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A thin, light wire 75.0 cm long having a circular cross section 0.557 mm in diameter has a 25.0 kg weight attached to it, causing it to stretch by 1.13 mm.

Option 1: True
Option 2: False

1 Answer

3 votes

Final answer:

The Young's modulus of the wire is 1.95 x 10^11 Pa.The answer to the question is true.

Step-by-step explanation:

Young's modulus can be calculated using the formula:

Y = (F * L) / (A * ∆L)

Where Y is the Young's modulus, F is the force applied, L is the original length of the wire, A is the cross-sectional area of the wire, and ∆L is the change in length of the wire.

In this case, the force applied is the weight attached to the wire, which is 25.0 kg * 9.8 m/s^2 = 245 N. The original length of the wire is 75.0 cm = 0.75 m. The cross-sectional area of the wire can be calculated using the formula:

A = (π/4) * (d^2)

Where A is the cross-sectional area, π is a constant approximately equal to 3.14159, and d is the diameter of the wire. Substituting the values, we get:

A = (3.14159/4) * (0.000557^2) = 9.623 x 10^-8 m^2.

The change in length of the wire is given as 1.13 mm = 0.00113 m.

Plugging in the values into the formula for Young's modulus, we get:

Y = (245 * 0.75) / (9.623 x 10^-8 * 0.00113) = 1.95 x 10^11 Pa.

Therefore, the correct option is True.

User Pierre Mage
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