The resistances in left and right gap of a meter brigdge are 20 `omega` and 30 `omega` respecitively when the resistance in the left gap is reduced to half its value then balance point shifts by

Options

A. 15 cm to the right
B. 15 cm to the left
C. 20 cm to the right
D. 20 cm to the left

Answer

B

Explanation

`(X)/(R )=(20)/(30)=(L_(X))/(L_(R )) therefore L_(X)=(2)/(3)L_( R)`

For the metre bridge, `L_(X)+L_(R )=100`,
`therefore L_(X)=40 cm and L_(R )=60 cm`
In the second case, `X=10 Omega and R= 30 Omega`
`therefore (L_(X))/(L_(R ))=(10)/(30)=(1)/(3) therefore L_(X)=(L_(R ))/(3)`
`therefore (L_(R ))/(3)+L_( R)=100 therefore (4)/(3)L_(R )=100`
`therefore L_(R )=75 cm and L_(X)=25 cm`
Initially the null point is at 40 cm from A and in the second case it is at 25 cm from A.
Thus it is shifted by (40-25)=15 cm to the left.