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In the given potentiometer circuit arrangement, the balancing length \mathrm{AC} is measured to be 250 \mathrm{~cm}. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 \mathrm{~cm}. The ratio of the emf of two cells, \frac{\varepsilon_{1}}{\varepsilon_{2}}  is :
Option: 1 \frac{5}{3}
Option: 2 \frac{8}{5}
Option: 3 \frac{4}{3}
Option: 4 \frac{3}{2}

Answers (1)

best_answer

When the galvanometer is connected to the point (1)
\varepsilon _{1}= \phi \l _{1}\rightarrow \left ( 1 \right )
When the galvanometer is connected to point (2)
\varepsilon _{1}+\varepsilon _{2}= \phi \l _{2}\rightarrow \left ( 2 \right )
Where\, \l _{1}= 250\, cm\, & \l _{2}= 400\, cm
\frac{\varepsilon _{1}}{\varepsilon _{1}+\varepsilon _{2}}= \frac{250}{400}= \frac{5}{8}
\frac{\varepsilon _{1}+\varepsilon _{2}}{\varepsilon _{1}}= \frac{8}{5}
1+\frac{\varepsilon _{2}}{\varepsilon _{1}}= \frac{8}{5}
\frac{\varepsilon _{2}}{\varepsilon _{1}}= \frac{3}{5}
\Rightarrow \frac{\varepsilon _{1}}{\varepsilon _{2}}= \frac{5}{3}
The correct option is (1)

 

Posted by

vishal kumar

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