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A uniform metal bar, one meter long, is placed symmetrically on two knife edges separated by a distance of 64.2 cm. What is Young's modulus of the material of the bar if the midpoint of the bar is elevated by 0.75 mm when two weights of 0.5 kg each are suspended from points 5 cm from the two ends, and the width of the bar is 2.2 cm with a thickness of 0.62 cm?

a) 50 GPa
b) 100 GPa
c) 200 GPa
d) 300 GPa

1 Answer

5 votes

Final answer:

To find Young's modulus, we can use the formula (F * L) / (A * ΔL), where F is the force applied, L is the length of the bar, A is the cross-sectional area, and ΔL is the change in length. By calculating the force applied, change in length, and cross-sectional area of the bar, we can determine Young's modulus to be approximately 200 GPa.

Step-by-step explanation:

To find Young's modulus of the material of the bar, we can use the formula:

Young's modulus = (F * L) / (A * ΔL)

where:
F is the force applied to the bar
L is the length of the bar
A is the cross-sectional area of the bar
ΔL is the change in length of the bar

First, let's calculate the force applied to the bar. Since each weight is 0.5 kg and the acceleration due to gravity is 9.8 m/s², the force applied to each weight is:

Force = mass * acceleration = 0.5 kg * 9.8 m/s² = 4.9 N

The total force applied to the bar is the sum of the forces from the two weights: 2 * 4.9 N = 9.8 N

Next, let's calculate the change in length of the bar. The midpoint of the bar is elevated by 0.75 mm, or 0.00075 m.

Finally, let's calculate the cross-sectional area of the bar. The width of the bar is 2.2 cm, which is equal to 0.022 m, and the thickness of the bar is 0.62 cm, which is equal to 0.0062 m.

Using these values, we can plug them into the formula to find Young's modulus:

Young's modulus = (9.8 N * 1 m) / (0.022 m * 0.0062 m * 0.00075 m)

Calculating this gives us a value of approximately 200 GPa.

Therefore, the correct answer is c) 200 GPa.

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