A rod of uniform cross-sectional area A and length L has a weight W. It is suspended vertically from a fixed. support. If Young's modulus for rod is Y, then elongation produced in rod is

Question

A rod of uniform cross-sectional area A and length L has a weight W. It is suspended vertically from a fixed. support. If Young's modulus for rod is Y, then elongation produced in rod is (A) (W L/Y A) (B) (W L/2 Y A) (C) (W L/4 Y A) (D) (3 W L/4 Y A)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9f3a230b582764a5a55615439884f813-.png" /> Center of mass is at (L/2) distance from top so it can be assumed for easy calculation that W weight is hanged to a (L/2) length string Now use (F L/A Y) ⋅ Δ L Δ L=(W × L/2 A Y)

Q. A rod of uniform cross-sectional area $A$ and length $L$ has a weight $W$ . It is suspended vertically from a fixed. support. If Young's modulus for rod is $Y$ , then elongation produced in rod is

$\frac{W L}{Y A}$

$\frac{W L}{2 Y A}$

$\frac{W L}{4 Y A}$

$\frac{3 W L}{4 Y A}$

Center of mass is at $\frac{L}{2}$ distance from top so it can be assumed for easy calculation that $W$ weight is hanged to a $\frac{L}{2}$ length string
Now use $\frac{F L}{A Y} \cdot Δ L$
$Δ L = \frac{W \times L}{2 A Y}$

Q. A rod of uniform cross-sectional area A and length L has a weight W. It is suspended vertically from a fixed. support. If Young's modulus for rod is Y, then elongation produced in rod is

YAWL​

2YAWL​

4YAWL​

4YA3WL​