Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

Concept: 

In order for both pendulums to react in the same phase, we must identify the vibrations created by the shorter pendulum.

Time period of motion = T = \(2\pi √{\frac{L}{g}} \)    ---- (1) 

where L = length of pendulum , g = acceleration due to gravity

 from equation (1) we can conclude that, T∝ √ L   ---- (2)

Calculation:

Given: length of long pendulum (L₁) = 121 cm = (121/100) m = 1.21m

 length of short pendulum (L₂) = 100cm = 100/100 m = 1m

From equation (2) we get: T∝ √ L 

we have L₁ = 1.21 m , L₂ = 1 m

T₁ ∝ √ L₁ =>  T₁ ∝ √ 1.21    ---- (3)

T2 ∝ √ L2 =>  T2 ∝ √ 1    ---- (4)

 diving equation 3 by equation 4 we get:

  =>  \(\frac{T_{1}}{T_{2}} = \sqrt{\frac{1.21}{1}} = \sqrt{1.21}\) = 1.1 

 => T₁ = 1.1 T₂  => 10T1 = 11 T2

When a longer pendulum vibrates 10 times than shorter pendulum have to vibrate 11 times to complete that path.

Hence option 2) is correct.