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 Omega`, as shown in the figure. The least count of the scale used in the metre bridge is 1mm. The unknown resistance is

Options

A. `60 +- 0.15 Omega`
B. `135 +- 0.56 Omega`
C. `60 +- 0.25 Omega`
D. `135 +- 0.23 Omega`

Answer

c

Explanation

`(R)/(90) = (40)/(60) or R = 60 Omega`
If balancing length of bridge wire zero end is x, then `(R)/(90) = (x)/((100-x))`
Taking log of both the sides we get
`log_(e)R- log_(e) 90 = log_(e) x- log_(e) (100- x)`
Diffrentiating it we get
`(Delta R)/(R) = (Delta x)/(x) - (Delta (100-x))/((100-x))`
Error in `R, Delta R = pm((Delta x)/(x)+(Delta (100-x))/(100-x))R`
`= +- ((Delta x)/(x)+(Deltax)/(100-x))R=pm((0.1)/(40) + (0.1)/(60))xx60`
`= +- 0.25 Omega`
`:. R = 60 +- 0.25 Omega`