ANS:C - 65%

Flow index = water content at 10 no of blows - water content at 100 no of blows.

So, flow index = W10-W100 = 70-20 = 50%. Using the formula of one point method.

WL= Wn(N/25)^0.1.
Where N should be between 15 to 35.
Using nearest value from data i.e. N=10 & Wn = 70.

WL = 0.70 * (10/25)^0.1 = 0.6387~0.65 ans.
10 -> 70%.
100 -> 20%.
25 -> ?.

Ans = [(70-20)/log (100/10)] * log25.
= 50*log25.
= 69.9.
Now, using that Flow Index of 50 and the same formula to find moisture content at 25 No. of blows is 50%.

Flow Index = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50.1%. LL = w*(N/25)^0.121.
LL = 70*(100/25)^0.121 = 82.78%.
LL = 20*(10/25)^0.121 = 17.90%.
LL = 83 - 18 = 65%. Take log(no of blows) only for finding flow index or for finding liquid limit using a flow index value.

If the graph is given and we want to find LL, then just do interpolate without taking the log of no of blows. Because we are plotting as such in semilog to get a straight line.

ie, (70-20)/(100-10) = (LL-20)/(100-25).
LL = 61.66. The formula of liquid limit =w* (N/25) ^0.121.

Find, LL for each condition and then subtract them. You will get 65% as answer.