A meter bridge is balanced by putting `20pi` resistance in the left gap and `40pi` in the right, gap, if `40pi` resistance is now shunted with `40pi` resistance the shift in the null point towards right is nearly

Options

A. 16.67 cm
B. 50 cm
C. 25 cm
D. 70.67 cm

Answer

A

Explanation

For the case of balanced bridge
we have `X/R=(l_(x))/(l_(r))`
`=20/20=(l_(x))/(l_(r))`
`:.2l_(x)=l_(r)=(100-l_(x))`
`impliesl_(x)=33.33cm`
Now `40Omega` resistance is shunted by `40Omega`.
Thus `R=(40xx40)/(40+40)=20Omega`
Thus, to balanced bridge again,
`20/20=(l_(x))/(l_(r))`
`:.l_(x)=l_(r)=50cm`
Thus, shift in null point is given by
`50-33.33=16.67cm`