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Two cells of emf `epsilon_(1)` and `epsilon_(2)(epsilon_(2) lt epsilon_(1))` are joined as shown in figure : When a potentiometer is connected between X and Y it balances for 300 cm length against `epsilon_(0)`. On conntecting the same potentiometer between X and Z it balances for 100 cm length against `epsilon_(1)` and `epsilon_(2)`. Then the ratio `(epsilon_(2))/(epsilon_(1))` is :A. `(1)/(3)`B. `(3)/(4)`C. `(1)/(4)`D. `(2)/(3)` |
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Answer» Correct Answer - D `epsilon_(1)=300 alpha`....(i) `-epsilon_(2)+epsilon_(1)=100 alpha` ....(ii) where `alpha` is the potential gradient `:. (epsilon_(2))/(epsilon_(1))=(2)/(3)` |
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