In balanced meter bridge, the resistance of bridge wire is `0.1Omega cm` . Unknown resistance X is connected in left gap and `6Omega` in right gap, null point divides the wire in the ratio 2:3 . Find the current drawn the battery of 5V having negligible resistance

Options

A. 1A
B. 1.5A
C. 2A
D. 5A

Answer

A

Explanation

According to question,
`(l_(1))/(l_(2))=(2)/(3) " "` (given)
For potentiometer, `(R_(1))/(R_(2))=(l_(1))/(l_(2))=(2)/(3)`
`(x)/(6)=(2)/(3)rArr x = 4 Omega m^(-1)`
Total resistance of known and unknown resistance
`= 6+4=10 Omega`
The resistance of wire is `0.1xx100=10 Omega`
So, effective resistance `R_("eff")` is given by
`(1)/(R_("wff"))=(1)/(10)+(1)/(10) rArr R_("eff")=5 Omega`
Hence, current drwan from the battery is given by Ohm's law, V = lR
`rArr " " l=(V)/(R )=(5)/(5)=1 A`