QJEEM24Ph08df: JEE Main 2024 Physics - Class 11 JEEM24Ph08df Solution: The density and breaking stress of a wire are and respectively. …

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 ( $\rho$ ), length ( $L$ ), and area ( $A$ ) as:

$m = \rho V = \rho A L$

We can substitute this expression in the equation for weight:

$W = \rho A L g$

On a planet where the acceleration due to gravity is $\frac{1}{3}$ rd of that on Earth, we substitute $g_{\text{planet}} = \frac{1}{3}g_{\text{Earth}} = \frac{1}{3} \times 10 \, \text{m/s}^2 = \frac{10}{3} \, \text{m/s}^2$ . The breaking force $F$ which is equivalent to the weight at the breaking point, is therefore given by:

$F = \rho A L g_{\text{planet}}$

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:

$\sigma = \frac{\rho A L g_{\text{planet}}}{A}$

$\sigma = \rho L g_{\text{planet}}$

Solving for $L$ :

$L = \frac{\sigma}{\rho g_{\text{planet}}}$

Given that $\sigma = 1.2 \times 10^8 \, \text{N/m}^2$ , $\rho = 6 \times 10^4 \, \text{kg/m}^3$ , and $g_{\text{planet}} = \frac{10}{3} \, \text{m/s}^2$ , we can substitute these values into the formula:

$L = \frac{1.2 \times 10^8}{6 \times 10^4 \times \left( \frac{10}{3} \right) }$

$L = \frac{1.2 \times 10^8}{2 \times 10^5}$

$L = 600 \, \text{m}$

Therefore, the maximum length of the wire that can be suspended from a rigid support on this planet without breaking is $600$ meters.