Q.
During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at $40.0cm$ using a standard resistance of $90 \, \Omega$ ,as shown in the figure. The least count of the scale used in the metre bridge is $1mm$ . The unknown resistance is
NTA AbhyasNTA Abhyas 2022
Solution:
The unknown resistance is given by, $R=\left(\frac{x}{100 - x}\right)90$
Putting $x=40.0cm$ , $R=60 \, \Omega$
For error analysis, $R=\left(\frac{40 . 0 \pm 0 . 1}{100 . 0 - \left(40 . 0 \pm 0 . 1\right)}\right)\times 90=\left(\frac{40 . 0 \pm 0 . 1}{60 . 0 \pm 0 . 1}\right)\times 90$
In multiplication and division, fractional errors get added up.
Therefore, $\frac{\Delta R}{R}=\frac{0 . 1}{40}+\frac{0.1}{60}$
$\Rightarrow \Delta R=\frac{100}{\left(40\right) \left(60\right)}\times 0.1\times 60$ $=0.25 \, \Omega$
Hence, $R=\left(60 \pm 0 . 25\right)\Omega$
Putting $x=40.0cm$ , $R=60 \, \Omega$
For error analysis, $R=\left(\frac{40 . 0 \pm 0 . 1}{100 . 0 - \left(40 . 0 \pm 0 . 1\right)}\right)\times 90=\left(\frac{40 . 0 \pm 0 . 1}{60 . 0 \pm 0 . 1}\right)\times 90$
In multiplication and division, fractional errors get added up.
Therefore, $\frac{\Delta R}{R}=\frac{0 . 1}{40}+\frac{0.1}{60}$
$\Rightarrow \Delta R=\frac{100}{\left(40\right) \left(60\right)}\times 0.1\times 60$ $=0.25 \, \Omega$
Hence, $R=\left(60 \pm 0 . 25\right)\Omega$