Determine the equation of the elastic curve for the cantilever beam of uniform cross section as shown in the figure due to the applied bending moment M1. (Consider Young’s modulus as E and Moment of inertia as I)

Correct option is C

​$$\begin{aligned}&\text{For a cantilever beam with a moment } M_1 \text{ applied at the free end:} \\&\bullet \text{The bending moment } M(x) \text{ is constant along the beam:} \\&M(x) = M_1 \quad \text{(for all } 0 \leq x \leq L). \\[1.5em]&\textbf{Differential Equation of the Elastic Curve} \\[0.5em]&\text{The elastic curve is governed by:} \\&EI \frac{d^2 y}{dx^2} = M(x). \\[0.5em]&\text{Substitute } M(x) = M_1: \\&EI \frac{d^2 y}{dx^2} = M_1. \\[1.5em]&\textbf{Integrate to Find Slope and Deflection} \\[0.5em]&\text{First Integration (Slope } \theta(x)\text{):} \\&EI \frac{dy}{dx} = M_1 x + C_1. \\[0.5em]&\text{Boundary Condition: At the fixed end } (x = 0), \text{ slope } \frac{dy}{dx} = 0: \\&0 = M_1 \cdot 0 + C_1 \Rightarrow C_1 = 0. \\[0.5em]&\text{Thus:} \\&\frac{dy}{dx} = \frac{M_1}{EI} x. \\[1.5em]&\text{Second Integration (Deflection } y(x)\text{):} \\&EI y(x) = \frac{M_1}{2} x^2 + C_2. \\[0.5em]&\text{Boundary Condition: At the fixed end } (x = 0), \text{ deflection } y = 0: \\&0 = \frac{M_1}{2} \cdot 0^2 + C_2 \Rightarrow C_2 = 0. \\[0.5em]&\text{Thus:} \\&y(x) = \frac{M_1}{2EI} x^2.\end{aligned}$$​​