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The figure

Shows the experimental set up of a meter bridge. The null point is found to be 60 cm away from the end A with X and Y in positions as shown. When a resistance of `15 Omega` is connected in series with Y, the null points is found to shift by 10 cm towards the end A of the wire. Find the position of the null point if resistance of `30 Omega` were connected in parallel with Y.

Text Solution

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In first casr, `X/Y = 60/40 = 3/2`....(i)
In second case, `X/Y+15 = 50/50 = 1`.....(ii)
Dividing (i) by (ii), we get
`X/Y = (Y + 15)/X = 3/2 xx 1` or `1 + 15/Y = 3/2`
or `Y=30 Omega`
From(i), `X= 3/2Y = 3/2 xx 30 = 45 Omega`
When a resistance of `30 Omega` is connected in parallel with Y, the total resistance in the right gap of bridge will be `Y' = ( 30 xx Y)/ (30 + Y) = (30 xx 30)/(30 +30) = 15 Omega`
If the null point is obtained at length l from end A of the meter bridge, then
`X/15 = l/(100-l) or 45/15 = l/(100-l)`
or `300 - 3 l = l or 4 l = 300 or l =75 cm`
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