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Question

The areas of cross section of two wires of the same material are in the ratio $$1:2$$ and their lengths are in the ratio $$2 : 1$$. If their ends have the same potential differences then currents through them will be in the ratio of


A
1:2
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B
2:1
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C
1:4
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D
4:1
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Solution

The correct option is D $$1 : 4$$
The resistance of a wire of length $$l$$ and cross section area $$A$$ is given by,
$$R=\rho(l/A)$$.
So
$$\dfrac{R_1}{R_2}=\dfrac{l_1}{l_2} \dfrac{A_2}{A_1}=\dfrac{4}{1}$$.
Now current through a wire is $$i \propto 1/R$$.
$$\therefore \dfrac{i_1}{i_2}=\dfrac{R_2}{R_1}=\dfrac{1}{4}$$
If their ends have the same potential differences then currents through them
will be in the ratio of $$1:4$$.

Physics

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