The circuit diagram given in the figure shown is the experimental setup for the measurement of unknown resistance by using a metre bridge. The wire connected between the points P and Q ( PQ = 100 cm )

Question

The circuit diagram given in the figure shown is the experimental setup for the measurement of unknown resistance by using a metre bridge. The wire connected between the points P and $Q (PQ = 100 cm)$ has uniform cross-sectional area and its resistivity is directly proportional to the distance from point P . Null point is obtained with the jockey J with $R_{1}$ and $R_{2}$ in the gaps. If $R_{1}$ and $R_{2}$ are interchanged, the jockey has to be displaced through a distance $Δ$ from the previous position along the wire to stablish the null point. If the ratio of $\frac{R _{1}}{R _{2}} = 3$ , the value of $Δ$ (in cm) is: [Ignore any end corrections] Screenshot 2026-09-07 210455.png

Options:

Answer:

D

Solution:

$From the given circuit diagram \frac{R _{1}}{R _{2}} = \frac{x ^{2}}{( 100 ) ^{2} - x ^{2}} \Rightarrow x = 503 Also \frac{R _{2}}{R _{1}} = \frac{y ^{2}}{( 100 ) ^{2} - y ^{2}} \Rightarrow y = 50; Δ = 50 (3 - 1) cm$

TestHub · Accepted Answer

​ From the given circuit diagram R 2 ​ R 1 ​ ​ = ( 100 ) 2 − x 2 x 2 ​ ⇒ x = 50 3 ​ ​ Also R 1 ​ R 2 ​ ​ = ( 100 ) 2 − y 2 y 2 ​ ⇒ y = 50 ; Δ = 50 ( 3 ​ − 1 ) cm ​

Physics - Current Electricity Question with Solution | TestHub

Options:

Doubts & Discussion