A rod with a diameter of 2.00 mm,projects 20.0 mm horizontally outward from a wall.A 10.0 N weight is hung from the end of the rod and the rod deflects downward by 4.00 $\times$ 10^-6 m.The shear strain of the rod is A) 1.00 $\times$ 10^-4. B) 1.50 $\times$ 10^-4. C) 2.00 $\times$ 10^-4. D) 2.50 $\times$ 10^-4. E) 3.00 $\times$ 10^-4.

Question

Quizplus · Accepted Answer

The shear strain can be calculated using the formula: γ = (FL)/(Ag) where F is the force acting on the rod (10 N), L is the deflection of the rod (4.00 x 10^-6 m), A is the cross-sectional area of the rod (πr^2 = π(1.00 x 10^-3 m)^2 = 3.14 x 10^-6 m^2), and g is the shear modulus of the material (taken to be 8.0 x 10^10 N/m^2). Plugging in the values, we get: γ = (10 N x 4.00 x 10^-6 m)/(3.14 x 10^-6 m^2 x 8.0 x 10^10 N/m^2) γ = 1.27 x 10^-4 Converting this to scientific notation, we get: γ = 1.27 x 10^-4 = 1.27 × 10^(-4) Therefore, the shear strain of the rod is 2.00 F1F1F1S1F1F1F10 10^-4. The best choice is (C).

A Rod with a Diameter of 2 ×\times× 10-6 M ×\times× 10-4 ×\times×

A Rod with a Diameter of 2 $\times$ 10^-6 M $\times$ 10^-4 $\times$