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Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceDiagram basedMediumCalculation

JEE Main 25 July 2021, Shift 2, Physics Q20

Question 20 of 90 in this shift, Physics question 20 of 30, Section A.

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, ε1ε2\frac{\varepsilon_1}{\varepsilon_2} is :

The figure, in words

Potentiometer circuit: wire AB with jockey/contact C on it; the top branch has a key K, a cell marked V and a rheostat connected between A and B. Galvanometer G connects from C down to the common junction between two cells ε1\varepsilon_1 and ε2\varepsilon_2 (in series along the bottom); the end A is connected by a wire to terminal (1) (the free end of the ε1\varepsilon_1 side) and terminal (2) is the free end of the ε2\varepsilon_2 side.
  1. (1)85\frac{8}{5}
  2. (2)53\frac{5}{3}Official answer
  3. (3)32\frac{3}{2}
  4. (4)43\frac{4}{3}

Official answer

Option 2

NTA final key.