On applying the load, the increment in length of a wire is 1 mm. On applying the same load on another wire of same length and material, but having half the radius, the increment will be:

Question

On applying the load, the increment in length of a wire is 1 mm. On applying the same load on another wire of same length and material, but having half the radius, the increment will be: (A) 0.25 mm (B) 4.0 mm (C) 0.5 mm (D) 2.0 mm

Tardigrade Admin · Accepted Answer

Δ l =(F L/A &gamma;)=(F L/π r2 &gamma;) Δ l =(1/r2) Δ l2=(1 m m/((1)2)2)=4 m m

Q. On applying the load, the increment in length of a wire is $1 mm$ . On applying the same load on another wire of same length and material, but having half the radius, the increment will be:

0.25 mm

4.0 mm

0.5 mm

2.0 mm

$Δ l = \frac{F L}{A γ} = \frac{F L}{π r ^{2} γ}$
$Δ l = \frac{1}{r ^{2}}$
$Δ l_{2} = \frac{1 mm}{( \frac{1}{2} ) ^{2}} = 4 mm$

Q. On applying the load, the increment in length of a wire is 1mm. On applying the same load on another wire of same length and material, but having half the radius, the increment will be: