b
(b,d)Since collision is perfectly inelastic so all the blocks will stick together one by one and move in a form of combined mass.
Time required to cover a distance \(‘L’\) by first block \( = \frac{L}{v}\)
Now first and second block will stick together and move with \(v/2 \) velocity (by applying conservation of momentum) and combined system will take time \(\frac{L}{{v/2}} = \frac{{2L}}{v}\) to reach up to block third.
Now these three blocks will move with velocity \(v/3 \) and combined system will take time \(\frac{L}{{v/3}} = \frac{{3L}}{v}\) to reach upto the block fourth.
So, total time \( = \frac{L}{v} + \frac{{2L}}{v} + \frac{{3L}}{v} + ...\frac{{(n - 1)L}}{v}\)\( = \frac{{n(n - 1)L}}{{2v}}\)
and velocity of combined system having \(n\) blocks as \(\frac{v}{n}\).