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Two opposite forcesF(1) = 120 N and F(2)...

Two opposite forces`F_(1) = 120 N` and `F_(2) = 80 N` act on an elastic plank of modulus of elasticity `Y= 2 x 10^(11) N//m^(2)` and length `l=1m` placed over a smooth horizontal surface. The cross-sectional area of the plank is `S = 0.5 m^(2)`. The change in length of the plank is `x xx 10^(-11)m`. Find the value of `x`.

A

1

B

1.5

C

1.4

D

1.1

Text Solution

Verified by Experts

The correct Answer is:
A
Consider an element of thickness dx, Change in the length of the element is `dl=(T)/(S)(dx)/(Y)` and `T = F_(1) - (F_(1) - F_(2)) (x)/(l)`
`underset(0) overset(Deltal)int dl underset(0)overset(l)int(F_(1)-((F_(1) -F_(2))dx)/(l)dx)/(SY)`

`Deltal = ((F_(1)+F_(2)))/(2SY) = (200xx 1)/(2xx0.5xx2xx10^(11)) = 1xx10^(-9)m`
`therefore x =1`
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