A potentiometer wire is $4 \mathrm{~m}$ long and potential difference of $3 \mathrm{~V}$ is maintained…

A potentiometer wire is $4 \mathrm{~m}$ long and potential difference of $3 \mathrm{~V}$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $100 \mathrm{~cm}$ of the potentiometer wire is
  1. $0.50\mathrm{V}$
  2. $0.60 \mathrm{~V}$
  3. $0 .75 \mathrm{~V}$
  4. $0 .25 \mathrm{~V}$

Solution

Potential gradient $=\frac{3}{4} \frac{\mathrm{V}}{\mathrm{m}}$ So, emf against $100 \mathrm{~cm}=\frac{3}{4} \times 1 \mathrm{~V}=0.75 \mathrm{~V}$

Asked in: MHT CET 2020 (16 Oct Shift 1)

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