Question

5.
A block of mass M is pulled with the help of a wire
of mass M and length L by applying F force on other
end. If area of cross section of wire is A and
Young's modulus is Y then elongation in wire will be​

Answer 1

Answer:

The elongation in the wire is $\eta = \frac{3FL}{4AY}$

Explanation:

As we know that the tension in the wire is given as

$F = (M + M) a$

$a = \frac{F}{2M}$

now we know that tension in the wire at a distance {x} from the block is given as

$T = (\frac{M}{L}x + M)a$

$T = (\frac{M}{L}x + M)\frac{F}{2M}$

now extension is given as

$d\eta = \int \frac{T}{AY} dx$

$d\eta = \frac{F}{2AY} \int(\frac{xdx}{L} + dx$

$\eta = \frac{F}{2AY} (\frac{L}{2} + L)$

$\eta = \frac{3FL}{4AY}$

#Learn

Topic : Young's Modulus and elasticity

https://brainly.in/question/5867736