Question

Figure shows the potentiometer arrangement to compare the emf of cells E₁ and E₂. Length of the resistance wire AB is 100 cm. If null point obtained for E₁ and E₂ are at distance 20 cm and 40 cm respectively from B then E₁/E₂ is –

• A. 1 : 2
• B. 4 : 5
• C. 3 : 2
• D. 4 : 3

Answer 1

Answer: (D)

4 : 3

Answer 2

E₁ = Q × l₁ ..... (1)

E₂ = Q × l₂ ...... (2)

$\frac { E _ { 1 } } { E _ { 2 } }$ = $\frac { \ell _ { 1 } } { \ell _ { 2 } }$ = $\frac { 80 } { 60 }$ = $\frac { 4 } { 3 }$

(l₁ & l₂ should be measured from A)