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 pm 0.15 Omega`
B. `135 pm 0.56 Omega`
C. `60 pm 0.25 Omega`
D. `135 pm 0.23 Omega`

Answer

C

Explanation

Let `l_(1)` is the distance of null point from left and `l_(2)` be the same from right end. Note that sun of `l_(1)" and "l_(2)` is 100 cm and that we can use later on. Using condition+N408 of balanced wheatstone.s bridge we can write the following equation :
`(R )/(90)=(l_1)/(l_2)`
`implies R=90xx(l_1)/(l_2)" ""........"(i)`
We can substitute `l_(1)" and "l_(2)` in above equation (i) to calculate value of R as follows :
`R= 90xx(40)/(60)=60 Omega`
From equation (i) we can write the following :
`(triangleR)/(R )=(triangle l_1)/(l_1)+(triangle l_2)/(l_2)`
`implies (triangleR)/(60)=(0.1)/(40)+(0.1)/(60) implies triangleR= 6((1)/(40)+(1)/(60))`
`implies triangleR= 6((6+4)/(240))=(1)/(4)=0.25 Omega`
Hence unknown resistance is : `(60 pm 0.25) Omega`. Option c is correct.