Now the rod is held by two strings and weight is suspended it from it.

Now the tension in the strings will be in vertically upward direction and the weight of block is acting in vertically downward direction now since rod is held stationary it must be in equilibrium i.e. all the forces on it must cancel each other and also net resultant torque on it due to all the forces must also be zero

Now tension in the strings is the force acting due to them on rod and same amount of force is also acting on them in opposite direction let the Force on Rod due to tension in string A and B be T_A and T_Brespectively now let the block be at a distance d from string end tied to String A, let the weight of block of mass m be w

Now we know weight is given by

w = m × g

Where m is mass of block and g is acceleration due to gravity

All the forces on the rod is represented by free body diagram as shown in the figure below

So Balancing all the forces for equilibrium condition i.e. net magnitude of forces in vertically upward direction must be equal to magnitude of force in downward direction so we get

T_A + T_B = w

Now rod is in rotational equilibrium also since it is not rotating so sum of torque due to all the forces acting on rod must be zero around any point

We know magnitude torque on body due to a force with respect to any point is given by

Where r_⊥ is the perpendicular distance of point from line of force

And F is the magnitude of force

Now let us consider o point where mass is attached for rotation equilibrium, here forces are perpendicular to rod so the perpendicular distance from any point to line of force is same as the distance from point o now force due to point

We can see torque due to force T_A will be in clockwise direction denoted by and due to T_Bin anticlockwise direction denoted by and there will be no torque to weight of mass m since it is acting on the same point about we are calculating torques, so r_⊥ or the perpendicular distance is zero so torque is also zero since

Now the direction of torques are shown in figure below

Now for force T_A

r_⊥ = d

(Distance from o to left most end where T_A is acting)

F = T_A

So torque is

Now for force T_B

r_⊥ = 1.05 – d

(Distance from o to Right most end where T_B is acting since total length of rod is 1.05m)

F = T_B

So torque is

Now both torque must cancel each other i.e. should be equal in magnitude

i.e.

or

so we get the relation

(a) Now we know stress in a wire is restoring force per unit area , restoring force is same as the Tension in the wire or the force applied by the wire

We know stress in in a wire is given as