## Four rods that obey Hooke’s law are each put under tension. (a) A rod 50.0 cm 50.0 cm long with cross-sectional area 1.00 mm 2 1.00 mm2 and

Four rods that obey Hooke’s law are each put under tension. (a) A rod 50.0 cm 50.0 cm long with cross-sectional area 1.00 mm 2 1.00 mm2 and with a 200 N 200 N force applied on each end. (b) A rod 25.0 cm 25.0 cm long with cross-sectional area 1.00 mm 2 1.00 mm2 and with a 200 N 200 N force applied on each end. (c) A rod 20.0 cm 20.0 cm long with cross-sectional area 2.00 mm 2 2.00 mm2 and with a 100 N 100 N force applied on each end.

## Answers ( )

Answer:

Four rods that obey Hooke’s law are each put under tension. (a) A rod 50.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end. A rod 25.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end. .00 mm2 and with a 100 N on each end. (c) A rod 20.0 cm long with cross-sectional area 2 force applied Order the rods according to the tensile stress on each rod, from smallest to largest.

Explanation:

Tensile stress is given as =

σ = Force/Area

A. Given that

Length of rod = 50cm

Cross-sectional Area A = 1mm²

A = 1 × 10^-6m²

Force applied F = 200N

Then, σ = F/A

σ = 200/1×10^-6

σ = 200 × 10^6 N/m²

σ = 200 Mpa

B. Given that,

Length of rod = 25cm.

Cross-sectional Area A = 1mm²

A = 1 × 10^-6m²

Force applied F = 200N

Then, σ = F/A

σ = 200/1×10^-6

σ = 200 × 10^6 N/m²

σ = 200 Mpa

C. Given that,

Length of rod = 20cm.

Cross-sectional Area A = 2mm²

A = 2 × 10^-6m²

Force applied F = 100N

Then, σ = F/A

σ = 100/2×10^-6

σ = 50 × 10^6 N/m²

σ = 50 Mpa

It is noticed that

50Mpa < 20Mpa =200Mpa

Then, C < B = A

The tensile strength of rod A and B are equal and their tensile strength is greater than that of Rod C