Question

One end of a uniform wire of length L and weight
Wis suspended from rigid support and a weight
2W is suspended from its lower end. If A is the
area of cross-section, and Y is Young's modulus.
Then elongation in wire is
(1)
WL
AY
(2) 3W
AY
2WL
5 WL
2 AY
AY​

Answer 1

force = tension = W+2W =3W

Y= FL/A ∆L

∆L ( enlogation ) = 3W L/ AY

option 2 hoga

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Answer 2

Given that,

Length of wire = L

Suspended weight from rigid support = W

Suspended weight from lower end = 2W

Area of cross section = A

Young modulus = Y

We need to calculate the force

Using formula of force

$F=W+2W$

$F=3W$

We need to calculate the elongation in wire

Using formula of young's modulus

$Y=\dfrac{F L}{A\Delta L}$

$\Delta L=\dfrac{FL}{AY}$

Put the value into the formula

$\Delta L=\dfrac{3WL}{AY}$

Hence, The elongation in wire is $\dfrac{3WL}{AY}$