In Fig., AB is a 2 m long rod of negligible weight. It is supported at the ends by wires X and Yofequal length. Young's modulus for wire X is `2 xx 10^(11) N//m^(2)` and its area of cross-section 3 `mm^(2)` . For wire Y, Young's modulus is `1.4 xx 10^(11)" N//m^(2)` and its area of cross-section is 6 `mm^(2)`. A mass M is suspended at any point along the rod.

Distance of a point on the rod from A, such that the mass M suspended from it will result in equal strains in wires X and Y, is

Options

A. `(8)/(5)` m
B. `(7)/(6)` m
C. `(4)/(3)`m
D. `(2)/(3)` m

Answer

B

Explanation

`Y . (F.L)/(Al) :. "Strain" (l)/(L) = (F)/(AY) `
`(F_(1))/(A_(1)Y_(1))= (F_(2))/(A_(2)Y_(2))`
`(F_(1))/(F_(2)) = (A_(1)Y_(1))/(A_(2)Y_(2))= (3xx2 xx 10^(11))/(6xx 1.4 xx 10^(11)) = (1)/(1.4)`
`(F_(1))/(F_(2))= (2-x)/(x) =(1)/(1.4)`
2.8 -1.4 x =x
2.8 = 2.4 x `:. x = (2.8)/(2.4 ) = (7)/(6)` m