a) mass m should be suspended close to wire A to have equal stresses in both the wires

b) mass m should be suspended close to B to have equal stresses in both the wires

c) mass m should be suspended at the middle of the wires to have equal stresses in both the wires

d) mass m should be suspended close to wire A to have equal strain in both wires

Answer:

The correct answers are

b) mass m should be suspended close to B to have equal stresses in both the wires

d) mass m should be suspended close to wire A to have equal strain in both wires

EXPLANATION :-
Let a small mass m be suspended to the rod .
So , Stress in the wire = Force / Area  =  F / a
If the two wires have equal stresses, then ,

it would be , F1 / a1  =  F2 / a2
Where , we know ,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminium wire
F1 / F2 = a1 / a2  =  1 / 2   ———–(i)
Now, Taking torque about the point of suspension y from wire ASo , we have ,
F1y = F2 (l – y)

or, F1 / F2 = (l – y) / y    ———-(ii)

We got two equations

By using the equations (i) and (ii) , we can write ,

(l – y) / y  = 1 / 2

or, 2(l – y)  =  y

or, y= 2l/3 ;

To have equivalent stresses in both wires, mass m should be suspended close to B. (option B correct)

Now, If the strains in the two wires are equal then we have the relation

(F1 / a1)/Y1 = (F2 / a2)/Y2

or, F1/F2 = (a1Y1/a2Y2)

or, F1/F2 = 10/7 ; (putting values of a1,a2,Y1 & Y2)   …..(iii)

Let the mass is suspended at a distance x from wire A

F1x = F2(l-x)

or, (l-x)/x = 10/7    ….using equation (iii)

or, 17x = 7l

or, x = 7l/17

To have equal strain in both wires, Mass m should be suspended near to wire A. (Option D is correct)