Question

A wire having mass per unit length 10 g/cm and density $$800\text{kg/}{\text{m}}^3$$ suspended rigid support. The stress in the wire if 10 kg-wt is attached to its free end is?

Answer 1

Hint:Use the relation between density, mass and volume to express the volume of the wire. Divide the relation by the length of wire to get the area of the wire. The stress is the ratio of applied force and area of cross-section.

Formula used:
Stress, $$\sigma ={\displaystyle \frac{F}{A}}$$
Here, F is the applied force and A is the area.

Complete step by step answer:
We have given the mass per unit length of the wire, $${\displaystyle \frac{m}{l}}=\left(10{\displaystyle \frac{\text{g}}{\text{cm}}}\right)\left({\displaystyle \frac{1\text{kg}}{1000\text{g}}}\right)\left({\displaystyle \frac{100\text{cm}}{1\text{m}}}\right)=1\text{kg/m}$$.
The density of the wire is, $$\rho =800\text{kg/}{\text{m}}^3$$.
We have the relation,
$$\rho ={\displaystyle \frac{m}{V}}$$
$$\Rightarrow V={\displaystyle \frac{m}{\rho }}$$
Here, m is the mass of the wire and V is the volume.

Dividing the above equation by $$l$$, we get,
$${\displaystyle \frac{V}{l}}={\displaystyle \frac{m}{\rho l}}$$
We know that the volume is given as, $$V=l^3$$. Therefore, the above expression becomes,
$${\displaystyle \frac{l^3}{l}}={\displaystyle \frac{m}{\rho l}}$$
$$\Rightarrow l^2={\displaystyle \frac{m}{\rho l}}$$
$$\Rightarrow A={\displaystyle \frac{m}{\rho l}}$$ ……. (Since $$Area=l^2$$) …… (1)
We know that the stress is the ratio of applied force per unit area of cross-section. Therefore,
$$\sigma ={\displaystyle \frac{F}{A}}$$
Here, F is the applied force and A is the area.

In this case, the applied force is the weight of the block attached to the free end of the wire. Therefore, we can write the above equation as,
 $$\sigma ={\displaystyle \frac{W}{A}}$$
Using equation (1) in the above equation, we get,
$$\sigma ={\displaystyle \frac{W}{{\displaystyle \frac{m}{\rho l}}}}$$
$$\Rightarrow \sigma ={\displaystyle \frac{W\rho }{m/l}}$$
Substituting$$W=10\text{N}$$, $$\rho =800\text{kg/}{\text{m}}^3$$ and $${\displaystyle \frac{m}{l}}=1\text{kg/m}$$ in the above equation, we get,
$$\sigma ={\displaystyle \frac{\left(10\right)\left(800\right)}{1}}$$
$$\therefore \sigma =8000\text{N/}{\text{m}}^2$$

Thus,the stress in the wire if 10 kg-wt is attached to its free end is $$8000\text{N/}{\text{m}}^2$$.

Note: Always convert the mass per unit length from $$\text{gm/c}{\text{m}}^3$$ to $$\text{kg/}{\text{m}}^3$$. We have taken the volume of the wire irrespective of its shape to express it in terms of area of the wire. The stress is also the pressure because the pressure is also the force per unit area of cross-section.