The total length of a sonometer wire fixed between two bridges is $110 \mathrm{cm}$. Now, two more bridges are placed to divide the length of the wire in the ratio $6:3:2$. If the tension in the wire is $400 \mathrm{N}$ and the mass per unit length of the wire is $0.01 \mathrm{kg} {\mathrm{m}}^{-1}$, then the minimum common frequency with which all the three parts can vibrate, is

• Correct answer
  A

  $1000Hz$
• B

  $1100Hz$
• C

  $100Hz$
• D

  $110Hz$

Hence, the correct answer is option: 'A'.