When the arrangement consisting of a horizontal thick copper wire of length 2L and radius 2R , one end of which is welded to an end of another horizontal thin copper wire of length L and radius R is stretched by applying forces at two ends. Ratio of the elongation in the thick wire to that in the thin wire at equilibrium is

Question

When the arrangement consisting of a horizontal thick copper wire of length 2L and radius 2R , one end of which is welded to an end of another horizontal thin copper wire of length L and radius R is stretched by applying forces at two ends. Ratio of the elongation in the thick wire to that in the thin wire at equilibrium is (A) 0.50 (B) 0.25 (C) 4.00 (D) 2.00

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-6i6tmzqazxwflquy.jpg" /> Δ ℓ =(F L/A Y) (Δ ℓ1/Δ ℓ2)=(L1/A1) (A2/L2)=(2 L/π(2 R)2) × (π R2/L)=(1/2)

$0.50$

$0.25$

$4.00$

$2.00$

$Δ ℓ = \frac{F L}{A Y}$
$\frac{Δ ℓ _{1}}{Δ ℓ _{2}} = \frac{L _{1}}{A _{1}} \frac{A _{2}}{L _{2}} = \frac{2 L}{π ( 2 R ) ^{2}} \times \frac{π R ^{2}}{L} = \frac{1}{2}$

0.50

0.25

4.00

2.00

Δℓ=AYFL​ Δℓ2​Δℓ1​​=A1​L1​​L2​A2​​=π(2R)22L​×LπR2​=21​

$0.50$

$0.25$

$4.00$

$2.00$

$Δ ℓ = \frac{F L}{A Y}$
$\frac{Δ ℓ _{1}}{Δ ℓ _{2}} = \frac{L _{1}}{A _{1}} \frac{A _{2}}{L _{2}} = \frac{2 L}{π ( 2 R ) ^{2}} \times \frac{π R ^{2}}{L} = \frac{1}{2}$