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
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$\mathrm{N} \sqrt{\ell_{\ell}}=(\mathrm{N}+1) \sqrt{\ell_{\mathrm{s}}}$
$\Rightarrow \mathrm{N} \sqrt{121}=(\mathrm{N}+1) \sqrt{100}$
$\Rightarrow \mathrm{N}(11)+(\mathrm{N}+1) 10$
$\Rightarrow 11 \mathrm{N}=10 \mathrm{N}+10$
$\Rightarrow \mathrm{N}=10$
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