with a certain resistance in the left gap of a metre bridge,the balance point is obtained when a resistance of 10 ohm is taken out of resistance box. on increasing the resistance from the resistance box by 12.5 ohm , the balance point shift towards by 20 cm . find the value of unknown resistance.

Dear Student,

Please find below the solution to the asked query:

The diagram of the meter bridge is as shown in the below figure.



Here let R is the known resistance and x is unknown resistance. As we know meter bridge works based on Wheatstone Bridge principle. So,

$\frac{R}{l}=\frac{x}{100-l}$

Here, two values for R are given and for those the difference between the balancing length is 20 cm. So,


$$\begin{gathered} \frac{10}{l}=\frac{x}{100-l} \& \frac{22.5}{l+20}=\frac{x}{80-l} \\ \Rightarrow 1000-10l=xl\Rightarrow l=\frac{1000}{10+x} \\ \therefore \left(22.5\right)\left(80-l\right)=\left(x\right)\left(l+20\right)\Rightarrow \left(22.5\right)\left(4-\frac{50}{10+x}\right)=\left(x\right)\left(\frac{50}{10+x}+1\right) \\ \Rightarrow \left(22.5\right)\left(\frac{40+4x-50}{10+x}\right)=\left(x\right)\left(\frac{50+10+x}{10+x}\right) \\ \Rightarrow \left(22.5\right)\left(4x-10\right)=\left(x\right)\left(60+x\right) \\ \Rightarrow 90x-225=60x+x^2 \\ \Rightarrow x^2-30x+225=0 \\ \Rightarrow x=15 \Omega \end{gathered}$$

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