In the given potentiometer circuit arrangement, the balancing length {AC} is measured to be 250 {cm}. When the galvanometer connection is shifted from point (1)

Q: In the given potentiometer circuit arrangement, the balancing length ${AC}$ is measured to be $250$ ${cm}$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram, the balancing length becomes $400\, {cm}$. The ratio of the emf of two cells, $\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

c $E_{1}=k l_{1} \ldots .$ (i) $E_{1}+E_{2}=k l_{2} \ldots$ (ii) $\frac{E_{1}}{E_{1}+E_{2}}=\frac{l_{1}}{l_{2}}=\frac{250}{400}=\frac{5}{8}$ $8 E_{1}=5 E_{1}+5 E_{2}$ $3 E_{1}=5 E_{2}$ $\frac{E_{1}}{E_{2}}=\frac{5}{3}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

In the given potentiometer circuit arrangement, the balancing length ${AC}$ is measured to be $250$ ${cm}$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram, the balancing length becomes $400\, {cm}$. The ratio of the emf of two cells, $\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

A$\frac{4}{3}$
B$\frac{3}{2}$
C$\frac{5}{3}$
D$\frac{8}{5}$

JEE MAIN 2021, Diffcult

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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