The density and breaking stress of a wire are 6 \times 10^4 ~kg / m^3 and 1.2 \times 10^8 ~N / m^2 respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is \frac13^ rd of the value on the surface of earth. The maximum length of the wire with breaking is \underline\hspace2cm m (take, g=10 ~m / s^2 ).

Properties of Matter PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q86 with Solution

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Correct answer: 600

For a wire hanging under its own weight, the maximum tension occurs at the topmost point.

If the wire is just about to break under its own weight, then

\text{stress at top} = \text{breaking stress}

Now,

\text{stress} = \frac{\text{force}}{\text{area}} = \frac{\rho A L g'}{A} = \rho L g'

where:

So, for maximum length,

\rho L g' = \sigma_b

where $\sigma_b$ is the breaking stress.

\rho = 6 \times 10^4\ \text{kg/m}^3

\sigma_b = 1.2 \times 10^8\ \text{N/m}^2

Gravity on planet is one-third of earth's gravity:

g' = \frac{g}{3} = \frac{10}{3}\ \text{m/s}^2

L = \frac{\sigma_b}{\rho g'}

Substitute values:

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

First simplify denominator:

6 \times 10^4 \times \frac{10}{3} = 2 \times 10^5

Therefore,

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

L = 0.6 \times 10^3 = 600

\boxed{600\ \text{m}}

Stored correct answer = $600$

Our derived answer matches the stored answer.