QJEEM23Ph85b0: JEE Main 2023 Physics - Class 11 JEEM23Ph85b0 Solution: The Young's modulus of a steel wire of length and…

The elongation of the wire can be calculated using the formula for stress and strain. The stress in the wire is given by:

$\sigma = \frac{mg}{A}$

where m is the mass of the block (4 kg), g is the acceleration due to gravity on the planet (1/4 of its value on the earth, or 2.5 m/s²), and A is the cross-sectional area of the wire (3 mm²).

The strain in the wire is given by:

$\epsilon = \frac{\Delta L}{L}$

where ΔL is the elongation of the wire and L is the original length of the wire (6 m).

Using Hooke's law, which states that stress is proportional to strain, we can find the elongation of the wire:

$\sigma = Y\epsilon$

where Y is the Young's modulus of the wire (2 $\times$ 10¹¹ N/m²).

Combining the above equations, we can find the elongation of the wire:

$\epsilon = \frac{\sigma}{Y} = \frac{mg}{A Y} = \frac{4 \times 2.5}{3 \times 10^{-6} \times 2 \times 10^{11}} = \frac{5}{3 \times 10^{-6} \times 10^{11}} = \frac{5}{3 \times 10^{5}}$

$\therefore$ $\frac{\Delta L}{L}$ $= \frac{5}{3 \times 10^{5}}$

$\Rightarrow$ $\Delta L = {{5 \times 6} \over {3 \times {{10}^5}}}$ = ${1 \over {{{10}^4}}}$ = 0.1 mm

So, the elongation of the wire is 0.1 mm.