A uniform wire of mass m, length L, area of cross-section A and Young's modulus Y hangs from the ceiling. Its elongation under its own weight will be

Question

A uniform wire of mass m, length L, area of cross-section A and Young's modulus Y hangs from the ceiling. Its elongation under its own weight will be (A) zero (B) (m g L/2 A Y) (C) (mgL/AY) (D) (2mgL/AY)

Tardigrade Admin · Accepted Answer

In case of elongation by its own weight mg will act at centre of gravity of the wire so that length of wire which is stretched will be ((L/2)). ∴ Δ L=((m g)((L/2))/A Y) =(m g L/2 A Y)

Q. A uniform wire of mass $m$ , length $L$ , area of cross-section $A$ and Young's modulus $Y$ hangs from the ceiling. Its elongation under its own weight will be

zero

$\frac{m gL}{2 A Y}$

$\frac{m gL}{A Y}$

$\frac{2 m gL}{A Y}$

In case of elongation by its own weight $m g$ will act at centre of gravity of the wire so that length of wire which is stretched will be $(\frac{L}{2})$ .
$∴ Δ L = \frac{( m g ) ( \frac{L}{2} )}{A Y}$
$= \frac{m gL}{2 A Y}$

Q. A uniform wire of mass m, length L, area of cross-section A and Young's modulus Y hangs from the ceiling. Its elongation under its own weight will be

zero

2AYmgL​

AYmgL​

AY2mgL​

In case of elongation by its own weight mg will act at centre of gravity of the wire so that length of wire which is stretched will be (2L​). ∴ΔL=AY(mg)(2L​)​ =2AYmgL​

Q. A uniform wire of mass $m$ , length $L$ , area of cross-section $A$ and Young's modulus $Y$ hangs from the ceiling. Its elongation under its own weight will be

$\frac{m gL}{2 A Y}$

$\frac{m gL}{A Y}$

$\frac{2 m gL}{A Y}$

In case of elongation by its own weight $m g$ will act at centre of gravity of the wire so that length of wire which is stretched will be $(\frac{L}{2})$ .
$∴ Δ L = \frac{( m g ) ( \frac{L}{2} )}{A Y}$
$= \frac{m gL}{2 A Y}$