The extension produced in a wire by the application of load is 5 mm. The extension in a wire of the same material and length but half the radius, by the same load is:

Question

The extension produced in a wire by the application of load is 5 mm. The extension in a wire of the same material and length but half the radius, by the same load is: (A) 20 mm (B) 30 mm (C) 40 mm (D) 50 mm

Tardigrade Admin · Accepted Answer

Y=(F × L/A × Δ L) ⇒ Δ L propto (1/ r 2) ∴ (Δ L1/Δ L2)=((r2/r1))2 ⇒ Δ L2=Δ L1 ×((r1/r2))2 =5 × 4=20 mm

Q. The extension produced in a wire by the application of load is 5 mm. The extension in a wire of the same material and length but half the radius, by the same load is:

20 mm

30 mm

40 mm

50 mm

$Y = \frac{F \times L}{A \times Δ L}$
$\Rightarrow Δ L \propto \frac{1}{r ^{2}}$
$∴ \frac{Δ L _{1}}{Δ L _{2}} = (\frac{r _{2}}{r _{1}})^{2}$
$\Rightarrow Δ L_{2} = Δ L_{1} \times (\frac{r _{1}}{r _{2}})^{2}$
$= 5 \times 4 = 20 mm$

Q. The extension produced in a wire by the application of load is 5 mm. The extension in a wire of the same material and length but half the radius, by the same load is:

20 mm

30 mm

40 mm

50 mm

Y=A×ΔLF×L​ ⇒ΔL∝r21​ ∴ΔL2​ΔL1​​=(r1​r2​​)2 ⇒ΔL2​=ΔL1​×(r2​r1​​)2 =5×4=20mm

$Y = \frac{F \times L}{A \times Δ L}$
$\Rightarrow Δ L \propto \frac{1}{r ^{2}}$
$∴ \frac{Δ L _{1}}{Δ L _{2}} = (\frac{r _{2}}{r _{1}})^{2}$
$\Rightarrow Δ L_{2} = Δ L_{1} \times (\frac{r _{1}}{r _{2}})^{2}$
$= 5 \times 4 = 20 mm$