This question was previously asked in

AAI JE (Technical) Official Paper 2020

Option 1 : 0.5 mg

CT 1: Indian History

20623

10 Questions
10 Marks
6 Mins

**Concept:**

At hinged point **'A'**, Two reactive forces are there Vertical Force 'A_{y}' and Horizontal Force 'A_{x}'. At Point 'B', Two Normal forces are acting, Vertical Force 'B_{y}' and Horizontal Force 'B_{x}', As the wall is frictionless, so B_{y} = 0.

Find out the Reactive and Normal Forces at point 'A' and 'B' respectively, then apply moment as 0 at any of the points as Rod is in equilibrium, and find A_{y} - A_{x}

**Calculation:**

**Given:**

Weight (mg) is acting vertically downward at the uniform rod of mass 'm' at its mid-point.

\(\theta\) =\(45^o\), B_{y} = 0.

From the figure,

\(\sum F_x\) = 0,

∴ Ax = Bx ...(1)

\(\sum F_y\) = 0,

A_{y} = mg (B_{y} = 0) ...(2)

Now, Applying moment at Hinged point,

\(\sum M_A\) = 0,

\(B_x \times lsin\theta = mg \times \frac{l}{2}cos\theta\)

\(B_x = \frac{mg}{2} = 0.5(mg)\)

From(1), Ax = Bx = 0.5(mg) - (3)

Now, According to question, the difference between magnitudes of the vertical and horizontal components of reactions force at point A

Ay - A_{x} = mg - 0.5(mg) = **0.5(mg)**

**∴ Option(1) is the correct answer.**