A rod of length 1.05 m having negliaible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in fig. The cross-sectional area of wire A and B are `1 mm^(2)` and 2` mm^(2)`, respectively . At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires. Given,
`Y_(steel) = 2 xx 10^(11) Nm^(-2) and Y-(alumi n i um) = 7.0 xx 10^(10)N^(-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. Mass m should be suspended close to wire A to have equal strain on both wires.

Answer

B, D

Explanation

Refer Fig. 7(EP).7,
let the mass m be suspended at distance 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)` …(i)
`s_(1) =(F_(1))/(A_(1))= (F_(1))/(10^(-6))` and `S_(2)=(F_(2))/(A_(2))=(F_(2))/(2xx10^(-6))`
For equal stress, `S_(1)=S_(2)` or `(F_(1))/(10^(-4))=(F_(2))/(2xx10^(6))` or `(F_(1))/(F_(2))=(1)/(2)` ....(ii)
From (i) and (ii), we get `(l-x)/(x)=(1)/(2)`
On solving, `x=2 l//3`
It means mass m is suspended close to wire B.
As, strain `=(strees)/(Y)`.
For equal strain, `(F_(1)//10^(-6))/(Y_(s))=(F_(2)//(2xx10^(-6)))/(Y_(Al))`
`(F_(1))/(F_(2))=(Y_(s))/(2Y_(Al))=((200xx10^(9)))/(2xx70xx10^(9))=(10)/(7)`
From (i) and (iii), `(l-x)/(x)=(10)/(7)`
On solving, `x=(7)/(17)l`
It means m is suspended close to wire A.