Question

One end of a uniform wire of length $L$ and of weight $W$ is attached rigidly to a point in the ceiling and a weight $W_1$ is suspended from its lower end. If $S$ is the area of cross–section of the wire, the stress in the wire at a height $3L/4$ from its lower end is

• A. $\frac{W_1}{S}$
• B. $\frac{1}{S}(W_1+\frac{W}{4})$
• C. $\frac{1}{S}(W_1+\frac{3W}{4})$
• D. $\frac{1}{S}(W_1+W)$

Answer 1

The correct option is C $\frac{1}{S}(W_1+\frac{3W}{4})$

Since the wire is uniform, i.e. its mass per unit length is constant over its entire length $L$,

the total downward weight = the weight of the suspended mass + weight of length $\frac{3L}{4}$ of the wire

$F=W_1+\frac{3W}{4}.$

$\therefore Stress=\frac{force}{area}=\frac{F}{S}=\frac{W_1+\frac{3}{4}W}{S}$