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)