A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m.By what distance will the wire stretch? Young's modulus for steel is 2.0 × 10¹¹ Pa. A)1.5 cm B)2.5 cm C)3.5 cm D)4.5 cm

Question

Quizplus · Accepted Answer

The stretch of a wire under tension can be found using the formula:

ΔL = (F * L) / (A * E)

where ΔL is the change in length, F is the force applied, L is the original length, A is the cross-sectional area, and E is the Young's modulus.

In this case, the force is the weight of the load, which is 50 kg * 9.8 m/s^2 = 490 N. The cross-sectional area can be found using the formula for the area of a circle: A = π * (d/2)^2 = π * (0.001 m / 2)^2 = 7.85 × 10^-7 m^2.

Plugging in the values, we get:

ΔL = (490 N * 11.2 m) / (7.85 × 10^-7 m^2 * 2.0 × 10^11 Pa) = 0.0035 m = 3.5 cm

Therefore, the wire will stretch by 3.5 cm. The answer is choice C.

A 50-Kg Load Is Suspended from a Steel Wire of Diameter