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If the human bone fractures with stress at [tex]120 \, \text{N/mm}^2[/tex], then the maximum tension on the bone with an area of [tex]5 \, \text{cm}^2[/tex] is:

A. 60 N
B. 60000 N
C. 24000 N
D. 2400 N
E. 600 N

Calculate the change in length of the upper leg bone when a [tex]75.0 \, \text{kg}[/tex] man supports his weight on one leg. Assume the bone is equivalent to a uniform rod that is [tex]40.0 \, \text{cm}[/tex] long and [tex]2.50 \, \text{cm}[/tex] in radius. Young's modulus for bones is [tex]9 \times 10^9 \, \text{N/m}^2[/tex]. Use [tex]\pi = 3.14[/tex].

A. 0.1665 mm
B. 1.665 mm
C. 0.01665 m
D. 0.1665 m
E. 0.01665 mm

Answer :

Given that:

Stress = 120 N/m²Area of bone = 5 cm² = 0.0005 m²

Maximum tension on the bone can be found out using the formula: Stress = Tension / Areaof boneTension = Stress × Area of bone= 120 N/m² × 0.0005 m²= 0.06 N = 60N. Therefore, the maximum tension on the bone with an area 5 cm² is 60N.

The change in length of the upper leg bone when a 75.0 kg man supported his weight on one leg can be found out using the formula:ΔL/L = F/((π × r²) × Y)where,ΔL = Change in length of the upper leg bone L = Length of the upper leg bone F = Force applied Y = Young's modulus = 9 × 10¹⁰ N/m²π = 3.14r = Radius of the upper leg bone = 2.50 cm = 0.025 mF = mg, where, m = Mass of the man = 75 kg g = Acceleration due to gravity = 9.8 m/s²F = 75 kg × 9.8 m/s²= 735 N. Substitute the given values in the above formula to find ΔL/L.ΔL/L = F/((π × r²) × Y)= 735 N/((π × (0.025 m)²) × (9 × 10¹⁰ N/m²))= 0.001665 m= 1.665 mm. Therefore, the change in length of the upper leg bone when a 75.0 kg man supported his weight on one leg is 0.001665 m or 1.665 mm.

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