This is a homework, so I won't ask for a complete solution, but can anyone tell me how should I approach this? I've been sitting now for hours without any progress, all that I know is the answer should be (0.01Ohm). If someone is willing to dedicate his/her time to truly solve this I'd really be grateful! Thanks.
First you should know that the book answer is only an approximation. The exact equations are complex and nonlinear.
If was going to solve this precisely, I would calculate Thevenin equivalent of both sides, which have a series resistance of 50||1000 on one side and (50 + a)|| 1000 on the other side, with 100 ohms between them. Then go and calculate the delta in Thevenin voltage to get the change in resistance for an exact 5uA through the galvo.
However we can quickly approximate if we assume constant current in the legs of the bridge of about 10mA and that gives you the change in Thevenin voltage. Simplify the total resistance and the book answer falls out. It's in error by maybe 5%, but close enough.
The approach is critical in this kind of question- it's almost a trick question because if you take a perfectionist approach you'll waste a lot of time and maybe have difficulty in an exam situation- but real life is like that too- if you take the wrong approach you need to recognize it as quickly as possible and abandon the overly complex approach.
The key question here is: How much must the bridge be out of balance in order to show on the galvanometer — i.e., put at least 0.5 µA through its 100 Ω resistance?
The approach would be to set up a Thévenin source for the left side of the bridge (based on 10 V, 1 kΩ and 50 Ω) and another Thévenin source for the right side of the bridge (based on 10 V, 1 kΩ and 50 Ω + ΔR). Put the 100 Ω resistance of the galvanometer between them and then solve for the value of ΔR that results in 0.5 µA of current.
Is this enough of a hint to get you going?
As long as the answer you get is greater than the actual resolution of the resistance standard, then that's the final answer; otherwise, you're limited to the resolution of the standard.