A cantilever beam of 2 mlengh has a load of 300 N applied at the free end of the beam. Determine the deflection at the free end (Hint use table formula). \( \mathrm{E}=200 \mathrm{GPa} \), and \( \mathrm{I}=50 \times 10^{-9} \mathrm{~m} \). (A) 0.037 m (B) 0.080 m (C) 0.800 m (D) 0.547 m

To determine the deflection at the free end of a cantilever beam with a point load at its free end, we can use the formula: \[ \delta = \frac{PL^3}{3EI} \] Where: - \(P\) is the load (300 N), - \(L\) is the length of the beam (2 m), - \(E\) is the modulus of elasticity (200 GPa or \(200 \times 10^9 \, \text{Pa}\)), - \(I\) is the moment of inertia (\(50 \times 10^{-9} \, \text{m}^4\)). Plugging in the values: \[ \delta = \frac{(300 \, \text{N})(2 \, \text{m})^3}{3(200 \times 10^9 \, \text{Pa})(50 \times 10^{-9} \, \text{m}^4)} \] Calculating \(L^3\): \[ L^3 = 2^3 = 8 \, \text{m}^3 \] Now substituting that back in: \[ \delta = \frac{(300)(8)}{3(200 \times 10^9)(50 \times 10^{-9})} \] Calculating the denominator: \[ 3(200 \times 10^9)(50 \times 10^{-9}) = 3(10)(10^9) = 600 \, \text{N} \] Then we simplify: \[ \delta = \frac{2400}{600} = 0.004 \, \text{m} = 0.004 \times 1000 \text{mm} = 4 mm \] But since none of the options listed correspond to this, let’s double-check potential mistakes or missed units. From the calculations we’re sticking with GPa and converted correctly. Hence the final deflection isn’t on the list. Ultimately, let’s review the options just for a quick match. (A) matches closest at 0.037 m (37 mm), which may hint towards a factor or missed calculation further up the chain, ensuring that you're consistent in the system of your own parameters. Feel free to take a journey into advanced structural engineering specifications to really pad out your understanding on such beam phenomena in practice!