A 1m long steel wire of cross sectional area 2.5×10^-6 is fixed at one end. A load of 15kg is suspended from the free end so that the wire hangs vertically the young modulus of steel 2.0×10^11nm² assuming that the elastic limit has not been exceeded

Answer 1

We can use the formula for stress to calculate the force exerted on the wire:

Stress = Force / Area

The force exerted on the wire is equal to the weight of the load:

Force = mass x gravity

Force = 15kg x 9.8 m/s² = 147 N

Now, we can calculate the stress on the wire:

Stress = 147 N / (2.5×10^-6 m²) = 5.88 x 10^7 N/m²

Next, we can use the Young's modulus formula to find the strain in the wire:

Young's Modulus = Stress / Strain

Strain = Stress / Young's Modulus

Strain = (5.88 x 10^7 N/m²) / (2.0 x 10^11 N/m²) = 2.94 x 10^-4

Finally, we can calculate the extension in the wire using the formula for strain:

Strain = Change in Length / Original Length

Change in Length = Strain x Original Length

Change in Length = (2.94 x 10^-4) x 1m = 2.94 x 10^-4 m

Therefore, the steel wire will extend by 2.94 x 10^-4 meters when a 15kg load is suspended from the free end.