MCQ
Two simple pendulum whose lengths are $1\,m$ and $121\, cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will the two be in phase again
  • A
    $11$
  • $10$
  • C
    $21$
  • D
    $20$

Answer

Correct option: B.
$10$
b
$\mathrm{N} \sqrt{\ell_{\ell}}=(\mathrm{N}+1) \sqrt{\ell_{\mathrm{s}}}$

$\Rightarrow \mathrm{N} \sqrt{121}=(\mathrm{N}+1) \sqrt{100}$

$\Rightarrow \mathrm{N}(11)+(\mathrm{N}+1) 10$

$\Rightarrow 11 \mathrm{N}=10 \mathrm{N}+10$

$\Rightarrow \mathrm{N}=10$

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