Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is (in $\Omega$)

  • A
    $10$
  • B
    $100$
  • C
    $110$
  • D
    $120$

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