Question 45: Current Electricity

Question

The resistance of a $$10 \mathrm{~m}$$ long wire is $$10 \Omega$$. Its length is increased by $$25 \%$$ by stretching the wire uniformly. The new resistance is

Accepted Answer

To calculate the new resistance after the wire is stretched, we need to use the fact that resistance is proportional to the length and inversely proportional to the cross-sectional area of the wire. The formula for resistance is given by:
$$ R = ho \frac{L}{A} $$
where $$ ho $$ is the resistivity of the material, $$ L $$ is the length, and $$ A $$ is the cross-sectional area.
Initially, the length of the wire is $$10 	ext{ m}$$ and the resistance is $$10 \Omega$$. When the length is increased by $$25 \%$$, the new length becomes:
$$ L_{	ext{new}} = 10 	ext{ m} + (0.25 	imes 10 	ext{ m}) = 12.5 	ext{ m} $$
LetÃ¢ÂÂs denote the initial resistance as $$ R_{	ext{initial}} = 10 \Omega $$ and the initial length as $$ L_{	ext{initial}} = 10 	ext{ m} $$.
The volume of the wire remains constant during stretching, so:
$$ L_{	ext{initial}} 	imes A_{	ext{initial}} = L_{	ext{new}} 	imes A_{	ext{new}} $$
This implies:
$$ 10 	ext{ m} 	imes A_{	ext{initial}} = 12.5 	ext{ m} 	imes A_{	ext{new}} $$
So,
$$ A_{	ext{new}} = \frac{A_{	ext{initial}} 	imes 10}{12.5} = 0.8 	imes A_{	ext{initial}} $$
The new resistance is calculated as:
$$ R_{	ext{new}} = ho \frac{L_{	ext{new}}}{A_{	ext{new}}} $$
Using the relationship between initial and final conditions, we get:
$$ R_{	ext{new}} = ho \frac{L_{	ext{new}}}{A_{	ext{new}}} = ho \frac{L_{	ext{new}}}{0.8 	imes A_{	ext{initial}}} = 1.25 	imes ho \frac{L_{	ext{new}}}{A_{	ext{initial}}} $$
Since $$ R_{	ext{initial}} = ho \frac{L_{	ext{initial}}}{A_{	ext{initial}}} $$, we can write:
$$ R_{	ext{new}} = 1.25 	imes R_{	ext{initial}} 	imes \frac{L_{	ext{new}}}{L_{	ext{initial}}} $$
Substitute the known values:
$$ R_{	ext{new}} = 1.25 	imes 10 \Omega 	imes \frac{12.5 	ext{ m}}{10 	ext{ m}} $$
Simplify:
$$ R_{	ext{new}} = 1.25 	imes 10 \Omega 	imes 1.25 $$
$$ R_{	ext{new}} = 1.5625 	imes 10 \Omega $$
$$ R_{	ext{new}} = 15.625 \Omega $$
Rounding it to one decimal place, $$ R_{	ext{new}} \approx 15.6 \Omega $$.
Therefore, the new resistance is
Option B: $$ 15.6 \Omega $$.

Answer

$$18.6 \Omega$$

Answer

$$15.6 \Omega$$

Answer

$$12.8 \Omega$$

Answer

$$14.9 \Omega$$

Current Electricity

The resistance of a $10 \mathrm{~m} long wire is 10 \Omega. Its length is increased by 25 \%$ by stretching the wire uniformly. The new resistance is

Choose the correct answer: