Arod of length I and negligible mass is suspended at its two ends by two wires of steel (wirc A) and aluminium (wire B) of equal lengths (figure). The cross-sectional areas of wires A and B are 1.0 `mm^(2)` and 2.0 `mm^(2)` respectively.
`(Y_(A) =70 xx 10^(9) Nm^(-2) ` and `Y_("steel " ) = 200xx 10^(9) Nm^(-2))`

Options

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. Massm should be suspended close to wire A to have equal strain in both wires

Answer

B, D

Explanation

Let the mass m be suspended at di stance x from left end of the rod for equal stress in the wires Let `F_(1)` and `F_(2)` be the tension in the wires . Then
`F_(1) x = F_(2) (l-x) "or" (F_(1))/(F_(2)) = (l-x)/(x) `
`S_(1) = (F_(1))/(A_(1)) = (F_(1))/(10^(-6)) ` and `S_(2) = (F_(2))/(A_(2)) = (F_(2))/(2xx 10^(-6))`
For equal stress `S_(1) =S_(2) "or " (F_(1))/(10^(-)) = (F_(2))/(2xx10^(-6)) "or" (F_(1))/(F_(2)) =(1)/(2)`
From Eqs (i) and (ii) we get `(l-x)/(x) =(1)/(2)`
On solving `x = (2l)/(3)`
It means mass m is suspended close to wire B .
As Strain = `(Stress)/(Y) `
For equal strain `((F_(1))/(10^(6)))/(Y_(s))=((F_(2))/((200xx10^(-6))))/T_(Al)`
`(F_(1))/(F_(2))=(Y_(s))/(2Y_(Al))=((200xx10^(9)))/(2xx70xx10^(9))=(10)/(7)`
From Eqs . (i) and (iii) `(l-x)/(x) = (10)/(7) , ` On solving ` x= (7)/(17)` l
It means mass m is suspended close to wire A.