{2} = 4{E}_{1}$

$\dfrac{3}{2} = \dfrac{{E}_{1}}{{E}_{2}}$

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