Electrical measuring instruments

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In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is $l = 25 c m$ . If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance $l'$ (in cm) will now be_______

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Solution

$\begin{array}{l} \frac{R}{S} = \frac{L}{100 - L} = \frac{25}{75} \\ \Rightarrow R = \frac{1}{3} S \\ A l s o, R = \frac{ρ L}{A} = \frac{ρ L}{π r^{2}} \\ N e w v a l u e = R^{1} = \frac{ρ . \frac{L}{2}}{π {(\frac{r}{2})}^{2}} = 2 R \\ N o w, \frac{2 R}{S} = \frac{L^{1}}{100 - L^{1}} \\ \Rightarrow \frac{2. S / 3}{S} = \frac{L^{1}}{100 - L^{1}} \\ \Rightarrow \frac{2}{3} = \frac{L^{1}}{100 - L^{1}} \\ \Rightarrow \frac{200 - 2 L^{1}}{1} = \frac{3 L^{1}}{1} \\ \Rightarrow L^{1} = 40 c m \end{array}$

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Electrical measuring instruments

Remember concepts with our Masterclasses.

In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is l=25cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance l'(in cm) will now be_______

RS=L100−L=2575 ​⇒R=13S ​Also, R=ρLA=ρLπr2New value=R1=ρ.L2πr22=2R ​Now , 2RS=L1100−L1 ⇒2.S/3S=L1100−L1​⇒23=L1100−L1 ⇒200−2L11=3L11 ​⇒L1=40cm

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In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is $l = 25 c m$ . If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance $l'$ (in cm) will now be_______