A spring of young's modulus 2 ×10¹¹ pa is suspended vertically and subjected to a load of 5 kg and elongation is 2 mm. when the load is doubled-match the following (g = 9.8 m/s²)
      Section - A                         Section - B
a) Elongating force                  e) 4 mm 

b) Stress                                    f) Unchanged 

c) Elongation                            g) 98 N 

d) Young's modulus                h) Doubled 

The formula for the elongating force (F) in a spring is given by Hooke's Law:

$\mathrm{F} = \mathrm{k} \times \mathrm{\Delta L}$

where k is the spring constant, and ΔL is the change in length. The stress (σ) is given by:

$\mathrm{\sigma } = \frac{\mathrm{F}}{\mathrm{A}}$

where A is the cross-sectional area of the spring. The Young's modulus (Y) can be calculated as:

$\mathrm{Y} = \frac{\mathrm{\sigma }}{\mathrm{\epsilon }}$

where ε is the strain, which is the ratio of the change in length to the original length.

Given that the Young's modulus of the spring is $2\times {10}^{11}$ Pa, the original elongation is 2 mm, and the original load is 5 kg (i.e., $5\times 9.8$N), we can calculate the original values for the other quantities as follows:

$\mathrm{F}=\mathrm{mg}$

Now, when the load is doubled, the new elongation is 4 mm, and the new force is $2\times 5\times 9.8 \mathrm{N} = 98 \mathrm{N}.$ 

Stress,

$\mathrm{S}=\frac{2\mathrm{mg}}{\mathrm{A}}=2\left(\mathrm{S}\right)$

which is doubled.

Strain,

$\mathrm{strain}=\frac{\mathrm{stress}}{\mathrm{Y}}$

$∆\mathrm{L}=\frac{\mathrm{mgL}}{\mathrm{AY}}$

$∆\mathrm{L}\propto \mathrm{m}$

$∆{\mathrm{L}}_{\mathrm{f}}=2∆\mathrm{L}$

Hence the correct answer is  a–g; b–h; c–e; d–f .