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Question

Physics Question on Current electricity

During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at 40.0cm40.0\, cm using a standard resistance of 90Ω90 \Omega, as show in the scale used in the meter bridge is 1mm1\, mm. The unknown resistance is

A

60±0.15Ω60 \pm 0.15\, \Omega

B

135±0.56Ω135 \pm 0.56\, \Omega

C

60±0.25Ω60 \pm 0.25\, \Omega

D

135±0.23Ω135 \pm 0.23 \,\Omega

Answer

60±0.25Ω60 \pm 0.25\, \Omega

Explanation

Solution

R=x100x90R =\frac{ x }{100- x } 90
R=60Ω\therefore R =60\, \Omega
dRR=100(x)(100x)dx\frac{ dR }{ R }=\frac{100}{( x )(100- x )} dx
dR=100(40)(60)0.1×60\therefore dR =\frac{100}{(40)(60)} 0.1 \times 60
=0.25Ω=0.25 \,\Omega
Alternatively
dRR=0.140+0.160\frac{ dR }{ R }=\frac{0.1}{40}+\frac{0.1}{60}
dR=0.25Ω\therefore dR =0.25 \,\Omega