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Question

Physics Question on Current electricity

In a meter bridge, the wire of length 1m1\, m has a non-uniform cross-section such that, the variation dRdl\frac{dR}{dl} of its resistance RR with length ll is dRdl1l\frac{dR}{dl} \propto \frac{1}{\sqrt{l}}. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point PP. What is the length APAP ?

A

0.25 m

B

0.3 m

C

0.35 m

D

0.2 m

Answer

0.25 m

Explanation

Solution

For the given wire : dR=CddR = C \frac{d \ell}{\sqrt{\ell}} , where C=constantC = constant.
Let resistance of part AP is R1R_1 and PB is R2R_2
RR=R1R2\therefore \frac{R'}{R'} = \frac{R_{1}}{R_{2}} or R1=R2R_{1} =R_{2} By balanced
WSB concept.
Now dR=cd\int dR = c \int \frac{d\ell}{\sqrt{\ell}}
R1=C01/2d=C.2.\therefore R_{1} = C \int^{\ell}_{0} \ell^{-1/2} d\ell = C.2 . \sqrt{\ell}
R2=C11/2d=C.(22)R_{2} = C \int^{1}_{\ell} \ell^{-1/2} d\ell =C. \left(2-2\sqrt{\ell}\right)
Putting R1=R2R_{1} =R_{2}
C2=C(22)C_{2} \sqrt{\ell} =C \left(2-2 \sqrt{\ell}\right)
2=1\therefore 2 \sqrt{\ell} = 1
=12\sqrt{\ell} = \frac{1}{2}
i.e. =14m0.25m\ell = \frac{1}{4} m \Rightarrow 0.25\, m