Two resistances are connected in two gaps of a meter bridge. The balance point is 20, cm from the zero end. A resistance of 15,

Q: Two resistances are connected in two gaps of a meter bridge. The balance point is $20\, cm$ from the zero end. A resistance of $15\, ohms$ is connected in series with the smaller of the two. The null point shifts to $40\, cm$. The value of the smaller resistance in $ohms$ is

c (c) Let $S$ be larger and $R$ be smaller resistance connected in two gaps of meter bridge. $S = \left( {\frac{{100 - l}}{l}} \right)R = \frac{{100 - 20}}{{20}}R = 4R$ .....$(i)$ When $15\,\Omega $ resistance is added to resistance $R$, then $S = \left( {\frac{{100 - 40}}{{40}}} \right)(R + 15) = \frac{6}{4}(R + 15)$ .... $(ii)$ From equations $(i)$ and $(ii)$ $R = 9\,\Omega $

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Two resistances are connected in two gaps of a meter bridge. The balance point is $20\, cm$ from the zero end. A resistance of $15\, ohms$ is connected in series with the smaller of the two. The null point shifts to $40\, cm$. The value of the smaller resistance in $ohms$ is

A$3$
B$6$
C$9$
D$12$

Diffcult

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

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