The resistance of a wire is 5 Ω. It's new resistance in ohm if stretched to 5 times of it's original length will be:

Question

The resistance of a wire is 5 Ω. It's new resistance in ohm if stretched to 5 times of it's original length will be: (A) 125 (B) 5 (C) 25 (D) 625

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fbd22b71847618b5862ec1ee03528432-.png" /> ∵ Volume of wire is constant in stretching V i = V f A i ℓ i = A f ℓ f A ℓ= A prime(5 ℓ) A prime=( A /5) R f =(ρ ℓ f / A f )=(ρ(5 ℓ)/(( A )5)) =25((ρ ℓ/ A )) =25 × 5=125 Ω

Q. The resistance of a wire is $5Ω$ . It's new resistance in ohm if stretched to 5 times of it's original length will be:

$125$

5

25

625

$∵$ Volume of wire is constant in stretching
$V_{i} = V_{f}$
$A_{i} ℓ_{i} = A_{f} ℓ_{f}$
$A ℓ = A^{'} (5 ℓ)$
$A^{'} = \frac{A}{5}$
$R_{f} = \frac{ρ ℓ _{f}}{A _{f}} = \frac{ρ ( 5 ℓ )}{( \frac{A}{5} )}$
$= 25 (\frac{ρ ℓ}{A})$
$= 25 \times 5 = 125Ω$

Q. The resistance of a wire is 5Ω. It's new resistance in ohm if stretched to 5 times of it's original length will be:

125

5

25

625

∵ Volume of wire is constant in stretching Vi​=Vf​ Ai​ℓi​=Af​ℓf​ Aℓ=A′(5ℓ) A′=5A​ Rf​=Af​ρℓf​​=(5A​)ρ(5ℓ)​ =25(Aρℓ​) =25×5=125Ω

Q. The resistance of a wire is $5Ω$ . It's new resistance in ohm if stretched to 5 times of it's original length will be:

$125$

$∵$ Volume of wire is constant in stretching
$V_{i} = V_{f}$
$A_{i} ℓ_{i} = A_{f} ℓ_{f}$
$A ℓ = A^{'} (5 ℓ)$
$A^{'} = \frac{A}{5}$
$R_{f} = \frac{ρ ℓ _{f}}{A _{f}} = \frac{ρ ( 5 ℓ )}{( \frac{A}{5} )}$
$= 25 (\frac{ρ ℓ}{A})$
$= 25 \times 5 = 125Ω$