An elastic spring under tension of 3 N has a length a. Its length is b under tension 2 N. For its length (3a − 2b), the value of tension will be ______ N.

For a spring with natural length $$L_0$$ and spring constant $$k$$:

Under tension 3N: $$a = L_0 + 3/k$$ ... (1)

Under tension 2N: $$b = L_0 + 2/k$$ ... (2)

From (1) - (2): $$a - b = 1/k$$, so $$k = 1/(a-b)$$

From (2): $$L_0 = b - 2(a-b) = 3b - 2a$$

For length $$3a - 2b$$: $$T = k(L - L_0) = \frac{(3a-2b)-(3b-2a)}{a-b} = \frac{5a-5b}{a-b} = 5$$ N

The answer is 5.