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Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueEasyData or graph

JEE Main 25 February 2021, Shift 1, Physics Q30

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

In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1E_1 and E2E_2 as well. For key K1K_1 (closed), the jockey is adjusted to touch the wire at point J1J_1 so that there is no deflection in the galvanometer. Now the first battery (E1E_1) is replaced by second battery (E2E_2) for working by making K1K_1 open and K2K_2 closed. The galvanometer gives then null deflection at J2J_2. The value of E1E2\frac{E_1}{E_2} is ab\frac{a}{b}, where a = __________ .

The figure, in words

Potentiometer circuit: the potentiometer wire AB is drawn as a zig-zag of parallel segments running from A at the top to B at the bottom, with total length labelled 1 m (arrow at the bottom from B). A battery E, a key K and a rheostat Rh (zig-zag resistor with an arrow contact) are connected in series across A and B. At the top, two cells are drawn in parallel branches: E1 with key K1 (drawn with a filled dot, closed) and E2 with key K2 (drawn open); they connect to point A by a slanted line and, through a galvanometer G, to a jockey contact. Jockey position J1 is marked at a distance labelled 20 cm from A (arrow pointing up at the wire), and J2 is marked at a distance labelled 60 cm (arrow pointing up, joined by a dashed line to the galvanometer lead).

Official answer

1

NTA final key.