Question 550: Current Electricity

Question

A wire of length $$3 \mathrm{~m}$$ connected in the left gap of a meter-bridge balances $$8 \Omega$$ resistance in the right gap at a point, which divides the bridge wire in the ratio $$3: 2$$. The length of the wire corresponding to resistance of $$1 \Omega$$ is

Accepted Answer

Let $$R_1$$ be the resistance of $$3 \mathrm{~m}$$ long wire connected in the left gap.
For meter-bridge, $$\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{l_1}{l_2}$$
$$\begin{aligned}
& 	herefore \quad \frac{\mathrm{R}_1}{8}=\frac{3}{2} \
& 	herefore \quad \mathrm{R}_1=\frac{3}{2} 	imes 8=12 \Omega
\end{aligned}$$
Length of the wire corresponding to the resistance of $$1 \Omega$$ is $$l=\frac{3}{12}=0.25 \mathrm{~m}$$

Answer

$$1 \mathrm{~m}$$

Answer

$$0.75 \mathrm{~m}$$

Answer

$$0.5 \mathrm{~m}$$

Answer

$$0.25 \mathrm{~m}$$

Current Electricity

A wire of length $3 \mathrm{~m} connected in the left gap of a meter-bridge balances 8 \Omega resistance in the right gap at a point, which divides the bridge wire in the ratio 3: 2. The length of the wire corresponding to resistance of 1 \Omega$ is

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