Question

A wire elongates by 1.0 mm when a load W is hung from it. If this wire goes over a a pulley and two weights W each are hung at the two ends, he elongation of he wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm

Answer 1

$\Large\underline{\underline{\sf \red{Given}:}}$

• Wire elongated by (l) = 1.0 when weight W is hung from it.

• Wire goes over a pulley and two weight =

$\Large\underline{\underline{\sf \red{To\:Find}:}}$

• W each are hung at the two ends , elongation in wire = ?

$\Large\underline{\underline{\sf \red{Solution}:}}$

Young's Modulus :-

$\large{\boxed{\sf Y=\dfrac{FL}{Al} }}$

Here ,

F = Force

A = Cross Section area

l = Elongation

L = Lenght of wire

Y = Young's Modulus

In first Condition :

$\implies{\sf Y=\dfrac{W×L}{1×A}}$

$\implies{\sf Y=\dfrac{WL}{A}}$

In second Condition :-

For one side of wire :

$\implies{\sf Y=\dfrac{W×L/2}{1/2×A}}$

$\implies{\sf Y=\dfrac{WL}{1×A}}$

$\implies{\sf Y=\dfrac{WL}{A}}$

No change in Young's Modulus.

$\Large\underline{\underline{\sf \red{Answer}:}}$

Here , there is no change in Young's Modulus

•°• Wire elongates by 1.0mm

Answer 2

A wire elongates by 1.0 mm when a load W is hung from it. If this wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be 1.0 mm.

Explanation:

• The rod's topmost part experiences maximum stress because of the entire rod's weight.
• The rod of uniform mass distribution and stretched by its own weight.
• In first case, the wire tension’s elongation in the wire ∝ and in the second case

        $T 2=2 W \times \frac{w}{w}+W=W$

       $\text { As } \frac{T 1}{T 2}=1 . \text { Therefore, }\frac{l1}{l2} =1$

       $L 2=l 1=1.0 \mathrm{mm}$.