Question 59: Current Electricity

Question

A resistor of wire $$24 \mathrm{~cm}$$ length and resistance $$8 \Omega$$ is stretched in to a uniform wire of $$48 \mathrm{~cm}$$ length, then the new resistance will be :

Accepted Answer

The resistance $$R$$ of a wire is given by the formula:
$$R = ho \frac{L}{A}$$
where:
$$ho$$ is the resistivity of the material,
$$L$$ is the length of the wire, and
$$A$$ is the cross-sectional area of the wire.
When the wire is stretched to double its original length (from $$24 \, 	ext{cm}$$ to $$48 \, 	ext{cm}$$), its volume ($$V = A 	imes L$$) remains constant because the material is conserved. Thus, if the length $$L$$ is doubled, the cross-sectional area $$A$$ must be halved to keep the volume constant.
The new resistance, $$R'$$, can then be calculated using the new length $$L' = 48 \, 	ext{cm}$$ and the new area $$A' = \frac{A}{2}$$, keeping in mind that the resistivity $$ho$$ does not change because it depends only on the material:
$$R' = ho \frac{L'}{A'} = ho \frac{2L}{\frac{A}{2}} = 4 ho \frac{L}{A} = 4R$$
Given that the original resistance is $$8 \Omega$$:
$$R' = 4 	imes 8 \Omega = 32 \Omega$$
Therefore, the correct answer is
Option A
: $$32 \Omega$$.

Answer

$$32 \Omega$$

Answer

$$8 \Omega$$

Answer

$$16 \Omega$$

Answer

$$4 \Omega$$

Current Electricity

A resistor of wire $24 \mathrm{~cm} length and resistance 8 \Omega is stretched in to a uniform wire of 48 \mathrm{~cm}$ length, then the new resistance will be :

Choose the correct answer: