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JEE Advanced 2014 · Paper 2, Physics question 3

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is

Metre bridge: unknown resistance R in the left gap and the standard 90 Ω in the right gap; a galvanometer connects the junction to the jockey, which touches the wire at 40.0 cm from the left end; a cell and key across the wire.
Choose one option
Answer and worked solution

Official answer: C

Balance. With the null point at l=40.0l=40.0 cm,
R90=4060  ⟹  R=60 Ω.\frac R{90}=\frac{40}{60}\implies R=60\ \Omega.

Error. R=90l100−lR=90\dfrac{l}{100-l}, and each length is read to Δl=0.1\Delta l=0.1 cm:
ΔRR=Δll+Δl100−l=0.140+0.160=0.00417.\frac{\Delta R}R=\frac{\Delta l}l+\frac{\Delta l}{100-l}=\frac{0.1}{40}+\frac{0.1}{60}=0.00417.
So ΔR=60×0.00417=0.25 Ω\Delta R=60\times0.00417=0.25\ \Omega.

(C)

Topics:Metre bridge and Ohm's law experimentsWheatstone bridge and metre bridgeErrors in measurement
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