Question 253: Properties of Matter

Question

The density and breaking stress of a wire are $$6 	imes 10^4 \mathrm{~kg} / \mathrm{m}^3$$ and $$1.2 	imes 10^8 \mathrm{~N} / \mathrm{m}^2$$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is $$\frac{1}{3}^{	ext {rd }}$$ of the value on the surface of earth. The maximum length of the wire with breaking is _______ $$\mathrm{m}$$ (take, $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$).

Accepted Answer

The breaking stress of a wire, denoted as $$\sigma$$, is the maximum tension per unit area it can withstand before it breaks. It is given as:
$ \sigma = \frac{F}{A} $
where $F$ is the breaking force and $A$ is the cross-sectional area of the wire. The weight of the wire, when it's on the verge of breaking, equals the maximum force $F$ it can sustain. The weight of an object is given by:
$ W = mg $
where $m$ is the mass of the object and $g$ is the acceleration due to gravity. Since the wire's mass, $m$, can also be expressed in terms of its density ($ho$), length ($L$), and area ($A$) as:
$ m = ho V = ho A L $
We can substitute this expression in the equation for weight:
$ W = ho A L g $
On a planet where the acceleration due to gravity is $\frac{1}{3}$rd of that on Earth, we substitute $g_{	ext{planet}} = \frac{1}{3}g_{	ext{Earth}} = \frac{1}{3} 	imes 10 \, 	ext{m/s}^2 = \frac{10}{3} \, 	ext{m/s}^2$. The breaking force $F$ which is equivalent to the weight at the breaking point, is therefore given by:
$ F = ho A L g_{	ext{planet}} $
Since the breaking stress $\sigma$ is also $F/A$, we can set the two expressions equal to find the maximum length $L$ of the wire before breaking:
$ \sigma = \frac{ho A L g_{	ext{planet}}}{A} $
$ \sigma = ho L g_{	ext{planet}} $
Solving for $L$:
$ L = \frac{\sigma}{ho g_{	ext{planet}}} $
Given that $\sigma = 1.2 	imes 10^8 \, 	ext{N/m}^2$, $ho = 6 	imes 10^4 \, 	ext{kg/m}^3$, and $g_{	ext{planet}} = \frac{10}{3} \, 	ext{m/s}^2$, we can substitute these values into the formula:
$ L = \frac{1.2 	imes 10^8}{6 	imes 10^4 	imes \left( \frac{10}{3} ight) } $
$ L = \frac{1.2 	imes 10^8}{2 	imes 10^5} $
$ L = 600 \, 	ext{m} $
Therefore, the maximum length of the wire that can be suspended from a rigid support on this planet without breaking is $600$ meters.

Properties of Matter

Integer / Numeric Answer Type