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The resistance in the two arms of the me...

The resistance in the two arms of the meter bridge are `5 Omega` and `R Omega`, respectively. When the resistance `R` is shunted with an equal resistance, the new balance point is `1.6 l_(1)`.
The resistance `R` is

A

`10 Omega`

B

`15 Omega`

C

`20 Omega`

D

`25 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B
`(5)/(R)=(l_(1))/(100-l_(1))`
`(5)/(R//2)=(1.6 l_(1))/(100-1.6l_(1))`
Solving `l_(1) = 25 cm and R = 15 Omega`.
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