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A soil sample has a moisture content of $20\%$ and a void ratio $($e$)$ of $(2/3).$ If the moisture content becomes two times the initial moisture content, how the saturation of this soil will vary? Specific gravity is $2\mathrm{.}5.$

Hint $1$: The moisture content at the saturated state $($i$.$e$.$ S$=1)$ for this soil is equal to $26\%(0\mathrm{.}26).$

Hint $2$: Use the following equation to find the answer:

Question $8$Answer

a$.$

Saturation becomes twice

b$.$

Saturation will not change

c$.$

Saturation becomes $(4/3)$ times the initial value

d$.$

Saturation becomes $(3/2)$ times the initial value

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The correct answer is:

Saturation becomes $(4/3)$ times the initial value