Given information:
- Length of the first pendulum, L1 = 100 cm
- Length of the second pendulum, L2 = 121 cm
- The two pendulums are at the mean position in the same phase at some instant

To find:
- After how many oscillations of the longer pendulum will the two be in the same phase at the mean position again

Solution:
Let T1 and T2 be the time periods of the first and second pendulums, respectively.

The time period of a simple pendulum is given by:
T = 2π√(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity.

Using this formula, we can find the time periods of the two pendulums:
T1 = 2π√(L1/g)
T2 = 2π√(L2/g)

Since the two pendulums start from the same mean position and in the same phase, their oscillations will be in sync if they complete an equal number of oscillations in the same time interval.

Let the number of oscillations of the second pendulum required to be in sync with the first pendulum be n.

After n oscillations of the second pendulum, the time taken will be:
nT2 = n(2π√(L2/g))

The first pendulum would have completed a certain number of oscillations in this time interval. Let that number be m.

The time taken by the first pendulum to complete m oscillations is:
mT1 = m(2π√(L1/g))

Since the two pendulums are in sync at the end of this time interval, we can equate the two expressions:

nT2 = mT1

Substituting the values of T1 and T2, we get:
n(2π√(L2/g)) = m(2π√(L1/g))

Simplifying the equation, we get:
n/√L2 = m/√L1

Since L2 > L1, we can say that n > m.

Also, we know that n and m are positive integers, because they represent the number of oscillations of the pendulums.

So, we need to find a pair of positive integers such that n > m and n/√L2 = m/√L1.

We can simplify the second expression as follows:
n/m = √(L2/L1)

Since L2/L1 = (121/100), we can write:
n/m = √(121/100) = 11/10

We need to find a pair of positive integers whose ratio is 11/10 and whose difference is 1.

The only such pair is 11 and 10.

Therefore, n = 11 and m = 10.

Hence, the second pendulum needs to complete 11 oscillations for the two pendulums to be in sync at the mean position again.

Therefore, the correct answer is option B (10).