Question 175: Current Electricity

Question

A wire of resistance $$160 ~\Omega$$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

Accepted Answer

Let the original length of the wire be L and its cross-sectional area be A. Then, its resistance R is given by:
$$R = \frac{\rho L}{A}$$
where $$\rho$$ is the resistivity of the material of the wire.
When the wire is melted and drawn into a wire of one-fourth of its length, its new length is L/4 and its new cross-sectional area is 4A (since the same amount of material is now spread over a longer length). Therefore, its new resistance R' is given by:
$$R' = \frac{\rho (L/4)}{4A} = \frac{R}{16}$$
Substituting the given value of R, we get:
$$R' = \frac{160}{16} = 10 ~\Omega$$
Therefore, the new resistance of the wire is $$10 ~\Omega$$.

Answer

$$640 ~\Omega$$

Answer

$$40 ~\Omega$$

Answer

$$16 ~\Omega$$

Answer

$$10 ~\Omega$$

Current Electricity

A wire of resistance $160 ~\Omega$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

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